(PDF) Abstract Algebra for Polynomial Operations

Abstract Algebra for Polynomial Operations

Maya Mohsin Ahmed

c

⃝Maya Mohsin Ahmed 2009

To my students

As we express our gratitude, we must never forget that the highest

appreciation is not to utter words, but to live by them.

- John F. Kennedy.

3

4

Contents

1 Polynomial Division. 9

1.1 Rings and Fields. . . . . . . . . . . . . . . . . . . . . . 9

1.2 Polynomial division. . . . . . . . . . . . . . . . . . . . 11

1.3 Gr¨obnerbases. ...................... 15

2 Solving Systems of Polynomial Equations. 27

2.1 Ideals and Varieties. . . . . . . . . . . . . . . . . . . . 27

2.2 Elimination Theory. . . . . . . . . . . . . . . . . . . . . 33

2.3 Resultants. ........................ 39

3 Finding Roots of polynomials in Extension Fields. 51

3.1 Modular Arithmetic and Polynomial irreducibility in Q. 51

3.2 Field Extensions. . . . . . . . . . . . . . . . . . . . . . 57

3.3 QuotientRings....................... 64

3.4 Splitting ﬁelds of polynomials. . . . . . . . . . . . . . . 69

4 Formulas to ﬁnd roots of polynomials. 81

4.1 Groups. .......................... 81

4.2 Cyclicgroups........................ 87

4.3 Normal Subgroups and Quotient Groups. . . . . . . . . 91

4.4 Basic properties of ﬁnite groups. . . . . . . . . . . . . . 95

4.5 Finite Abelian Groups. . . . . . . . . . . . . . . . . . . 101

4.6 Galoistheory........................ 107

4.7 Proof of Galois’ Criterion for solvability. . . . . . . . . 119

5 Constructing and Enumerating integral roots of systems

of polynomials. 135

5.1 MagicSquares. ...................... 135

5.2 Polyhedral cones. . . . . . . . . . . . . . . . . . . . . . 138

5

5.3 Hilbert bases of Polyhedral cones . . . . . . . . . . . . 141

5.4 ToricIdeals......................... 145

5.5 Hilbert Functions. . . . . . . . . . . . . . . . . . . . . . 149

5.6 Ehrhart Polynomials. . . . . . . . . . . . . . . . . . . . 154

6 Miscellaneous Topics in Applied Algebra. 159

6.1 Counting Orthogonal Latin squares. . . . . . . . . . . . 159

6.2 Chinese Remainder Theorem. . . . . . . . . . . . . . . 163

6.3 Cryptology ........................ 166

6.4 Algebraic codes. . . . . . . . . . . . . . . . . . . . . . . 171

A 189

A.1 The Euclidean Algorithm. . . . . . . . . . . . . . . . . 189

A.2 Polynomial irreducibility. . . . . . . . . . . . . . . . . . 192

A.3 Generating Functions. . . . . . . . . . . . . . . . . . . 194

A.4 Algorithms to compute Hilbert bases. . . . . . . . . . . 197

A.5 Algorithms to compute toric ideals. . . . . . . . . . . . 198

A.6 Algorithms to compute Hilbert Poincar´e series. . . . . . 201

6

Foreword

To forget one’s purpose is the commonest form of stupidity - Nietzsche.

I have been asked, time and again, what the purpose is of learning

Abstract Algebra. I wrote this book to answer this perennial question.

Traditionally, Algebra books begin with deﬁnitions and theorems and

applications might appear as examples. Many students are not inclined

to learn without a purpose. The beautiful subject of Algebra closes

doors on them. The responses of many students to Abstract Algebra

remind me of Gordan’s reaction to the proof of the Hilbert’s basis

Theorem - This is not Mathematics. This is Theology.

The focus of this book is applications of Abstract Algebra to poly-

nomial systems. The ﬁrst ﬁve chapters explore basic problems like

polynomial division, solving systems of polynomials, formulas for roots

of polynomials, and counting integral roots of equations. The sixth

chapter uses the concepts developed in the book to explore coding the-

ory and other applications.

This book could serve as a textbook for a beginning Algebra course,

a student takes immediately after a Linear Algebra course. Linear Al-

gebra is not a prerequisite but will provide the basis for the natural

progression to nonlinear Algebra. This book could also be used for

an elective course after an Abstract Algebra course to focus on appli-

cations. This book is suitable for third or fourth year undergraduate

students.

Maya Mohsin Ahmed

7

8

Chapter 1

Polynomial Division.

Judge a man by his questions rather than by his answers – Voltaire.

If someone asks you whether you know how to divide polynomials

your ﬁrst answer would be sure you do. You learned that in high

school or earlier. But now if the question is rephrased and you are

asked whether you know how to divide polynomials in more than one

variable, then to your surprise, you ﬁnd you do not know the answer

unless you have taken a couple of courses in Abstract Algebra. In this

chapter we introduce Rings and Fields which are algebraic objects that

allow you to solve such problems.

1.1 Rings and Fields.

Deﬁnition 1.1.1. Aring is a nonempty set Requipped with two oper-

ations (usually written as addition and multiplication) that satisfy the

following axioms.

1. Ris closed under addition: if a∈Rand b∈Rthen a+b∈R.

2. Addition is associative: if a, b, c ∈R, then a+ (b+c) = (a+b)+c.

3. Addition is commutative: if a, b ∈R, then a+b=b+a

4. There is an additive identity (or zero element) 0Rin Rsuch that

a+ 0R=a= 0R+afor every a∈R.

5. For each a∈Rthere is an additive inverse (denoted by -a) in

R, that is the equation a+x= 0Rhas a solution in R. For

convenience we write b+ (−a)as b−afor a, b ∈R.

9

6. Ris closed under multiplication: if a∈R, and b∈Rthen a·b∈

R.

7. Multiplication is associative: if a, b, c ∈R, then a·(b·c) = (a·b)·c.

8. Distributive laws of multiplication hold in R: if a, b, c ∈R, then

a·(b+c) = a·b+a·cand

(a+b)·c=a·c+b·c.

Example 1.1.1.

1. The set of integers Z={. . . , −2,−1,0,1,2, . . . }is a ring.

2. The set of rational numbers Qis a ring.

3. The set of complex numbers Cis a ring.

4. Let kbe a ring. The set of all polynomials in nvariables with coef-

ﬁcients in k, denoted by k[x1, x2, . . . , xn], with the usual operation

of addition and multiplication of polynomials, is a ring. Conse-

quently, C[x1, x2, . . . , xn], Q[x1, x2, . . . , xn], and Z[x1, x2, . . . , xn]

are rings.

A ring in which the operation of multiplication is commutative is

called a commutative ring. A ring with identity is a ring Rthat contains

an element 1Rsatisfying the axiom:

a·1R=a= 1R·afor all a∈R.

Deﬁnition 1.1.2. An integral domain is a commutative ring Rwith

identity 1R̸= 0Rthat satisﬁes the condition:

Whenever a, b ∈Rand ab = 0R,then a= 0Ror b= 0R.

Example 1.1.2.

The sets Zand Qare integral domains.

Deﬁnition 1.1.3. Aﬁeld is a commutative ring with identity in which

every nonzero element has an inverse.

Note that in a ﬁeld Fdivision is closed, i.e., if a, b ∈F, then

a/b =ab−1∈F.

10

Example 1.1.3.

1. The sets Qand Care ﬁelds.

2. The set Zis not a ﬁeld.

3. The set k[x1, x2, . . . , xn] is not a ﬁeld.

Many results from elementary algebra are also true for rings.

Example 1.1.4. Let Rbe a ring. If a, b ∈R, then a−(−b) = a+b.

Proof. Since b−b=b+ (−b) = 0R, we get that the inverse of (−b)

−(−b) = b.

Therefore

a−(−b) = a+b.

Similar properties of rings are explored in the exercises.

1.2 Polynomial division.

We ﬁrst look at polynomial divisions that involve only one variable x.

The monomial of a polynomial with the highest degree is called the

leading monomial and the coeﬃcient of the leading monomial is called

the leading coeﬃcient. The leading term of a polynomial is the product

of the leading coeﬃcient and the leading monomial. The degree of the

leading term is also the degree of the polynomial. The nonzero constant

polynomials have degree zero. The constant polynomial 0 does not have

a degree.

Theorem 1.2.1 (The Division Algorithm).Let f(x)and g(x)be poly-

nomials with real coeﬃcients such that g(x)̸= 0. Then there exists

unique polynomials q(x)and r(x)such that

f(x) = g(x)q(x) + r(x)

and degree r(x)<degree g(x).

The polynomial q(x)is called the quotient and the polynomial r(x)

is called the remainder.

11

The proof of the division algorithm is dealt with in the exercises.

Example 1.2.1. If we divide f=x4+x+ 1 by g=x2−1, we get

r= 2x+ 1 as remainder. Observe that the degree of ris less than the

degree of g.

But the story changes when we work with polynomials involving

more than one variable. For example, determining which is the leading

term of the polynomial x2+xy +y2is not as straightforward as the

one variable case. Consequently, we need to establish an ordering of

terms for multivariable polynomials.

Let Zn

≥0denote the set of n-tuples with nonnegative integer coordi-

nates and let kbe a ﬁeld. Consider the ring of polynomials k[x1, x2, . . . xn].

Observe that we can reconstruct the monomial xα=xα1

1···xαn

1

from the n-tuple of exponents (α1, . . . , αn)∈Zn

≥0. In other words, there

is a one-to-one correspondence between monomials in k[x1, . . . , xn] and

Zn

≥0. This correspondence allows us to use any ordering >we establish

on the space Zn

≥0as an ordering on monomials, that is,

α > β in Zn

≥0implies xα> xβin k[x1, . . . , xn].

Deﬁnition 1.2.1. AMonomial ordering on k[x1, . . . , xn]is any re-

lation >on Zn

≥0, or equivalently, any relation on the set of monomials

xα, α ∈Zn

≥0, satisfying:

1. >is a total (or linear) ordering on Zn

≥0, which means that, for

every pair α, β ∈Zn

≥0exactly one of the three statements

α > β, α =β, β > α

should be true.

2. If α > β ∈Zn

≥0, then α+γ > β +γ, whenever γ∈Zn

≥0.

3. >is a well-ordering in Zn

≥0, that is, every nonempty subset of Zn

≥0

has a smallest element under >.

We now look at some common monomial orderings.

Deﬁnition 1.2.2 (Lexicographic (or Lex) ordering). Let α=

(α1, . . . , αn)and β= (β1, . . . , βn)∈Zn

≥0. We say α >lex βif, in the

vector diﬀerence α−β∈Zn

≥0, the left-most nonzero entry is positive.

And we write xα>lex xβif α >lex β.

12

Example 1.2.2. 1. Consider the polynomial f=x2+xy +y2. We

have x2>lex xy because (2,0) >lex (1,1): check that in the vector

diﬀerence (2,0) −(1,1) = (1,−1), the leftmost entry is positive.

Similarly, x2>lex y2since (2,0) >lex (0,2). Therefore, the lead-

ing term of the polynomial fwith respect to the lexicographic

ordering is x2.

2. The leading term of the polynomial x+y4with respect to the lex

ordering is x.

Diﬀerent monomial orderings give diﬀerent leading terms for the

same polynomial and we make the choice of monomial ordering that

serves our purpose best.

Deﬁnition 1.2.3 (Graded lex order). Let α, β ∈Zn

≥0and let

|α|=

n



i=1

αi,|β|=

n



i=1

βi.

We say α >glex βif

|α|>|β|or |α|=|β|and α >lex β.

Example 1.2.3. 1. The leading term of the polynomial x2+xy +y2

with respect to graded lex order is still x2. This is because the

degrees of all other terms being the same, the condition x >lex y

determines the leading term.

2. The leading term of the polynomial x+y4is y4with respect to

the graded lex ordering.

We refer the reader to Chapter 2 in [17] for other monomial order-

ings and also for a detailed study of the same. Now that we have a

notion of monomial orderings, can we satisfactorily divide polynomials

with more than one variable? The answer still is no because there is

one more problem we must discuss. We do this with an example.

Example 1.2.4. Let us divide f=x2+xy + 1 with the polynomial

g=xy −xwith respect to the graded lex ordering.

The leading term of fis x2and is not divisible by the leading term

xy of g. In the one variable case this would imply that fis not divisible

13

by g. But in the case of multivariable polynomials fis still divisible

by gbecause the second term of fis divisible by the leading term of

g. So we ignore the leading term of fand perform division as shown

below.

q: 1

xy −xx2+xy + 1

xy −x

x2+x+ 1

The quotient q= 1 and the remainder r=x2+x+ 1 and we write

f=qg +r. So the idea is to continue dividing till none of the terms of

fis divisible by the leading term of g. Observe that

lead term r >glex lead term g.

Recall that this cannot happen in one variable polynomial division.

To conclude, we list the two steps involved in dividing a multivari-

able polynomial fby a multivariable polynomial g:

1. Choose a monomial ordering.

2. Divide until none of the terms of the remainder is divisible by the

leading term of g.

Sometimes we need to divide a polynomial fby a set of polynomials

F={f1, . . . , fn}, that is, write fas

f=

n



i

qifi+rwhere qiare quotients and ris the remainder.

For example, we want to know whether the solutions of a system

of polynomials in F={f1, . . . , fn}are also roots of a polynomial f

(this question is formalized in Section 2.1). To answer this question,

we divide fby the set {f1, . . . , fn}to write f=n

iqifi+r. If the

remainder r= 0, then the solutions of the system Fare roots of f.

In the following example, we demonstrate the dependence of the

remainder on the order of division. The remainder is diﬀerent when

the order of division is diﬀerent.

14

Example 1.2.5. Let F={f1, f2}where f1=xy −1 and f2=y2−1,

and let f=xy2−y3+x2−1. We divide the polynomial fﬁrst by f1

and then by f2with respect to the graded lex ordering:

q1:y

q2:−y

xy −1

y2−1xy2−y3+x2−1

xy2−y

−y3+x2+y−1

−y3+y

x2−1

Therefore,

f=q1f1+q2f2+rwhere r=x2−1, q1=y, q2=−y.

Now we change the order of division and divide fby f2ﬁrst and

then f1:

q1: 0

q2:x−y

y2−1

xy −1xy2−y3+x2−1

xy2−x

−y3+x2+x−1

−y3+y

x2+x−y−1

This gives us

f=q1f1+q2f2+rwhere r=x2+x−y−1, q1= 0, q2=x−y.

Since the remainder is not unique, we cannot say at this point, whether

r= 0 for some q1and q2. To get a unique remainder for a given

monomial ordering, no matter what the order of division is, we use

Gr¨obner bases which are discussed in the next section.

1.3 Gr¨obner bases.

Subsets of a ring need not be rings. For example, the set of even

integers is a ring whereas the set of odd integers is not (the sum of two

15

odd integers is not odd). A subset of a ring that is also a ring is called

asubring.

Deﬁnition 1.3.1. A subring Iof a ring Ris an ideal provided:

Whenever r∈Rand a∈I, then r·a∈Iand a·r∈I.

Ideals bring the generalized notion of being closed under scalar mul-

tiplication we ﬁnd in vector spaces to rings.

Example 1.3.1.

1. {0R}and Rare ideals for every ring R.

2. The only ideals of a ﬁeld Rare {0R}and R. See Exercise 5.

3. The set of even integers is an ideal of the ring Z.

We now prove a result that is handy while proving a subset of a

ring is an ideal and help skip the many checks of the deﬁnition.

Proposition 1.3.1. A nonempty subset Iof a ring Ris an ideal if

and only if it has the following two properties:

1. if a, b ∈I, then a−b∈I;

2. if r∈Rand a∈I, then r·a∈Iand a·r∈I.

Proof. Every ideal has these two properties by deﬁnition. Con-

versely suppose Ihas properties (1) and (2). Since Iis a subset of R,

addition is associative and commutative, multiplication is associative,

and the distributive laws of multiplication hold in Ias well. Therefore,

to prove Iis a subring we only need to prove that Iis closed under

addition and multiplication, 0R∈I, and that the additive inverse of

every element of Iis also in I. Since Iis nonempty there is some

element a∈I. Applying (1), we get a−a= 0R∈I. Now if a∈I,

then again by (1), 0r−a=−a∈I. Now, let a, b ∈I. Since −b∈I,

a−(−b) = a+b∈I. Thus Iis closed under addition. If a, b ∈I, then

a, b ∈Rsince Iis a subset of R. Consequently, Property (2) implies

that a·b∈I. Hence Iis closed under multiplication. Thus, Iis an

ideal.

In many cases, ideals tend to be inﬁnite sets. So it is convenient to

describe ideals in terms of a ﬁnite set, whenever possible.

16

Proposition 1.3.2. Let Rbe a ring and let F={f1, . . . , fs}be a

subset of R. Then the set I={s

i=1 ai·fi:ai∈R}is an ideal.

Iis called the ideal generated by the set Fand is denoted I=<

f1, . . . , fs>.

Proof. We use Proposition 1.3.1 to prove Iis an ideal. Let a, b ∈I

such that a=s

i=1 ai·fiand b=s

i=1 bi·fiwhere ai, bi∈Rfor

i= 1 to s. Then a−b=s

i=1(ai−bi)·fi∈Ibecause (ai−bi)∈R

for all isince Ris a ring. Thus Isatisﬁes property (1) in Proposition

1.3.1. Again, since Ris a ring, for r∈R,r·ai∈Rfor i= 1 to s.

Therefore, r·a=s

i=1(rai)·fi∈Iby deﬁnition of I. Similarly we

prove that a·r∈I. Thus Ialso satisﬁes property (2) of Proposition

1.3.1. Therefore, Iis an ideal.

Example 1.3.2.

1. The zero ideal is generated by a single element: I=<0R>={0R}

for every ring R.

2. An ideal Ican have diﬀerent sets of generators. Let R=Q[x1, . . . , xn]

be the polynomial ring with rational coeﬃcients. Then the ideal

I=< xy −1, y2−1>=< x −y, y2−1>(see Exercise 8).

Is every ideal of ring Rﬁnitely generated? Not always, but in the

case of Noetherian rings this is true.

Deﬁnition 1.3.2. A ring Ris a Noetherian ring if every ideal Iof

Ris ﬁnitely generated, i.e., I=< f1, . . . , fs>such that fi∈Rfor

i= 1 to s.

Theorem 1.3.1 (Hilbert’s Basis Theorem).If Ris a Noetherian ring

then so is the polynomial ring R[x].

The Proof of the Hilbert’s basis Theorem is given in [7, 17] and is

beyond the scope of this book. An ideal that is generated by one ele-

ment is called a principal ideal. A principal ideal domain is an integral

domain in which every ideal is principal.

Example 1.3.3. The ﬁeld kis ﬁnitely generated as an ideal (k=<

1>). The only other ideal of kis <0>. In fact, both the ideals

of kare principal ideals and hence ﬁnitely generated. Thus, ﬁelds

are Noetherian. Therefore, Theorem 1.3 implies k[x1] is Noetherian

whenever kis a ﬁeld. Applying the theorem subsequently we derive

k[x1, x2, . . . , xn] is Noetherian whenever kis a ﬁeld

17

A Gr¨obner basis of an ideal Iis a set of generators of I, and we

now proceed to deﬁne it.

Let I⊂k[x1, . . . , xn] be an ideal other than {0}. Let LT(I) denote

the set of leading terms of elements of I, that is,

LT(I) = {cxα: there exists f∈Iwith LT(f) = cxα}.

We denote <LT(I)>to be the ideal generated by the elements of

LT(I).

Deﬁnition 1.3.3. Fix a monomial order. A ﬁnite subset G={g1, . . . , gt}

of an ideal Iis said to be Gr¨obner basis if

<LT(g1), . . . , LT(gt)>=<LT(I)> .

In other words, a set {g1, . . . , gt}is a Gr¨obner basis of Iif and only

if the leading term of any element of Iis divisible by one of the LT(gi)

because the ideal <LT(I)>is generated by LT(gi).

In order to compute Gr¨obner bases, we deﬁne S-polynomials. For a

ﬁxed monomial ordering, let LM(f) denote the leading monomial of a

polynomial fand let LT(f) denote the leading term of f.

Deﬁnition 1.3.4. 1. Let the leading monomials of polynomials f

and gbe

LM(f) =

n



i=1

xiαiand LM(g) =

n



i=1

xiβi.

We call xγthe least common multiple (LCM) of LM(f)and

LM(g), if γ= (γ1, . . . , γn)such that γi=max (αi, βi)for each i.

2. The S-polynomial of fand gis the combination

S(f, g) = xγ

LT (f)·f−xγ

LT (g)·g.

Observe that we construct a S-polynomial of the polynomials fand

gby eliminating the lead terms of fand g, and that the S-polynomial

always has a smaller lead term than the lead terms of fand g.

Example 1.3.4. We now return to Example 1.2.5. Consider the

graded lex ordering, then

LM(f1) = xy and LM(f2) = y2.

18

The least common multiple of LM (f1) and LM(f2) is

xγ=xy2.

Therefore

S(f1, f2) = xy2

xy f1−xy2

y2f2=yf1−xf2(1.1)

=y(xy −1) −x(y2−1) = x−y.

In his 1965 Ph.D. thesis, Bruno Buchberger created the theory of

Gr¨obner bases and named these objects after his advisor Wolfgang

Gr¨obner. We now provide his algorithm to compute a Gr¨obner basis

of an ideal.

Algorithm 1.3.1. (Buchberger’s Algorithm.)

•Input: A set of polynomials F={f1, . . . , fs}

•Output: A Gr¨obner basis G={g1, . . . , gt}associated to F.

•Method:

Choose a monomial ordering.

Start with G:= F.

Repeat G′:= G

1. For each pair {p, q}, p ̸=qin G′ﬁnd S-polynomial S(p, q).

2. Divide S(p, q) by the set of polynomials G′.

3. If S̸= 0 then G:= G∪ {S}

Until G=G′.

Observe that for each pair {p, q}, p ̸=qin the Gr¨obner basis G

the remainder after dividing the S-polynomial S(p, q) by Gis always

zero. Gr¨obner bases for the same set of polynomials diﬀer according

to the monomial order we choose in our algorithm. The proof of the

Buchberger’s Algorithm is found in [17].

Given a monomial ordering can we ﬁnd a unique Gr¨obner basis?

The answer is yes and this basis also has the smallest number of poly-

nomials and is called reduced.

19

Deﬁnition 1.3.5. Areduced Gr¨obner basis for a set of polynomials

Fis a Gr¨obner basis Gof Fsuch that:

1. The leading coeﬃcient is 1for all p∈G.

2. For all p∈G, none of the terms of pis divisible by the leading

term of qfor each q∈G− {p}.

To ﬁnd the reduced Gr¨obner basis we need to modify Algorithm

1.3.1 a little. We now add one more step before repeating the loop.

Algorithm 1.3.2. (Computing a reduced Gr¨obner basis.)

•Input: A set of polynomials F={f1, . . . , fs}

•Output: The reduced Gr¨obner basis G={g1, . . . , gt}of F.

•Method:

Choose a monomial ordering.

Start with G:= F.

Repeat G′:= G

1. For each pair {p, q}, p ̸=qin G′, ﬁnd S-polynomial S(p, q).

2. Divide S(p, q) by the set of polynomials G′.

3. If S̸= 0 then G:= G∪ {S}

4. Divide each p∈Gby G− {p}to get p′. If p′̸= 0, replace p

by p′in G. If p′= 0 then G=G− {p}.

Until G=G′.

Example 1.3.5. We return to Example 1.2.5 and compute the reduced

Gr¨obner basis of the ideal generated by Fwith respect to the graded

lex ordering.

Initially the Gr¨obner basis G=F. We go to Step 1 in Algo-

rithm 1.3.2 and compute S(f1, f2). We have from Equation 1.1 that

S(f1, f2) = x−y. Let f3=S(f1, f2). The remainder after dividing f3

by Gis also f3. Since f3̸= 0, in accordance with Step 3, we add f3to

G, that is G={f1, f2, f3}. Now proceed to Step 4. The remainder is

zero when f1is divided by {f2, f3}. Therefore G={f2, f3}. Verify that

more polynomials cannot be eliminated from Gand go back to the be-

ginning of the loop with G={f2, f3}. In Step 1, f4=S(f2, f3) = y3−x

20

whose remainder is zero when we divide it by G. We now can exit

the loop and conclude that the reduced Gr¨obner basis with respect to

Graded lex ordering is

G={x−y, y2−1}.

Gr¨obner bases can be computed using mathematical softwares like

Singular ( http://www.singular.uni-kl.de), CoCoA (http://cocoa.dima.unige.it),

and Macaulay2( http://www.math.uiuc.edu/Macaulay2). Here, we demon-

strate how to compute Gr¨obner bases using Singular.

Example 1.3.6. We use Singular to compute the reduced Gr¨obner

basis Gof the ideal (xy −1, y2−1) with respect to the graded lex

ordering. The command to compute Gr¨obner basis of an ideal Iis

std(I). We get G={x−y, y2−1}. A sample input output session of

Singular to compute a Gr¨obner basis is given below.

> ring r = 0, (x,y), Dp;

> ideal I = xy-1, y^2-1;

> std(I);

_[1]=x-y

_[2]=y2-1

> exit;

Auf Wiedersehen.

Lemma 1.3.1. Let rbe the remainder we get when we divide fby a

Gr¨obner basis Gof the ideal I=< F >. Then, ris also a remainder

when fis divided by F.

Proof. The S-polynomials are at ﬁrst monomial combinations of

polynomials in F. Later, in the Buchberger’s algorithm, S-polynomials

include polynomials from G. But gi∈Gare either S-polynomials

or remainders when S-polynomials are divided by polynomials in G.

Therefore, from the expression f=gi∈Gaigi+rwe get from dividing

fby G, we are always able to write f=fi∈Fqifi+rsuch that qiare

polynomials. And rremains the same.

Now we have all the tools to perform polynomial divisions by a set.

We demonstrate the process with an example. The Gr¨obner basis used

in the process is not required to be reduced, in general.

Example 1.3.7. Going back to Example 1.2.5, we divide f=xy2−

y3+x2−1 by F.

21

From Example 1.3.5, we know that the Gr¨obner basis with respect

to the glex ordering of the ideal I=< F > is G={x−y, y2−1}.

By Lemma 1.3.1, the remainder we get by dividing fby Gis also a

remainder when fis divided by F.

We now show that the order of division do not matter when fis

divided by G.

When we divide fby x−yﬁrst and then by y2−1, we get the

remainder r= 0 as described below.

q1 : y2+x+y

q2 : 1

x−y

y2−1xy2−y3+x2−1

xy2−y3

x2−1

x2−xy

xy −1

xy −y2

y2−1

0

We now change the order of division, that is, we divide fby g2

ﬁrst and then g1to demonstrate that the remainder remains the same.

When we divide a polynomial with a set of polynomials, just like in the

case of dividing a polynomial with a single polynomial, the remainder

has to be a polynomial such that none of its terms are divisible by any

polynomial in the set. For example after dividing fby g2and g1once,

we get a remainder y2−1. We need to divide y2−1 again with g2to

get the actual remainder 0. The details are given below. Observe that

the quotients, unlike the remainder, depend on the order of division.

22

q1 : x+y+ 1

q2 : x−y+ 1

y2−1

x−yxy2−y3+x2−1

xy2−x

−y3+x2+x−1

−y3+y

x2+x−y−1

x2−xy

xy +x−y−1

xy −y2

y2+x−y−1

x−y

y2−1

0

Consequently, we get

f=xy2−y3+x2−1 = (y2+x+y)f3+f2.(1.2)

We also know from Equation 1.1 that f3=S(f1, f2) = yf1−xf2.

Therefore,

f=q1f1+q2f2+ 0 where

q1=y(y2+x+y) and

q2=−x(y2−x−y) + 1.(1.3)

A zero remainder implies that the solutions of Fare roots of f.

It is easy to check that f, indeed, vanishes at the two solutions of F,

namely, (1,1) and (−1,−1).

We leave it as an exercise to prove that f∈Iif and only if the

remainder we get when fis divided by Gis zero.

In conclusion, the strategy we follow to divide a polynomial fby a

set of polynomials Fto get a unique remainder is as follows:

23

1. Compute Gr¨obner basis G={g1, . . . , gt}of the ideal I=< F >.

2. Divide fby Gto get a unique remainder r. Note that none of

the terms of rare divisible by any polynomial in G.

3. Trace the quotients qi,i= 1 to nfrom the S-polynomials to write

f=q1f1+· ·· +qnfn+r.

In this chapter, we saw that replacing a set of polynomials with

a Gr¨obner basis gave us a unique remainder. We will see some more

applications of Gr¨obner bases in later chapters.

Exercises.

1. Prove that the set of all n×nmatrices with the usual opera-

tions of matrix multiplication and addition over real numbers is

a noncommutative ring with identity.

2. Prove that the set Tof all continuous functions from Rto Ris a

ring with identity where addition and multiplication is deﬁned as

follows. Let f, g ∈T, the

(f+g)(x) = f(x) + g(x) and (fg)(x) = f(x)g(x).

3. Let Rand Sbe rings. Deﬁne addition and multiplication on the

Cartesian product R×Sby

(r, s)+(r′, s′) = (r+r′, s +s′)

(r, s)·(r′, s′) = (r·r′, s ·s′).

Prove that R×Sis a ring. Also prove that if Rand Sare

commutative, then so is R×S, and that if Rand Seach have an

identity, then so does R×S.

4. Let Rbe a ring. Prove that for any element a, b, c ∈R

(a) the equation a+x= 0Rhas a unique solution;

(b) a+b=a+cimplies b=c;

(c) a·0R= 0R= 0R·a;

(d) (−a)·(−b) = a·b;

24

(e) −(−a) = a.

5. Prove that the only ideals of a ﬁeld Rare <0R>and R.

6. Prove that every ideal in Zis principal (Hint: show that I=<

c >, where cis the smallest integer in I).

7. If kis a ﬁeld, show that k[x] is a principal ideal domain.

8. Prove that the ideals < xy −1, y2−1>and < x −y, y2−1>

are the same. (Hint: Prove that both the ideals have the same

minimal Gr¨obner basis).

9. Let Ibe an ideal, prove that f∈Iif and only if the remainder

we get when fis divided by a Gr¨obner basis of Iis zero.

10. Use the principle of induction to prove the division algorithm

(Theorem 1.2.1).

11. Show that the remainder is zero when the polynomial x2y−xy2−

y2+ 1 is divided by the set {xy −1, y2−1}.

12. Compute the Gr¨obner basis of the ideal < x −z4, y −z5>with

respect to the lex and graded lex orderings.

13. Write a computer program to ﬁnd the Gr¨obner basis of an ideal

w.r.t the lex ordering.

25

26

Chapter 2

Solving Systems of

Polynomial Equations.

The greatest challenge to any thinker is stating the problem in a way

that will allow a solution – Bertrand Russell.

In this chapter, we look at solutions to systems of polynomial equa-

tions. Systems of polynomials are solved by eliminating variables. In

Linear Algebra, where all the polynomials involved are of degree one,

eliminating variables involved matrix operations. For systems of higher

order polynomials we use Gr¨obner bases to do the same.

2.1 Ideals and Varieties.

Let kbe a ﬁeld, and let f1, . . . , fsbe polynomials in k[x1, . . . , xn].

In this section, we will consider two fundamental questions about the

system of equations deﬁned by F={f1, . . . , fs}:

1. Feasibility - When does the system deﬁned by Fhave a solution

in kn?

2. Which are the polynomials that vanish on the solution set of F?

Solution sets of ﬁnite sets of polynomials are commonly known as

varieties:

Deﬁnition 2.1.1. Let kbe a ﬁeld. and let f1, . . . , fsbe polynomials in

k[x1, . . . , xn]. The set

V(f1, . . . , fs) = {(a1, . . . , an)∈kn:fi(a1, . . . , an) = 0 for all 1≤i≤s}

27

is called the aﬃne variety deﬁned by f1, . . ., fs.

Example 2.1.1.

1. V(x2+y2−1) is the circle of radius 1 centered at the origin in C.

2. V(xy −1, y2−1) = {(1,1),(−1,−1)}in C.

3. Observe that a variety depends on the coeﬃcient ﬁeld: let f=

x3y−x2y−x3+x2−2xy + 2y+ 2x−2, then

V(f) = 





{(√2, y),(−√2, y),(x, 1),(1, y)}in R,

{(x, 1),(1, y)}in Q.

Now we look at solutions of all the polynomials in an ideal I.

Deﬁnition 2.1.2. Let I⊂k[x1, . . . , xn]be an ideal. We denote by

V(I)the set

V(I) = {(a1, . . . , an)∈kn:f(a1, . . . , an) = 0 for all f∈I}.

Though Iis usually inﬁnite for inﬁnite ﬁelds, computing V(I) is

equivalent to ﬁnding the roots of a ﬁnite set of polynomials. We prove

this fact next.

Theorem 2.1.1. V(I)is an aﬃne variety. In particular, if I=<

f1, . . . , fs>, then V(I) = V(f1, . . . , fs).

Proof. By Hilbert’s Basis Theorem 1.3.1, I=< f1, . . . , fs>for

some generating set {f1, . . . , fs}. We now show that V(I) = V(f1, . . . , fs).

Let (a1, . . . , an)∈V(I), then since fi∈I,fi(a1, . . . , an) = 0 for all

i= 1 to s. Therefore,

V(I)⊂V(f1, . . . , fs).(2.1)

Now let (a1, . . . , an)∈V(f1, . . . , fs) and let f∈I. Since I=<

f1, . . . , fs>, we can write f=s

i=1 hififor some hi∈k[x1, . . . , xn].

But then

f(a1, . . . , an) =

s



i=1

hi(a1, . . . , an)fi(a1, . . . , an)

=

s



i=1

hi(a1, . . . , an)·0 = 0.

28

Therefore,

V(f1, . . . , fs)⊂V(I).(2.2)

Equations 2.1 and 2.2 prove that V(I) = V(f1, . . . , fs).

Theorem 2.1.1 implies that the solutions of a given set of polynomi-

als Fare the same as the solutions of an ideal Igenerated by F. The

biggest advantage of passing from Fto I=< F >, as we shall see, is

that we can replace Fby a Gr¨obner basis for all practical purposes.

A ﬁeld kis algebraically closed if every non-constant polynomial

in k[x] has a root in k. For example, Ris not algebraically closed

because x2+ 1 has no roots in R. on the other hand, Cis an alge-

braically closed ﬁeld because of the fundamental theorem of algebra

(every non-constant polynomial in C[x] has a root in C). The next

theorem answers the feasibility question for algebraically closed ﬁelds.

Theorem 2.1.2 (The Weak Nullstellensatz).Let kbe an algebraically

closed ﬁeld and let I⊂k[x1, . . . , xn]be an ideal such that V(I)is empty,

then I=k[x1, . . . , xn].

The proof of this Theorem is beyond the scope of this book and we

refer the reader to [17] for a proof. The Weak Nullstellensatz implies

that every proper ideal has a solution in an algebraically closed ﬁeld.

If the ﬁeld is not algebraically closed, the Weak Nullstellensatz holds

one way, that is, if I=k[x1, . . . , xn], then V(I) is empty. The next

lemma is useful while checking whether I=k[x1, . . . , xn].

Lemma 2.1.1. Let kbe a ﬁeld, then I=k[x1, . . . , xn]if and only if

1∈I.

Proof. If I=k[x1, . . . , xn] then 1 ∈I. This is because k⊂

k[x1, . . . , xn] and 1 ∈kbecause kis a ﬁeld.

Conversely, if 1 ∈I, then a·1∈Ifor every a∈k[x1, . . . , xn] by def-

inition of an ideal. Therefore, k[x1, . . . , xn]⊂I. But I⊂k[x1, . . . , xn].

Thus, I=k[x1, . . . , xn].

Consequently, if we want to check whether a given system of polyno-

mials F={f1, . . . , fs}has a solution, we compute the reduced Gr¨obner

basis Gof the ideal I= (f1, . . . , fs). If G={1}we conclude that Fhas

29

no solution. We leave it as an exercise to prove that if I=k[x1, . . . , xn]

then the reduced Gr¨obner basis of Iis {1}(Exercise 3).

In Section 1.2, we talked about how being able to write a polyno-

mial fas f=s

i=1 qifi(that is, remainder is zero when fis divided

by {f1, . . . , fs}) meant that the fvanished on the solution set of the

system of equations fi= 0, i = 1..s. This is because f=s

i=1 qifi

implies that fbelongs to the ideal I=< f1, . . . , fs>. Moreover, by

Theorem 2.1.1, V(I) = V(f1, . . . , fs). Consequently, f∈Ithen fvan-

ishes on V(f1, . . . , fs). Are these the only polynomials that vanish on

V(f1, . . . , fs)? Now, we explore this question.

The next lemma proves that the set of all polynomials that vanish

on a given variety V, denoted by I(V), is an ideal.

Lemma 2.1.2. Let V⊂knbe an aﬃne variety, and let

I(V) = {f∈k[x1, . . . , xn] : f(a1, . . . , an) = 0 for all (a1, . . . , an)∈V},

then I(V) is an ideal of R=k[x1, . . . , xn].

Proof. We use Proposition 1.3.1 to prove I(V) is an ideal. Let

f, g ∈I(V) and let (a1, . . . , an)∈V. Then

f(a1, . . . , an)−g(a1, . . . , an) = 0 −0 = 0.(2.3)

Therefore f−g∈I(V). For every h∈Rand f∈I(V),

h(a1, . . . , an)f(a1, . . . , an) = h(a1, . . . , an)·0 = 0.(2.4)

This implies that hf ∈I(V). Properties 2.3 and 2.4 implies I(V)

is an ideal.

From the discussion above Lemma 2.1.2, we know that I⊂I(V(I)).

Is I(V(I)) = I? The answer in general is no. It is usually a bigger

ideal that contains I. We now compute I(V(I)) for algebraically closed

ﬁelds.

Theorem 2.1.3 (Hilbert’s Nullstellensatz).Let kbe an algebraically

closed ﬁeld, and let f, f1, . . . , fs∈k[x1, . . . , xn]. Then f∈I(V(f1, . . . , fs))

if and only if there exists an integer m≥1such that

fm∈< f1, . . . fs> .

30

Proof. If fm∈< f1, . . . fs>, then fm=s

i=1 Aififor some

Ai∈k[x1, . . . , xn]. Consequently, fvanishes at every common zero

of polynomials f1, . . . , fsbecause fmvanishes at these zeroes. There-

fore f∈I(V(f1, . . . , fs)). Conversely, assume that fvanishes at every

common zero of the polynomials f1, . . . , fs. We must show that there

exists an integer m≥1 and polynomials Ai, . . . , Assuch that

fm=

s



i=1

Aifi.(2.5)

To do this we introduce a new variable yand then consider the ideal

˜

I=< f1, . . . fs,1−fy >∈k[x1, . . . , xn, y].

We claim that V(˜

I) is empty. To see this let (a1, . . . , an, an+1)∈

kn+1. There are only two possibilities. Either

1. (a1, . . . , an) is a common zero of f1, . . . , fsor

2. (a1, . . . , an) is not a common zero of f1, . . . , fs

In the ﬁrst case, f(a1, . . . , an) = 0 by our assumption that fvan-

ishes at every common zero of f1, . . . , fs. Therefore, the polynomial

1−yf takes the value 1 −an+1f(a1, . . . , an)=1̸= 0. This implies

(a1, . . . , an, an+1)̸∈ V(˜

I).

In the second case, for some t, 1 ≤t≤s,ft(a1, . . . , an)̸= 0.

We treat ftas a function of n+ 1 variables that does not depend on

the last variable to conclude that ft(a1, . . . , an, an+1)̸= 0. Therefore,

(a1, . . . , an, an+1)̸∈ V(˜

I). Since (a1, . . . , an, an+1) was arbitrary, we

conclude that V(˜

I) is empty. This implies, by the Weak Nullstellensatz,

that 1 ∈˜

I. Therefore, for some polynomials pi, q ∈k[x1, . . . , xn, y],

1 =

s



i=1

pi(x1, . . . , xn, y)fi+q(x1, . . . , xn, y)(1 −yf).(2.6)

Now let 1 −yf = 0, that is y= 1/f(x1, . . . , xn). Then Equation 2.6

implies that

1 =

s



i=1

pi(x1, . . . , xn,1/f)fi.

31

Multiply both sides of the equation by fmwhere mis chosen large

enough to clear denominators to get Equation 2.5, thereby proving the

theorem.

The Hilbert’s Nullstellensatz motivates the next deﬁnition.

Deﬁnition 2.1.3. Let I⊂k[x1, . . . , xn]be an ideal. The radical of I

denoted √Iis the set

{f:fm∈Ifor some integer m≥1}.

Theorem 2.1.4 (The Strong Nullstellensatz).Let kbe an algebraically

closed ﬁeld. If Iis an ideal in k[x1, . . . , xn], then

I(V(I)) = √I.

Proof. f∈√Iimplies that fm∈Ifor some m. Hence fmvanishes

on V(I), which implies fvanishes on V(I). Consequently, f∈I(V(I)).

Therefore √I⊂I(V(I)) (2.7)

Conversely, suppose that f∈I(V(I)). Then, by deﬁnition, fvan-

ishes on V(I). By Hilbert’s Nullstellenatz, there exists an integer m≥1

such that fm∈I. But this implies that f∈√I. Thus, we prove

I(V(I)) ⊂√I(2.8)

Equations 2.7 and 2.8 imply

I(V(I)) = √I.

Exercise 2 shows that √Iis an ideal in k[x1, . . . , xn] containing I.

We do not discuss algorithms to compute radical ideals in this text. It

is a diﬃcult problem nevertheless. We now illustrate how to compute

radical ideals using the Software Singular.

Example 2.1.2.

We compute (J), where J=< xy −1, y2−1>. An input-output

Singular session for doing this is given below. For this computation we

load a Singular library called primdec.lib.

32

> LIB "primdec.lib";

> ring r = 0, (x,y), Dp;

> ideal J = x*y -1, y^2-1;

> radical(J);

_[1]=y2-1

_[2]=xy-1

_[3]=x2-1

> exit;

Auf Wiedersehen.

In the next examples we compare Jand √J.

Example 2.1.3. 1. In Example 2.1.2, we saw that when

J=< xy −1, y2−1>, √J=< x2−1, y2−1, xy −1> .

The reduced Gr¨obner basis of √Jw.r.t the graded lex ordering

is {x−y, y2−1}. And we know from Example 1.3.6 that the

Gr¨obner basis of Jis also {x−y, y2−1}. Therefore, √J=J.

So, in this example, I(V(J)) = J.

2. Let J=< x2, y2>, then the variety V(J) = {(0,0)}. We compute

I(V(J)) = √J=< x, y >. Note that < x, y > is strictly larger

than J, for instance, x̸∈< x2, y2>. Hence, J⊂√J.

2.2 Elimination Theory.

As we know, solving systems of polynomial equations involves elimi-

nating variables. We begin by eliminating all the polynomials involving

variables x1, . . . , xlfrom the ideal I.

Deﬁnition 2.2.1. Given I= (f1, . . . , fs)⊂k[x1, . . . , xn], the lth

elimination ideal Ilis the ideal of k[xl+1, . . . , xn]deﬁned by

Il=I∩k[xl+1, . . . , xn].

We check that Ilis an ideal of k[xl+1, . . . , xn] in Exercise 4. Note

that I=I0is the 0 th elimination ideal.

For a ﬁxed integer lsuch that 1 ≤l≤n, we say a monomial

order >on k[x1, . . . , xn] is of l- elimination type provided that any

33

monomial involving one of x1, . . . , xlis greater than all other monomials

in k[xl+1, . . . , xn]. For example, the lex monomial ordering, where x1>

x2··· > xn, is a l- elimination type ordering. In the next theorem we

extract a Gr¨obner basis for the lth elimination ideal Ilfrom a Gr¨obner

basis of I.

Theorem 2.2.1 (The Elimination Theorem).Let I⊂k[x1, . . . , xn]be

an ideal and let Gbe a Gr¨obner basis of Iwith respect to a l- elimination

type monomial ordering. Then, for every 0≤l≤n, the set

Gl=G∩k[xl+1, . . . , xn]

is a Gr¨obner basis of the lth elimination ideal Il.

Proof. Since Gl⊂Ilby construction, to show that Glis a Gr¨obner

basis, it suﬃces to prove that

<LT(Il)>=<LT(Gl)> .

It is obvious that <LT(Gl)>⊂<LT(Il)>. To prove the other

inclusion <LT(Il)>⊂<LT(Gl)>, we show that if f∈Il, then

LT(f) is divisible by LT(g) for some g∈Gl. Since f∈I, and Gis a

Gr¨obner basis of I, LT(f) is divisible by some g∈G. But f∈Ilmeans

that LT(g) only involves variables xl+1, . . . , xn. Consequently, since the

monomial ordering is of l-elimination type, g∈k[xl+1, . . . , xn].

In section 2.1, we saw that the solutions of a set of polynomials

Fare the same as the solutions of an ideal Igenerated by F. The

advantage of passing from a set to an ideal is that we can replace F

by any set of generators of I, to get the solution set of F. In the

next example, we demonstrate how to solve a system of polynomial

equations using l-elimination ideals.

Example 2.2.1. In this example, we solve the system of equations

x2+y+z= 1,

x+y2+z= 1,

x+y+z2= 1,

x2+y2+z2= 1.

Let

F={x2+y+z−1, x +y2+z−1, x +y+z2−1, x2+y2+z2−1},

34

and let Ibe the ideal generated by F, that is,

I=< x2+y+z−1, x +y2+z−1, x +y+z2−1, x2+y2+z2−1> .

The reduced Gr¨obner basis Gof Iwith respect to the lex ordering

x > y > z is

G={z2−z, 2yz +z4+z2−2z, y2−y−z2+z, x +y+z2−1}.

By Theorem 2.2.1 the Gr¨obner basis of elimination ideals I1and I2

are

G1=G∩k[y, z] = (z2−z, 2yz +z4+z2−2z, y2−y−z2+z),

and

G2=G∩k[z] = (z2−z),

respectively.

The Gr¨obner basis of I2involves only the variable z. By Exercise

7, k[z] is a principal ideal domain. Therefore I2is generated by one

element.

We now perform a backward substitution to solve the given system

of equations deﬁned by G2. Solving z2−z= 0, we get z= 0 or z= 1.

Next we solve the equations deﬁned by the polynomials in the set

G2−G1, that is,

2yz +z4+z2−2z= 0

y2−y−z2+z= 0.

When z= 0, the above equations imply y= 0 or y= 1, on the other

hand, when z= 1, we get y= 0.

Finally, we solve the system of equations deﬁned by G−G1, namely,

x+y+z2−1 = 0.(2.9)

Consequently, when we substitute y= 0, z = 0 in Equation 2.9, we

get x= 1; when we substitute y= 1, z = 0 in Equation 2.9, we get

x= 0; and when we substitute y= 0, z = 1 in Equation 2.9, we get

x= 0.

Observe that the process leads us to the solutions of G. Recall that

V(G) = V(I) = V(F). Therefore, the solution set of the given system

of equations is {(1,0,0),(0,1,0),(0,0,1)}.

35

Can we always extend a partial solution to the complete one? Not

always, but the next theorem tells us when such an extension is possible

for the ﬁeld of complex numbers.

Theorem 2.2.2 (The Extension Theorem).Let I=< f1, . . . , fs>⊂

C[x1, . . . , xn]and let I1be the ﬁrst elimination ideal of I. For each

1≤i≤s, write fiin the form

fi=gi(x2, . . . , xn)xNi

1+terms in which x1has degree < Ni,

where Ni≥0and gi∈C[x2, . . . , xn]is nonzero. Suppose that

we have a partial solution (a2, . . . , an)∈V(I1). If (a2, . . . , an)̸∈

V(g1, . . . , gs), then there exists a1∈Csuch that (a1, . . . , an)∈V(I).

We will prove this theorem in Section 2.3. We illustrate this theorem

with an example.

Example 2.2.2.

In the case of the ideal

I=f1=x2+y+z−1,

f2=x+y2+z−1,

f3=x+y+z2−1,

f4=x2+y2+z2−1,

the coeﬃcients giof the highest powers of xin all the polynomials fi

are 1. By the Weak Nullstellensatz Theorem, V(g1, g2, g3, g4) = V(1)

is empty. Consequently, by Theorem 2.2.2, all the partial solutions can

be extended to a complete solution.

We look at another example where such an extension is not possible.

Example 2.2.3. Consider the ideal

I=< f1=xy −1, f2=xz −1>⊂k[x, y, z].

The reduced Gr¨obner basis Gof Iwith respect to the graded lex or-

dering is G={y−z, xz −1}. Thus G1={y−z}. A partial solution is

y=z= 0. But, observe that coeﬃcients of xof the polynomials f1, f2

simultaneously vanish at y=z= 0, that is, (0,0) ∈V(y, z). Therefore,

by the extension theorem this partial solution cannot be extended to

a complete solution of the system of equations F={f1= 0, f2= 0}.

On the other hand, every partial solution (c, c) such that c̸= 0 can be

extended to a complete solution (1/c, c, c) of F.

36

Apart from solving systems of equations, elimination ideals are also

used to ﬁnd implicit equations of a surface from its parametrization.

We present, without proof, a theorem that describes the method to do

this. The proof of this theorem is given in [17] and requires concepts

not discussed in this book.

Theorem 2.2.3 (Implicitization).1. Let kbe an inﬁnite ﬁeld. Let

f1, . . . , fn∈k[t1, . . . , tm]and let

x1=f1(t1, . . . , tm)

.

xn=fn(t1, . . . , tm)

be a polynomial parametrization. Let Ibe the ideal

I=< x1−f1, . . . , xn−fn>⊂k[t1, . . . , tm, x1, . . . , xn]

and let Im=I∩k[x1, . . . , xn]be the mth elimination ideal. Then

V(Im)is the smallest variety in kncontaining the parametriza-

tion.

2. Let

x1=f1(t1, . . . , tm)

g1(t1, . . . , tm)

.

xn=fn(t1, . . . , tm)

gn(t1, . . . , tm)

be a rational parametrization, where f1, . . . , fn, g1, . . . , gnare in

k[t1, . . . , tm]. Let Ibe the ideal

< g1x1−f1,˙,gnxn−fn,1−g1g2··· gnY >⊂k[Y, t1, . . . , tm, x1, . . . , xn]

and let Im+1 =I∩k[x1, . . . , xn]be the (m+ 1) elimination ideal.

Then, V(Im+1)is the smallest variety containing this parametriza-

tion.

Example 2.2.4. In this example, we show that the surface deﬁned by

the following parametric equations

37

x=1−t2

1 + t2,

y=2t

1 + t2.(2.10)

lie on the circle

x2+y2= 1.

Let

I=<(1 + t2)x−(1 −t2),(1 + t2)y−2t, 1−(1 + t2)2Y > .

Then, the Gr¨obner basis Gof Iw.r.t the Lex ordering t > Y > x >

yis

G={x2+y2−1,4Y−2x+y2−2, ty +x−1, tx +t−y}

The Gr¨obner basis of I2is {x2+y2−1}which is also the equation

of the circle. Therefore, Theorem 2.2.3 implies V(x2+y2−1) is the

smallest variety containing the Parametrization 2.10. Observe that the

above Parametrization do not describe the whole circle because the

point (−1,0) on the circle is not covered by this parametrization.

Example 2.2.5. In this example, we show that the surface deﬁned by

the following polynomial parametrization

x=t1t2,

y=t1t2

2,

z=t2

1.(2.11)

lie on surface x4−y2z.

The Gr¨obner basis Gof the ideal I=< x −t1t2, y −t1t2

2, z −t2

1>

with respect to the lex ordering t1> t2> x > y is

G={x4−y2z, t2yz−x3, t2x−y, t2

2z−x2, t1y−t2

2z, t1x−t2z, t1t2−x, t12−z}.

This implies I2=< x4−y2z >. Therefore, by Theorem 2.2.3, the

smallest variety containing the Parametrization 2.11 is x4−y2z.

38

2.3 Resultants.

In this section, we introduce resultants which are used to determine

whether two polynomials have a common factor without having to fac-

torize the polynomials involved. We also use resultants to prove the

Extension Theorem from Section 2.2.

We begin with a lemma that discusses a key property of two poly-

nomials that have a common factor.

Lemma 2.3.1. Let f, g ∈k[x1, . . . , xn]be of degrees l > 0and m > 0,

respectively, in x1. Then fand ghave a common factor with positive

degree in x1if and only if there are polynomials A, B ∈k[x1, x2, . . . , xn]

such that

1. Aand Bare not both zero.

2. Ahas degree at most m−1and Bhas degree at most l−1in x1.

3. Af + Bg = 0.

Proof. First assume fand ghave a common factor h∈k[x1, . . . , xn]

with positive degree in x1. Then f=hf1and g=hg1, where f1, g1∈

k[x1, . . . , xn]. Note that f1has degree at most l−1 in x1and g1has

degree at most m−1 in x1. Then

g1·f+ (−f1)·g=g1·hf1−f1·hg1= 0.

Thus A=g1and B=−f1have the required properties.

Conversely, suppose that Aand Bhave the above three properties.

By Property 1, we may assume B̸= 0. Let

k(x2, . . . , xn) = {f

g;f, g ∈k[x2, . . . , xn], g ̸= 0}.

Check that k(x2, . . . , xn) is a ﬁeld. If fand ghave no common

factor of positive degree in x1, in k(x2, . . . , xn)[x1], then we use the

Euclidean Algorithm (see Section A.1) to ﬁnd polynomials A′, B′∈

k(x2, . . . , xn)[x1] such that A′f+B′g= 1 . Now multiply by Band

use Bg =−Af to get

B= (A′f+B′g)B=A′Bf +B′Bg =A′Bf −B′Af = (A′B−B′A)f.

39

Since Bis nonzero and the degree of fis l, this equation shows that

Bhas degree at least lin x1, which contradicts Property 2. Hence there

must be a common factor of fand gin k(x2, . . . , xn)[x1]. By Exercise

7, fand ghave a common factor in k[x1, . . . , xn] of positive degree in

x1, if and only if, they have a common factor in k(x2, . . . , xn)[x1] of

positive degree in x1. This proves the theorem.

To show that Aand Bin Lemma 2.3.1 actually exist, we write f

and gas polynomials in x1with coeﬃcients ai, bi∈k[x2, . . . , xn]:

f=a0x1l+···+al, a0̸= 0,

g=b0x1m+· ·· +bm, b0̸= 0.(2.12)

Our goal is to ﬁnd coeﬃcients ci, di∈k[x2, . . . , xn] such that

A=c0x1m−1+· ·· +cm−1,

B=d0x1l−1+···+dl−1,(2.13)

and

Af +Bg = 0.(2.14)

Consequently, comparing coeﬃcients of x1in Equation 2.14, we get

the following system of equations

aoc0+b0d0= 0 (coeﬃcient of x1l+m−1)

a1c0+a0c1+b1d0+b0d1= 0 (coeﬃcient of xl+m−2

1)

.

alcm−1+bmdl−1= 0 (coeﬃcient of x0

1) (2.15)

Since there are l+mlinear equations and l+munknowns, there

is a nonzero solution if and only if the coeﬃcient matrix has a zero

determinant. This leads to the following deﬁnition.

Deﬁnition 2.3.1. Given polynomials f, g ∈k[x1, . . . , xn]of positive

degree in x1, write them in the form 2.12. Then the Sylvester matrix

of fand gwith respect to x1denoted Syl(f, g, x1)is the coeﬃcient

40

matrix of the system of equations given in 2.15. Thus, Syl(f, g, x1)is

the following (l+m)×(l+m)matrix:

Syl(f, g, x1) =







a0b0

a1a0b1b0

a1...b1...

.

....a0

.

....b0

.

.a1

.

.b1

albm

al

.

.bm

.

......

albm







,

where the ﬁrst mcolumns contain the coeﬃcients of f, such that the

ﬁrst i−1entries of the ith column are zeroes, 1≤i≤m; the last l

columns contain the coeﬃcients of g, such that the ﬁrst j−1entries

of the m+jth column are zeroes, 1≤j≤l; and the empty spaces are

ﬁlled by zeros.

The resultant of fand gwith respect to x1denoted Res(f, g, x1)

is the determinant of the Sylvester matrix. Thus,

Res(f, g, x1) = det(Syl(f, g, x1)).

The resultant is deﬁned in such a way that its vanishing detects

the presence of common factors. We prove this fact in the following

theorem.

Theorem 2.3.1. Let f, g ∈k[x1, . . . , xn]have positive degree in x1,

then Res(f, g, x1) = 0 if and only if fand ghave a common factor in

k[x1, . . . , xn]which has positive degree in x1.

Proof. The resultant is zero means that the determinant of the

coeﬃcient matrix of Equations 2.15 is zero. This happen if and only if

there exists a nonzero solution to the system of equations 2.15. This

is equivalent to existence of polynomials Aand Bsuch that Aand B

are not both zero, degree of Ais less than degree of fand the degree

of Bis less than the degree of g, in x1, and Af +Bg = 0. By Lemma

2.3.1, this happens if and only if fand ghave a common factor in

k[x1, . . . , xn] which has positive degree in x1.

41

Example 2.3.1. Consider the polynomials

f=x2y+x2−3xy2−3xy and g=x3y+x3−4y2−3y+ 1.

To compute Res(f, g, x), write fand gas

f= (y+ 1)x2+ (−3y2−3y)x,

g= (y+ 1)x3+ (−4y2−3y+ 1).

Res(f, g, x) = det





y+ 1 0 0 y+ 1 0

−3y2−3y y + 1 0 0 y+ 1

0−3y2−3y y + 1 0 0

0 0 −3y2−3y−4y2−3y+ 1 0

0 0 0 0 −4y2−3y+ 1







=−108y9−513y8−929y7−738y6−149y5+ 112y4+ 37y3

−14y2−3y+ 1 ̸= 0.

Res(f, g, x)̸= 0 implies that fand ghave no common factor with

positive degree in x, by Theorem 2.3.1.

To compute Res(f, g, y), write fand gas

f= (−3x)y2+ (x2−3x)y+x2,

g=−4y2+ (x3−3)y+ (x3+ 1).

Res(f, g, y) = det





−3x0−4 0

x2−3x−3x x3−3−4

x2x2−3x x3+ 1 x3−3

0x20x3+ 1





= 0.

Res(f, g, y) = 0 implies that fand ghave a common factor with

positive degree in y, by Theorem 2.3.1. To verify this, we factorize f

and gto get f=x(y+ 1)(−3y+x) and g= (y+ 1)(−4y+1+x3). We

see that (y+ 1) is indeed a common factor of fand gwith a positive

degree in y.

Resultants can be computed using the Software Singular. A sample

input-output session is provided below.

42

> ring r = 0, (x,y), dp;

> poly f = x^2*y-3*x*y^2+x^2-3*x*y;

> poly g = x^3*y+x^3-4*y^2-3*y+1;

> resultant(f,g,x);

-108y9-513y8-929y7-738y6-149y5+112y4+37y3-14y2-3y+1

> resultant(f,g,y);

0

>quit;

Auf Wiedersehen.

In the case of polynomials fand gwith only one variable x, the

resultant Res(f, g, x) is usually denoted as Res(f , g).

Example 2.3.2. Let

f=x2+xand g=x2+ 4x+ 4.

Res(f, g) = det 





1010

1141

0144

0004







= 4 ̸= 0.

Therefore the polynomials fand gare relatively prime.

Lemma 2.3.2. Let f, g ∈k[x1, . . . , xn]be of positive degree in x1with

coeﬃcients ai, bi∈k[x2, . . . , xn], then Res(f, g, x1)∈k[x2, . . . , xn].

Proof. Since Res(f, g, x1) is a determinant involving only aiand bi,

it follows that Res(f, g, x1)∈k[x2, . . . , xn].

Lemma 2.3.3. Let f, g ∈k[x1, . . . , xn]of positive degree in x1with

coeﬃcients ai, bi∈k[x2, . . . , xn]. Then

Af +Bg =Res(f , g, x1),

where Aand Bare polynomials in x1whose coeﬃcients are integer

polynomials in aiand bi.

Proof. The lemma is true when Res(f, g, x1) = 0, because we can

choose A=B= 0. Assume that Res(f, g, x1)̸= 0. Write fand gin

the form of Equations 2.12. Let

A′=c0x1m−1+· ·· +cm−1,

B′=d0x1l−1+· ·· +dl−1,(2.16)

43

where the coeﬃcients ci, di∈k[x2, . . . , xn], such that

A′f+B′g= 1.

Comparing coeﬃcients we get

aoc0+b0d0= 0 (coeﬃcient of x1l+m−1)

a1c0+a0c1+b1d0+b0d1= 0 (coeﬃcient of xl+m−2

1)

.

alcm−1+bmdl−1= 1 (coeﬃcient of x0

1) (2.17)

These equations are the same as 2.15 except for the 1 on the right

hand side of the last equation. Thus, the coeﬃcient matrix is the

Sylvester matrix of fand g. Therefore, Res(f, g, x1)̸= 0 guarantees

that the System 2.17 has a unique solution. We use Cramer’s rule to

ﬁnd this unique solution. Recall that the Cramer’s rule states that the

i-th unknown is a ratio of two determinants, where the denominator

is the determinant of the coeﬃcient matrix and the numerator is the

determinant of the matrix where the i-th column of the coeﬃcient

matrix has been replaced by the right hand side vector of the system.

For example, the ﬁrst unknown c0is given by

c0=1

Res(f, g, x1)det







0b0

0a0b1b0

a1...b1...

.

....a0

.

....b0

.

.a1

.

.b1

0albm

.

.....

.

.....

.

1albm







.

Since a determinant is an integer polynomial in its entries, it follows

that

c0=an integer polynomial in ai, bi

Res(f, g, x1).

Similarly, we conclude that the denominator for ckand dkfor ev-

ery kis always Res(f, g, x1) and the numerator is always an integer

polynomial in aiand bi.

44

Since A′=c0x1m−1+··· +cm−1, we can pull out the common

denominator Res(f, g, x1) and write

A′=1

Res(f, g, x1)A,

where A∈k[x1, . . . , xn], and the coeﬃcients of Aare integer poly-

nomials in ai, bi. Similarly, we can write

B′=1

Res(f, g, x1)B,

where B∈k[x1, . . . , xn], and the coeﬃcients of Bare integer polyno-

mials in ai, bi.

Since A′f+B′g= 1, we can multiply through by Res(f, g, x1) to

obtain

Af +Bg = Res(f, g, x1).

Theorem 2.3.2. Let f, g ∈k[x1, . . . , xn]have positive degree in x1,

then Res(f, g, x1)is in the ﬁrst elimination ideal < f, g > ∩k[x2, . . . , xn].

Proof. By Lemma 2.3.3,

Af +Bg =Res(f , g, x1),

where A, B ∈k[x1, . . . , xn]. Hence Res(f, g, x1)∈< f , g >. Apply-

ing Lemma 2.3.2, we get Res(f, g, x1)∈k[x2, . . . , xn]. Consequently,

Res(f, g, x1)∈< f, g > ∩k[x2, . . . , xn].

Over the complex numbers, two polynomials in C[x] have a common

factor if and only if fand ghave a common root by Theorems A.2.2

and A.2.8. Thus, we get the following corollary.

Corollary 2.3.3. If f, g ∈C[x], then Res(f, g, x) = 0 if and only if f

and ghave a common root in C.

To prove the Extension Theorem, we ﬁrst need to prove it for the

case of two polynomials, and then extend the result to the general case.

We begin by proving the following theorem which is used in the proof

of the Extension Theorem for two polynomials.

45

Theorem 2.3.4. Given f, g ∈C[x1, . . . , xn], write fand gin the form

of Equations 2.12, so that ai, bi∈C[x2, . . . , xn]. If Res(f, g, x1)van-

ishes at (c2, . . . cn)∈Cn−1, then either a0or b0vanishes at (c2, . . . , cn),

or there is a c1∈Csuch that fand gvanish at (c1, c2, . . . cn)∈Cn.

Proof. Let c= (c2, . . . , cn) and let f(x1,c) = f(x1, c2, . . . , cn). It

suﬃces to show that f(x1,c) and g(x1,c) have a common root when

a0(c) and b0(c) are both nonzero. To prove this, write

f(x1,c) = ao(c)xl

1+· ·· +al(c), ao(c)̸= 0,

g(x1,c) = bo(c)xm

1+· ·· +bm(c), bo(c)̸= 0.

By hypothesis h= Res(f, g, x1) vanishes at c. Therefore

0 = h(c) = Res(f(x1,c), g(x1,c), x1).

Then Corollary 2.3.3 implies that f(x1,c) and g(x1,c) have a com-

mon root.

Theorem 2.3.5. [The Extension Theorem for two polynomials.] Let

I=< f , g >⊂C[x1, . . . , xn]and let I1be the ﬁrst elimination ideal

of I. Write fand gin the form of Equations 2.12, so that ai, bi∈

C[x2, . . . , xn]. Suppose we have a partial solution c= (c2, . . . , cn)∈

V(I1), and if (c2, . . . , cn)̸∈ V(a0, b0), then there exists c1∈Csuch that

(c1, . . . , cn)∈V(I).

Proof. By Theorem 2.3.2, we know that Res(f, g, x1)∈I1, so that

the resultant vanishes at the partial solution c. If neither a0nor b0

vanishes at c, then the required c1exists by Theorem 2.3.4.

Now suppose a0(c)̸= 0 but b0(c) = 0. Since xN

1f∈< f, g +xN

1f >

and g=g+xN

1f−xN

1f, we conclude that g∈< f, g+xN

1f >. Therefore

< f, g >⊂< f, g +xN

1f >. Clearly < f, g +xN

1f >⊂< f, g >. Hence

< f, g >=< f, g +xN

1f > . (2.18)

We choose Nlarge enough so that x1Nfhas larger degree in x1than g.

The leading coeﬃcient of g+x1Nfis a0, which is nonzero at c. This

allows us to use Theorem 2.3.4 to conclude that there is a c1∈Csuch

that (c1,c)∈V(f, g +xN

1f), and hence (c1,c)∈V(f, g) by 2.18.

Let f1, . . . , fs∈C[x1, . . . , xn], then the resultant for f1, . . . , fs,

s≥3 is deﬁned by introducing new variables u2, . . . , usand encoding

46

f2, . . . , fsin to a single polynomial u2f2+···+usfs∈C[u2, . . . , us, x1, . . . , xn].

By Theorem 2.3.2, Res(f1, u2f2+···usfs, x1) lies in C[u2, . . . , us, x2, . . . , xn].

Therefore, to get polynomials in x2, . . . , xn, we expand the resultant in

terms of powers of u2, . . . , us, that is, we write

Res(f1, u2f2+···usfs, x1) = 

α

hα(x2, . . . , xn)uα,

where uα=u2α2···usαs. The polynomials hαare called the gen-

eralized resultants of f1, . . . , fs. The generalized resultants are not of

much practical use, but we use it to prove the Extension Theorem.

Finally, we have the necessary tools to prove the Extension Theo-

rem, that is, a partial solution acan be extended if the leading terms

of f1, . . . , fsdo not simultaneously vanish at a.

Proof of the Extension Theorem 2.2.2. Let a= (a2, . . . , an). We

seek a common root a1of f1(x1,a), f2(x1,a), . . . , fs(x1,a). The case

s= 2 was proved in Theorem 2.3.5, which also covers the case s= 1

since V(f1) = V(f1, f1). It remains to prove the theorem when s≥3.

Since a̸∈ V(g1, . . . , gs), we may assume that g1(a)̸= 0. Let hα∈

C[x2, . . . , xn] be the generalized resultants of f1, . . . , fs, that is,

Res(f1, u2f2+· ·· +usfs, x1) = 

α

hαuα.(2.19)

By Lemma 2.3.3,

Af1+B(u2f2+· ·· +usfs) = Res(f1, u2f2+·· · +usfs, x1),(2.20)

for some polynomials A, B ∈C[u2, . . . , us, x1, . . . , xn].

Write A=αAαuαand B=βBβuβ, where Aα, Bβ∈C[x1, . . . , xn].

Set e2= (1,0, . . . , 0), . . . , es= (0, . . . , 0,1), so that u2f2+·· · +usfs=

47

i≥2ueifi. Then Equation 2.19 can be written as

αhαuα= (αAαuα)f1+βBβuβi≥2ueifi

=α(Aαf1)uα+i≥2,β Bβfiuβ+e1

=α(Aαf1)uα+α



i≥2

β+ei=α

Bβfi



uα

=α



Aαf1+i≥2

β+ei=α

Bβfi



uα.

If we equate the coeﬃcients of uα, we obtain

hα=Aαf1+

i≥2

β+ei=α

Bβfi,

which proves that hα∈I, and hence in I1, for all α. Since a∈V(I1),

it follows that hα(a) = 0 for all α. Therefore, by 2.19, the resultant

h= Res(f1, u2f2+···+usfs, x1) vanishes at a, that is,

h(a, u2, . . . , un) = 0.

Suppose we can assume about f2that

g2(a)̸= 0 and f2has degree in x1greater than f3, . . . , fs.(2.21)

Then, since

Res(f1(x1,a), u2f2(x1,a) + · ·· +usfs(x1,a)) = 0,

the polynomials f1(x1,a), and u2f2(x1,a) + · ·· +usfs(x1,a) have

a common factor d∈C[x1] of positive degree in x1by Theorem 2.3.4.

Check that since ddivides u2f2(x1,a) + ··· +usfs(x1,a), ddivides

fi(x1,a) for i= 2 to s. Consequently, dis a common factor for all

48

the polynomials f1, . . . , fs. Let a1be a root of d(a1exists because we

are working with complex numbers), then a1is a common root of all

fi(x1,a). This proves the Extension Theorem when we can assume the

condition 2.21 to be true.

Finally, if 2.21 is not true for f2, . . . , fs, then we have to use a

diﬀerent basis for Iso that the condition 2.21 is true. Replace f2by

f2+xN

1f1, where Nis such that xN

1f1has a higher degree in x1than

f2, f3, . . . , fsso that the leading coeﬃcient of f2+xN

1f1is g1. Check

that

I=< f1, f2+xN

1f1, f3, . . . , fs> .

Then, the previous argument gives us a1as a common root of f1(x1,a)

and f2(x1,a) + xN

1f1(x1,a), f3(x1,a),··· , fs(x1,a). Consequently, a1

is a common root of f1(x1,a), f2(x1,a), f3(x1,a),··· , fs(x1,a). This

completes the proof of the Extension Theorem.

Exercises.

1. Let Vand Wbe aﬃne varieties. Prove that V⊂Wif and only

if I(W)⊂I(V).

2. If Iis an ideal in k[x1, . . . , xn], prove that √Iis an ideal in

k[x1, . . . , xn] containing I. Further prove that

√I=√I.

3. Prove that if I=k[x1, . . . , xn] then the reduced Gr¨obner basis of

Iis {1}.

4. Let Ibe an ideal of k[x1, . .. , xn].Prove that Il=I∩k[xl+1, . . . , xn]

is an ideal of k[xl+1, . . . , xn].

5. Solve the following system of equations.

x2+y+z= 1,

x+y2+z= 1,

x+y+z2= 1.

6. Find the implicit equations of the following parametrizations.

49

(a) The tangent surface to the twisted cubic.

x=t+u,

y=t2+ 2tu,

z=t3+ 3t2u.

(b) The Enneper surface.

x= 3u+ 3uv2−u3,

y= 3v+ 3u2v−v3,

z= 3u2−3v2.

(c) The Folium of Descartes.

x=3t

1+t3,

y=3t2

1+t3.

7. Suppose f, g ∈k[x1, . . . , xn] have positive degree in x1. Then

prove that fand ghave a common factor in k[x1, . . . , xn] of pos-

itive degree in x1if and only if they have a common factor of

positive degree in x1in k(x2, . . . , xn)[x1].

8. Find the resultant of the following polynomials. Do they have a

common factor?

(a) f=x3+ 11x2+ 36x+ 28 and g=x3−17x2−25x+ 1001.

(b) f=x3+ 13x2+ 48x+ 38 and g=x3−21x2+ 71x+ 429.

9. Find Res(f, g, x), Res(f, g, y), and Res(f, g, z), when

(a)

f=x2+xy +xz −x−y−z,

g=x2z2−y2z2+xz3−yz3+x2y−y3+xyz −y2z.

(b)

f=xy +y2+xz + 2yz +z2−2x−2y−2z,

g=xz2−yz2+xy −y2.

50

Chapter 3

Finding Roots of polynomials

in Extension Fields.

In the book of life, the answers aren’t in the back - Charles M. Schulz.

The fundamental theorem of algebra says that every polynomial

with real coeﬃcients has a root in the ﬁeld of complex numbers C. In

this chapter, we prove that for any polynomial with coeﬃcients in an

arbitrary ﬁeld, there is always an extension ﬁeld which contains all the

roots of this polynomial.

3.1 Modular Arithmetic and Polynomial irreducibil-

ity in Q.

If Ais a set, then any subset of A×Ais called a relation of A. The

operation of division deﬁnes a relation among integers deﬁned as below.

Deﬁnition 3.1.1. Let a, b, n be integers with n > 0. Then ais con-

gruent to bmodulo n[written a≡b(mod n)], provided that ndivides

a−b.

Example 3.1.1. 17 ≡2 (mod 5) because 5 divides 17 −2 = 15.

Similarly, we check that 4 ≡28 (mod 6) and 3 ≡ −9 (mod 4).

Deﬁnition 3.1.2. Let aand nbe integers with n > 0. The congru-

ence class of amodulo n(denoted [a]) is the set of all those integers

that are congruent to amodulo n, that is,

[a] = {b|b∈Zand b≡a(mod n)}.

51

We denote ndivides aas n|a. Note that if n|a, then there is an

integer ksuch that a=kn. Therefore a≡bimplies a=b+kn for

some k∈Z. In other words,

[a] = {a+kn|k∈Z}.

Example 3.1.2. 1. When n= 5,

[17] = {17+5k|k∈Z}={. . . , −13,−8,−3,2,7,12,17,22,27,32, . . . }.

2. When n= 7,

[17] = {17 + 7k|k∈Z}={. . . , −11,−4,3,10,17,24,31,38, . . . }.

We now look at several properties of the congruence modulo nre-

lation of integers.

Theorem 3.1.1. Let nbe a positive integer. For all a, b, c ∈Z,

1. a≡a(mod n) (≡is reﬂexive);

2. if a≡b(mod n), then b≡a(mod n) (≡is symmetric);

3. if a≡b(mod n) and b≡a(mod n), then a≡c(mod n) (≡is

transitive).

Proof.

1. Since a−a= 0 and n|0, we have a≡a(mod n).

2. a≡b(mod n) implies n|(a−b) by deﬁnition. But that means

n|(b−a). Hence b≡a(mod n).

3. if a≡b(mod n) and b≡a(mod n) then there are integers kand

tsuch that a−b=nk and b−c=nt. Therefore

(a−b)+(b−c) = nk +nt

(a−c) = n(k+t).

Thus n|a−cand therefore a≡c(mod n).

Theorem 3.1.2. a≡c(mod n) if and only if [a] = [c].

52

Proof. Assume a≡c(mod n). To show that [a] = [c], we ﬁrst show

[a]⊂[c]. Let b∈[a] then by deﬁnition b≡a(mod n). Since we assume

a≡c(mod n), we have b≡c(mod n) by transitivity. Thus b∈[c]

and we prove that [a]⊂[c]. Observe that the assumption a≡c(mod

n) implies c≡a(mod n) by symmetry. Therefore, to prove [c]⊂[a],

we just reverse the role of aand cin the above argument.

Conversely, assume [a] = [c]. Since a≡a(mod n) by reﬂexivity we

have a∈[a] = [c]. Therefore a∈[c] and hence a≡c(mod n).

Example 3.1.3. Since, 17 ≡2 (mod 5) we get [17] = [2].

Corollary 3.1.3. Two congruence classes modulo nare either disjoint

or identical.

Proof. If [a] and [c] are disjoint there is nothing to prove. Assume

that [a]∩[c] is nonempty. Let b∈[a]∩[c], then b≡a(mod n) and

b≡c(mod n). By symmetry we ﬁrst get a≡b(mod n) and then by

transitivity a≡c(mod n). Finally, Theorem 3.1.2 implies [a] = [c].

Corollary 3.1.4. There are exactly ndistinct congruence classes mod-

ulo n, namely, [0],[1],··· ,[n−1].

Proof. We ﬁrst prove that no two of 0,1,2, . . . , n −1 are congruent

modulo n. Let sand tbe integers such that 0 ≤s<t<n. Then

0< t −s < n and therefore, ndoes not divide t−s, that is t̸≡ s

(mod n). Since no two of 0,1,2, . . . , n −1 are congruent modulo n

we have that [0],[1],··· ,[n−1] are all distinct. Next we show that

a∈Zis one of these nclasses. By division algorithm, a=qn +rsuch

that 0 ≤r < n. Therefore a≡r(mod n) or in other words a∈[r].

Therefore, ais in one of the classes [0],[1],··· ,[n−1].

Deﬁnition 3.1.3. The set of all congruence classes modulo nis de-

noted Zn.

Example 3.1.4. Z5={[0],[1],[2],[3],[4]}where

[0] = {. . . , −15,−10,−5,0,5,10,15, . . . },[1] = {. . . , −14,−9,−4,1,6,11,16, . . . },

[2] = {. . . , −13,−8,−3,2,7,12,17, . . . },[3] = {. . . , −12,−7,−2,3,8,13,18, . . . },

[4] = {. . . , −11,−5,−1,4,9,14,19, . . . }.

53

Deﬁnition 3.1.4. Addition and multiplication in Znare deﬁned by

[a] + [b] = [a+b]and [a]·[b] = [a·b].

Example 3.1.5. In Z5we have [3]+[4] = [7] = [2] = {. . . , −8,−3,2,7,12, . . . , }

and [3] ·[2] = [6] = [1] = {. . . , −9,−4,1,6,11, . . . }.

Theorem 3.1.5. The set Znwith the addition and multiplication of

classes is a commutative ring with identity.

Proof. It is easily veriﬁed that [0] is the additive identity, [1] is the

multiplicative identity in Znand that the additive inverse of a class

[a] is [−a]. All other properties are derived from the fact that Zis a

commutative ring.

Thus, sets transform to number-like objects on which we can per-

form arithmetic operations. Therefore, from now on, throughout the

book, brackets are dropped in the notation of congruence classes when-

ever the context is clear. For example, [a]·[b] is written as a·b.

Theorem 3.1.6. Zpis a ﬁeld whenever pis a prime.

Proof. By Theorem 3.1.5, we know that Zpis a commutative ring

with identity. To show that Zpis a ﬁeld we need to prove that if a∈Zp

such that a̸= 0, then ahas a multiplicative inverse x. Now, a̸= 0

implies a̸≡ 0 (mod p), that is, ais not divisible by p. Therefore,

the greatest common divisor (gcd) of aand pis 1. We use Euclid’s

algorithm to write ax +py = 1 (see Section A.1). This implies p

divides ax −1. In other words, ax ≡1 (mod n). Therefore xis the

inverse of ain Zp. And the proof is now complete.

Given f∈Q[x], we can clear denominators and get cf ∈Z[x] for

some nonzero integer c, such that cf(x) has the same degree as f(x).

This allows us to reduce factorization problems in Q[x] to factorization

problems in Z[x].

Theorem 3.1.7. Let f(x)∈Z[x], then f(x)factors as a product of

polynomials of degrees mand nin Q[x]if and only if f(x)factors as

a product of polynomials of degrees mand nin Z[x].

Proof. Clearly, if f(x) factorizes in Z[x], then f(x) factors in Q[x].

Conversely, suppose f(x) = g(x)h(x) in Q[x]. Let aand bbe inte-

gers such that ag(x) and bh(x) have integer coeﬃcients. Therefore,

abf(x)=(ag(x))(bh(x)) ∈Z[x]. Now let pbe a prime that divides

54

ab, that is let ab =pt. Then by Exercise 4, pdivides every coeﬃcient

of ag(x) or pdivides every coeﬃcient of bh(x). Let us say pdivides

every coeﬃcient of ag(x). Then ag(x) = pk(x) such that k(x)∈Z[x].

Thus, we get ptf(x) = (pk(x))(bh(x)). Canceling pfrom both sides we

have tf(x) = k(x)bh(x). Now we repeat the argument with any prime

divisor of t. Continuing thus, we cancel every prime factor of ab till

the left side of the equation is ±f(x) and the right side is the product

of two polynomials in Z[x], one with the same degree as g(x) and the

other with the same degree as h(x).

Example 3.1.6. Let

f= (1/2)x2−(5/4)x+ (1/2).

Then

4f= 2x2−5x+ 2 = (2x−1)(x−2) ∈Z[x].

Hence

f=1

4(2x−1)(x−2) ∈Q[x].

A polynomial f(x)∈k[x], where kis a ring, is said to be an associate

of g(x)∈k[x] if f(x) = cg(x) for some nonzero c∈k.

Deﬁnition 3.1.5. Let kbe a ﬁeld. A non-constant polynomial p(x)∈

k[x]is said to be irreducible if its only divisors are its associates and

nonzero constant polynomials. A non-constant polynomial that is not

irreducible is said to be reducible.

Example 3.1.7. The polynomial x2+ 1 is irreducible in R(apply

Corollary A.2.4) but is reducible in C.

We use the ﬁelds Zpto determine irreducibility of polynomials in

Q. Let f(x) = anxn+an−1xn−1+· ·· +a1x+a0∈Z[x], then f(x)

denotes the polynomial [an]xn+ [an−1]xn−1+··· + [a1]x+ [a0]∈Zp[x].

Theorem 3.1.8. Let f(x) = anxn+an−1xn−1+·· · +a1x+a0be

a polynomial with integer coeﬃcients, and let pbe a positive prime

that does not divide an. If f(x)is irreducible in Zp[x], then f(x)is

irreducible in Q[x].

Proof. Suppose, on the contrary, that f(x) is irreducible in Zp[x]

and that f(x) is reducible in Q[x]. By Theorem 3.1.7, f(x) factors in

55

Z[x]. Let f(x) = h(x)g(x) such that h(x) and g(x) are non-constant

polynomials in Z[x]. Since pdoes not divide an, it cannot divide the

leading coeﬃcients of h(x) or g(x) (their product is an). Therefore, de-

gree of g(x) is the same as degree of g(x) and degree of h(x) is the same

as degree of h(x). In particular, g(x) and h(x) are not constant polyno-

mial in Zp[x]. By Exercise 6, we have f(x) = g(x)h(x) in Z[x] implies

that f(x) = g(x)h(x) in Zp[x]. This contradicts the irreducibility of

f(x) in Zp[x]. Therefore f(x) is irreducible in Q[x].

The advantage of using this theorem for proving irreducibility is that

for each nonnegative integer nthere are only ﬁnitely many polynomials

of degree nin Zp[x]. In fact, there are pn+1 −pnpolynomials of degree n

in Zp[x] (see Exercise 7). So we determine whether a given polynomial

is irreducible by checking the ﬁnite number of possible factors.

Example 3.1.8. To show that f(x) = x5+ 8x4+ 3x2+ 4x+ 7 is

irreducible in Q[x], we reduce f(x) mod 2 and we get f(x) = x5+x2+1

in Z2[x]. f(x) has no roots in Z2[x] because f(0) ̸= 0 and f(1) ̸= 0 (see

Theorem A.2.1). Therefore f(x) has no linear factors (see Theorem

A.2.2). The only quadratic polynomials in Z2[x] are x2, x2+x, x2+

1, x2+x+ 1. We use long division to show none of these polynomials

divide f(x). f(x) cannot have factors of degree 3 or 4 because then the

other factor has to be either linear or quadratic which is not possible.

Therefore f(x) is irreducible in Z2[x]. This implies f(x) is irreducible

in Q[x].

If a polynomial f(x) is reducible mod p, then it does not imply that

f(x) is reducible in Q[x]. Consequently, application of Theorem 3.1.8

can be time consuming because we need to ﬁnd the right pto prove

irreducibility.

Example 3.1.9. To prove that f(x) = 7x3+ 6x2+ 4x+ 6 is irre-

ducible in Q[x], we use p= 5. Check that f(x) is reducible in Z2[x]

and Z3[x]. Now f(x) = 2x3+x2+ 4x+ 1 has no roots in Z5[x] be-

cause f(0), f (1), f(2), f (3), f(4) do not evaluate to zero. Thus, f(x) is

irreducible in Z5[x] (by Corollary A.2.4) and hence in Q[x].

The number of irreducible polynomials of a given degree nin Zp[x]

is also known.

56

Proposition 3.1.1. The number of irreducible polynomials of degree

nin Zp[x]is

1

n

d/n

µ(d)pn/d

where

µ(d) = 





1for d= 1

0if dhas a square factor

(−1)rif dhas rdistinct prime factors.

The proof of Proposition 3.1.1 is available in [19].

Example 3.1.10. In Z2[x], there is exactly 1 irreducible polynomial

of degree 2 because

1

2

d/2

µ(d)p2/d =1

2µ(1)22+µ(2)21=1

2(4 −2) = 1.

Note that x2+x+ 1 is irreducible because it has no roots by Corollary

A.2.4. Thus x2+x+ 1 is the only irreducible polynomial of degree 2

in Z2.

In Section A.2 we list other irreducibility tests for polynomials. In

the next section we use irreducible polynomials to construct extension

ﬁelds.

3.2 Field Extensions.

Let kbe a ﬁeld. Given a polynomial fin k[x] our goal is to ﬁnd

a ﬁeld containing kin which fhas a root. To do this we need to

study congruence relations in the polynomial ring k[x]. Congruency is

a recurring theme in this chapter that allows us to construct new ﬁelds.

Deﬁnition 3.2.1. Let kbe a ﬁeld and f(x), g(x), p(x)∈k[x], and let

p(x)be a nonzero polynomial. Then f(x)is congruent to g(x)modulo

p(x), written as

f(x)≡g(x)(mod p(x)),

provided that p(x)divides f(x)−g(x).

57

Example 3.2.1. It is easy to verify that x2≡ −1 (mod x2+ 1),

x3+ 2x+ 1 ≡x+ 1 (mod x2+ 1), and x4−1≡0 (mod x2+ 1).

We state some properties of this congruence modulo relation with-

out proof. The proofs of Theorems 3.2.1, 3.2.2, 3.2.3, 3.2.6, 3.2.7, and

Corollary 3.2.4 are similar to proofs in the previous section, and are

assigned as exercises.

Theorem 3.2.1. Let kbe a ﬁeld and let p(x)be a nonzero polynomial

in k[x]. Then the relation of congruence modulo p(x)is

1. reﬂexive: f(x)≡f(x)(mod p(x));

2. symmetric: if f(x)≡g(x)(mod p(x)), then g(x)≡f(x)(mod p(x));

3. transitive: if f(x)≡g(x)(mod p(x)) and g(x)≡h(x)(mod p(x)),

then f(x)≡h(x)(mod p(x)).

Theorem 3.2.2. Let kbe a ﬁeld and p(x)a nonzero polynomial in

k[x]. If f(x)≡g(x)(mod p(x)) and h(x)≡k(x)(mod p(x)), then

1. f(x) + h(x) = g(x) + k(x)(mod p(x)),

2. f(x)h(x) = g(x)k(x)(mod p(x)).

Example 3.2.2. Since x2≡ −1 (mod x2+ 1) and x3+ 2x+ 1 ≡x+ 1

(mod x2+ 1) we get

(x2)+(x3+ 2x+ 2) ≡ −1 + (x+ 1) = x(mod x2+ 1)

and

(x2)(x3+ 2x+ 2) ≡(−1)(x+ 1) = −x−1(mod x2+ 1).

Deﬁnition 3.2.2. Let kbe a ﬁeld and f(x), p(x)∈k[x]such that p(x)

is a nonzero polynomial. The congruence class of f(x)modulo p(x)is

denoted [f(x)] and consists of all polynomials in k[x]that are congruent

to f(x)modulo p(x), that is

[f(x)] = {g(x); g(x)∈k[x]and g(x)≡f(x)(mod p(x))}.

In other words

[f(x)] = {f(x) + q(x)p(x); q(x)∈k[x]}.

58

Example 3.2.3. The congruence class of x+ 1 modulo x2+ 1 is the

set

[x+ 1] = {(x+ 1) + q(x)(x2+ 1); q(x)∈k[x]}.

Note that the set [x+ 1] contains all the polynomials that has the

remainder x+ 1 when divided by x2+ 1.

Theorem 3.2.3. f(x)≡g(x)(mod p(x)) if and only if [f(x)] = [g(x)].

Corollary 3.2.4. Two congruence classes modulo p(x)are either dis-

joint or identical.

Corollary 3.2.5. Let kbe a ﬁeld and let p(x)be a nonzero polynomial

of degree nin k[x]. Consider the set Ssuch that

S={r(x) : r(x)∈k[x]and degree of r(x)is less than n}.

Then, if f(x)∈k[x],[f(x)] = [r(x)] for some r(x)∈S. Moreover the

congruence classes of diﬀerent polynomials in Sare distinct.

Proof. Two diﬀerent polynomials in Scannot be congruent mod-

ulo p(x) because their diﬀerence has degree less than nand hence

is not divisible by p(x). Therefore diﬀerent polynomials in Smust

be in diﬀerent congruence classes by Theorem 3.2.3. Now given a

polynomial f(x)∈k[x] we can use the division algorithm to write

f(x) = q(x)p(x) + r(x) where r(x) has degree less than n. Note that

f(x)≡r(x)(mod p(x)). Therefore, f(x)∈k[x] implies [f(x)] = [r(x)]

for some r(x)∈S.

The set of all congruence classes modulo p(x) is denoted by k[x]/(p(x)).

Example 3.2.4. Consider R[x]/(x2+ 1). The possible remainders on

division by x2+ 1 are polynomials of the form a+bx where a, b ∈R.

R[x]/(x2+ 1) = {[a+bx] : a, b ∈R}={[0],[x],[2x+ 5],[1/5x+ 3], . . . }.

Consequently, R[x]/(x2+ 1) is an inﬁnite set.

Example 3.2.5. The possible remainders on division by the polyno-

mial x2+x+1 ∈Z2[x] are polynomials of the form ax+bwith a, b ∈Z2.

There are only four possible remainders (see Exercise 14). Therefore

Z2[x]/(x2+x+ 1) = {[0],[1],[x],[x+ 1]}.

59

Deﬁnition 3.2.3. Let kbe a ﬁeld and let p(x)be a non-constant poly-

nomial in k[x]. Addition and multiplication in k[x]/(p(x)) are deﬁned

by

[f(x)] + [g(x)] = [f(x) + g(x)],

[f(x)][g(x)] = [f(x)g(x)].

Example 3.2.6. In R[x]/(x2+ 1)

[x+1]+[x−1] = [2x].

[x+ 1][x−1] = [x2−1] = [−2].

Theorem 3.2.6. Let kbe a ﬁeld and let p(x)be a non-constant poly-

nomial in k[x]. Then the set k[x]/(p(x)) of congruence classes modulo

p(x)is a commutative ring with identity.

Theorem 3.2.7. Let kbe a ﬁeld and let p(x)be an irreducible polyno-

mial in k[x]. Then k[x]/(p(x)) is a ﬁeld.

Example 3.2.7. The polynomial p(x) = x2+ 1 is irreducible in R[x]

because it has no roots in R(see Theorem A.2.7). Therefore, by The-

orem 3.2.7, R[x]/(x2+ 1) is a ﬁeld.

If Fand Kare ﬁelds such that F⊆K, we say that Kis an

extension ﬁeld of F. Next, we prove that if kis a ﬁeld and p(x) is an

irreducible polynomial in k[x], then k[x]/(p(x)) is an extension ﬁeld of

kthat contains a root of p(x). To do this we introduce the concept of

isomorphisms.

Deﬁnition 3.2.4. Let fbe a function from a set Xto a set Y. Then

1. fis surjective (or onto) if for every y∈Ythere is a x∈Xsuch

that f(x) = y.

2. fis injective (or one-to-one) if x̸=x′implies f(x)̸=f(x′).

3. fis a bijection if it is both injective and surjective.

Deﬁnition 3.2.5. Let Rand Sbe rings. A function f:R→Sis

called a homomorphism if it satisﬁes the condition

f(a+b) = f(a) + f(b)and f(ab) = f(a)f(b)for all a, b ∈R.

60

Deﬁnition 3.2.6. Let Rand Sbe rings. A function f:R→Sis

called an isomorphism if fis a bijective homomorphism. The ring

Ris said to be isomorphic to S(in symbols R∼

=s) if there is an

isomorphism from Rto S.

What is the purpose of isomorphisms? Two isomorphic sets are

considered essentially same for all practical purposes.

Example 3.2.8. 1. Z6is not isomorphic to Z12 because the orders

of the two rings are diﬀerent.

2. Consider the ﬁeld Kof 2 ×2 matrices of the form

a b

−b a 

We prove that Kis isomorphic to the ﬁeld Cof complex numbers.

Deﬁne a function f:K→Cby the rule

fa b

−b a =a+bi.

To prove that fis injective suppose that

fa b

−b a =fr s

−s r .

Then a+bi =r+si in C. By the rules of equality in Cwe must

have a=rand b=s. Therefore

a b

−b a =r s

−s r .

Consequently, fis injective. The function is surjective because

any complex number a+bi is the image under fof the matrix

a b

−b a 

in K. Finally

61

f a b

−b a +c d

−d c =fa+c b +d

−b−d a +c

= (a+c)+(b+d)i

= (a+bi)+(c+di)

=fa b

−b a +fc d

−d c 

and

f a b

−b a  c d

−d c =fac −bd ad +bc

−ad −bc ac −bd 

= (ac −bd)+(ad +bc)i

= (a+bi)(c+di)

=fa b

−b a fc d

−d c .

Therefore, fis an isomorphism.

3. An element ain a ring Rwith identity is called a unit if there

exists u∈Rsuch that au = 1R=ua. In the ring Z8has four

units 1,3,5,7. The ring Z4×Z2has only two units, namely (1,1,)

and (3,1). Therefore Z8is not isomorphic to Z4×Z2.

In the next theorem we show that the ﬁeld k[x]/(p(x)) contains an

isomorphic copy of the ﬁeld k. Though we do not prove that k[x]/(p(x))

contains the ﬁeld kitself it is mathematically correct to conclude that

k[x]/(p(x)) is an extension ﬁeld of k. As we explore this ﬁeld of math-

ematics further we realize that most theorems here are proved up to

isomorphisms.

Theorem 3.2.8. Let kbe a ﬁeld and let p(x)be an irreducible polyno-

mial in k[x]. Then k[x]/(p(x)) is an extension ﬁeld of kthat contains

a root of p(x).

Proof. By Theorem 3.2.7, k[x]/(p(x)) is a ﬁeld. Let k∗be the subset

of k[x]/(p(x)) consisting of the congruence classes of all the constant

62

polynomials, that is k∗={[c]; c∈k}. Deﬁne a map ϕ:k→k∗by

ϕ(c) = [c]. Clearly ϕis surjective by deﬁnition. Since

ϕ(a+b) = [a+b] = [a] + [b] = ϕ(a) + ϕ(b) and

ϕ(ab) = [ab] = [a][b] = ϕ(a)ϕ(b)

ϕis a homomorphism. To see that ϕis injective suppose ϕ(a) = ϕ(b).

Then [a]=[b] which implies p(x) divides a−b. But the degree of

p(x)≥1 and degree of a−bis zero. Therefore, a−b= 0. Thus a=b

and ϕis injective. Therefore ϕis an isomorphism. Hence k[x]/(p(x))

is an extension ﬁeld of k.

Let p(x) = anxn+·· · +a1x+a0. Recall, that k[x]/(p(x)) denotes

all the remainders possible when divided by p(x). Therefore, p(x)∈[0]

and if a∈kthen a∈[a] in k[x]/(p(x)). Now

p([x]) = an[x]n+···+a1[x] + a0

= [an][x]n+· ·· + [a1][x]+[a0]

= [anxn+· ·· +a1x+a0]

= [p(x)]

= [0k]

Therefore, [x] is a root of p(x) in k[x]/(p(x)).

Example 3.2.9. By Theorem 3.2.8 we get that R[x]/(x2+ 1) is a ﬁeld

that contains a root [x] (denoted usually by i) of x2+ 1.

Next we show that R[x]/(x2+1) is the same as the ﬁeld of complex

numbers C.

Theorem 3.2.9. The ﬁeld R[x]/(x2+ 1) is isomorphic to the ﬁeld of

complex numbers C.

We know from Example 3.2.4 that R[x]/(x2+ 1) = {[a+bx] : a, b ∈

R}. Let f:R[x]/(x2+ 1) →Csuch that f([a+bx]) = a+bi. We

show that fis an isomorphism. Suppose f([a+bx]) = f([c+dx]), then

a+bi =c+di. Consequently, a=cand b=d. Therefore fis injective.

If a+bi ∈C, then f([a+bx]) = a+bi. Therefore fis surjective. Next

63

we show that fis a homomorphism.

f([a+bx]) + f([c+dx]) = (a+bi)+(c+di) = (a+c) + (b+d)i

=f([(a+c)+(b+d)x])

=f([a+bx] + [c+dx]).

f([a+bx])f([c+dx]) = (a+bi)(c+di)

= (ac −bd)+(bc +ad)i

=f([(ac +bdx2)+(bc +ad)x]) since [x2] = [−1]

=f([a+bx][c+dx]).

Thus R[x]/(x2+ 1) ∼

=C.

3.3 Quotient Rings.

Deﬁnition 3.3.1. Let Ibe an ideal in a ring Rand let a, b ∈R. Then

ais congruent to bmodulo I[written a≡b(mod I)], provided a−b∈I.

Congruence in Zand polynomial rings are speciﬁc examples of con-

gruence modulo an ideal.

Example 3.3.1.

1. a≡b(mod n) is the same as a≡b(mod I), where I=<n>is

the principal ideal generated by nin Z. Note that a−b∈< n >

if and only if ndivides a−b.

2. Similarly, x3+ 2x+ 1 ≡x+ 1 (mod x2+ 1) is the same as

x3+ 2x+ 1 ≡x+ 1 (mod I) where I=< x2+ 1 >is the principal

ideal generated by x2+ 1 in the polynomial ring Q[x].

Theorem 3.3.1. Let Ibe an ideal in a ring R. Then the relation of

congruence modulo Iis

1. reﬂexive: a≡a(mod I) for every a∈R;

2. symmetric: if a≡b(mod I), then b≡a(mod I);

3. transitive: if a≡b(mod I) and b≡c(mod I), then a≡c(mod

I).

64

Theorem 3.3.2. Let Ibe an ideal in a ring R. If a≡b(mod I) and

c≡d(mod I), then

1. a+c≡b+d(mod I);

2. ac ≡bd (mod I).

Let Ibe an ideal in a ring Rand if a∈R, then the congruence

class of amodulo Iis the set of all elements of Rthat are congruent

to amodulo I, that is, the set

{b∈R:b≡a(mod I)}

={b∈G:b−a∈I}

={b∈G:b=a+i, for some i∈I}

={i+a:i∈I}.

As a consequence the congruence class of amodulo Iis denoted

a+Iand is called a coset of Iin R. The set of all cosets of Iis

denoted by R/I.

Theorem 3.3.3. Let Ibe an ideal in a ring Rand let a, c ∈R. Then

a≡c(mod I) if and only if a+I=c+I.

Corollary 3.3.4. Let Ibe an ideal in a ring R. Then two cosets of I

are either disjoint or identical.

Theorem 3.3.5. Let Ibe an ideal in a ring R. If a+I=b+Iand

c+I=d+Iin R/I, then

(a+c) + I= (b+d) + Iand ac +I=bd +I.

Theorem 3.3.6. Let Ibe an ideal in a ring R, then R/I is a ring with

addition and multiplication of cosets as deﬁned above.

Proofs of Theorems 3.3.1, 3.3.2, 3.3.3, 3.3.5, 3.3.6, and Corollary

3.3.4 are similar to the proofs we provided for Zin Section 3.1 and are

assigned as exercises.

The ring R/I is called a quotient ring.

65

Example 3.3.2. 1. If R=Z8and I=<2>, then

R/I ={0 + I, 1 + I}.

2. If R=Z2[x] and I=< x2+x+ 1 >, then

R/I ={0 + I, 1 + I, x +I, (x+ 1) + I}.

A quotient ring preserves many properties of the original ring R.

Theorem 3.3.7. Let Ibe an ideal in a ring R. Then

1. If Ris commutative, then R/I is a commutative ring.

2. If Rhas an identity, then so does the ring R/I.

Proof.

1. If Ris commutative and a, c ∈R, then ac =ca. Consequently, in

R/I we have (a+I)(c+I) = ac +I=ca +I= (c+I)(a+I).

Hence R/I is commutative.

2. The identity in R/I is the coset 1R+Ibecause (a+I)(1R+I) =

a1R+I=a+Iand similarly (1R+I)(a+I) = a+I.

Let f:R→Sbe a homomorphism of rings, then the kernel of fis

the set K={r∈R|f(r) = 0S}.

Theorem 3.3.8. Let f:R→Sbe a homomorphism of rings, then

the kernel Kis an ideal in R.

Proof. If a, b ∈K, then f(a−b) = f(a)−f(b)=0S−0S= 0S.

Therefore a−b∈K. If r∈Rand a∈K, then f(ra) = f(r)f(a) =

f(r)0S= 0Sand f(ar) = f(a)f(r) = 0Sf(r) = 0S. Therefore ra ∈K

and ar ∈K. Thus, by Proposition 1.3.1, Kis an ideal of R.

Theorem 3.3.9. Let f:R→Sbe a homomorphism of rings with

kernel K. Then K= (0R)if and only if fis injective.

Proof. Suppose K= (0R) and f(a) = f(b). Then since fis a

homomorphism, f(a−b) = f(a)−f(b) = 0S. Hence a−bis in the

kernel K. Consequently, a−b= 0Rwhich implies a=b. Therefore

fis injective. Conversely, let fbe injective and let f(c) = 0S. Since

f(0R)=0S(see Exercise 10), we get f(c) = f(0R). Therefore c= 0R

by injectivity. Hence the kernel consists of the single element 0R.

66

Theorem 3.3.10. Let Ibe an ideal in a ring R. Then the map π:

R→R/I given by π(r) = r+Iis a surjective homomorphism with

kernel I.

Proof. The map πis surjective because given any coset r+I∈R/I,

π(r) = r+I.πis a homomorphism because

π(r+s) = (r+s) + I= (r+I)+(s+I) = π(r) + π(s) and

π(rs) = rs +I= (r+I)(s+I) = π(r)π(s).

Now π(r) = 0R+Iif and only if r+I= 0R+Iwhich occurs if only if

r≡0R(mod I), that is, if and only if r∈I. Therefore Iis the kernel

of π.

We now prove the First Isomorphism Theorem which is a very useful

tool to prove isomorphism of rings.

Theorem 3.3.11. (First Isomorphism Theorem) Let f:R→Sbe a

surjective homomorphism of rings with kernel K. Then the quotient

ring R/K is isomorphic to S.

Proof. Consider the map ϕ:R/K →Ssuch that ϕ(r+K) = f(r).

If r+K=t+Kthen r−t∈Kby Theorem 3.3.3. Therefore f(r−t) =

0S. Since fis a homomorphism, f(r−t) = f(r)−f(t)=0S, which

implies f(r) = f(t). Hence ϕis a well deﬁned function independent of

how the coset is written. Since fis surjective, for s∈Sthere is some

r∈Rsuch that f(r) = s. Thus ϕis surjective because s=f(r) =

ϕ(r+K). If ϕ(r+K) = ϕ(c+K) then f(r) = f(c) which implies

0S=f(r)−f(c) = f(r−c). Hence r−c∈K, which implies that

r+K=c+K(again by Theorem 3.3.3). Therefore ϕis injective.

Finally ϕis a homomorphism because

ϕ[(c+K) + (d+K)] = ϕ[(c+d) + K] = f(c+d) = f(c) + f(d)

=ϕ(c+K) + ϕ(d+K)

and

ϕ[(c+K)(d+K)] = ϕ(cd +K) = f(cd) = f(c)f(d)

=ϕ(c+K)ϕ(d+K).

Therefore, ϕ:R/K →Sis an isomorphism.

67

Example 3.3.3. We use the First Isomorphism to show that Z[x]/ <

x >∼

=Z. Let f:Z[x]→Zbe such that each polynomial p(x) is

mapped to its constant term cp. If c∈Zthen f(x+c) = c. Therefore

fis surjective. Verify that the constant term of p(x) + q(x) is cp+

cqand the constant term of p(x)q(x) is cpcq. Therefore f(p+q) =

f(p) + f(q) and f(pq) = f(p)f(q). Hence fis a homomorphism. The

polynomials with a zero constant term are precisely those that have x

as a factor. Therefore kernel of fis the ideal < x >. Applying the

First Isomorphism we derive that Z[x]/ < x >∼

=Z.

Like before we use quotient rings to construct new ﬁelds.

Deﬁnition 3.3.2. An ideal Min a ring Ris said to be maximal if

M̸=Rand whenever Jis an ideal such that M⊆J⊆R, then M=J

or J=R.

Example 3.3.4. We prove that (3) is a maximal ideal in Z. Suppose

Jis an ideal such that (3) ⊆J⊆Z. If J̸= (3) then there exists a∈J

such that 3 does not divide a, that is 3 and aare relatively prime.

Therefore the greatest common divisor of aand 3 is 1. Hence by the

Euclidean Algorithm (see Section A.1) there are u, v ∈Zsuch that

3u+av = 1. Since 3, a ∈J, it follows that 1 ∈J. Therefore J=Z

proving that Jis maximal.

Theorem 3.3.12. Let Mbe an ideal in a commutative ring Rwith

identity. Then Mis a maximal ideal if and only if the quotient ring

R/M is a ﬁeld.

Proof. Suppose R/M is a ﬁeld and M⊆J⊆Rfor some ideal J.

If M̸=J, then there exists a∈Jwith a̸∈ M. By Theorem 3.3.3,

a+M= 0R+M, if and only if, a∈M. Hence a+M̸= 0R+M. Since

R/M is a ﬁeld, a+Mhas inverse b+Msuch that (a+M)(b+M) =

ab +M= 1R+M. This implies ab ≡1R(mod M) which means that

ab −1R=mfor some m∈M. Since a, b ∈Jit follows that 1R∈J.

Consequently, J=R. Therefore Mis a maximal ideal.

Conversely, suppose that Mis a maximal ideal. R/M is a commu-

tative ring with identity by Theorems 3.3.6 and 3.3.7. Consequently,

R/M is a ﬁeld if every nonzero element of R/M has a multiplicative

inverse. If a+Mis a nonzero element in R/M, then by Theorem

3.3.3, a̸∈ M. The set J={m+ra :r∈Rand m∈M}is an ideal

in Rthat contains Mby Exercise 12. Furthermore, a= 0R+ 1Rais

68

in Jso that M̸=J. By maximality we must have J=R. Hence

1R∈Jwhich implies that 1R=m+ca for some m∈Mand c∈R.

Note that ca −1R=m∈Mwhich implies ca ≡1R(mod M). Hence

ca +M= 1R+M. Consequently the coset c+Mis the inverse of

a+Min R/M:

(c+M)(a+M) = ca +M= 1R+M.

Therefore R/M is a ﬁeld.

Example 3.3.5. Now we can prove that (3) is a maximal ideal in Zby

a diﬀerent method than the one used in Example 3.3.4. By Theorem

3.1.6, Z/(3) = Z3is a ﬁeld. Hence, Theorem 3.3.12 proves that (3) is

a maximal ideal in Z.

3.4 Splitting ﬁelds of polynomials.

In this section, given a polynomial p∈F[x] such that Fis a ﬁeld, we

show that an extension ﬁeld K⊇Fexists such that psplits completely

as linear factors. We also classify all the ﬁnite ﬁelds up to isomorphism.

Let Rbe a ring with identity. Then Ris said to have characteristic

nif nis the smallest positive integer such that n1R= 0R.

Example 3.4.1. The ring Z5has characteristic 5.

Theorem 3.4.1. Let Rbe a ring with identity.

1. The set P={k1R|k∈Z}is a subring of R.

2. If Rhas characteristic 0then P∼

=Z.

3. If Rhas characteristic n > 0then P∼

=Zn.

Proof. Deﬁne f:Z→Rby f(k) = k1R. Then fis a homomor-

phism because

f(k+t) = (k+t)1R=k1R+t1R=f(k) + f(t);

and

f(kt) = (kt)1R= (k1R)(t1R) = f(k)f(t).

The image of fis the set Ptherefore Pis a ring (see Exercise 13).

Consequently fcan be considered as a surjective homomorphism from

69

Zto P. Then by the First Isomorphism Theorem we get P∼

=Z/kerf .

If Rhas characteristic 0 then the only integer ksuch that k1R= 0 is

k= 0. So that the kernel of fis the ideal <0>in Zand

P∼

=Z/ < 0>∼

=Z.

If Rhas characteristic n > 0 then we prove that Kernel of fis the

principal ideal < n >. Suppose that k1R= 0R. Divide kby nto write

k=nq +rwhere 0 ≤r < n. Then

r1R=r1R+ 0R

=r1R+n1R,since n1R= 0R

=r1R+nq1R

= (r+nq)1R

=k1R

= 0R.

Since r < n and nis the smallest positive integer such that n1R= 0R

(by deﬁnition of the characteristic) we must have r= 0. Therefore

k=nq implying that k∈< n >. Therefore Ker f=< n >. Therefore

P∼

=Z/ < n >=Zn.

If a ﬁeld Fhas characteristic zero then Theorem 3.4.1 implies that

Fhas a copy of Zand therefore is inﬁnite.

Corollary 3.4.2. Every ﬁnite ﬁeld Fhas characteristic pfor some

prime p.

Proof. Suppose the characteristic of Fis nand nis not a prime

number. Then n=kt where kand tare positive integers such that

k < n and t < n. Then

0F= (kt)1R= (k1R)(t1R).

This implies either (k1R) = 0 or (t1R) = 0 (see Exercise 19) con-

tradicting the fact that nis the smallest integer such that n1R= 0R.

Therefore, the characteristic of Fis a prime number.

Let Kbe an extension ﬁeld of F. Let w, u1, . . . , unbe elements of

K. If w∈Kcan be written in the form w=a1u1+a2u2+· ·· +anun

with each ai∈F, we say that wis a linear combination of u1, . . . , un.

If every element of Kis a linear combination of u1, . . . , un, we say that

the set (u1, . . . , un)spans Kover F.

70

Example 3.4.2. The set {1, i}spans Cover R.

A subset {u1, . . . , un}of Kis said to be linearly independent over

Fprovided that whenever

c1u1+c2u2+· ·· +cnun= 0F

with each ci∈F, then ci= 0Ffor every i. A set that is not linearly

independent is said to be linearly dependent. A set {u1, . . . , um}is

linearly dependent over Fif there exists elements b1, . . . , bmin Fnot

all zero such that b1u1+·· · +bmum= 0F.

Example 3.4.3. 1. The set {1 + i, 2i, 2 + 8i}is linearly dependent

over Rsince

2(1 + i) + 3(2i)−(2 + 8i) = 0.

2. The set {1, i}is linearly independent over R.

A subset {u1, . . . , un}of Kis said to be a basis of Kover Fif it

spans Kand is linearly independent over F.

Example 3.4.4. The set {1, i}is a basis of Cover R.

If Khas a ﬁnite basis over Fthen Kis said to be ﬁnite dimensional

over F. The dimension of Kover Fis the number of elements in any

basis of Kand is denoted [K:F]. In the exercises you will show that

if S={u1, . . . , un}spans Kover Fthen some subset of Sis a basis of

Kover F. The order of a ﬁeld is the number of elements in the ﬁeld.

We now look at the order of a ﬁeld.

Theorem 3.4.3. A ﬁnite ﬁeld Fhas order pn, where pis the charac-

teristic of Fand n= [F:Zp].

Proof. By Theorem 3.4.1, since Fhas characteristic p,Zp⊂F.

Hence, there is certainly a ﬁnite set of elements that spans Fover

Zp(the set Fitself for example). Consequently Fhas a ﬁnite basis

(u1, . . . , un) over Zp(see Exercise 20). Every element of Fcan be

uniquely written in the form

c1u1+c2u2+· ·· +cnun(3.1)

with each ci∈Zp. Since there are ppossibilities for each cithere are

precisely pndistinct linear combinations of the form 3.1. So the order

of Fis pn.

71

If u1, u2, . . . , unare elements of an extension ﬁeld Kof F, then we

denote F(u1, u2. . . , un) to be smallest subﬁeld of Kthat contains Fand

all the ui.F(u1, u2. . . , un) is said to be a ﬁnitely generated extension

of Fgenerated by u1, . . . , un. An extension ﬁeld F(u) generated by

one element is called a simple extension.

An element uof an extension ﬁeld Kover Fis algebraic over Fif

it is the root of a nonzero polynomial in F[x].

Deﬁnition 3.4.1. The minimal polynomial of an element u∈Kover

Fis an irreducible monic polynomial p(x)such that p(u) = 0F. More-

over if uis a root of g(x)∈F[x], then p(x)divides g(x).

Example 3.4.5. The minimal polynomial of i∈Cis x2+ 1 over R.

In the exercises you will show that a minimal polynomial of an

algebraic element over a ﬁeld Falways exist and is unique.

Theorem 3.4.4. Let Kbe an extension ﬁeld of Fand u∈Kan

algebraic element over Fwith minimal polynomial p(x)of degree n.

Then {1F, u, u2, . . . , un−1}is a basis of F(u)over Fand therefore

[F(u) : F] = n.

Proof. Let ϕ:F[x]→F(u) be such that ϕ(f(x)) = f(u). Every

constant polynomial cis mapped to itself by ϕand ϕ(x) = u. So Image

of ϕ(Imϕ) is a ﬁeld that contains both Fand u. But since F(u) is

the smallest ﬁeld that contains both Fand u,F(u)⊆Imϕ. But by

the deﬁnition of ϕand since F(u) is a ﬁeld we have that Imϕ⊆F(u).

Therefore Imϕ=F(u). Therefore every nonzero element in F(u) is of

the form f(u) for some f(x)∈F[x]. Dividing f(x) by p(x) we write

f(x) = q(x)p(x) + r(x) such that degree of r(x) is less than n. Conse-

quently f(u) = q(u)p(u) + r(u) = q(u)0F+r(u) = r(u). Hence f(u)

has degree less than n. Therefore the set {1F, u, u2, . . . , un−1}spans

F(u) over F. To show that this set is linearly independent suppose

that c0+c1u+···+cn−1un−1= 0Fwith each ci∈F. Then uis a root

of this polynomial and therefore p(x) divides this polynomial which has

degree less than n. This is possible only when c0+c1u+···+cn−1un−1

is the zero polynomial, that is, each ci= 0F. Thus, {1F, u, u2, . . . , un−1

is a basis of F(u).

In the Exercises you will prove that F(u)∼

=F[x]/(p(x)) by showing

that ϕin Theorem 3.4.4 is an isomorphism. As a consequence if uand

vare roots of the same minimal polynomial then F(u)∼

=F(v).

72

Let E, F be ﬁelds and let σ:F→Ebe an isomorphism. Then

it can be easily veriﬁed that the map that sends a polynomial p(x) =

c0+c1x+···+cnxnin F[x] to σ(p(x) = σ(c0) + σ(c1)x+· ··+σ(cn)xn

is an isomorphism. That is σextends F∼

=Eto F[x]∼

=E[x]. If p(x) is

irreducible, then σ(p(x) is also irreducible (see Exercise 32). The next

step is to show that σextends to an isomorphism between extension

ﬁelds.

Theorem 3.4.5. Let σ:F→Ebe an isomorphism of ﬁelds. Let ube

an algebraic element in some extension ﬁeld of Fwith minimal poly-

nomial p(x)∈F[x]. Let σ(p(x)) be the irreducible polynomial obtained

by applying σto the coeﬃcients of p(x)and let vbe a root of σ(p(x)).

Then σextends to an isomorphism of ﬁelds F(u)and E(v).

Proof. By Exercise 25, F[x]/(p(x)) ∼

=F(u) and E[x]/(σ(p(x))) ∼

=

E(v). Since σis an isomorphism, the maximal ideal (p(x)) gets mapped

to the maximal ideal σ(p(x)). Therefore the Kernel of the composition

of the surjective functions

F[x]→E[x]→E[x]/(σ(p(x))) →E(v).

is (p(x)). By the First Isomorphism Theorem F[x]/p(x)∼

=E(v).

Thus F(u)∼

=E(v).

If f(x) factors in K[x] as

f(x) = c(x−u1)(x−u2). . . (x−un)

then we say that f(x)splits over the ﬁeld K. In other words, K

contains all the roots of f(x).

Deﬁnition 3.4.2. If Fis a ﬁeld and f(x)∈F[x], then an extension

ﬁeld Kof Fis said to be a splitting ﬁeld of f(x)over Fprovided

that

1. f(x)splits over K, say f(x) = c(x−u1)(x−u2)···(x−un)and

2. K=F(u1, u2, . . . , un).

Example 3.4.6. 1. The polynomials f(x) = 2x4+x3−21x2−14x+

12 factorizes as (x+ 3)(x−1

2)(2x2−4x−8) over Q. The roots of

the factor 2x2−4x−8 are 1 ±√5 (apply quadratic formula). So

the splitting ﬁeld of f(x) over Qis Q(√5).

73

2. The splitting ﬁeld of f(x) = x2+ 1 over Ris R(i) = C(see

Exercise 18), where i=√−1. But the splitting ﬁeld of f(x) over

Qis Q(i) which is a much smaller ﬁeld than C.

By Theorem A.2.7 f(x) is irreducible in R[x] if and only if f(x) is a

ﬁrst degree polynomial or a second degree polynomial such that its

discriminant is negative. Consequently the splitting ﬁeld of f(x)

is either Ror R(i) = C. This gives us the Fundamental Theorem

of Algebra, that is, every polynomial with real coeﬃcients has a

root in C.

Next we prove that splitting ﬁelds always exist.

Theorem 3.4.6. Let Fbe a ﬁeld and let f(x)be a non-constant poly-

nomial of degree nin F[x]. Then there exits a splitting ﬁeld Kof f(x)

over Fsuch that [K:F]≤n!.

Proof. The proof is by induction on the degree of f(x). If f(x) has

degree 1 then Fis the splitting ﬁeld of f(x) and [F:F] = 1 <1!. Sup-

pose the theorem is true for all polynomials of degree less than nand

that f(x) has degree n. Every polynomial is a product of irreducible

factors therefore f(x) has an irreducible factor in F[x]. Multiplying

this factor by the inverse of its leading coeﬃcient we get a monic ir-

reducible factor p(x) of f(x). By Theorem 3.2.8 there is an extension

ﬁeld that contains a root uof p(x) and hence of f(x). Moreover p(x)

is necessarily the minimal polynomial of u. Consequently by Theorem

3.4.4 [F(u) : F] = deg p(x)≤deg f(x) = n. Now f(x) factorizes

as f(x) = (x−u)g(x) for some g(x)∈F(u)[x]. Since g(x) has de-

gree n−1, the induction hypothesis gives us a splitting ﬁeld Kof

g(x) over F(u) such that [K:F(u)] ≤(n−1)!. In K[x], g(x) = c(x−

u1)···(x−un−1) and hence f(x) = c(x−u)(x−u1)···(x−un−1). Since

K=F(u)(u1, . . . , un−1) = F(u, u1, . . . , un−1), Kis a splitting ﬁeld of

f(x) over Fsuch that [K:F] = [K:F(u)][F(u) : F]≤n(n−1)! = n!.

This completes the inductive step and hence the proof of the Theo-

rem.

Two splitting ﬁelds of a polynomial are isomorphic. The standard

way to prove this fact is by proving a stronger result that an isomor-

phism σbetween ﬁelds Fand Eextends to an isomorphism of splitting

ﬁelds. Then by setting F=Eand σto be the identity map we get

that any two splitting ﬁelds of a polynomial are isomorphic.

74

Theorem 3.4.7. Let σ:F→Ebe an isomorphism of ﬁelds, f(x)a

non-constant polynomial in F[x]and σf (x)the corresponding polyno-

mial in E[x]. If Kis a splitting ﬁeld of f(x)over Fand Lis a splitting

ﬁeld of σf (x)over E, then σextends to an isomorphism K∼

=L.

Proof. The proof is by induction on the degree of f(x). If deg

f(x) = 1, then K=F.σ(f(x)) also has degree 1 and therefore

E=L. Thus σprovides the isomorphism of the splitting ﬁelds too.

Now suppose the Theorem is true for polynomials of degree n−1 and

f(x) has degree n. As in Theorem 3.4.6, f(x) has a monic irreducible

factor p(x). Let ube a root of p(x) and vbe a root of σ(p(x)). Then

by Theorem 3.4.5 F(u)∼

=E(v). Now f(x) = (x−u)g(x) and degree of

g(x) = n−1. Therefore by the induction hypothesis the isomorphism

F(u)∼

=E(v) can be extended to an isomorphism K∼

=Lwhere K

is the splitting ﬁeld of g(x) over F(u) and Lis the splitting ﬁeld of

σ(g(x)) over E(v). Consequently Kand Lare also splitting ﬁelds of

f(x) and σ(f(x)) and this proves the Theorem.

A polynomial f(x) is said to be separable if it has no repeated roots

in any splitting ﬁeld. The derivative of

f(x) = c0+c1x+c2x2+···+cnxn∈F[x]

is

f′(x) = c1+ 2C2x+ 3C3x2+···ncnxn−1∈F[x].

When F=Rthis is the usual derivative of calculus.

Lemma 3.4.1. Let Fbe a ﬁeld and f(x)∈F[x]. If f(x)and f′(x)

are relatively prime in F[x]then f(x)is separable.

Proof. Let Kbe a splitting ﬁeld of f(x) and suppose on the contrary

f(x) is not separable. Then f(x) must have a repeated root uin K.

Hence f(x) = (x−u)2g(x) for some g(x)∈K[x] and by Exercise 26

f′(x) = (x−u)2g′(x) + 2(x−u)g(x).

Therefore f′(u) = 0Fand uis a root of f′(x). Consequently, the

minimal polynomial of udivides both f(x) and f′(x). Therefore f(x)

and f′(x) are not relatively prime which is a contradiction. Hence f(x)

is separable.

Theorem 3.4.8. Let Fbe a ﬁeld of characteristic zero. Then every

irreducible polynomial in F[x]is separable.

75

Proof. An irreducible polynomial p(x)∈F[x] is nonconstant and

hence

p(x) = cxn+ (lower degree terms), withc̸= 0Fand n≥1.

Then

p′(x) = (nc)xn−1+ (lower degree terms), withnc ̸= 0F.

Therefore p′(x) is a nonzero polynomial of lower degree than p(x).

Since p(x) is irreducible, p(x) and p′(x) are relatively prime. Hence

p(x) is separable by Lemma 3.4.1.

The Theorem is false if Fdoes not have characteristic 0.

Example 3.4.7. Consider the polynomial f(x) = x2−yin Z2(y)where

yis an indeterminate. Then f(x)is irreducible because it has no roots

in Z2(y). Since f′(x) = 0,f(x)is not separable by Lemma 3.4.1.

Corollary 3.4.9. Let Fbe a ﬁeld. Then an irreducible polynomial

f(x)∈F[x]is separable if f′(x)̸= 0.

Proof. The proof is similar to the proof of Theorem 3.4.8.

Theorem 3.4.10. Let Kbe an extension ﬁeld of Zpand na positive

integer. Then Khas order pnif and only if Kis a splitting ﬁeld of

xpn−xover Zp.

Proof. Assume Kis a splitting ﬁeld of xpn−x∈Zp[x]. Since

f′(x) = pnxpn−1−1 = −1, f(x) is separable by Lemma 3.4.1. Moreover,

the set Econsisting of the pndistinct roots of f(x) is a subﬁeld of K

by Exercise 27. Since Kis a splitting ﬁeld, Kis the smallest ﬁeld

containing the set Eof roots. Hence, K=E, which implies Khas

order pn.

Conversely, suppose Khas order pn. Theorem 4.5.8 implies that

every nonzero element cof Ksatisﬁes cpn−1= 1K. Therefore cis a root

of xpn−x. 0Kis also a root of xpn−x. Hence, the pnelements of K

are all the possible roots of xpn−x. Therefore Kis the splitting ﬁeld

of xpn−x.

Corollary 3.4.11. For each positive prime pand positive integer n,

there exists a ﬁeld of order pn.

76

Proof. A splitting ﬁeld of xpn−xover Zpexists by Theorem 3.4.6

It has order pnby Theorem 3.4.10.

Example 3.4.8. Let p= 2, n = 2 in Corollary 3.4.11. Since

x4−x=x(x+ 1)(x2+x+ 1) ∈Z2,

the splitting ﬁeld of x4−xis Z2/(x2+x+ 1) = {[0],[1],[x],[x+ 1]}.

Corollary 3.4.12. Two ﬁnite ﬁelds of the same order are isomorphic.

Proof. If Kand Lare ﬁelds of order pn, then both are splitting ﬁelds

of xpn−xover Zp, by Theorem 3.4.10. Hence they are isomorphic by

Theorem 3.4.7.

Finite ﬁelds have many applications in many areas including combi-

natorics, cryptography, projective geometry, and experimental design.

We use ﬁnite ﬁelds to count mutually orthogonal Latin squares and to

generate algebraic codes in Chapter 6.

Exercises.

1. A relation T⊂A×Aon a set Ais called an equivalence relation

provided that Tis reﬂexive ((a, a)∈T, for every a∈A), sym-

metric (if (a, b)∈T, then (b, a)∈T), and transitive (if (a, b)∈T

and (b, c)∈T, then (a, c)∈T). Let ∼be an equivalence relation

on a set A. Then the equivalence class of a∈A, denoted [a], is

the set

[a] = {b|b∈Aand b∼a}.

Prove that if a, b ∈Athen a∼bif and only if [a] = [b] and

that any two equivalence classes are either disjoint or identical.

Note that the congruence modulo relations in this chapter are

equivalence relations.

2. Show that

39 mod 181 = 39,181 mod 39 = 25,39 mod 39 = 0,

−17 mod 55 = 38,0 mod 39 = 0,25 mod 5 = 0,

−13 mod 5 = 2,1 mod 39 = 1,39 mod 13 = 0.

77

3. Prove the Freshman’s dream: Let pbe a prime and Ra commuta-

tive ring with identity of characteristic p. Then for every a, b ∈R

and every positive integer n,

(a+b)pn=apn+bpn.

4. Let f(x), g(x), h(x)∈Z[x] with f(x) = g(x)h(x). If pis a prime

that divides every coeﬃcient of f(x), then either pdivides every

coeﬃcient of g(x) or pdivides every coeﬃcient of h(x).

5. Prove that f(x) is an associate of g(x) if and only if g(x) is an

associate of f(x).

6. Verify that f(x) = g(x)h(x) in Z[x] implies that f(x) = g(x)h(x)

in Zp[x].

7. Prove that there are pn+1 −pnpolynomials of degree nin Zp[x].

8. Determine whether the two rings are isomorphic.

(a) Qand R.

(b) R×Rand C.

(c) Z4×Z4and Z16.

(d) Z6and Z2×Z3.

9. Let f:C→Cbe the complex conjugation map given by f(a+

bi) = a−bi. Show that fis an isomorphism.

10. Let f:R→Sbe a homomorphism of rings. Prove that f(0R) =

0S. Also prove that f(−a) = −f(a) for every a∈R.

11. Prove that f, g :R→Rgiven by f(x) = x+ 1 and g(x) = 2xare

not isomorphisms.

12. Let Rbe a commutative ring with identity and let Mbe an ideal

of R. Prove that the set J={m+ra|r∈Rand m∈M}is an

ideal in Rthat contains M.

13. If Rand Sare rings and f:R→Sis a homomorphism, prove

that f(R) = {f(a)∈S|a∈R}is a subring of S.

14. Let p(x)∈Zn[x] be a polynomial of degree k. Prove that there

are nkdistinct congruence classes in Zn[x]/(p(x)).

78

15. Let I={0,3}in Z6. Verify that Iis an ideal and show that

Z6/I ∼

=Z3.

16. Let Ibe an ideal in a noncommutative ring Rsuch that ab−ba ∈I

for all a, b ∈R. Prove that R/I is commutative.

17. Use the First Isomorphism Theorem to show that Z20/ < 5>∼

=

Z5.

18. Prove that the ﬁeld R(i) is C, where i=√−1.

19. Let Fbe a ﬁeld and let a, b ∈F. If ab = 0Fprove that either

a= 0 or b= 0.

20. Prove that if S={u1, . . . , un}spans Kover Fthen some subset

of Sis a basis of Kover F.

21. Let Kbe an extension ﬁeld of F. Prove that any two ﬁnite bases

of Kover Fhave the same number of elements.

22. Let F, K , and Lbe ﬁelds such that F⊆K⊆L. If [K:F]

and [L:K] are ﬁnite, then prove that Lis a ﬁnite dimensional

extension of Fand [L:F] = [L:K][K:F].

23. Let Kand Lbe ﬁnite dimensional extension ﬁeld of Fand let

f:K→Lbe an isomorphism such that f(c) = cfor every c∈F.

Prove that [K:F] = [L:F].

24. Prove that a minimal polynomial of an algebraic element over a

ﬁeld Falways exist and is unique.

25. In Theorem 3.4.4 show that ϕis an isomorphism between F(u)

and F[x]/(p(x)).

26. Let kbe a ﬁeld and let f, g ∈k[x]. Prove that the following rules

hold for derivatives: (f+g)′(x) = f′(x) + g′(x) and (fg)′(x) =

f(x)g′(x) + g(x)f′(x)

27. Let Kbe a splitting ﬁeld of xpn−x∈Zp[x]. Prove that the set

Econsisting of all the pndistinct roots of the polynomial xpn−x

is a subﬁeld of K.

28. Prove that if Kis a ﬁnite dimensional extension ﬁeld of F, then

Kis an algebraic extension of F.

79

29. Prove that if Kis a ﬁnitely generated separable extension ﬁeld of

F, then K=F(u) for some u∈K.

30. Prove that if K=F(u1, . . . , un) is a ﬁnitely generated extension

ﬁeld of Fand each uiis algebraic over F, then Kis a ﬁnite

dimensional algebraic extension of F.

31. Let f(x) be an irreducible polynomial in Zp[x] such that degree

of f(x) divides n. Show that the polynomial f(x) is a factor of

xpn−xin Zp[x].

32. Let σ:F→Ebe an isomorphism of ﬁelds, and let σ(p(x)) denote

the polynomial obtained by applying σto the coeﬃcients of p(x).

Show that σ(p(x)) is irreducible.

80

Chapter 4

Formulas to ﬁnd roots of

polynomials.

There is something to complete in this demonstration. I do not have

the time - Evariste Galois.

Most of us know how to solve a polynomial of degree 2 using the

quadratic formula. It is natural to ask whether there are such formulas

for polynomials of degrees greater than 2. In this chapter, we provide

formulas for ﬁnding roots of polynomials of degrees 3 and 4, and prove

that no formulas can exist for polynomials of degrees greater than 4.

4.1 Groups.

In this section, we introduce groups which are algebraic structures sim-

ilar to rings but with only a single operation. We use groups later in

the chapter to analyze roots of polynomial equations.

Deﬁnition 4.1.1. A group is a nonempty set Gequipped with an op-

eration ∗that satisﬁes the following properties.

1. Closure: If a∈Gand b∈G, then a∗b∈G.

2. Associativity: a∗(b∗c) = (a∗b)∗c, for all a, b, c ∈G.

3. There is an element e∈G(called the identity element) such that

a∗e=a=e∗afor every a∈G.

81

4. For each a∈G, there is an element a−1∈G(called the inverse

of a) such that a∗a−1=e=a−1∗a.

A group Gis said to abelian if its operation ∗is commutative, that

is,

a∗b=b∗afor all a, b ∈G.

Generally, for groups the multiplicative notation is used. Whenever

the operation is addition we switch to suitable notation. For example

we replace −aas inverse of ainstead of a−1and so on.

Example 4.1.1. 1. We prove that the set G={1,−1, i, −i} ∈ Cis

a group under multiplication by checking the four axioms in the

deﬁnition of a group. From the operation table for Ggiven below

we verify that 1 is the multiplicative identity, every element has

an inverse and that closure and associativity holds in G. Thus G

is a group. We also check that Gis commutative from the same

table.

·1 -1 i -i

1 1 -1 i -1

-1 -1 1 -i i

i i -i -1 1

-i -i i 1 -1

Table 4.1: The operation table of G.

2. It is easy to verify that every ring is an abelian group under

addition. Also check that the nonzero elements of a ﬁeld form an

abelian group under multiplication.

3. Let G1, G2, . . . , Gnbe groups. We deﬁne a coordinate-wise oper-

ation on the Cartesian product G1×G2× · ·· × Gn:

(a1, a2, . . . , an)(b1, b2, . . . , bn) = (a1b1, a2b2, . . . , anbn).

Check that G1×G2× ·· · × Gnis a group under this operation.

4. From Example 3, we know that in the ring Z8, the set of units

U8={1,3,5,7}.U8is a group under multiplication (see operation

table in Example 4.1.2).

82

Just like in the case of rings, isomorphisms play a critical role and

isomorphic groups are considered to be essentially the same.

Deﬁnition 4.1.2. Let Gand Hbe groups. A function f:G→H

is a homomorphism if f(a∗b) = f(a)∗f(b)for all a, b ∈G. The

group Gis said to be isomorphic to the group Hif there is a bijective

homomorphism from Gto H.

Example 4.1.2. We show that the multiplicative group U8={1,3,5,7}

of units in Z8is isomorphic to the additive group Z2×Z2. Let the func-

tion f:U8→Z2×Z2be such that

f(1) = (0,0), f (3) = (1,0), f(5) = (0,1), f (7) = (1,1).

fis bijective by its deﬁnition. We determine that fis a ho-

momorphism from the operation tables of the two groups, that is,

f(ab) = f(a)f(b) for a, b ∈U8. Thus U8∼

=Z2×Z2.

U8Z2×Z2

◦1 3 5 7

1 1 3 5 7

3 3 1 7 5

5 5 7 1 3

7 7 5 3 1

+ (0,0) (1,0) (0,1) (1,1)

(0,0) (0,0) (1,0) (0,1) (1,1)

(1,0) (1,0) (0,0) (1,1) (0,1)

(0,1) (0,1) (1,1) (0,0) (1,0)

(1,1) (1,1) (0,1) (1,0) (0,0)

Next, we look at groups of permutations.

Deﬁnition 4.1.3. A permutation of the set Gof nelements is an

ordered arrangement of the nelements.

Let Sndenote the set of all permutations of the set {1,2, . . . , n}.

Example 4.1.3. The set S3of permutations of the set S={1,2,3}is

S3={123,231,312,213,321,132}.

We now describe a recursive algorithm to generate all the permu-

tations of {1,2, . . . , n}.

Algorithm 4.1.1 (Generating permutations).1. Write down each

permutation of {1,2, . . . , n −1},ntimes.

83

2. Interlace nwith these permutations from left to right to get Sn.

Example 4.1.4. We derive the permutations of the set {1,2}from the

permutation of the set {1}using Algorithm 4.1.1.

12

21

Again, applying Algorithm 4.1.1, we get that the permutations of

the set {1,2,3}are

1 2 3

132

31 2

2 1 3

231

32 1

Observe that a permutation is a bijective function ffrom the set G

to itself. We now introduce the cycle notation of permutations which

we use henceforth. Let a1, a2, . . . , ak,k≥1 be distinct elements of the

set {1,2, . . . , n}. Then (a1, a2, . . . , ak) denotes the permutation in Sn

that maps a1to a2,a2to a3,...,akto a1and maps every other element

of {1,2, . . . , n}to itself. (a1, a2, . . . , ak) is called a cycle of length kor

ak-cycle.

Example 4.1.5. In the cycle notation the identity permutation 123 ∈

S3can be written either as (1),(2),or (3), but the usual convention is

to denote the identity by (1) or e. The permutation 213 = (12), and

so on. Thus, in the cycle notation,

S3={(1),(123),(132),(12),(13),(23)}.

The product of permutations is the composition of permutations as

functions.

Example 4.1.6. In S4the product (243)(1243) is (1423) and (123)(12) =

(13).

Two cycles are said to be disjoint if they have no elements in com-

mon. We leave it as an exercise to show that every permutation in Sn

is a product of disjoint cycles.

84

Example 4.1.7. In S8the permutation 51724638 is the same as (1542)(37).

Lemma 4.1.1. Every permutation in Snis a product of transpositions.

Proof. Every permutation is a product of cycles by Exercise 6. Any

cycle (a1a2···ak) is a product of transpositions:

(a1a2···ak) = (a1ak)(a1ak−1)· ··(a1a3)(a1a2).

There are n!=1·2· · ·· · nelements in Snand Snis a nonabelian

group with the operation of product of permutations (see Exercise 2).

Check that the set of all permutations of a set Gwith nelements is

isomorphic to Sn. Shortly, we prove that every group is isomorphic to

a group of permutations.

Deﬁnition 4.1.4. A subset Kof a group Gis a subgroup of Gif Kis

itself a group under the operation in G.

Example 4.1.8. 1. Since every ring Ris a group under addition,

every subring is a subgroup of R. In particular, every ideal Ris

a subgroup of R.

2. The six subgroups of the group S3are

{e},{e, (12)},{e, (13)},{e, (23)},{e, (123),(132)},and S3.

3. A permutation is said to be even if it can be written as a product

of even number of transpositions. Otherwise it is called an odd

permutation. The set of all even permutations of Sn, denoted by

An, is a subgroup.

The next result helps us skip a couple of steps while checking

whether a subset of a group is a subgroup.

Theorem 4.1.1. A nonempty subset Hof a group Gis a subgroup of

Gprovided that

1. if a, b ∈H, then ab ∈Hand

2. if a∈Hthen a−1∈H.

85

Proof. By deﬁnition H⊂Gis a subgroup of Gif His a group.

Now Properties 1 and 2 are the closure and inverse axioms for a group.

Associativity holds in Hbecause His a subset of G. So we only have

to prove that the identity e∈H. Since His nonempty, there exists

an element c∈H. Now c−1∈Hby Property 2 and cc−1=e∈Hby

Property 1. Therefore His a group and hence a subgroup of G.

Note that to prove that a ﬁnite subset is a subgroup you need to

only check for closure (see Exercise 31).

Theorem 4.1.2. Let Gand Hbe groups and let f:G→Hbe a

homomorphism. Then Im fis a subgroup of H. If fis injective then

G∼

=Im f.

Proof. The identity eHis in Im fbecause

f(eG)f(eG) = f(eGeG) = f(eG) = eHf(eG).(4.1)

Since His a group, f(eG)−1exists. Multiplying Equation 4.1 by

f(eG)−1on both sides, we get f(eG) = eH. Therefore Im fis nonempty.

Since fis a homomorphism, f(a)f(b) = f(ab). Hence Im fis closed.

Now

f(a−1)f(a) = f(a−1a) = f(eG) = eH.

Similarly, we prove that f(a)f(a−1) = eH. Therefore, f(a−1) = f(a)−1.

Thus the inverse of f(a) is also in Im f. Therefore Im fis a subgroup

of Hby Theorem 4.1.1. Now fis a surjective function from Gto Im

f. Consequently, if fis also an injective homomorphism, then fis an

isomorphism.

The number of elements in a group is called the order of the group.

We denote the order of a group Gas |G|. An element ain a group is

said to have ﬁnite order if ak=efor some positive integer k. The order

of an element ais the smallest positive integer nsuch that an=e. The

order of ais denoted by |a|. The element ais said to have inﬁnite order

if ak̸=efor every positive integer k.

Example 4.1.9. 1. |Sn|=n!.

2. In the group G={±1,±i}under multiplication of complex num-

bers, |G|= 4. The order of iis 4 because i2=−1, i3=−i, i4= 1.

Similarly −ihas order 4. Whereas −1 has order 2. Finally, 1,

which is the multiplicative identity, has order 1.

86

3. In the additive group Z5, 3 has order 5 because:

3 + 3 = 1,3 + 3 + 3 = 4,3 + 3 + 3 + 3 = 2,3 + 3 + 3 + 3 + 3 = 0.

But in the additive group of integers Z, 3 has inﬁnite order.

Now we are ready to show that every group is isomorphic to a

permutation group.

Theorem 4.1.3 (Cayley’s Theorem).Every group is isomorphic to a

group of permutations. Moreover, every ﬁnite group Gof order nis

isomorphic to a subgroup of the symmetric group Sn.

Proof. Let A(G) be the set of all permutations of the set G. By

Exercise 12, A(G) is a group with composition as the group operation.

A(G) is also the set of all bijective functions from Gto G. Let a∈G

and let the map ϕa:G→Gbe such that ϕa(x) = ax. Then ϕa∈A(G)

by Exercise 26. Now deﬁne f:G→A(G) by f(a) = ϕa. Now

f(ab)(x) = ϕab(x) = ab(x). On the other hand f(a)◦f(b) = (ϕa◦

ϕb)(x) = ϕa(ϕb(x)) = ϕa(bx) = abx. Therefore f(ab) = f(a)◦f(b).

Thus fis a homomorphism. Consequently, Im fis a subgroup of A(G)

by Theorem 4.1.2. Suppose f(a) = f(b), then ϕa(x) = ϕb(x) for all

x∈G. Consequently, a=ae =ϕa(e) = ϕb(e) = be =b. Hence fis

injective. Therefore G∼

=Im fby Theorem 4.1.2.

If Ghas nelements, then A(G) is isomorphic to Snby Exercise 2.

But since Gis isomorphic to a subgroup of A(G) it follows that Gis

isomorphic to a subgroup of Sn.

Thus, in eﬀect, permutation groups are the only groups up to iso-

morphism. This representation of a group is sometimes useful because

permutations are concrete objects and calculations are straightforward.

But usually other isomorphic representations of a group lead to a bet-

ter understanding about the basic underlying structure of the group as

we shall see in following sections.

4.2 Cyclic groups.

In this section we study groups that are generated by a single element.

The next theorem deals with the properties of the order of an element

in a group. These properties are useful in determining the inherent

structure of the group.

87

Theorem 4.2.1. Let Gbe a group and let a∈G.

1. If ahas inﬁnite order, then the elements ak, with k∈Z, are all

distinct.

2. If ahas ﬁnite order nthen ak=eif and only if ndivides k.

Moreover, ai=ajif and only if i≡j(mod n).

3. If ahas order nand n=td with d > 0, then athas order d.

Proof.

1. Suppose ai=ajwith i>j. The multiplying both sides by a−j

shows that ai−j=e. Since i−j > 0 we get ahas ﬁnite order

which is a contradiction. Therefore the elements ak, with k∈Z,

are all distinct.

2. If ndivides k, say k=nt, then ak=ant = (an)t=et=e.

Conversely suppose that ak=e. Then divide kby nto get

k=nq +rsuch that 0 ≤r≤n. Consequently

e=ak=anq+r= (an)qar=eqar=ear=ar.

By the deﬁnition of order, nis the smallest positive integer with

an=e. Therefore r= 0 implying k=nq. Hence ndivides k.

Like before, ai=ajif and only if ai−j=e. And ai−j=eif and

only if ndivides i−j, that is, if and only if i≡j(mod n).

3. Now (at)d=atd =an=e. Consequently to show that dis the

order of atwe need to show that dis the smallest integer such

that (at)d=e. Let kbe any positive integer such that (at)k=e,

then atk =e, Since nis the order of a, by Part 2, ndivides

tk. Therefore tk =nr = (td)rfor some integer r. This implies

k=dr. Since kand dare positive integers and ddivides kwe get

d≤k. Thus, we conclude that athas order d.

Theorem 4.2.2. Let Gbe a group and let a∈G. Let <a>denote

the set of all powers of a, that is

< a >={an|n∈Z}={. . . , a−2, a−1, a0, a1, a2, . . . }.

Then, < a > is a subgroup of G.

88

Proof. The product of any two elements of < a > is in < a >

because aiaj=ai+j. The inverse of akis a−k, and a−kis also in < a >.

Therefore < a > is a subgroup by Theorem 4.1.1.

The group < a > is called the cyclic subgroup generated by a. If the

subgroup < a > is the entire group G, we say that Gis a cyclic group.

Observe that cyclic groups are necessarily abelian.

Example 4.2.1. 1. In S3, the cyclic subgroup <(123) >is

<(123) >={e, (123),(132)}.

2. In the additive group Z8, the cyclic subgroup <2>={2,4,6,0}.

The cyclic subgroup <1>is the entire group Z8and therefore

Z8is a cyclic group. Generalizing, Zn=<1>is cyclic.

3. The group Z=<1>and therefore is a cyclic group.

4. We prove that the group Zm×Znis cyclic if and only if gcd(m, n) =

1. Observe that the order of Zm×Znis mn. Let gcd(m, n) =

d > 1. Then m=dr and n=ds for some integers rand s. Thus

drs < d2rs =mn. If (a, b)∈Zm×Zn, then

drs(a, b) = (drsa, drsb) = (msa, nrb) = (0,0).

Thus the order of (a, b) is a divisor of drs and hence is strictly less

than mn. Thus Zm×Znis not cyclic when Let gcd(m, n)̸= 1.

When the gcd(m, n) = 1,

Zm×Zn=<(1,1) > .

Theorem 4.2.3. Let Gbe a group and let a∈G.

1. If ahas inﬁnite order, then < a > is an inﬁnite subgroup consist-

ing of the distinct elements akwith k∈Z.

2. If ahas ﬁnite order n, then < a > is a subgroup of order nand

< a >={e=a0, a1, . . . , an−1}.

Proof.

1. This follows from Part 1 of Theorem 4.2.1.

89

2. Part 2 of Theorem 4.2.1 says that ai=ajif and only if i≡j

(mod n). Every integer is in the congruency class of one of the

integers in {0,1, . . . , n −1}(see Section 3.1). Since no two in-

tegers 0,1, . . . , n −1 are congruent modulo n,ai̸=ajif i, j ∈

{0,1, . . . , n −1}. Therefore < a >={a0, a1, . . . , an−1}. Conse-

quently, < a > is a subgroup of order n.

The next theorem shows that cyclic groups have a nice classiﬁcation

up to isomorphism.

Theorem 4.2.4. Every inﬁnite cyclic group is isomorphic to Z. Every

ﬁnite cyclic group of order nis isomorphic to Zn.

Proof. Let G=< a > be an inﬁnite cyclic group. Deﬁne f:Z→G

by f(i) = ai. The map fis surjective by deﬁnition of a cyclic group. f

is injective by Part 1 of Theorem 4.2.3. fis a homomorphism because

f(i+j) = ai+j=aiaj=f(i)f(j). Thus fis an isomorphism.

Now suppose G=< a > and ahas ﬁnite order n. Then G=

{a0, a1, . . . , an−1}by Part 2 of Theorem 4.2.3. Let f:Zn→Gbe

such that f(i) = ai.fis injective by deﬁnition and fis a surjective

homomorphism just like above. Therefore, fis an isomorphism from

Znto G.

Subgroups can be generated by more than one element. Let Gbe

a group and a1, . . . , an∈G. Consider the set

< a1, a2, . . . , an>={

n



i=1

airi:ri∈Z, ri≥0}.

We leave it as an exercise to verify that < a1, a2, . . . , an>is a

subgroup of G.

Example 4.2.2. 1. The subgroup <(12),(123) >is the entire group

S3because

(123)2= (132),(123)3=e, (123)(12) = (13),(123)2(12) = (23).

2. The 6 transpositions of S4can be generated by the three trans-

positions (12),(13), and (14) as shown below.

(13)−1(12)(13) = (13)(12)(13) = (23)

(14)−1(12)(14) = (14)(12)(14) = (24)

(14)−1(13)(14) = (14)(13)(14) = (34)

90

Since every permutation is a product of transpositions (Lemma

4.1.1), we get <(12),(13),(14) >=S4.

4.3 Normal Subgroups and Quotient Groups.

In this section, we prove the First Isomorphism Theorem for groups.

We begin with congruence relations in a group.

Deﬁnition 4.3.1. Let Kbe a subgroup of a group Gand let a, b ∈G.

Then ais congruent to bmodulo K[written a≡b(mod K)] provided

that ab−1∈K.

Example 4.3.1. 1. In Z8, 3 ≡1 (mod 2) because 3−1 = 2 ∈<2>.

2. In S3, (12) ≡(13) (mod <(123) >) because (12)(13)−1= (12)(13) =

(132) ∈<(123) >.

Theorem 4.3.1. Let Kbe a subgroup of a group G. Then the relation

of congruence modulo Kis

•reﬂexive: a≡a(mod K) for all a∈G;

•symmetric: if a≡b(mod K), then b≡a(mod K);

•transitive: if a≡b(mod K) and b≡c(mod K), then a≡c(mod

K).

If Kis a subgroup of Gand if a∈G, then the congruence class

of amodulo Kis the set of all elements of Gthat are congruent to a

modulo K, that is, the set

{b∈G:b≡a(mod K)}={b∈G:ba−1∈K}

={b∈G:b=ka, for some k∈K}

={ka :k∈K}.

As a consequence the congruence class of amodulo Kis denoted

Ka and is called a right coset of Kin G. The set of all congruence

classes modulo Kis denoted G/K. A left coset of Kis denoted by aK

and is deﬁned as aK ={ak :k∈K}. If Gis abelian, then Ka =aK.

Example 4.3.2. 1. In S3

<(123) >(12) = {e(12),(123)(12),(132)(12)}={(12),(13),(23)}.

Check that the only right cosets of the subgroup <(123) >are

<(123) > e and <(123) >(12).

91

2. In Z8,

<2>+1 = {0 + 1,2+1,4+1,6+1}={1,3,5,7}.

Similarly, <2>+2 = {0,2,4,6}. Check that the only right

cosets of the subgroup <2>are <2>+0 and <2>+1.

Theorem 4.3.2. Let Kbe a subgroup of a group Gand let a, c ∈G.

Then a≡c(mod K) if and only if Ka =Kc.

Corollary 4.3.3. Let Kbe a subgroup of a group G. Then two right

cosets of Kare either disjoint or identical.

Proofs of Theorems 4.3.1, 4.3.2, and Corollary 4.3.3 are similar to

the proofs provided for congruence classes in Zin Section 3.1 and we

do not discuss it further.

Theorem 4.3.4. Let Kbe a subgroup of a group G.

1. Gis union of the right cosets of K.

2. If Kis ﬁnite, any two right cosets of Khave the same number of

elements.

Proof.

1. Let a∈G, then a∈Ka. Therefore, every element of Gis in one

of the cosets of K. Moreover, every coset of Kcontains elements

of G. Hence G=∪a∈GKa.

2. Deﬁne f:K→Ka by f(x) = xa. Let y∈Ka, then y=xa for

some x∈K. Therefore, f(x) = y. Consequently, fis surjective.

If f(x) = f(y), then xa =ya and therefore x=y. Thus, fis

injective. Consequently fis a bijection. Therefore |K|=|Ka|

for every right coset Ka of K.

Recall from Section 3.3 that the set of cosets of an ideal is a ring.

But the set of cosets of a subgroup need not be a group. Let Nbe a

subgroup of a group G. The set of right cosets G/N is called a quotient

group if G/N is a group. We prove, shortly, that G/N is a group if and

only if Nis a normal subgroup.

Deﬁnition 4.3.2. A subgroup Nof a group Gis said to be normal if

Na =aN for every a∈G.

92

Example 4.3.3. 1. Let N=<(123) >be the cyclic group gener-

ated by (123) in S3. Then the only two right cosets of Nin S3are

Ne and N(12). Therefore Nis a normal subgroup of S3because

Ne ={e, (123),(132)}=eN

N(12) = {(12),(13),(23)}= (12)N.

2. Every subgroup of an abelian group is normal.

3. < e > is a normal subgroup for every group.

4. Let H=Anbe the subgroup of even permutations of Sn. Then,

Ha =H=aH if ais an even cycle. By Exercise 11, |An|=1

2|Sn|.

Therefore, by Theorem 4.3.4, Hhas exactly two right cosets. Let

abe an odd cycle. Then the two right cosets of Hare He and

Ha. Similarly, the two left cosets are eH and aH. Consequently,

Ha =aH for all a∈Sn. Thus Anis a normal subgroup of Sn.

Lemma 4.3.1. If Nis a normal subgroup of Gthen for each a∈G,

a−1Na =N.

Proof. We ﬁrst show that a−1Na ⊆N. Let x∈a−1N a, then

x=a−1na for some n∈N. Since Nis normal Na =aN for every

a∈G. Therefore na =an′for some n′∈N. Consequently,

x=a−1na =a−1an′=n′∈N.

Therefore a−1Na ⊆N.

Next we need to show N⊆a−1Na. Let n∈N. Since Nis normal,

na−1=a−1n′for some n′∈N. Therefore

n=na−1a=a−1n′a.

Hence n∈a−1Na. This implies N⊆a−1Na. Thus, a−1Na =N.

Theorem 4.3.5. Let Nbe a normal subgroup of G. If a≡b(mod N),

and c≡d(mod N), then ac ≡bd (mod N).

Proof. Since a≡b(mod N), ab−1∈N. Therefore ab−1=n1for

some n1∈N. Similarly cd−1=n2for some n2∈N. By Exercise 1,

(bd)−1=d−1b−1. Consequently, ac(bd)−1=acd−1b−1=an2b−1. The

element an2is in aN. Since Nis normal aN =N a. Therefore an2=

n3afor some n3∈N. Thus ac(bd)−1=an2b−1n3ab−1=n3n1∈N.

Consequently ac ≡bd (mod N).

93

Theorem 4.3.6. Let Nbe a normal subgroup of a group G. If N a =

Nb and Nc =Nd in G/N, then Nac =Nbd.

Proof. By Theorem 4.3.2, Na =Nb implies a≡b(mod N) and

Nc =Nd implies c≡d(mod N). Consequently, ac ≡bd (mod

N) by Theorem 4.3.5. Hence, applying Theorem 4.3.2 again, we get

Nac =Nbd.

Theorem 4.3.7. If Nis a normal subgroup of G, then G/N is a group

under the operation deﬁned by (Na)(N c) = N ac. If Gis an abelian

group then so is G/N.

Proof. The operation in G/N is well deﬁned by Theorem 4.3.6.

Since NaNe =Nae =Nea =N eN a, the coset N=N e is the iden-

tity element in G/N. The inverse of Na is N a−1because N aN a−1=

Naa−1=Ne =Na−1a=Na−1N a. Associativity in G/N follows

from associativity in G: (Na)(NbNc) = NaN bc =N abc =N(ab)c=

(NaNb)Nc. Therefore G/N is a group. If Gis abelian, then commu-

tativity follows in G/N from the commutativity in G:NaN b =N ab =

Nba =NbNa.

Example 4.3.4. Examples of Quotient groups:

1.

Z8/ < 2>={<2>+0, < 2>+e}.

2.

S3/ < 123 >={<(123) > e, < (123) >(12)}.

The next theorem shows that there is a surjection between sub-

groups of a group Gand the subgroups of its quotient group G/N .

Theorem 4.3.8. Let Nbe a normal subgroup of a group G. If Tis any

subgroup of G/N, then there is a subgroup Hof Gsuch that N⊂H

and T=H/N.

Proof. Let H={a∈G|Na ∈T}, then His a subgroup of Gby

Exercise 17. Let a∈N, then Na =Ne ∈T, so that a∈H. Therefore

N⊆H. Now the quotient group H/N consists of all cosets Na such

that a∈H. Therefore T=H/N by the deﬁnition of H.

Deﬁnition 4.3.3. Let f:G→Hbe a homomorphism of groups.

Then the kernel of fis the set {a∈G|f(a) = eH}.

94

Theorem 4.3.9. Let f:G→Hbe a homomorphism of groups with

kernel K. Then Kis a normal subgroup of G.

Proof. If c, d ∈K, then f(c) = eHand f(d) = eHby deﬁnition of

the kernel. Hence f(cd) = f(c)f(d) = eHeH=eH. Therefore cd ∈K

and Kis closed. If c∈Kthen f(c−1) = f(c)−1=eH−1=eH.

Therefore c−1∈K. It follows that Kis a subgroup by Theorem

4.1.1. To show Kis a normal subgroup of G, we must prove that

for each a∈G, a−1Ka =K. Let a∈Gand c∈K. Then f(a−1ca) =

f(a−1)f(c)f(a) = f(a−1)eHf(a) = f(a)−1f(a) = eH. Thus a−1ca ∈K.

Consequently, Kis normal.

Theorem 4.3.10. If Nis a normal subgroup of a group G, then the

map π:G→G/N given by π(a) = Na is a surjective homomorphism

with Kernel N.

Proof. Translate the proof of Theorem 3.3.10 to this case.

Theorem 4.3.11. [First Isomorphism Theorem] Let f:G→Hbe a

surjective homomorphism of groups with kernel K. Then the quotient

group G/K is isomorphic to H.

Proof. Deﬁne ϕ:G/K →Hby ϕ(Ka) = f(a) and Show that

ϕis an isomorphism. The proof is similar to the proof of the First

Isomorphism Theorem for rings (see Theorem 3.3.11).

Example 4.3.5. Let R∗denote the multiplicative group of nonzero

real numbers and let R∗∗ denote the multiplicative group of positive

real numbers. Let f:R∗→R∗∗ be such that f(x) = x2. Then the

kernel of fis <1,−1>. Let y∈R∗∗, then f(√y) = y. Therefore

fis surjective. Hence by the First Isomorphism Theorem we get that

R∗/ < −1,1>∼

=R∗∗.

4.4 Basic properties of ﬁnite groups.

In this section we relate the order of a ﬁnite group to the orders of its

subgroups and elements.

If His a subgroup of a group Gthen the number of distinct right

cosets of Hin Gis called the index of Hin Gand is denoted by [G:H].

If Gis a ﬁnite group then [G:H] is ﬁnite. If Gis inﬁnite then [G:H]

can be either ﬁnite or inﬁnite.

95

Example 4.4.1. 1. Under addition, the group Zis a normal sub-

group of the abelian group Q. If 0 <c<a<1, then a−cis

not an integer. Therefore Z+aand Z+care distinct elements of

Q/Zby Theorem 4.3.2. Since there are inﬁnitely many rational

numbers between 0 and 1, the index [Q:Z] is inﬁnite. But the

order of Z+m

nis nbecause n(Z+m

n) = Z+m=Z=e. Thus

every element of Q/Zhas ﬁnite order.

2. Consider the subgroup N=<(123) >of S3. The index [S3:

N] = 2 by Exercise 4.3.3.

Theorem 4.4.1 (Lagrange’s Theorem).If His a subgroup of a ﬁnite

group G, then the order of Hdivides the order of G; in particular

|G|= [G:H]|H|.

Proof. Let [G:H] = n. Let H a1, . . . , Hanbe the ndistinct cosets

of H. By Theorem 4.3.4, G=Ha1∪Ha2∪ · ·· ∪ Han. Therefore

|G|=|Ha1|+|Ha2|+···+|Han|. Again, by Theorem 4.3.4, |Hai|=|H|

for every i. Therefore |G|=n|H|= [G:H]|H|.

Corollary 4.4.2. Let Gbe a ﬁnite group.

1. If a∈G, then the order of adivides the order of G.

2. If |G|=k, then ak=efor every a∈G.

3. If Nis a normal subgroup of G, then |G/N |=|G|/|N|.

Proof.

1. If a∈Ghas order nthen the cyclic subgroup < a > of Ghas

order nby Theorem 4.2.3. Consequently, by Lagrange’s Theorem,

ndivides |G|.

2. If ahas order n, then by Part 1, ndivides k. Therefore k=nt

for some t∈Z. Then ak=ant = (an)t=et=e.

3. |G/N|is the number of distinct right cosets of Nin G. Hence

|G/N|= [G:N].

By Lagrange’s Theorem |G|= [G:N]|N|. Therefore |G/N|=

|G|/|N|.

96

We use Lagrange’s theorem to show that every group of prime order

is cyclic.

Theorem 4.4.3. Let pbe a positive prime integer. Every group of

order pis cyclic and isomorphic to Zp.

Proof. If Gis a group of order pand ais any nonidentity element

of G, then the cyclic subgroup < a > is a group of order greater than

1. Since the order of the group < a > must divide pby Theorem 4.4.1,

and pis prime, order of < a >=p. Thus < a >=G. Since Gis a

cyclic group of order p,G∼

=Zpby Theorem 4.2.4.

If a prime pdivides |G|for a group G, then does Ghave an element

of order p? Cauchy’s Theorem says that there is always such an ele-

ment. We prove Cauchy’s Theorem in two steps, ﬁrst for ﬁnite abelian

groups, and then for all ﬁnite groups.

Theorem 4.4.4 (Cauchy’s Theorem for Abelian Groups.).If Gis a

ﬁnite abelian group and if pis a prime that divides the order of G.

Then Ghas an element of order p

Proof. The proof is by induction on the order of G. The the-

orem is true for |G|= 2 because in this case the nonidentity ele-

ment must have order 2. Assume the theorem is true for all abelian

groups of order less than nand suppose that |G|=n. Let abe any

nonidentity element of G, then |a|is divisible by some prime q, say

|a|=qt, then |at|=q. Therefore if q=pthe theorem is proved.

Let q̸=pand let Nbe the cyclic subgroup < at>.Nis normal

because Gis abelian. Consequently, since Nhas order q, by Corollary

4.4.2 the quotient group G/N has order |G|/|N|=n/q < n. Conse-

quently by the induction hypothesis the theorem is true for G/N. Now

|G|=|N||G/N|=q|G/N|by Theorem 4.4.1. Since pdivides |G|, and

q̸=p,pdivides |G/N|. Therefore G/N contains an element of order

p, say Nc. Since Ncp= (Nc)p=Ne,cp∈N. Because Nhas order q,

(cp)q=cpq =e. Therefore the order of cdivides pq. Now order of c̸= 1

because otherwise N c would have order 1 instead of pin G/N. The or-

der of cis not qbecause then (N c)q=Ncq=Ne in G/N which means

pwhich is the order of Nc divides q. This is not possible since qia

prime and p̸=q. Therefore the order of cis either por pq: in the later

case cqhas order p. Therefore the theorem is true for abelian groups

of order nand hence by induction for all ﬁnite abelian groups.

97

To prove Cauchy’s theorem for all ﬁnite groups, we need to develop

some additional concepts. Let Gbe a group and a, b ∈G. We say ais

conjugate to bif there exists x∈Gsuch that b=x−1ax.

Example 4.4.2. (12) is conjugate to (23) in S3because

(132)−1(12)(132) = (123)(12)(132) = (23).

Let Gbe a group, The conjugacy class of an element a∈Gconsists

of all the elements in Gthat are conjugate to a. We leave it as an

exercise to show that Gis a union of its distinct conjugacy classes.

Example 4.4.3. 1. For any x∈S3,x−1(12)xis either (12),(13), or

(23):

e−1(12)e=e(12)e= (12),

(12)−1(12)(12) = (12)(12)(12) = (12),

(23)−1(12)(23) = (23)(12)(23) = (13),

(13)−1(12)(13) = (13)(12)(13) = (23),

(132)−1(12)(132) = (123)(12)(132) = (23),

(123)−1(12)(123) = (132)(12)(123) = (13).

Therefore the conjugacy class of (12) in S3is {(12),(13),(23)}.

Verify that there are three distinct conjugacy classes in S3:

{e},{(123),(132)},and {(12),(13),(23)}.

Observe that

S3={e} ∪ {(123),(132)} ∪ {(12),(13),(23)}.

2. Verify that the distinct conjugacy classes of S4are

{e}

{(1234),(1243),(1324),(1342),(1423),(1432)}

{(12)(34),(13)(24),(14)(23)}

{(12),(13),(14),(23),(24),(34)}

{(123),(132),(124),(142),(134),(143),(234),(243)}

The centralizer of an element ain a group Gis denoted by C(a)

and consists of all elements in Gthat commute with a, that is,

C(a) = {g∈G|ga =ag}.

98

Example 4.4.4.

C((123)) = {(1),(123),(132)}in S3.

C(a) is a subgroup of G(see Exercise 33).

Theorem 4.4.5. Let Gbe a group and a∈G. The number of elements

in the conjugacy class of ais [G:C(a)], and divides |G|.

Proof. We ﬁrst show that xand yproduce the same conjugate of a

if and only if xand yare in the same coset of c(a):

x−1ax =y−1ay ⇔a=xy−1ayx−1

⇔a= (yx−1)−1a(yx−1)

⇔(yx−1)a=a(yx−1)

⇔yx−1∈C(a)

⇔C(a)y=C(a)x.

Therefore the number of distinct conjugates of ais the same as the

number of distinct cosets of C(a), namely [G:C(a)], which divides |G|

by the Lagrange’s Theorem 4.4.1.

Let Gbe a group and let C1, C2, . . . , Crbe the distinct conjugacy

classes of G. Then

|G|=|C1∪C2∪ · ·· ∪ Ct|=|C1|+|C2|+·· · +|Ct|.(4.2)

Let aibe an element in Cithen by Theorem 4.4.5

|G|= [G:C(a1)|+ [G:C(a2)] + · ·· + [G:C(at)].(4.3)

The equation (in either version 4.2 or 4.3) is called the class equation

of the group G.

Example 4.4.5. The class equation for the group S3is

|S3|=|{e}| +|{(123),(132)}| +|{(12),(13),(23)}|.

The center of a group Gis the set Z(G) consisting of those elements

of Gthat commute with every element of G, that is,

Z(G) = {c∈G|cx =xc for every x∈G}.

Verify that Z(G) is a subgroup of G.

99

Example 4.4.6. 1. If Gis an abelian group then the center of G,

Z(G) = G.

2. Check that Z(S3) =< e >.

3. Consider the Dihedral subgroup of S4

D4=ρ= (1234), ρ2= (13)(24), ρ3= (1432), ρ4=e,

τ= (12)(34), τ ρ = (24), τ ρ2= (14)(23), τρ3= (13) .

Every element of D4is of the form τmρnwhere mand nare

integers such that m, n ≥0. Therefore to show that ρ2commutes

with every element of D4, it suﬃces to show that it commutes with

ρand τ. Now ρρ2=ρ3=ρ2ρ. Since the inverse of ρ2is itself,

(ρ2)−1τρ2=ρ2τρ2= (13)(24)(12)(34)(13)(24) = (12)(34) = τ,

that is τρ2=ρ2τ. Consequently, ρ2∈Z(D4). Verify that no

other nonidentity element of D4is in Z(D4). Therefore Z(D4) =

{e, ρ2}.

Note that Z(G) is the union of one-element conjugacy classes and

the class equation can be written as

|G|=|Z(G)|+|C1|+|C2|+· ·· +|Cr|,(4.4)

where C1, . . . , Crare the distinct conjugacy classes of Gthat contain

more than one element. Moreover, |Ci|divides |G|, for i= 1 to r.

Theorem 4.4.6. If Nis a subgroup of Z(G), then Nis a normal

subgroup of G.

Proof. Let a∈Gand n∈N, then na =an because n∈Z(G).

Thus Na = aN for all a∈Gwhich implies Nis normal.

Theorem 4.4.7 (First Sylow Theorem).Let Gbe a ﬁnite group. If p

is a prime and pkdivides |G|, then Ghas a subgroup of order pk.

Proof. The proof is by induction on the order of G. If |G|= 1, then

p0is the only prime power that divides |G|, and Gitself is a subgroup

of order p0. Suppose that |G|>1 and assume inductively that the

theorem is true for all groups of order less than |G|. Combining the

forms of the class Equation 4.3 and 4.4, we get

|G|=|Z(G)|+ [G:C(a1)] + [G:C(a2)] + · ·· + [G:C(ar)],

100

where [G:C(ai)] >1 for each i. Moreover, |Z(G)| ≥ 1 because

e∈Z(G) and |C(ai)|<|G|otherwise [G:C(ai)] = 1.

Suppose pdoes not divide [G:C(aj)] for some j. Then since pk

divides |G|,pkmust divide |C(aj)|because, by Lagrange’s Theorem,

|G|=|C(aj)|[G:C(aj)]. Since the subgroup C(aj) has order less than

|G|, the induction hypothesis implies that C(aj), and hence G, has a

subgroup of order pk.

On the other hand, if pdivides [G:C(ai)] for every ithen since

pdivides |G|,pmust divide |Z(G)|because |Z(G)|=|G| − r

i=1[G:

C(ai)]. Since Z(G) is abelian, Z(G) contains an element cof order p

by Theorem 4.4.4. Let Nbe the cyclic group generated by Cthen N

is normal in Gby Theorem 4.4.6. Consequently |G/N |=|G|/p is less

than |G|and divisible by pk−1. By the induction hypothesis G/N has

a subgroup Tof order pk−1. By Theorem 4.3.8, there is a subgroup H

of Gsuch that N⊆Hand T=H/N. Now by Lagrange’s Theorem

|H|=|N||H/N|=|N||T|=ppk−1=pk. So Ghas a subgroup of order

pkin this case too.

Corollary 4.4.8 (Cauchy’s Theorem).If Gis a ﬁnite group whose

order is divisible by a prime p, then Gcontains an element of order p.

Proof. Since pdivides |G|, Theorem 4.4.7 implies that |G|has a

subgroup Kof order p. Since Kis cyclic by Theorem 4.4.3, Khas a

generator which is an element of order pin G.

4.5 Finite Abelian Groups.

A major goal of group theory is to classify all ﬁnite groups up to iso-

morphism. We do not cover the group classiﬁcation problem in great

detail in this book. The interested reader may refer to [19], [20], and

the references therein for a detailed study. However, in this section, we

classify all ﬁnite abelian groups up to isomorphism.

If Gis an abelian group and if pis a prime, then G(p) denotes the

set of elements in Gwhose order is some power of p:

G(p) = {a∈G:|a|=pnfor some n≥0}.

Lemma 4.5.1. G(p)is a subgroup of G.

Proof.

101

Let a, b ∈G(p) and let the order of aand bbe pnand pmrespectively.

Let n>mand let n=m+rwhere r≥0, then pn=pmpr. Now

(ab)pn=apnbpn=eG(bpm)pr= (eG)pr=eG. Thus the order of ab

divides pnby Theorem 4.2.1. Therefore the order is some power of

pand hence ab ∈G(p). Hence G(p) is closed. If a∈G(p), then

a−1∈G(p), because apn=eGimplies (a−1)pn=eG. Therefore G(p) is

a subgroup of Gby Theorem 4.1.1.

Theorem 4.5.1. Let Gbe an abelian group and let a∈Gbe an element

of ﬁnite order. Then a=a1a2·· ·akwith ai∈G(pi)where p1,··· , pk

are distinct primes that divide the order of a.

Proof. The proof is by induction on the number of distinct primes

that divide the order of a. If |a|is divisible only by the single prime

p1, then the order of ais a power of p1and hence a∈G(p1). So the

theorem is true for k= 1. Assume inductively that the theorem is

true for all elements whose order is divisible by at most k−1 distinct

primes and that |a|is divisible by the distinct primes p1, . . . pk. Then

|a|=pr1

1···prk

kwith each ri>0. Let m=pr1

2···prk

kand n=pr1

1so

that |a|=mn. Since the gcd (m, n) = 1, by Theorem A.1.1 there are

integers u, v such that 1 = mu +nv. Consequently

a=a1=a(mu+nv)=amuanv.

Since (amu)pr1

1= (amn)u=eu

G=eG, order of amu divides pr1

1. There-

fore amu ∈G(p1). Similarly, (anv)m=eG. Therefore the order of anv

divides m. But mhas only k−1 distinct prime divisors. Therefore by

the induction hypothesis anv =a2···akwith ai∈G(pi). Let a1=amu .

Then a=a1···akwith ai∈G(pi).

Theorem 4.5.2. If N1, . . . , Nkare normal subgroups of a group Gsuch

that every element of Gcan be written uniquely in the form a1a2. . . ak

with ai∈Ni, then G∼

=N1×N2× · ·· × Nk.

Proof. Let f:N1×N2×···×Nk→Gbe such that f(a1, a2, . . . , ak) =

a1a2···ak. Then fis an isomorphism between N1×N2×···×Nkand

G(see Exercise 20).

Theorem 4.5.3. If Mand Nare normal subgroups of a group Gsuch

that G=MN and M∩N=< eG>, then G∼

=M×N.

102

Proof. By hypothesis every element of Gis of the form mn with m∈

Mand n∈N. Now suppose that an element had two representations,

say m1n1=m2n2, with m1, m2∈Mand n1, n2∈N. Then multiplying

on the left by m−1

2and on the right by n−1

1, that is, m−1

2m1n1n−1

1=

m−1

2m2n2n−1

1shows that m−1

2m1=n2n−1

1. But m−1

2m1∈Mand

n2n−1

1∈Nand M∩N=< eG>. Hence m−1

2m1=eG=n2n−1

1. This

implies m1=m2and n1=n2. Therefore every element of Gcan be

written uniquely in the form mn such that m∈Mand n∈N. Hence,

by Theorem 4.5.2, G∼

=M×N.

Theorem 4.5.4. If Gis a ﬁnite abelian group, then

G∼

=G(p1)×G(p2)× · ·· × G(pt),

where p1, . . . , ptare the distinct primes that divide the order of the

group.

Proof. If a∈G, then |a|divides |G|, by Corollary 4.4.2. By The-

orem 4.5.1, a=a1a2··· atwith ai∈G(pi) (aj= 1 if a prime pj

does not divide |a|). To prove this expression is unique, suppose that

a1···at=b1· ··bt, with ai, bi∈G(pi). Since Gis abelian

a1b−1

1=b2a−1

2b3a−1

3···bta−1

t.

For each i,bia−1

i∈G(pi) and hence has order pri

iwith ri≥0. If

m=pr2

2···prt

t, then (bia−1

i)m=eGfor i≥2 so that

(a1b−1

1)m= (b2a−1

2)m(b3a−1

3)m···(bta−1

t)m=eG.

Consequently the order of a1b−1

1must divide m. Since a1b−1

1∈

G(p1), this is possible only if the order of a1b−1

1is 1, that is a1b1−1 = eG.

Therefore a1=b1. Similar arguments for i= 2, . . . , t show that ai=bi

for every i. Therefore every element can be uniquely written in the

form a=a1a2···atwith ai∈G(pi). Consequently, by Theorem 4.5.2,

G∼

=G(p1)×G(p2)× · ·· × G(pt).

An element aof a p-group Gis called an element of maximal order

if |g| ≤ |a|for every g∈G. In other words, if |a|=pn, and g∈G, then

|g|=pjwith j≤n. Since pn=pn−jpj,gpn= (gpj)pn−j

=efor every

g∈G. Elements of maximal order always exist in a ﬁnite p-group.

Lemma 4.5.2. Let Gbe a ﬁnite abelian p-group and let abe an element

of maximal order in G. Then there is a subgroup Kof Gsuch that

G∼

=< a > ×K.

103

Proof. Consider those subgroups Hof Gsuch that <a>∩H=<

eG>. There is at least one such subgroup H=< eG>and since G

is ﬁnite there is a largest subgroup Kwith this property. To show

that G∼

=< a > ×K, we need only show that G=< a > K by

Theorem 4.5.3. Suppose this is not the case, then there exists b∈G

such that b̸=eGand b̸∈< a > K. Let qbe the smallest integer

such that bpq∈< a > K. Such a qexists because Gis a p-group and

bpj=eG=eGeG∈< a > K for some j > 0. Then

c=bpq−1̸∈< a > K (4.5)

and cp=bpq∈< a > K. Let

cp=atkwhere t∈Zand k∈K. (4.6)

If ahas order pnthen xpn=eGfor all x∈Gbecause ahas maximal

order. Consequently by Equation 4.6

eG=cpn= (cp)pn−1= (atk)pn−1= (at)pn−1kpn−1.

Therefore (at)pn−1=k−pn−1∈<a>∩K=< eG>and thus

(a)tpn−1=eG. Consequently pn(order of a) divides tpn−1and it follows

that pdivides t. Let t=mp for some mthen cp=ampk. Therefore

k=cpa−pm = (ca−m)p. Let

d=ca−m,(4.7)

then dp∈Kbut d̸∈ K(otherwise c∈< a > K, which is a

contradiction to Equation 4.5). Verify that H={xdz|x∈K, z ∈Z}

is a subgroup of Gwith K⊆H. Since d=eGd∈Hand d̸∈ K,H

is larger than K. But Kis the largest group such that < a > ∩K=<

eG>, therefore < a > ∩H̸=< eG>. Let w̸=eG∈< a > ∩H, then

w=as=k1drsuch that k1∈Kand r, s ∈Z.(4.8)

Now pdoes not divide r, for if r=py the eG̸=w=as=k1dpy ∈<

a > ∩Kwhich is a contradiction. Consequently gcd (p, r) = 1 and by

Theorem A.1.1 there are integers u, v such that pu +rv = 1. Hence

c=c1=cpu+rv = (cp)u(cr)v

= (atk)u((dam)r)vby Equations 4.6 and 4.7

= (atk)u(dramr)v

= (atk)u((ask−1

1)amr)vby Equation 4.8

=a(tu+vs+mr)kuk−v

1∈< a > K.

104

This contradicts Equation 4.5. Therefore G=< a > K and hence

G=< a > ×Kby Theorem 4.5.3.

Theorem 4.5.5 (The fundamental theorem of ﬁnite abelian groups).

Every ﬁnite abelian group Gis a product of cyclic groups each of prime

power order.

Proof. By Theorem 4.5.4, Gis the product of its subgroups G(p),

one for each prime pthat divides |G|. Each G(p) is a p-group. So to

complete the proof it suﬃces to show that every ﬁnite abelian p-group

His a product of cyclic groups each of prime power order. We prove

this by induction on the order of H. The assertion is true when |H|= 2

by Theorem 4.2.3. Assume inductively that it is true for all groups

whose order is less than |H|and let abe an element of maximal order

pnin H. Then H∼

=<a>×Kby Lemma 4.5.2. By induction Kis a

direct sum of cyclic groups, each of prime power order. Consequently,

the same is true of < a > ×K. Hence, His a product of cyclic groups

each of prime power order.

Lemma 4.5.3. If (m, k) = 1, then Zm×Zk∼

=Zmk.

Proof. The order of (1,1) in Zm×Zkis the smallest positive integer

tsuch that 0 = t(1,1) = (t, t). Thus t≡0 (mod m) and t≡0

(mod k) so that m|tand k|t. But gcd (m, k) = 1 implies that mk|t.

Therefore mk ≤t. Since mk(1,1) = (mk, mk) = (0,0), we must have

mk =t=|(1,1)|. Therefore, Zm×Zkwhich is a group of order mk,

is a cyclic group generated (1,1). Consequently, by Theorem 4.2.4,

Zm×Zkis isomorphic to Zmk.

Theorem 4.5.6. Let n=pn1

1pn2

2···pnt

tbe such that p1, . . . ptare dis-

tinct primes, then Zn∼

=Zp1n1× · ·· × Zptnt.

Proof. The theorem is true for groups of order 2. Assume induc-

tively that it is true for groups of order less than n. Apply Lemma 4.5.3,

with m=pn1

1and k=pn2

2···pnt

tto get Zn∼

=Zpn1

1×Zk. Consequently,

the induction hypothesis shows that Zk=Zp2n2× ·· · × Zptnt.

Combining Theorems 4.5.5 and 4.5.6 yields a diﬀerent way of writing

a ﬁnite abelian group as a product of cyclic groups.

Example 4.5.1. Consider the group

Z2×Z2×Z4×Z8×Z3×Z3×Z5×Z25 ×Z125.

105

Arrange the prime power orders of the cyclic factors by size, with one

row for each prime:

222223

3 3

5 5253

Now rearrange the cyclic factors of Gusing the columns of this

array and apply Theorem 4.5.6.

Z2×(Z2×Z5)×(Z4× ×Z3×Z25)×(Z8×Z3×Z125 ).

That is

G∼

=Z2×Z10 ×Z300 ×Z3000

Observe that the order of each factor divides the order of the next one.

Generalizing Example 4.5.1 we get

Theorem 4.5.7. Every ﬁnite abelian group is the product of cyclic

groups of orders m1, m2, . . . , mt, where

m1|m2, m2|m3, . . . , mt−1|mt.

We now look at ﬁnite abelian groups related to ﬁelds.

Theorem 4.5.8. Let Fbe a ﬁeld and Ga ﬁnite subgroup of the mul-

tiplicative group F∗of nonzero elements. Then Gis cyclic.

Proof. Since Gis a ﬁnite abelian group, Theorem 4.5.7 implies

that G=Zm1× ··· × Zmtwhere each midivides mt. Consequently

every element gof Gmust satisfy gmt= 1Fand hence is a root of the

polynomial xmt−1F. Since Ghas order m1m2··· mtand xmt−1Fhas

at most mtroots (see Corollary A.2.3) we must have t= 1. Therefore

G∼

=Zmt.

Theorem 4.5.9. Let Kbe a ﬁnite ﬁeld and Fa subﬁeld. Then Kis

a simple extension of F.

Proof. By Theorem 4.5.8, the multiplicative group of nonzero el-

ements of Kis cyclic. If uis the generator of this group, then the

subﬁeld F(u) contains 0Fand all powers of uand hence contains every

element of K. Therefore K=F(u).

106

Theorem 4.5.10. Let pbe a positive prime. For each positive integer

n, there exists an irreducible polynomial of degree nin Zp[x].

Proof. There is an extension ﬁeld Kof Zpof order pnby Corollary

3.4.11. By Theorem 4.5.9, K=Zp(u) for some u∈K. By Theorem

3.4.4, the minimal polynomial of uin Zp[x] is irreducible of degree

[K:Zp]. Finally, Theorem 3.4.3 shows that [K:Zp] = n.

4.6 Galois theory.

Asimple radical extension of a ﬁeld Fis the extension ﬁeld we obtain

by adjoining the nth root of an element a∈F.

Deﬁnition 4.6.1. An element uwhich is algebraic over Fcan be solved

for in terms of radicals if uis an element of a ﬁeld Kwhich can be

obtained by a succession of simple radical extensions, that is,

F=K0⊂K1⊂ ··· ⊂ Ki⊂Ki+1 ⊂ ·· · ⊂ Ks=K(4.9)

where Ki+1 =Ki(ni

√ai)for some ai∈Ki,i= 0,1, . . . , s −1. Here

ni

√aidenotes a root of the polynomial xni−ai. Such a ﬁeld Kis called

a root extension of F.

Deﬁnition 4.6.2. A polynomial f(x)can be solved by radicals if all

its roots can be solved for in terms of radicals.

In other words f(x) is solvable by radicals if each of its roots is

obtained by successive ﬁeld operations (addition, subtraction, multi-

plication, and division) and root extractions. Consequently, if f(x) is

solvable by radicals, then there are formulas to ﬁnd roots of f(x). We

prove every polynomial of degree less than or equal to four is solvable

by radicals. We also prove this is not true for polynomials of degrees 5

or higher using theory developed by Evariste Galois and hence called

Galois theory.

Let Kbe an extension ﬁeld of F. An F-automorphism of Kis an

isomorphism σ:K→Kthat ﬁxes Felement wise (that is, σ(c) = c

for c∈F). The set of all F-automorphisms of Kis denoted by GalFK.

Theorem 4.6.1. If Kis an extension ﬁeld of F, then GalFKis a group

under the operation of composition of functions. GalFKis called the

Galois group of Kover F.

107

Proof. If σ, τ ∈GalFKthen σ◦τis an isomorphism from Kto K, by

Exercise 43. For each c∈F, (σ◦τ)(c) = σ(τ(c)) = σ(c) = c. Therefore

σ◦τ∈GalFK. Hence GalFKis closed. Composition of functions is

associative and the identity function is the identity element of GalFK.

If σ∈GalFK, then σ−1is an isomorphism from Kto K, by Exercise

44. Moreover, σ−1(c) = cfor every c∈F. Therefore σ−1∈GalFK.

Thus GalFKis a group.

Example 4.6.1. The complex conjugation map σ:C→Cgiven by

σ(a+bi) = a−bi is an automorphism of Cby Exercise 9 in Section

3.4. For every real number a,σ(a) = a. Consequently σ∈GalRC.

Theorem 4.6.2. Let Kbe an extension ﬁeld of Fand f(x)∈F[x]. If

u∈Kis a root of f(x)and σ∈GalFK, then σ(u)is a root of f(x).

Proof. If f(x) = c0+c1x+···+cnxn, then c0+c1u+···+cnun= 0F.

Since σis a homomorphism and σ(ci) = cifor each ci∈F,

0F=σ(0F) = σ(c0+c1u+· ·· +cnun)

=σ(c0) + σ(c1)σ(u) + · ·· +σ(cn)σ(un)

=c0+c1σ(u) + · ·· +cnσ(u)n=f(σ(u)).

Therefore σ(u) is a root of f(x).

Theorem 4.6.3. Let Kbe a splitting ﬁeld of some polynomial over F

and let u, v ∈K. Then there exists σ∈GalFKsuch that σ(u) = vif

and only if uand vhave the same minimal polynomial in F[x].

Proof. If uand vhave the same minimal polynomial over F, then

by Theorem 3.4.5 there is an isomorphism σ:F(u)→F(v) such that

σ(u) = vand σﬁxes Felement wise. Since Kis a splitting ﬁeld of some

polynomial over F, it is a splitting ﬁeld of the same polynomial over

both F(u) and F(v). Therefore σextends to an F-automorphism of K

by Theorem 3.4.7. That is σ∈GalFKand σ(u) = v. The converse is

an immediate consequence of Theorem 4.6.2.

Example 4.6.2. By Example 4.6.1, we have GalRChas at least two

elements, the identity map e, and the complex conjugation map σ. We

prove that these are the only elements of GalRC. Let τ∈GalRC. Since

iis a root of x2+ 1, τ(i) = ±iby Theorem 4.6.2. If τ(i) = i, then

τ(a+bi) = τ(a) + τ(b)τ(i) = a+bi.

108

Therefore τ=e. On the other hand, if τ(i) = −i, then

τ(a+bi) = τ(a) + τ(b)τ(i) = a+b(−i) = a−bi.

Consequently, τ=σ. Thus GalRC={e, σ}is a group of order 2 and

hence is isomorphic to Z2by Theorem 4.4.3.

Theorem 4.6.4. Let K=F(u1, . . . , un)be an algebraic extension ﬁeld

of F. If σ, τ ∈GalFKand σ(ui) = τ(ui)for each i= 1,2, . . . , n, then

σ=τ. In other words, an automorphism in GalFKis completely

determined by its action on u1, . . . , un.

Proof. Let β=τ−1◦σ, then β∈GalFK. The theorem is proved if

we show that βis the identity map ebecause β=e=τ−1◦σimplies

τ=σ. Since σ(ui) = τ(ui) for every i,

β(ui) = (τ−1◦σ)(ui) = τ−1(σ(ui)) = τ−1(τ(ui)) = e(ui) = ui.

Let v∈F(u1). By Theorem 3.4.4 there exist ci∈Fsuch that v=

c0+c1u1+··· +cm−1um−1

1, where mis the degree of the minimal

polynomial of u1over F. Since βis a homomorphism that ﬁxes u1and

every element of F,

β(v) = β(c0+c1u1+· ·· +cm−1um−1

1)

=β(c0) + β(c1)β(u1) + · ·· +β(cm−1)β(u1)m−1

=c0+c1u1+· ·· +cm−1um−1

1=v.

Thus β(v) = vfor every v∈F(u1). Repeating this argument by

replacing Fwith F(u1) and u1with u2, we show that β(v) = vfor

every v∈F(u1, u2). After a ﬁnite number of such repetitions we prove

that β(v) = vfor every v∈F(u1, . . . , un). Therefore βis the identity

function.

Corollary 4.6.5. If Kis the splitting ﬁeld of a separable polynomial

f(x)of degree nin F[x], then GalFKis isomorphic to a subgroup of

Sn.

Proof. By separability f(x) has ndistinct roots in K, say u1, . . . un.

Consider snto be the group of permutations of the set R={u1, . . . un}.

If σ∈GalFK, then σ(u1), . . . , σ(u2) are roots of f(x) by Theorem

4.6.2. Moreover, since σis injective, σ(ui) are all distinct, and hence is

109

a permutation of the set R. In other words, the restriction of σto the

set (denoted σ|R) is a permutation of R. Deﬁne a map θ:GalFK→Sn

by θ(σ) = σ|R. It is easily veriﬁed that σis a homomorphism of

groups. Since Kis the splitting ﬁeld of F,K=F(u1, . . . un). If σ|R=

τ|R, then σ(ui) = τ(ui) for every i, hence σ=τby Theorem 4.6.4.

Therefore, θis an injective homomorphism. Consequently GalFKis

isomorphic to Im θwhich is a subgroup of Sn.

Lemma 4.6.1. If f(x)∈F(x)and Kis a splitting ﬁeld of f, then the

order of GalFK= [K:F].

Proof. This result follows from the Fundamental Theorem of Galois

theory (Theorem 4.7.6) which is proved in Section 4.7.

Deﬁnition 4.6.3. If f(x)∈F(x), then the Galois group of the poly-

nomial f(x)is GalFK, where Kis the splitting ﬁeld of f(x)over F.

If f(x) is irreducible, then given any two roots of f(x) there is an

automorphism in the Galois group Gof f(x) that maps the ﬁrst root

to the second by Theorem 4.6.3. Such a group is said to be transitive

on roots of f(x), that is you can get from any given root to another

by applying some element of G. The fact that the Galois group of a

polynomial f(x) must be transitive on the roots of irreducible factors

of f(x) often helps in determining the structure of the Galois group.

Example 4.6.3. Let f(x) = (x2−3)(x2−5). The splitting ﬁeld of

f(x) is Q(√3,√5). The roots of the minimal polynomial x2−3 are

θ1=√3 and θ2=−√3. Consequently, any automorphism σ∈Gtakes

√3 to either √3 or −√3 by Theorem 4.6.2. Similarly, σtakes √5 to

either θ3=√5 or θ4=−√5, the roots of x2−5. Since σis completely

determined by its action on √3 and √5 by Theorem 4.6.4, there are at

most four choices for σ:

√3e

−→ √3

√5−→ √5

√3(12)

−→ −√3

√5−→ √5

√3(34)

−→ √3

√5−→ −√5

√3(12)(34)

−→ −√3

√5−→ −√5

Consequently G={e, (12),(34),(12)(34)} ⊂ S4. Check that G∼

=

Z2×Z2.

Example 4.6.4. Let f(x) = (x3−2). The roots of f(x) are 3

√2, ω 3

√2,

and ω23

√2, where ωis a root of the equation x3−1. The minimal

110

polynomial of ωis x2+x+ 1. Consequently, the splitting ﬁeld of f(x),

Q(3

√2, ω), has degree 6. Let σand τbe automorphisms deﬁned by

3

√2σ

−→ ω3

√2

ω−→ ω

3

√2τ

−→ 3

√2

ω−→ ω2=−ω−1

The elements of Q(3

√2, ω) are linear combinations of the basis

{1,3

√2,(3

√2)2, ω, ω 3

√2, ω(3

√2)2}.

Like before, the action of σand τon Q(3

√2, ω) can be determined

completely by their action on the basis elements.

For example:

σ(ω3

√2) = σ(ω)σ(3

√2) = ω(ω3

√2)) = (−ω−1) 3

√2.

Verify that

σ3=τ2=e, and στ =τ σ2.

Hence the Galois group of f(x) is S3by Exercise 10.

Deﬁnition 4.6.4. A group Gis said to be solvable if it has a chain

of subgroups

G=G0⊇G1⊇G2⊇ ··· ⊇ Gn−1⊇Gn=< e > (4.10)

such that each Giis a normal subgroup of the preceding group Gi−1

and the quotient group Gi−1/Giis abelian.

Example 4.6.5. In this example, we prove that S3is a solvable group.

Consider the chain

S3⊃<(123) >⊃(e).

The subgroup <e>is normal in <(123) >, and <(123) >is

normal in S3(see Example 4.3.3). The group <(123) > /e has order

3 by Corollary 4.4.2. Since 3 is a prime number, <(123) > /e is

isomorphic to Z3by Theorem 4.4.3, and hence is abelian. Similarly,

the group S3/ < (123) >has order 2, and is therefore isomorphic to

Z2. Thus S3/ < (123) >is abelian. Hence S3is a solvable group.

Theorem 4.6.6. Let Nbe a normal subgroup of a group G. Then

G/N is abelian if and only if aba−1b−1∈Nfor all a, b ∈G.

111

Proof

G/N is abelian if and only if

Nab =NaNb =NbN a =N ba for all a, b ∈G.

Now, Nab =Nba implies ab(ba)−1∈N. Since ab(ba)−1=aba−1b−1,

the result follows.

Theorem 4.6.7. For n≥5the group Snis not solvable.

Proof Suppose on the contrary that Snis solvable and that

Sn=G0⊇G1⊇G2⊇ ··· ⊇ Gn−1⊇Gt=<(1) >

is a chain of subgroups such that each Giis a normal subgroup of

Gi−1and the quotient group Gi−1/Giis abelian.

Let (rst) be any 3-cycle in Snand let u, v be any elements of the

set {1,2, . . . , n}other than r, s, and t.u,vexist because n≥5. Since

Sn/G1is abelian, Theorem 4.6.6 (with a= (tus), b= (srv)) shows

that G1must contain (tus)(srv)(tus)−1(srv)−1. Since(tus)−1=tsu

and (srv)−1=svr, we get

(tus)(srv)(tus)−1(srv)−1= (tus)(srv)(tsu)(svr) = (rst).

Therefore G1contains all the 3-cycles of Sn. We can repeat this argu-

ment to conclude that Gicontains all the 3-cycles for i= 0, . . . , t. This

means the identity subgroup Gtcontains all the 3-cycles which leads

to a contradiction. Therefore Snis not solvable.

Theorem 4.6.8. 1. Homomorphic images and quotient groups of

solvable groups are solvable.

2. Subgroups of a solvable group are solvable.

Proof.

1. Let Gbe a solvable group. Then Ghas a chain of subgroups

G=G0⊇G1⊇G2⊇ ··· ⊇ Gn−1⊇Gn=< e > (4.11)

such that each Giis a normal subgroup of the preceding group

Gi−1and the quotient group Gi−1/Giis abelian. Let f:G→H

112

be a homomorphism of groups and let Hi=f(Gi). Consider the

chain of subgroups

H=H0⊇H1⊇H2⊇ ··· ⊇ Hn−1⊇Hn=< e > . (4.12)

Verify that Hiis a normal subgroup of Hi−1for each i. To see that

Hi−1/Hiis abelian, let a, b ∈Hi−1. Then there exist c, d ∈Gi−1

such that f(c) = aand f(d) = b. Since Gi−1/Giis abelian,

cdc−1d−1∈Giby Theorem 4.6.6. Therefore

aba−1b−1=f(c)f(d)f(c−1)f(d−1) = f(cdc−1d−1)∈f(Gi) = Hi.

Consequently, Hi−1/Hiis abelian by Theorem 4.6.6. Thus His

solvable. A Quotient group of Gis homomorphic to Gby Theorem

4.3.10, and hence is solvable.

2. Let Hbe a subgroup of a solvable group Gand let

G=G0⊇G1⊇G2⊇ ··· ⊇ Gn−1⊇Gn=< e > (4.13)

be a solvable series for G. Consider the groups Hi=H∩Giand

the chain

H=H0⊇H1⊇H2⊇ ··· ⊇ Hn−1⊇Hn=< e > . (4.14)

Verify that Hiis a normal subgroup of Hi−1for each i. To show

that Hi−1/Hiis abelian, consider the map f:Hi−1/Hi→Gi−1/Gi

given by f(Hix) = Gix. Suppose Hix=Hiy, then xy−1∈Hi.

Since Hi=H∩Gi,xy−1∈Gi. Consequently, Gix=Giy

which implies f(Hix) = f(Hiy). Thus fis well deﬁned. Sup-

pose f(Hix) = f(Hiy), then Gix=Giywhich implies xy−1∈Gi.

Since Hix, Hiy∈Hi−1/Hi,x, y ∈Hi−1⊆H. Consequently,

since His a subgroup, xy−1∈H. Thus xy−1∈H∩Gi=Hi.

Therefore Hix=Hiy. Hence fis an injective map. Verify that

fis a homomorphism. Finally, since Gi−1/Giis abelian, and

Hi−1=H∩Gi−1, we get Hi−1/Hiis abelian. Thus His solvable.

Therefore subgroups of a solvable group are solvable.

Finally, we state Galois’ criterion for solvability of a polynomial by

radicals. We prove this theorem in Section 4.7.

113

Theorem 4.6.9. (Galois’ criterion) Let Fbe a ﬁeld of characteristic

zero and f(x)∈F[x]. Then f(x) = 0 is solvable by radicals if and only

if the Galois group of f(x)is solvable.

Example 4.6.6. Consider the equation f(x) = x6−4x3+ 4. Since

f(x) = x6−4x3+ 4 = (x3−2)2, the roots of f(x) are θ1=3

√2,

θ2=3

√2ω, and θ3=3

√2ω2, where ω= (−1 + √3i)/2 is a complex root

of 1 (ω3= 1). Clearly, f(x) is solvable by radicals. We will now verify

that the Galois group Gis solvable by showing that Gis S3which is

solvable (see Example 4.6.5).

Check that Q(3

√2, ω) is the splitting ﬁeld of f(x). By Theorem

4.6.3 there is an automorphism σ∈Gsuch that σ(θ1) = θ2. A root of

f(x) is mapped to another root by Gby Theorem 4.6.2. Therefore σ

takes θ3to itself or to θ1. Therefore σcan be either the permutation

(12) or (123) in S3. Thus Gcontains the permutations (12) and (123).

Therefore Gis S3by Exercise 8.

Example 4.6.7. By Example 4.6.3, we know that the Galois group G

of the polynomial f(x)=(x2−3)(x2−5) is isomorphic to Z2×Z2.

Hence Gis abelian. Consequently, the chain e⊂Gshows that Gis a

solvable group.

Example 4.6.8. In this example, we prove that f(x) = 2x5−10x−5

is not solvable by radicals. Eisenstein’s criterion (Theorem A.2.6) with

p= 5 implies that the polynomial f(x) is irreducible. The splitting

ﬁeld of f(x) has degree divisible by 5 by Theorem 3.4.4. Consequently

the order of the Galois group Gof fis divisible by 5 by Lemma 4.6.1.

Therefore Ghas an element of order 5 by Corollary 4.4.8. The only

elements of order 5 in S5are the 5-cycles. Therefore Gcontains a

5-cycle.

The roots of the derivative f′(x) = 10x4−10 are ±1,±i. If f(x)

had 4 real roots, then by the mean value theorem, f′(x) must have 3

real roots. Consequently, since f′(x) has only two real roots, f(x) has

at most 3 real roots. f(x) has real roots in the intervals (−2,0),(0,1),

and (1,2) because f(−2) <0, f (0) >0, f(1) <0, and f(2) >0, that

is, f(x) has exactly three real roots. Let τ∈Gdenote the automor-

phism of complex conjugation. Then τﬁxes the three real roots and

interchanges the two complex roots of f(x). Thus τis a transposition.

Exercise 8 shows that the only subgroup of S5that contains both a

5-cycle and a transposition is S5itself. Therefore G∼

=S5. Since S5is

114

not a solvable group by Theorem 4.6.7, Galois’ criterion implies that

f(x) is not solvable by radicals.

Deﬁnition 4.6.5. Let r1, r2, . . . rnbe the roots of a polynomial f(x).

Then the discriminant of fis i<j (ri−rj)2.

Observe that the discriminant vanishes if and only if there is a

repeated root.

Consider a general polynomial f(x) = anxn+an−1xn−1+an−2xn−2+

. . . +a1x+a0. We leave it as an exercise to show that the discriminant

D(f) of f(x) is

D(f) = (−1)1

2n(n−1) 1

an

R(f, f ′, x),

where R(f, f ′, x) is the resultant of f(x) and its derivative f′(x).

Example 4.6.9. The discriminant of the polynomial f(x) = x5−x−1

is 100050000

010005000

001000500

000100050

−1 0 0 0 −1 0 0 0 5

−1−1 0 0 0 −1 0 0 0

0−1−1 0 0 0 −1 0 0

0 0 −1−1 0 0 0 −1 0

000−10000−1

= 2869.

Let f(x)∈Q[x]. In determining the Galois group of f(x), we

may assume f(x)∈Z[x] and f(x) is separable. Therefore the the

discriminant Dof f(x) is not zero. For a prime p, consider the reduction

f(x)≡f(x) (mod p). If pdivides Dthen f(x) has discriminant D= 0

in Zp. Therefore f(x) is not separable. If pdoes not divide D, then f(x)

is a separable polynomial and can factored in to distinct irreducibles.

Theorem 4.6.10. Let f(x)∈Z[x]be separable polynomial, and let p

be a prime. Consider the reduction f(x)≡f(x)(mod p). If f(x)is

separable, that is, pdoes not divide the discriminant of f(x), then the

Galois group of f(x)over Zpis a permutation group isomorphic to a

subgroup of the Galois group of f(x)over Q.

115

Corollary 4.6.11. Let f(x)∈Z[x]be separable polynomial, and let p

be a prime. Consider the reduction f(x)≡f(x)(mod p). If f(x)is

separable, that is, pdoes not divide the discriminant of f(x), then the

Galois group of f(x)over Qcontains an element with cycle decompo-

sition (n1, n2, . . . nk)where n1, . . . , nkare the degrees of the irreducible

factors of f(x).

The proofs of Theorem 4.6.10 and Corollary 4.6.11 are a conse-

quence of Corollary 4.6.5 and some elementary number theory. The

interested reader may refer to [25] for proofs.

Example 4.6.10. By Example 4.6.3, the discriminant of f(x) = x5−

x−1 is 2869 = 19 ×151. To apply Corollary 4.6.11, we reduce f(x)

mod p, where pis a prime and p̸∈ {19,151}. Since x5−x−1≡

(x2+x+ 1)(x3+x2+ 1) (mod 2), by Corollary 4.6.11, the Galois group

of f(x) over Q,G, has a (2,3) cycle. Cubing this element we see that

Ghas a transposition. The polynomial f(x) has no roots mod 3 and

therefore has no linear factors. Consequently, if f(x) is a reducible

polynomial, then it has an irreducible quadratic factor. There are 3

irreducible polynomials of degree 2 in Z3[x], namely, x2+ 1, x2+x+ 2,

and x2+ 2x+ 2, none of which divide f(x). Thus f(x) is an irreducible

polynomial in Z3[x]. Hence there is a 5-cycle in G. Since S5is generated

by a 5-cycle and any transposition (see Exercise 9), G=S5which is

not solvable. Therefore f(x) is not solvable by radicals.

Proposition 4.6.1. There exist inﬁnitely many polynomials f(x)∈

Z[x]with Snas the Galois group.

Proof. By Theorem 4.5.10, for each positive integer n, there exists

an irreducible polynomial of degree nin Zp[x]. Consequently, let f1(x)

be an irreducible polynomial of degree nin Z2[x]. Let f2(x)∈Z3[x] be

a polynomial of degree n, such that, f2(x) is a product of an irreducible

polynomial of degree 2, say g(x), and irreducible polynomials of odd

degree. For example, if nis odd then f2(x) can be the product of

g(x), x, and an irreducible polynomial of degree n−3. If nis even,

f2(x) can be a product of g(x) and an irreducible polynomial of degree

n−2. Similarly, let f3∈Z5[x] be the product of xwith an irreducible

polynomial of degree n−1. Finally, let f(x)∈Z[x] be any polynomial

with f(x)≡f1(x)(mod 2)

≡f2(x)(mod 3)

≡f3(x)(mod 5).

116

By the Chinese Remainder Theorem, such an f(x) exists (see Exercise

2 in Section 6.4).

We now apply Corollary 4.6.11. The reduction of f(x) mod 2 shows

that f(x) is irreducible in Z[x], hence the Galois group is transitive on

the nroots of f(x). Raising the element given by the factorization of

f(x) mod 3 to a suitable odd power shows that the Galois group con-

tains a transposition. The factorization of f(x) mod 5 shows that the

Galois group contains an n−1 cycle. By Exercise 9 the only transitive

subgroup of Snthat contains an n−1 cycle and a transposition is Sn.

Therefore, it follows that the Galois group is Sn.

By Theorem 4.6.7, Snis not solvable for n≥5. Consequently,

Proposition 4.6.1 shows that there can be no general formulas for poly-

nomials with degrees greater than 4. We now demonstrate that Galois

groups of polynomials with coeﬃcients in ﬁelds with characteristic zero,

and degrees less than 5, are always solvable. We also provide formulas

to ﬁnd their roots.

Let kbe a ﬁeld with characteristic zero. Let f(x)∈k[x] and let G

be its Galois group.

1. Let f(x) be linear of the form

f(x) = x−a.

Then x=ais the only root of f(x) and Gis trivial.

2. Let f(x) be a quadratic polynomial of the form

f(x) = x2+bx +c.

If the discriminant of f(x), namely √b2−4c, is a perfect square

(f(x) is reducible), then Gis trivial. If f(x) is irreducible, then G

is Z2. The quadratic formula is given by

x=−b±√b2−4c

2.

3. Let f(x) be a polynomial of degree 3.

(a) Let f(x) be reducible. If f(x) splits in to three linear factors,

then Gis trivial. If f(x) splits in to a linear factor and a

quadratic factor, then Gis Z2.

117

(b) Let f(x) be irreducible, then Gis either A3or S3.

Let f(x) be of the form

f(x) = x3+ax2+bx +c. (4.15)

Let

p=1

3(3b−a2), q =1

27(2a3−9ab + 27c),and D=−4p3−27q2.

(4.16)

Then the roots of the Equation 4.15 are

x1=A+B−a

3,

x2=t2A+tB −a

3,

x3=tA +t2B−a

3.(4.17)

where

A=3

−27

2q+3

2√−3D, B =3

−27

2q−3

2√−3D, and t=−1

2+1

2√−3.

(4.18)

Example 4.6.11. (a) For the equation x3−x2+ 3x+ 5 = 0,

p= 2.66, q= 5.92, D=−1023.99, A= 1.46, and B=−5.46

(see Equations 4.16 and 4.18). Finally, Equations 4.17 imply

that the roots are x1=−1, x2= 1 −2i, and x3= 1 + 2i.

(b) Similarly, p=−9.33, q= 5.92, D= 2303.99, A= 4 + 3.46i,

and B= 4 −3.46ifor the equation x3+ 5x2−x−5 = 0 and

its roots are x1= 1, x2=−1, and x3=−5.

4. Let f(x) be a polynomial of degree 4 of the form

f(x) = x4+ax3+bx2+cx +d. (4.19)

The resolvent cubic equation,g(y) of Equation 4.19 is

y3−2py2+ (p2−4r)y+q2(4.20)

118

where

p=−3a2+ 8b

8, q =a3−4ab + 8c

8,

r=−3a4+ 16a2b−64ac + 256d

256 .

(a) Let g(y) be reducible. If g(y) splits in to three linear factors,

then G=< e, (12)(34),(13)(24),(14)(23) >. If g(y) splits in

to a linear factor and a quadratic factor, then Gis either D4

or the cyclic group {e, (1234),(13)(24),(1432)}.

(b) If g(y) is irreducible, then Gis either A4or S4.

To solve the quartic equation 4.19, we ﬁrst compute the roots y1,

y2, and y3, of the resolvent cubic equation 4.20. Then the roots

of the Equation 4.19 are

x1=√−y1+√−y2+√−y3

2,

x2=√−y1−√−y2−√−y3

2,

x3=−√−y1+√−y2−√−y3

2,

x4=−√−y1−√−y2+√−y3

2.(4.21)

Example 4.6.12. To solve the quartic equation x4−4x3+8.25x2−

8.5x+ 3.25 = 0, we ﬁrst solve the cubic equation x3−4.5x2+

5.0625x= 0. We use the cubic formula to ﬁnd the roots y1= 0,

y2= 2.25, and y3= 2.25. Consequently, by Equations 4.21,

the roots of the quartic equation are x1= 1, x2= 1, x3=

1−1.5i, and x4= 1 + 1.5i.

We refer the reader to [19] or [25] for details of these computations.

4.7 Proof of Galois’ Criterion for solvability.

In this section we present a proof of Galois’ Criterion for solvability of

polynomials by radicals.

119

Deﬁnition 4.7.1. An algebraic extension ﬁeld Kof Fis normal pro-

vided that whenever an irreducible polynomial in f(x)has one root in

K, then it splits over K, that is, f(x)has all its roots in K.

The next theorem proves that a splitting ﬁeld of a polynomial is

always a normal extension.

Theorem 4.7.1. The ﬁeld Kis a splitting ﬁeld over the ﬁeld Fof some

polynomial in F[x]if and only if Kis a ﬁnite dimensional, normal

extension of F.

Proof. If Kis the splitting ﬁeld of f(x)∈F[x], then K=F(u1, . . . , un)

where uiare roots of f(x). Consequently, [K:F] is ﬁnite by Exer-

cise 30 in Chapter 3. Let p(x) be an irreducible polynomial in F[x]

with a root v∈K. Let Lbe the splitting ﬁeld of p(x) over K. To

prove that p(x) splits over K, we need to show that every root of p(x)

in Lis actually in K. Let w̸=v∈Lbe any root of p(x). Then

there is a σ∈GalFKsuch that σ(v) = wby Theorem 4.6.3, that

is , F(v)∼

=F(w). Consequently, since Kis a splitting ﬁeld of the

polynomial f(x) over F(v) and K(w) is a splitting ﬁeld of f(x) over

F(w), σextends to an isomorphism between Kand K(w) by Theorem

3.4.7, such that, vis mapped to wand the elements of Fremain ﬁxed.

Therefore [K:F]=[K(w) : F] by Exercise 23 in Chapter 3. By The-

orem 3.4.4, [K(w) : K] is ﬁnite. Consequently, since [K:F] is ﬁnite,

Exercise 22 in Chapter 3 implies

[K:F] = [K(w) : F] = [K(w) : K][K:F].

Canceling [K:F] from both sides we get [K(w) : K] = 1, that is,

K(w) = K. Thus every root of p(x) is in Kwhich means that Kis

normal over F.

Conversely, assume Kis ﬁnite dimensional, normal extension of

Fwith basis {u1, . . . , un}. Then K=F(u1, . . . , un). Each uiis al-

gebraic over Fby Exercise 28 in Chapter 3. Let the minimal poly-

nomial of uibe pi(x). Since each pi(x) splits over Kby normality,

f(x) = p1(x)···pn(x) also splits over K. Therefore Kis the splitting

ﬁeld of f(x).

An element uin an extension ﬁeld Kof Fis said to be separable

over Fif uis a root of a separable polynomial in F[x]. The extension

ﬁeld Kis said to be a separable extension if every element of Kis

separable over F.

120

Theorem 4.7.2. Let Fbe a ﬁeld of characteristic zero, then every

algebraic extension ﬁeld Kof Fis a separable extension.

Proof. By Theorem 3.4.8, the minimal polynomial of each u∈K

is separable. Hence uis separable. Consequently, Kis a separable

extension.

Deﬁnition 4.7.2. A ﬁeld Kis said to be Galois over Fif Kis a ﬁnite

dimensional, normal, separable extension ﬁeld of F.

Let Kbe an extension ﬁeld of F. A ﬁeld Esuch that F⊆E⊆K

is called an intermediate ﬁeld of the extension. Since Kis also an

extension of Ethe Galois group GalEKconsists of all automorphisms

of Kthat ﬁx Eelement wise. Since F⊆E, every automorphism in

GalEKautomatically ﬁxes each element of F. Therefore, GalEKis a

subset (and hence subgroup) of GalFK.

Theorem 4.7.3. Let Kbe an extension ﬁeld of F. If His a subgroup

of GalFK, let

EH={k∈K|σ(k) = kfor every σ∈H}

Then EHis an intermediate ﬁeld of the extension. The ﬁeld EHis

called the ﬁxed ﬁeld of the subgroup H.

Proof. If c, d ∈EHand σ∈H, then

σ(c+d) = σ(c) + σ(d) = c+dand σ(cd) = σ(c)σ(d) = cd.

Therefore EHis closed under addition and multiplication. Since σ(0F) =

0Fand σ(1F) = 1Ffor every automorphism, 0Fand 1Fare in EH. For

any nonzero c∈EHand any σ∈H,

σ(−c) = −σ(c) = −cand σ(c−1) = σ(c)−1=c−1.

Consequently, EHcontains the inverses of all the nonzero elements.

Hence EHis a subﬁeld of K. Since His a subgroup of GalFK,σ(c) = c

for every c∈Fand σ∈H. Therefore F⊆EH.

Lemma 4.7.1. Let Kbe a ﬁnite dimensional extension ﬁeld of F. If

His a subgroup of the Galois group GalFKand EHis the ﬁxed ﬁeld

of H, then Kis a simple, normal, separable extension of EH.

121

Proof. Each u∈Kis algebraic over Fby Exercise 28 in Chapter 3

and hence algebraic over E. Every automorphism in Hmust map uto

some root of the minimal polynomial of u. Let u1, . . . utbe the distinct

images of uunder automorphisms in Hand let f(x)=(x−u1)(x−

u2)···(x−ut). Since uiare distinct, f(x) is a separable polynomial.

Since every automorphism σ∈Hpermutes u1, . . . , ut,

σf (x) = (x−σ(u1))(x−σ(u2)) ·· ·(x−σ(ut)) = f(x).

Consequently, every automorphism ﬁxes the coeﬃcients of f(x), hence

the coeﬃcients are in EH. Since uis a root of f(x), uis separable over

EH. Hence Kis a separable extension of EH. Since f(x) splits in K[x],

Kis normal over EHby Theorem 4.7.1. Since Kis ﬁnitely generated

over F,Kis ﬁnitely generated over EH. Hence K=EH(u) for some

u∈Kby Exercise 29 in Chapter 3. Therefore Kis simple.

Theorem 4.7.4. Let Kbe a ﬁnite-dimensional extension ﬁeld of F.

If His a subgroup of the Galois group GalFKand Eis a ﬁxed ﬁeld

of H, then H=GalEKand |H|= [K:E]. Therefore the Galois

correspondence is surjective.

Proof. Lemma 4.7.1 shows that K=E(u) for some u∈K. If the

minimal polynomial p(x) of uover Ehas degree n, then [K:E] = n

by Theorem 3.4.4. The Galois group GalEKis completely determined

by its action on uby Theorem 4.6.4 and uis always mapped to another

root of p(x) by an automorphism in GalEKby Theorem 4.6.2. This

implies that the number of distinct automorphisms in GalEKis at

most n, that is, |GalEK| ≤ n. Now H⊆GalEKby deﬁnition of ﬁxed

ﬁeld E. Therefore

|H| ≤ |GalEK| ≤ n= [K:E].

Let f(x) be as in Lemma 4.7.1. Then Hcontains at least tauto-

morphisms (the number of distinct images of uunder H). Since uis a

root of f(x), p(x) divides f(x). Hence

|H| ≥ t= deg f(x)≥deg p(x) = n= [K:E].

Combining the inequalities, we get

|H| ≤ |GalEK| ≤ [K:E]≤ |H|.

Therefore |H|=|GalEK|= [K:E], and hence H=GalEK.

122

Theorem 4.7.5. Let Kbe a Galois extension of Fand Ean interme-

diate ﬁeld. Then Eis a ﬁxed ﬁeld of the subgroup GalEK. Therefore

the Galois correspondence is injective for Galois extensions.

Proof. The ﬁxed ﬁeld E0of GalEKcontains Eby deﬁnition. To

show that E0⊆Ewe prove the contra positive: If u̸∈ Ethen u̸∈ E0.

Kis a Galois extension of the intermediate ﬁeld by Exercises 34 and

35. Kis an algebraic extension of Eby Exercise 28 in Chapter 3.

Consequently uis algebraic over Ewith minimal polynomial p(x)∈

E[x] of degree ≥2 (if degree p(x) = 1, then u∈E). The roots of

p(x) are distinct by separability and all of then are in Kby normality.

Let vbe a root of p(x) diﬀerent from u. Then there exists σ∈GalEk

such that σ(u) = vby Theorem 4.6.3. Therefore u∈E0and hence

E0=E.

Lemma 4.7.2. Let Kbe a ﬁnite dimensional normal extension ﬁeld

of Fand Ean intermediate ﬁeld which is normal over F. Then there

is a surjective homomorphism of groups θ:GalFK→GalFEwhose

kernel is GalEK.

Proof. Let σ∈GalFKand u∈E. Then uis algebraic over F

with minimal polynomial p(x). Since Eis a normal extension of F,

p(x) splits in E[x], that is, all the roots of p(x) are in E. Since σ(u)

is a root of p(x) by Theorem 4.6.2, σ(u)∈E. Therefore σ(E)⊆Efor

every σ∈GalFK. Thus the restriction of σto Eis an F-isomorphism

from Eto σ(E). Hence [E:F] = [σ(E) : F] by Exercise 23 in Chapter

3. Since F⊆σ(E)⊆E, [E:F]=[E:σ(E)][σ(E) : F] by Exercise

22 in Chapter 3. Thus [E:σ(E)] = 1. Therefore E=σ(E) and σ

restricted to Eis an automorphism in GalFE. Denote σrestricted to

Eby σ|E. Let θ:GalFK→GalFEbe such that θ(σ) = σ|E. Check

that θis a homomorphism of groups with kernel GalEK. To show

that θis surjective, note that Kis a splitting ﬁeld of a polynomial

f(x) by Theorem 4.7.1. Kis also the splitting ﬁeld of f(x) over E.

Consequently every τ∈GalFEcan be extended to an F-automorphism

σ∈GalFKby Theorem 3.4.7. This means that σ|E=τ, that is ,

θ(σ) = τ. Therefore θis surjective.

Theorem 4.7.6. [Fundamental Theorem of Galois Theory] If Kis a

Galois extension ﬁeld of F, then

1. There is a bijection between the set Sof all intermediate ﬁelds of

the extension and the set Tof all subgroups of the Galois group

123

GalFK, given by assigning each intermediate ﬁeld Eto the sub-

group Gal(K/E). Furthermore,

[K:E] = |GalEK|and [E:F] = [GalFK:GalEK].

2. An intermediate ﬁeld Eis a normal extension of Fif and only if

the corresponding group GalEKis a normal subgroup of GalFK,

and in this case Gal(E/F ) = GalFK/GalEK.

Proof. There is a bijection between the set Sof all intermediate

ﬁelds of the extension and the set Tof all subgroups of the Galois group

GalFK, given by assigning each intermediate ﬁeld Eto the subgroup

GalEKby Theorems 4.7.4 and 4.7.5. By Theorem 4.7.4, [K:E] =

|GalEK|. In particular if F=E, then [K:F] = |GalFK|. By

Exercise 22 in Chapter 3, [K:F]=[K:E][E:F]. Consequently, by

applying Lagrange’s Theorem 4.4.1, we get

[K:E][E:F] = [K:F] = |GalFK|=|GalEK|[GalFK:GalEK].

Dividing the equation by [K:E] = GalEKshows that

[E:F] = [GalFK:GalEK].

To prove part 2, assume that GalEKis a normal subgroup of

GalFK. Let p(x) be an irreducible in F[x] with a root uin E. To

show that Eis a normal extension ﬁeld we must show that p(x) splits

in E[x]. Since Kis normal over F,p(x) splits in K[x]. So we need

only show that each root vof p(x) is in E. There is an automorphism

σ∈GalFKsuch that σ(u) = vby Theorem 4.6.3. If τ∈GalEK, then

since GalEKis normal, τ◦σ=σ◦τ1for some τ1∈GalEK. Since

u∈E,τ(v) = τ(σ(u)) = σ(τ1(u)) = σ(u) = v. Hence vis ﬁxed by

every element τ∈GalEKand therefore is in E(see Theorem 4.7.5).

Thus Eis a normal extension of F.

Conversely, assume that Eis a normal extension of F. Then E

is ﬁnite dimensional over Fby part 1. By Lemma 4.7.2, there is a

surjective homomorphism of groups θ:GalFK→GalFEwith kernel

GalEK. Then GalEKis a normal subgroup of GalFKby Theorem

4.3.9, and GalFK/GalEK∼

=GalFEby the First Isomorphism Theo-

rem 4.3.11.

124

Example 4.7.1. Let f(x) = (x2−3)(x2−5). The splitting ﬁeld of

f(x) is Q(√3,√5). By Example 4.6.3 we know that

Gal(Q(√3,√5)/Q) = {e, σ, τ, στ},

such that

√3e

−→ √3

√5−→ √5

√3σ

−→ −√3

√5−→ √5

√3τ

−→ √3

√5−→ −√5

√3στ

−→ −√3

√5−→ −√5

By the Fundamental Theorem, corresponding to each subgroup of

Gal(Q(√3,√5)/Q), there is a ﬁxed subﬁeld of Q(√3,√5).

For example, the subﬁeld corresponding to the subgroup {e, σ}is

the set of elements ﬁxed by the map

σ:a+b√3 + c√5 + d√5→a−b√3 + c√5−d√5

which is the set of elements a+c√5, that is, the ﬁeld Q(√5). Similarly,

we can determine the ﬁxed ﬁelds for other subgroups of Gal(Q(√3,√5)/Q):

Subgroup Fixed Field

{e}Q(√3,√5)

{e, σ}Q(√5)

{e, τ }Q(√3)

{e, στ }Q(√15)

{e, σ, τ, στ}Q

See Figure 4.1.

Deﬁnition 4.7.3. The extension K/F is said to be cyclic if it is Galois

with a cyclic Galois group.

Deﬁnition 4.7.4. Let K1and K2be two subﬁelds of a ﬁeld K. Then

the composite ﬁeld of K1and K2, denoted K1K2is the smallest subﬁeld

of Kcontaining both K1and K2.

Note that K1K2is the intersection of all the subﬁelds of Kcontain-

ing both K1and K2.

Proposition 4.7.1. Let K1and K2be Galois extensions of a ﬁeld F,

then the composite K1K2is Galois over F.

125

{e, σ, τ, στ }Q

Q(√5)Q(√3) Q(√15)

Q(√3,√5)

{e}

{e, τ } {e, στ} {e, σ }

Figure 4.1: The Galois correspondence of subgroups and subﬁelds.

Proof. If K1is the splitting ﬁeld of the separable polynomial f1(x)

and K2is the splitting ﬁeld of the separable polynomial f2(x) then the

composite is the splitting ﬁeld for the square free part of the polynomial

f1(x)f2(x), hence is Galois over F.

Proposition 4.7.2. Let Fbe a ﬁeld of characteristic not dividing n

such that Fcontains all the n-th roots of unity. Then the extension

F(n

√a), for a∈F, is cyclic over Fof degree dividing n.

Proof. The extension K=F(n

√a) is Galois over Fif Fcontains

the n-th roots of unity since it is the splitting ﬁeld for xn−a. For any

σ∈GalFK,σ(n

√a) is another root of xn−a. Hence σ(n

√a) = ωσn

√a

where ωσis some n-th root of unity. Let Gndenote the group of n-th

roots of unity. Since Fcontains Gn, every n-th root of unity is ﬁxed

by GalFK. Hence for τ, σ ∈GalFK,

στ (n

√a) = σ(ωτn

√a) = ωτσ(n

√a) = ωτωσn

√a=ωσωτn

√a

which shows that ωστ =ωσωτ. Therefore the map f:GalFK→Gn

such that f(σ) = ωσis a homomorphism. The kernel of fis precisely

the identity and hence fis injective. Consequently, since Gnis cyclic

GalFKis cyclic. Since the image of fis a subgroup, |GalFK|divides

n. Consequently, by Theorem 4.7.6, Khas degree dividing n.

126

Deﬁnition 4.7.5. Let Fbe a ﬁeld of characteristic not dividing n

such that Fcontains all the n-th roots of unity. Let Kbe any cyclic

extension of degree nover F. Let σbe the generator of the cyclic group

GalFK. For u∈Kand any n-th root of unity ω, deﬁne the Lagrange

resolvent (u, ω)∈Kby

(u, ω) = u+ωσ(u) + ω2σ2(u) + ···+ωn−1σn−1(u).

Proposition 4.7.3. Let Fbe a ﬁeld of characteristic not dividing n

such that Fcontains all the n-th roots of unity. Let Kbe a cyclic

extension of F, then Kis of the form F(n

√a)for some a∈F.

Proof. Let σbe the generator of the cyclic group GalFKand let

u∈Kand ωbe a n-th root of unity. Since ω∈F, if we apply σto

the Lagrange resolvent (u, ω) we get

σ((u, ω)) = σ(u) + ωσ2(u) + ω2σ3(u) + ·· · +ωn−1σn(u).

Since σn= 1 in GalFKand ωn= 1 in Gn(the group of n-th roots

of unity), we get

σ((u, ω)) = σ(u) + ωσ2(u) + ω2σ3(u) + ·· · +ωn−1σn(u)

=ω−1(ωσ(u) + ω2σ2(u) + ·· · +ωn−1σn−1(u) + wnσn(u))

=ω−1(ωσ(u) + ω2σ2(u) + ·· · +ωn−1σn−1(u) + u)

=ω−1(u, ω).(4.22)

Therefore

σ(u, ω)n= (ω−1)n(u, ω)n= (u, ω)n.

Since (u, ω)nis ﬁxed by GalFK, (u, ω)n∈Ffor any u∈K. By

the linear independence of the automorphisms 1, σ, σ2, . . . , σn−1, there

is an element u∈Kwith (u, σ)̸= 0. Iterating Equation 4.22 we get

σi((u, ω)) = (ω−i)(u, ω) and we see that σidoes not ﬁx (u, ω) for any

i < n. Hence (u, ω) cannot lie in any proper subﬁeld of K, so K=

F((u, ω)). Since (u, ω)n=a∈Fwe have F(n

√a) = F((u, ω)) = K.

The Galois closure Kof a ﬁeld Fis the minimal Galois extension

of Fin the sense that if Lis a Galois extension of Fthen K⊆L.

Theorem 4.7.7. If uis contained in a root extension Kas in Equation

4.9, then uis contained in a root extension which is Galois over Fand

where each intermediate extension is cyclic.

127

Proof. Let Lbe the Galois closure of Kover F. For any σ∈GalFL,

we derive the chain of subﬁelds from Equation 4.9

F=σK0⊂σK1⊂ ··· ⊂ σKi⊂σKi+1 ⊂ ··· ⊂ σKs=σK.

Since σ(ni

√ai) is a root of xni−σ(ai), it follows that σKi+1 =

σKi(σ(ni

√ai)), that is, σKi+1 is a simple radical extension of σKi.

Therefore σ(K) is solvable by radicals. Hence Lwhich is the com-

posite of all the ﬁelds σ(K) such that σ∈GalFLis also solvable by

radicals (see Exercises 36 and 37). Therefore uis contained in a Galois

root extension Land there are subﬁelds Liof L

F=L0⊂L1⊂ ··· ⊂ Li⊂Li+1 ⊂ ·· · ⊂ Lr=L(4.23)

such that Li+1 is a simple radical extension of Li.

We now adjoin the ni-th roots of unity to Fto obtain a ﬁeld F′. This

extension is derived as a chain of subﬁelds such that each individual

extension is cyclic (adjoin one root at a time).

Form the composite of F′with the root extension 4.23

F⊆F′=F′L0⊂F′L1⊂ ··· ⊂ F′Li⊂F′Li+1 ⊂ ·· · ⊂ F′Lr=F′L.

Since F′and Lare Galois over F, the composite F′Lis Galois

over Fby Theorem 4.7.1. F′Li+1 is a simple radical extension of F′Li

and since F′Licontains the roots of unity F′Li+1 is also cyclic by

Proposition 4.7.2. Therefore F′Lis a root extension of Fwhere each

intermediate extension is cyclic.

Proposition 4.7.4. Suppose K/F is a Galois extension and F′/F is

any extension. Then K F ′/F ′is a Galois extension, with Galois group

Gal(K F ′/F ′)∼

=Gal(K/K ∩F′)

isomorphic to a subgroup of Gal(K/F ).

Proof. If K/F is Galois, then Kis the splitting ﬁeld of some sep-

arable polynomial f(x)∈F[x]. Then K F ′/F ′is the splitting ﬁeld

of f(x) viewed as a polynomial in F′(x), hence this extension is Ga-

lois. Consider the map ϕ:Gal(K F ′/F ′)→Gal(K/F ) such that

ϕ(σ)→σ|K. Check that this map deﬁned by restricting an automor-

phism σto the subﬁeld Kis a well deﬁned homomorphism. Since an

element in Gal(KF ′/F ′) acts as the identity on F′), the elements in the

128

kernel of ϕare trivial on both Kand F′) and hence on their composite.

So Ker ϕ={σ∈Gal(K F ′/F ′)|σ|K= 1}, contains only the identity

automorphism. Hence ϕis injective.

Let Hdenote the image of ϕin Gal(K/F ) and let KHdenote the

corresponding ﬁxed subﬁeld of Kcontaining F. Since every element

in Hﬁxes F′,K∩F′⊆KH. Since, any σ∈Gal(K F ′/F ′) ﬁxes

F′and acts on KH⊆Kvia its restriction σ|K∈H, ﬁxes KHby

deﬁnition. Therefore, the KHF′is ﬁxed by Gal(K F ′/F ′). By the

Fundamental Theorem, KHF′=F′. Consequently, KH⊆F′, which

gives the reverse inclusion KH⊆K∩F′. Hence KH =K∩F′. By

Fundamental Theorem, H=Gal(K F ′/F ′).

Theorem 4.7.8. Let Gbe a ﬁnite solvable group. Then Ghas a chain

of subgroups

G=G0⊇G1⊇G2⊇ ··· ⊇ Gn−1⊇Gn=< e > (4.24)

such that each Giis a normal subgroup of the preceding group Gi−1and

the quotient group Gi−1/Giis cyclic.

Proof. Proof is by induction on the order of G. The theorem is

true when |G|= 1. Let |G|>1. Assume the theorem holds for all

solvable groups of order less than |G|. Let Nbe a normal subgroup

of Gsuch that N̸=<e>. Such a subgroup exists because Gis a

solvable group of order greater than 1. Theorem 4.6.8 implies G/N is a

solvable group. By Lagrange’s Theorem 4.4.1, |G/N|<|G|. Hence the

induction hypothesis applies on G/N and there is a chain of subgroups

Tiof G/N such that

G/N =T0⊇T1⊇T2⊇ ··· ⊇ Tr−1⊇Tr=N(4.25)

such that Tiis a normal subgroup of the preceding group Ti−1and the

quotient group Ti−1/Tiis cyclic. By Theorem 4.3.8, for each Ti, there

is a subgroup Giof Gsuch that N⊂Giand Ti=Gi/N . Thus we get

a chain of subgroups Giof G

G=G0⊇G1⊇G2⊇ ··· ⊇ Gr−1⊇Gr=N. (4.26)

Appending the subgroup < e > to the end gives us a chain of

subgroups

G=G0⊇G1⊇G2⊇ ··· ⊇ Gr−1⊇N⊇< e > (4.27)

129

such that each Giis a normal subgroup of the preceding group Gi−1.

By Exercise 19, the quotient group Gi−1/Giis isomorphic to Ti−1/Ti,

and hence is cyclic. Therefore by induction the theorem holds for all

solvable groups.

Finally, we can prove Galois’ criterion for solvability of polynomials,

that is, for a polynomial f(x)∈F[x], where Fis a ﬁeld of characteristic

zero, f(x) is solvable by radicals if and only if the Galois group Gof

f(x) is solvable.

Proof of Theorem 4.6.9. Suppose ﬁrst that f(x) can be solved by

radicals. Then each root of f(x) is contained in an extension as in

Theorem 4.7.7. The composite Lof such extensions is also Galois

by Proposition 4.7.1. Let Gibe the subgroups corresponding to the

subﬁelds Ki,i= 0,1, . . . , s −1. Since Gal(Ki+1/Ki) = Gi/Gi+1 for

each iit follows that the Galois Group G=Gal(L/F ) is a solvable

group. The ﬁeld Lcontains the splitting ﬁeld of f(x) so the Galois

group of f(x) is a quotient group of a solvable group Gand hence is

solvable by Theorem 4.6.8.

Suppose now that the Galois group Gof f(x) is a solvable group

and let Kbe the splitting ﬁeld of f(x). Taking the ﬁxed ﬁelds of the

subgroups in the Chain 4.24 for Ggives a chain

F=K0⊂K1⊂ ··· ⊂ Ki⊂Ki+1 ⊂ ·· · ⊂ Ks=K

where Ki+1/Kifor each iis a cyclic extension of degree ni. Let

F′be an extension ﬁeld over F, that contains all the roots of unity of

order ni,i= 0, . . . , s −1. Form the composite ﬁelds K′

i=F′Ki. We

obtain a sequence of extensions

F⊆F′=F′K0⊆F′K1⊆ ··· ⊆ F′Ki⊆F′Ki+1 ⊆ ··· ⊆ F′Ks=F′K.

The extension F′Ki+1/F ′Kiis cyclic of degree dividing ni,i= 0, . . . , s−

1 by Proposition 4.7.4.

Since we now have appropriate roots of unity in the base ﬁelds, each

of these cyclic extensions is a simple radical extension by Proposition

4.7.3. Each of these roots of f(x) is therefore contained in the root

extension F′Kso that f(x) can be solved by radicals.

Exercises.

1. Let Gbe a group and let a, b ∈G. Prove that (ab)−1=b−1a−1.

130

2. Prove that Snis a nonabelian group with the operation of product

of permutations, and that the order of Snis n!. Also Prove that

the set of all permutations of a set Gwith nelements is isomorphic

to Sn.

3. Find the inverse of (1324) ∈S4.

4. Find the inverse of (15342) ∈S5.

5. Find the order of (12)(345) in S5.

6. Find the order of (123)(456) in S6.

item Prove that every permutation in Snis the product of disjoint

cycles.

7. Prove that (12) and (1234) generate S4.

8. Prove that the only subgroup Gof Snthat contains both a n-

cycle and a transposition is Snitself. (Hint: Relabel to show that

(12 ···n) is in G. Then show that Gcontains all the transposi-

tions. Finally use Lemma 4.1.1).

9. Prove that the only transitive subgroup of Snthat contains both

an−1-cycle and a transposition is Snitself.

10. Let Dn⊆Snbe deﬁned by

Dn=< r, s|rn=s2=e, rs =sr−1> .

(a) Show that D3=S3.

(b) Compute the orders of D4and D5.

11. Show that the order of the group of even permutations, An, is

n!/2.

12. Show that the set A(G) of all bijective functions from Gto Gis

a group with composition as the group operation.

13. Prove that the set of units U8in Z8is a group under multiplica-

tion.

14. Show that the group U15 is generated by the elements 7 and 11.

15. Show that the group U18 is cyclic.

131

16. Show that the additive group Z2×Z3is cyclic.

17. Let Nbe a normal subgroup of a group Gand let Tbe a subgroup

of G/N. Prove that H={a∈G|N a ∈T}is a subgroup of G.

18. Prove that a subgroup with index 2 is a normal subgroup.

19. Let Kand Nbe normal subgroups of a group Gwith N⊆K⊆G.

Then K/N is a normal subgroup of G/N, and the quotient group

(G/N)/(K/N) is isomorphic to G/K.

20. Let N1, . . . , Nkbe normal subgroups of a group Gsuch that ev-

ery element of Gcan be written uniquely in the form a1a2. . . ak

with ai∈Ni. Let f:N1×N2× ··· × Nk→Gbe such that

f(a1, a2, . . . , ak) = a1a2···ak. Then prove that fis an isomor-

phism between N1×N2× ·· · × Nkand G.

21. Prove that N={1,17}is a normal subgroup of U32.

22. Prove that U32/N is isomorphic to U16.

23. Consider S4, the group of permutations of the set {1,2,3,4}.

Show that K={e, (12)(34),(13)(24),(14)(23)}is a normal sub-

group of S4.

24. Write the operation table for S4/K.

25. Let Gbe a group such that all its subgroups are normal. If

a, b ∈G, Show that there is an integer ksuch that ab =bak.

26. Let Gbe a group. For a∈Glet the map ϕa:G→Gbe such

that ϕa(x) = ax. Then prove that ϕais a bijection from Gto G.

27. Prove that every abelian group of order pq is isomorphic to Zpq,

where pand qare distinct primes.

28. Prove that every group of order 4 is isomorphic to either Z4or

Z2×Z2.

29. Prove that every group of order 6 is isomorphic to either S3or

Z6.

30. Explain why the two groups are not isomorphic:

132

(a) Z6and S3

(b) Zand R

(c) Z4×Z2and D4

(d) Z4×Z2and Z2×Z2×Z2

31. Let Hbe a nonempty ﬁnite subset of a group G. If His closed

under the operation in Gprove that His a subgroup of G.

32. Let Gbe a group.

(a) Show that the conjugacy relation on Gis reﬂexive, symmet-

ric, and transitive.

(b) Two conjugacy classes are either disjoint or identical.

(c) The group Gis a union of its distinct conjugacy classes.

33. If Gis a group and a∈G, prove that the centralizer of ais a

subgroup of G.

34. Let Kbe a splitting ﬁeld of f(x) over F. If Eis a ﬁeld such that

F⊆E⊆K, show that Kis a splitting ﬁeld of f(x) over E.

35. If Kis separable over Fand Eis a ﬁeld such that F⊆E⊆K,

show that Kis separable over E.

36. Prove that the composite of two root extensions is also a root

extension.

37. Prove that the Galois closure Lof a ﬁeld Kis the composite of

all the ﬁelds σ(K) where σ∈GalFL.

38. Use the cubic formula to ﬁnd the roots of the following equations.

(a) x3−3x2+ 28x−26

(b) x3−7.75x2+ 18.375x−13.5

39. Use the quartic formula to ﬁnd the roots of the following equa-

tions.

(a) x4−3x3+ 11x2−27x+ 18

(b) x4−8x3+ 22.75x2−27x+ 11.25

133

40. Prove that the subgroup < e, (12)(34),(13)(24),(14)(23) ⊂S4is

isomorphic to Z2×Z2(Hint: every element has order 2).

41. Prove that the group S4is solvable. (Hint: use the chain of

subgroups < e >⊂< e, (12)(34),(13)(24),(14)(23) >⊂A4⊂S4).

42. Prove that the Galois group of a polynomial f(x)∈F[x] is a

subgroup of Anif and only if the discriminant D∈Fis a square

of an element of F.

43. If σ, τ ∈GalFK, then prove that σ◦τis an isomorphism from K

to K.

44. If σ∈GalFK, then prove that σ−1is an isomorphism from Kto

K.

45. Determine the Galois group Gof the polynomial f(x) = (x2−

2)(x2−3). Draw the Galois correspondence of the subgroups of

Gand the subﬁelds of the splitting ﬁeld of f(x).

46. Draw the Galois correspondence of the subgroups of the Galois

group of f(x) = x3−2 and the subﬁelds of Q(3

√2, ω), where ωis

a root of x3−1.

134

Chapter 5

Constructing and

Enumerating integral roots

of systems of polynomials.

Either write something worth reading or do something worth writing.

Benjamin Franklin

Solving linear systems of equations is dealt with in Linear Algebra.

Abstract Algebra techniques come in to play when we restrict our so-

lutions to be integral, that is, every coordinate of a solution vector is

an integer. Finding only integral solutions of a linear system is a much

more complex problem than ﬁnding all its solutions. In this chapter, we

describe how to construct and enumerate integral roots of systems of

linear equations as lattice points inside polyhedral cones. We illustrate

this method by constructing and enumerating magic squares.

5.1 Magic Squares.

Amagic square is a square matrix whose entries are nonnegative inte-

gers, such that the sum of the numbers in every row, in every column,

and in each diagonal is the same number called the magic sum. See Fig-

ure 5.1 for examples of some ancient magic squares. We refer the reader

to [4] or [6] to read more about the history of magic squares. Con-

structing and enumerating magic squares and other variations of magic

squares are classical problems of interest. The well-known squares in

135

4 9 2

3 5 7

8 1 6

A

7 12 1 14

2 13 8 11

16 3 10 5

9 6 15 4

B C

16 3 2 13

5 10 11 8

9 6 7 12

4 15 14 1

Figure 5.1: (A) Loh-Shu (China, 2858-2738 B.C.), (B) Jaina (India, 12 th

century), and (C) the D¨urer (Germany, 1514) Magic squares.

Figure 5.2 were constructed by Benjamin Franklin. In a letter to Peter

Collinson he describes the properties of the 8 ×8 square F1 as follows:

1. The entries of every row and column add to a common sum called

the magic sum.

2. In every half-row and half-column the entries add to half the

magic sum.

3. The entries of the main bent diagonals (see Figure 5.4) and all

the bent diagonals parallel to it (see Figure 5.5) add to the magic

sum.

4. The four corner entries together with the four middle entries add

to the magic sum.

Henceforth, when we say row sum, column sum, bent diagonal sum,

and so forth, we mean that we are adding the entries in the correspond-

ing conﬁgurations. Franklin mentions that the square F1 has ﬁve other

curious properties but fails to list them. He also says, in the same let-

ter, that the 16×16 square F3 has all the properties of the 8×8 square,

but that in addition, every 4 ×4 subsquare adds to the common magic

sum. More is true about this square F3. Observe that every 2 ×2

subsquare in F3 adds to one-fourth the magic sum. The 8 ×8 squares

have magic sum 260 while the 16×16 square has magic sum 2056. For

a detailed study of these three “Franklin” squares, see [2], [6], and [28].

We deﬁne 8 ×8Franklin squares to be squares with nonnegative

integer entries that have the properties (1) - (4) listed by Benjamin

Franklin and the additional property that every 2×2 subsquare adds to

one-half the magic sum (see Figure 5.3). The 8×8 squares constructed

136

F1

64 2 51 13 60 6 55 9

16 50 3 61 12 54 7 57

F2

58 39 26 7 250 231 218 199 186 167 154 135 122 103 90 71

198 219 230 251 6 27 38 59 70 91 102 123 134 155 166 187

60 37 28 5 252 229 220 197 188 165 156 133 124 101 92 69

201 216 233 248 9 24 41 56 73 88 105 120 137 152 169 184

55 42 23 10 247 234 215 202 183 170 151 138 119 106 87 74

203 214 235 246 11 22 43 54 75 86 107 118 139 150 171 182

53 44 21 12 245 236 213 204 181 172 149 140 117 108 85 76

205 212 237 244 13 20 45 52 77 84 109 116 141 148 173 180

51 46 19 14 243 238 211 206 179 174 147 142 115 110 83 78

207 210 239 242 15 18 47 50 79 82 111 114 143 146 175 178

49 48 17 16 241 240 209 208 177 176 145 144 113 112 81 80

196 221 228 253 4 29 36 61 68 93 100 125 132 157 164 189

62 35 30 3 254 227 222 195 190 163 158 131 126 99 94 67

194 223 226 255 2 31 34 63 66 95 98 127 130 159 162 191

64 33 32 1 256 225 224 193 192 161 160 129 128 97 96 65

F3

53 60 5 12 21 28 37 44

200 217 232 249 8 25 40 57 72 89 104 121 136 153 168 185

52 61 4 13 20 29 36 45

14 3 62 51 46 35 30 19

11 6 59 54 43 38 27 22

55 58 7 10 23 26 39 42

9 8 57 56 41 40 25 24

16 1 64 49 48 33 32 17

50 63 2 15 18 31 34 47

32 34 19 45 28 38 23 41

17 47 30 36 21 43 26 40

1 63 14 52 5 59 10 56

33 31 46 20 37 27 42 24

48 18 35 29 44 22 39 25

49 15 62 4 53 11 58 8

Figure 5.2: Squares constructed by Benjamin Franklin.

by Franklin have this extra property (this might be one of the unstated

curious properties to which Franklin was alluding in his letter). It is

worth noticing that the fourth property listed by Benjamin Franklin

becomes redundant with the assumption of this additional property.

Similarly, we deﬁne 16 ×16 Franklin squares to be 16 ×16 squares

that have nonnegative integer entries with the property that all rows,

columns, and bent diagonals add to the magic sum, the half-rows and

half-columns add to one-half the magic sum, and the 2 ×2 subsquares

add to one-fourth the magic sum. The 2×2 subsquare property implies

that every 4 ×4 subsquare adds to the common magic sum.

The property of the 2 ×2 subsquares adding to a common sum and

the property of bent diagonals adding to the magic sum are “continuous

properties.” By this we mean that, if we imagine the square as the

surface of a torus (i.e., if we glue opposite sides of the square together),

then the bent diagonals and the 2 ×2 subsquares can be translated

without eﬀect on the corresponding sums (see Figure 5.5).

137

= Magic sum

= Magic sum= Magic sum

= Magic sum = Magic sum

= half the Magic sum

= Magic sum= half the Magic sum = half the Magic sum

= Magic sum

Figure 5.3: Deﬁning properties of the 8 ×8 Franklin squares [6].

When the entries of a n×nmagic square (or Franklin square) are

1,2,3, . . . , n2, it is called a natural square. Observe that the squares

in Figures 5.1 and 5.2 are natural squares. Nevertheless, in this chap-

ter, our study is not restricted to natural squares. In the following

sections, we develop algebraic methods to construct and enumerate all

such squares.

5.2 Polyhedral cones.

A set Pof vectors in Rnis called a polyhedron if P={y:Ay ≤b}for

some matrix Aand vector b. A bounded polyhedron is called a polytope.

A nonempty set Cof points in Rnis a cone if au +bv belongs to C

whenever uand vare elements of Cand aand bare nonnegative real

numbers. A cone is pointed if the origin is its only vertex (or minimal

face; see [32]). A cone Cis polyhedral if C={y:Ay ≤0}for some

matrix A, i.e, if Cis the intersection of ﬁnitely many half-spaces. If,

138

Figure 5.4: The four main bent diagonals [28].

52 61 4 13 20 29 36 45 52 61 4

14 3 62 51 46 35 30 19 14 3 62

53 60 5 12 21 28 37 44 53 60 5

11 6 59 54 43 38 27 22 11 6 59

9 8 57 56 41 40 25 24 9 8 57

50 63 2 15 18 31 34 47 50 63 2

16 1 64 49 48 33 32 17 16 1 64

52 61 4 13 20 29 36 45 52 61 4

14 3 62 51 46 35 30 19 14 3 62

11 6 59 54 43 38 27 22 11 6 59

55 58 7 10 23 26 39 42 55 58 7

52 61 4 13 20 29 36 45 52 61 4

14 3 62 51 46 35 30 19 14 3 62

16 1 64 49 48 33 32 17 16 1 64

36 45

30 19

37 44

27 22

39 42

25 24

34 47

32 17

50 63 2 15 18 31 34 47 50 63 2

55 58 7 10 23 26 39 42 55 58 7

4536

53 60 5 12 21 28 37 44 53 60 5

50 63 2 15 18 31 34 47 50 63 2

9 8 57 56 41 40 25 24 9 8 57

16 1 64 49 48 33 32 17 16 1 64

Figure 5.5: Continuous properties of Franklin squares.

in addition, the entries of the matrix Aare rational numbers, then C

is called a rational polyhedral cone. A point yin the cone Cis called

an integral point if all its coordinates are integers.

For the purposes of constructing and enumerating magic squares, we

regard n×nmagic squares as either n×nmatrices or vectors in Rn2and

apply the normal algebraic operations to them. We also consider the

entries of an n×nmagic square as variables yij (1 ≤i, j ≤n). If we set

the ﬁrst row sum equal to all other mandatory sums, then magic squares

become nonnegative integral solutions to a system of linear equations

Ay = 0, where Ais an (2n+ 1) ×n2matrix each of whose entries is

0, 1, or -1. It is easy to verify that the sum of two magic squares is a

magic square and that nonnegative integer multiples of magic squares

are magic squares. Therefore, the set of magic squares is the set of all

integral points inside a polyhedral cone CMn={y:Ay = 0, y ≥0}in

139

Rn2, where Ais the coeﬃcient matrix of the deﬁning linear system of

equations. Observe that CMnis a pointed cone.

Like in the case of magic squares, we consider the entries of an n×n

Franklin square as variables yij (1 ≤i, j ≤n) and set the ﬁrst row sum

equal to all other mandatory sums. Thus, Franklin squares become

nonnegative integral solutions to a system of linear equations Ay = 0,

where Ais an (n2+ 8n−1) ×n2matrix each of whose entries is 0, 1,

or -1. The cone of Franklin squares is also pointed.

Example 5.2.1. 1. The equations deﬁning 3 ×3 magic squares are:

y11 +y12 +y13 =y21 +y22 +y23

y11 +y12 +y13 =y31 +y32 +y33

y11 +y12 +y13 =y11 +y21 +y31

y11 +y12 +y13 =y12 +y22 +y32

y11 +y12 +y13 =y13 +y23 +y33

y11 +y12 +y13 =y11 +y22 +y33

y11 +y12 +y13 =y13 +y22 +y31

Therefore, 3 ×3 magic squares are nonnegative integer solutions

to the system of equations Ay = 0 where:

A=







111−1−1−1000

1 1 1 0 0 0 −1−1−1

011−100−1 0 0

1 0 1 0 −1 0 0 −1 0

1 1 0 0 0 −100−1

0 1 1 0 −1 0 0 0 −1

1 1 0 0 −1 0 −1 0 0







and y=







y11

y12

y13

y21

y22

y23

y31

y32

y33







2. In the case of 4 ×4 magic squares, there are three linear relations

equating the ﬁrst row sum to all other row sums and four more

equating the ﬁrst row sum to column sums. Similarly, equating

the two diagonal sums to the ﬁrst row sum generates two more

linear equations. Thus, there are a total of 9 linear equations that

deﬁne the cone of 4 ×4 magic squares. The coeﬃcient matrix A

140

has rank 8 and therefore the cone CM4of 4 ×4 magic squares has

dimension 16 −8 = 8.

3. In the case of the 8 ×8 Franklin squares, there are seven linear

relations equating the ﬁrst row sum to all other row sums and

eight more equating the ﬁrst row sum to column sums. Similarly,

equating the eight half-row sums and the eight half-column sums

to the ﬁrst row sum generates sixteen linear equations. Equating

the four sets of parallel bent diagonal sums to the ﬁrst row sum

produces another thirty-two equations. We obtain a further sixty-

four equations by setting all the 2×2 subsquare sums equal to the

ﬁrst row sum. Thus, there are a total of 127 linear equations that

deﬁne the cone of 8 ×8 Franklin squares. The coeﬃcient matrix

Ahas rank 54 and therefore the cone of 8 ×8 Franklin squares

has dimension 10.

5.3 Hilbert bases of Polyhedral cones

In 1979, Giles and Pulleyblank introduced the notion of a Hilbert basis

of a cone [21]. For a given cone C, its set SC=C∩Znof integral

points is called the semigroup of the cone C.

Deﬁnition 5.3.1. A Hilbert basis for a cone Cis a ﬁnite set of points

HB(C)in its semigroup SCsuch that each element of SCis a linear

combination of elements from HB(C)with nonnegative integer coeﬃ-

cients.

Example 5.3.1. The integral points inside and on the boundary of

the parallelepiped in R2with vertices (0,0),(3,2),(1,3) and (4,5) in

Figure 5.6 form a Hilbert basis of the cone generated by the vectors

(1,3) and (3,2).

The minimal Hilbert basis of a cone is deﬁned to be the smallest

ﬁnite set Sof integral points with the property that any integral point

can be expressed as a linear combination with nonnegative integer coef-

ﬁcients of the elements of S. An integral point of a cone Cis irreducible

if it is not a linear combination with integer coeﬃcients of other inte-

gral points. The cone generated by a set Xof vectors is the smallest

cone containing Xand is denoted by cone X; so

cone X={λ1x1+.... +λkxk|k≥0; x1, . . . , xk∈X;λ1, . . . , λk≥0}.

141

(4,5)

(3,2)

(1,3)

(0,0)

Figure 5.6: A Hilbert Basis of a two dimensional cone.

Theorem 5.3.1. Each rational polyhedral cone Cis generated by a

Hilbert basis. If Cis pointed, then there is a unique minimal integral

Hilbert basis generating C(minimal relative to taking subsets).

Proof. Let Cbe a rational polyhedral cone, generated by b1, b2, ..., bk.

Without loss of generality b1, b2, ..., bkare integral vectors. Let a1, a2, ..., at

be all the integral vectors in the polytope P:

P={λ1b1+.... +λkbk|0≤λi≤1 (i= 1, .., k)}

Then a1, a2, ..., atgenerate Cas b1, b2, ..., bkoccur among a1, a2, ..., at

and as Pis contained in C. We will now show that a1, a2, ..., atalso

form a Hilbert basis. Let bbe an integral vector in C. Then there are

µ1, µ2, ..., µk≥0 such that

b=µ1b1+µ2b2+· ·· +µkbk.(5.1)

Let ⌊µi⌋denote the ﬂoor of µi, then

b=⌊µ1⌋b1+⌊µ2⌋b2+···+⌊µk⌋bk+(µ1−⌊µ1⌋)b1+(µ2−⌊µ2⌋)b2+·· ·+(µk−⌊µk⌋)bk.

Now the vector

b− ⌊µ1⌋b1− · ·· − ⌊µk⌋bk= (µ1− ⌊µ1⌋)b1+·· · + (µk− ⌊µk⌋)bk(5.2)

occurs among a1, a2, ..., atas the left side of the Equation 5.2 is

clearly integral and the right side belong to P. Since also b1, b2, ..., bk

occur among a1, a2, ..., at, it follows that 5.1 decomposes bas a non-

negative integral combination of a1, a2, ..., at. So a1, a2, ..., atform a

Hilbert basis.

142

Next suppose Cis pointed. Consider Hthe set of all irreducible

integral vectors. Then it is clear that any Hilbert basis must contain

H. So His ﬁnite because it is contained in P. To see that Hitself

is a Hilbert basis generating C, let bbe a vector such that bx > 0 if

x∈C\{0}(bexists because Cis pointed). Suppose not every integral

vector in Cis a nonnegative integral combination of vectors in H. Let

cbe such a vector, with bc as small as possible (this exists, as cmust

be in the set P). As cis not in H,c=c1+c2for certain nonzero

integral vectors c1and c2in C. Then bc1< bc and bc2< bc. Therefore

c1and c2are nonnegative integral combinations of vectors in H, and

therefore cis also.

The minimal Hilbert basis of a pointed cone is unique and hence-

forth, when we say the Hilbert basis, we mean the minimal Hilbert

basis. All the elements of the minimal Hilbert basis are irreducible.

Since magic squares are integral points inside a cone, Theorem 5.3.1

implies that every magic square is a nonnegative integer linear combi-

nation of irreducible magic squares.

We use the software 4ti2 to compute Hilbert bases (see [26]; software

implementation 4ti2 is available from http://www.4ti2.de). Algorithms

to compute Hilbert bases are discussed in Appendix A.

Example 5.3.2. 1. The minimal Hilbert basis of the 3 ×3 magic

squares is given in Figure 5.7. A Hilbert basis construction of the

Loh-shu magic square is given in Figure 5.8.

2. The minimal Hilbert basis of the polyhedral cone of 4 ×4 magic

squares is given in Figure 5.9. Two diﬀerent Hilbert basis con-

structions of the Jaina magic square is given in Figures 5.10 and

5.11. Thus, Hilbert basis constructions are not unique.

1 0 2

2 1 0

0 2 1

2 0 1

0 1 2

1 2 0

0 2 1

2 1 0

1 0 2

1 2 0

0 1 2

2 0 1

1 1 1

Figure 5.7: The minimal Hilbert Basis of 3 ×3 Magic squares.

Example 5.3.3. Let Sndenote the group of n×npermutation matrices

acting on n×nmatrices. Let (ri, rj) denote the operation of exchanging

143

+3 + =

1 2 0

0 1 2

2 0 1

0 2 1

2 1 0

1 0 2

1 1 1

4 9 2

3 5 7

8 1 6

Figure 5.8: A Hilbert basis construction of the Loh-Shu magic square.

0 0 1 0

0 1 0 0

0 0 0 1

1 0 0 0

0 0 1 0

1 0 0 0

0 1 0 0

0 0 0 1

1 0 0 0

1 0 0 00 0 1 0

0 0 1 0

0 1 0 0 0 1 0 0

0 0 0 1

0 0 0 1 0 1 0 0

0 1 0 0

0 0 0 1

1 0 0 0

1 0 0 0 1 0 0 0

0 0 1 0

0 0 1 0 0 0 1 0

0 0 1 0

1 0 1 0

0 0 0 2

0 1 1 0

1 1 0 0

0 0 2 0

0 1 0 1

1 1 0 0

1 0 0 1

0 0 1 1

0 1 0 1

1 0 1 0

1 1 0 0

0 1 1 0

0 0 0 2

1 0 1 0

1 0 0 1

1 1 0 0

0 1 0 1

0 0 2 0

0 1 0 1

1 1 0 0

0 0 1 1

1 0 1 0

1 0 0 1

0 0 1 1

0 2 0 0

1 0 1 0

0 2 0 0

1 0 1 0

0 0 1 1

1 0 0 1

1 1 0 0

1 0 1 0

0 1 0 1

0 0 1 1

0 1 0 1

0 1 1 0

0 0 1 1

2 0 0 0

1 0 1 0

0 0 1 1

1 1 0 0

0 1 0 1

0 0 1 1

0 1 1 0

2 0 0 0

0 1 0 1

h1

h13

h2 h3 h4 h5

h6 h7 h8 h9 h10

h11 h12 h14 h15

h16 h17 h18 h19 h20

Figure 5.9: The minimal Hilbert Basis of 4 ×4 Magic squares.

rows iand jof a square matrix, and let (ci, cj) denote the analogous

operation on columns. Let Gbe the subgroup of S8generated by

{(c1, c3),(c5, c7),(c2, c4),(c6, c8),(r1, r3),(r5, r7),(r2, r4),(r6, r8)}.

The Hilbert basis of the polyhedral cone of 8 ×8 Franklin squares

is generated by the action of the group Gon the three squares T1, T2,

and T3 in Figure 5.12 and their counterclockwise rotations through

90 degree angles. Not all squares generated by these operations are

distinct. Let Rdenote the operation of rotating a square 90 degrees in

the counterclockwise direction. Observe that R2·T1 is the same as T1

and R3·T1 coincides with R·T1. Similarly, R2·T2 is just T2, and R3·T2

144

1 0 0 0

0 0 1 0

0 0 0 1

0 1 0 0

0 0 0 1

0 0 1 0

1 0 0 0

0 0 0 1

0 1 0 0

0 0 1 0

0 0 0 1

0 1 0 0

1 0 0 0

0 0 1 0

0 1 0 0

0 0 1 0

1 0 0 0

0 0 0 1

7 12 1 14

2 13 8 11

16 3 10 5

9 6 15 4

=

+4 +2 +8

+3 +12 +4

0 0 1 0

0 1 0 0

0 0 0 1

1 0 0 0

0 0 0 1

1 0 0 0

0 0 1 0

0 1 0 0

h1 h3 h4 h5

h6 h7 h8 Jaina magic square

Figure 5.10: A Hilbert basis construction of the Jaina magic square.

1 0 0 0

0 0 1 0

0 0 0 1

0 1 0 0

0 0 0 1

0 0 1 0

1 0 0 0

0 0 0 1

0 1 0 0

0 0 1 0

0 0 0 1

0 1 0 0

1 0 0 0

0 0 1 0

0 1 0 0

0 0 1 0

1 0 0 0

0 0 0 1

7 12 1 14

2 13 8 11

16 3 10 5

9 6 15 4

=

1 1 0 0

0 1 1 0

0 0 0 2

1 0 1 0

1 0 0 1

0 0 1 1

1 0 1 0

0 2 0 0

1 1 0 0

1 0 1 0

0 1 0 1

0 0 1 1

0 1 1 0

2 0 0 0

0 1 0 1

+2 + +2 +8

++ +11 +

h3h15 h17 h5

h6 h7 h8 Jaina magic squareh20

h14

Figure 5.11: Another Hilbert basis construction of the Jaina magic square.

is the same as R·T2. Also T1 and R·T1 are invariant under the action

of the group G. Therefore the Hilbert basis of the polyhedral cone of

8×8 Franklin squares consists of the ninety-eight Franklin squares: T1

and R·T1; the thirty-two squares generated by the action of Gon T2

and R·T2; the sixty-four squares generated by the action of Gon T3

and its three rotations R·T3, R2·T3, and R3·T3.

Two diﬀerent Hilbert basis constructions of the Franklin squares F2

are provided in Figures 5.13 and 5.14.

5.4 Toric Ideals.

In this section, we demonstrate with the example of magic squares how

to avoid repetitions while enumerating integer solutions of equations.

We map integral points to monomials and then apply algebraic methods

145

T1 T2 T3

0 1 0 1 0 1 0 1

1 0 1 0 1 0 1 0

0 1 0 1 0 1 0 1

1 1 0 1 1 1 0 1

0 1 1 1 0 1 1 1

1 0 2 0 1 0 2 0

0 1 1 1 0 1 1 1

1 0 2 0 1 0 2 0

0 1 0 1 0 1 0 1

1 0 1 0 1 0 1 0

0 1 0 1 0 1 0 1

1 0 1 0 1 0 1 0

0 1 0 1 0 1 0 1

1 0 1 0 1 0 1 0

0 1 0 1 0 1 0 1

1 0 1 0 1 0 1 0

Figure 5.12: Generators of the Hilbert basis of 8 ×8 Franklin squares.

to eliminate duplicate solutions.

Let A={a1, a2, ..., ar}be a subset of Zn,ai= (ai1, ai2, . . . , ain), and

ϕbe the unique ring homomorphism between the rings k[x1, x2, . . . , xr]

and k[t±1

1, t±1

2, . . . , t±1

n] such that ϕ(xi) = tai, the monomial deﬁned by

tai=

j=1,...,n

taij

j.

The kernel of ϕis an ideal of k[x1, x2, . . . , xr] called the toric ideal

of Aand is denoted by IA.

We now demonstrate how to use toric ideals while enumerating

magic squares. Diﬀerent combinations of the elements of a Hilbert

basis sometimes produce the same magic square. Figures 5.10 and 5.11

exhibit two diﬀerent Hilbert basis constructions of the Jaina magic

square. This is due to algebraic dependencies among the elements

of the Hilbert basis. Repetitions have to be avoided when counting

squares. We solve this problem by using toric ideals of the Hilbert

bases.

Let HB(CMn) = {h1, h2, . . . hr}be a Hilbert basis for the cone of

n×nmagic squares. Denote the entries of the square hpby yp

ij, and let k

be any ﬁeld. Let ϕbe the ring homomorphism between the polynomial

rings k[x1, x2, . . . , xr] and k[t11, t12 , . . . , t1n, t21, t22, . . . t2n, . . . , tn1, tn2, . . . , tnn]

such that ϕ(xp) = thp, the monomial deﬁned by

thp=

i,j=1,...,n

typ

ij

ij .

Since the entries of hiare all nonnegative, we are dealing with only

polynomial rings in this case. Observe that, in general, the deﬁnition

146

0 1 1 0 1 1 0 0

1 0 0 1 0 0 1 1

0 1 1 0 1 1 0 0

1 0 0 1 0 0 1 1

0 1 0 1 0 1 0 1

1 0 1 0 1 0 1 0

1 1 0 0 0 0 1 1

0 0 1 1 1 1 0 0

1 1 0 0 0 0 1 1

0 0 1 1 1 1 0 0

1 0 1 0 1 0 1 0

0 1 0 1 0 1 0 1

1 0 1 0 1 0 1 0

0 1 0 1 0 1 0 1

1 0 1 0 1 0 1 0

0 1 0 1 0 1 0 1

5+ 16 +4

+

+2

+3 +2

+32

0 1 1 0 0 1 1 0

1 0 0 1 1 0 0 1

0 1 1 0 0 1 1 0

1 0 0 1 1 0 0 1

0 1 1 0 0 0 1 1

1 0 0 1 1 1 0 0

0 1 1 0 0 0 1 1

1 0 0 1 1 1 0 0

0 0 1 1 0 0 1 1

1 1 0 0 1 1 0 0

0 0 1 1 0 0 1 1

1 1 0 0 1 1 0 0

1 0 1 0 1 0 1 0

0 1 0 1 0 1 0 1

=

17 47 30 36 21 43 26 40

32 34 19 45 28 38 23 41

33 31 46 20 37 27 42 24

48 18 35 29 44 22 39 25

49 15 62 4 53 11 58 8

64 2 51 13 60 6 55 9

1 63 14 52 5 59 10 56

16 50 3 61 12 54 7 57

h1 h2 h3

h4 h5 h6

h7 h8 F2

Figure 5.13: Constructing Benjamin Franklin’s 8 ×8 square F2.

of the toric ideal is not restricted to polynomial rings alone. See [1],

[9], or [39] for a detailed study of toric ideals.

Monomials in k[x1, x2, . . . , xr] correspond to magic squares under

this map, and multiplication of monomials corresponds to addition of

magic squares. For example, the monomial x5

1x200

3corresponds to the

magic square 5h1+ 200h3. Diﬀerent combinations of Hilbert basis ele-

ments that give rise to the same magic square can then be represented

as polynomial equations. Thus, from the two diﬀerent Hilbert basis

constructions of the Jaina magic square represented in Figures 5.10

and 5.11, we learn that

h1 + 4 ·h3+2·h4+8·h5 + 3 ·h6 + 12 ·h7+4·h8 =

h3 + 8 ·h5 + h6 + 11 ·h7 + h8 + h14 + 2 ·h15 + 2 ·h17 + h20

In k[x1, x2, . . . , xr], this algebraic dependency of Hilbert basis elements

translates to

x1x4

3x2

4x8

5x3

6x12

7x4

8−x3x8

5x6x11

7x8x14x2

15x2

17x20 = 0.

Consider the set of all polynomials in k[x1, x2, . . . , xr] that are mapped

to the zero polynomial under ϕ. This set, which corresponds to all the

147

0 1 1 1 1 1 0 1

2 0 1 0 1 0 2 0

0 1 1 1 1 1 0 1

1 1 0 1 0 1 1 1

2 0 1 0 1 0 2 0

0 1 1 1 1 1 0 1

1 1 0 1 0 1 1 1

1 0 2 0 2 0 1 0

1 1 0 1 0 1 1 1

0 1 1 1 1 1 0 1

1 1 0 1 0 1 1 1

1 0 2 0 2 0 1 0

1 1 0 1 0 1 1 1

0 1 1 1 1 1 0 1

1 1 0 1 0 1 1 1

1 0 2 0 2 0 1 0

0 1 1 0 1 1 0 0

1 0 1 0 1 0 1 0

0 1 1 0 1 1 0 0

1 0 1 0 1 0 1 0

2+ + +

0 1 0 1 0 1 0 1

1 0 1 0 1 0 1 0

0 1 0 1 0 1 0 1

1 1 1 0 1 1 1 0

1 0 1 1 1 0 1 1

0 2 0 1 0 2 0 1

1 0 1 1 1 0 1 1

1 1 1 0 1 1 1 0

0 1 1 1 0 1 1 1

2 0 1 0 2 0 1 0

0 1 1 1 0 1 1 1

2 0 1 0 2 0 1 0

0 1 1 1 0 1 1 1

1 1 0 1 1 1 0 1

0 1 1 0 0 1 1 0

1 0 0 1 1 0 0 1

0 1 1 0 0 1 1 0

1 0 0 1 1 0 0 1

0 1 1 0 0 1 1 0

1 0 0 1 1 0 0 1

0 1 1 0 0 1 1 0

1 0 0 1 1 0 0 1

0 1 1 0 0 0 1 1

1 0 0 1 1 1 0 0

0 1 1 0 0 0 1 1

1 0 0 1 1 1 0 0

+ 32 + 12

0 1 0 1 0 1 0 1

1 0 1 0 1 0 1 0

0 1 0 1 0 1 0 1

=

17 47 30 36 21 43 26 40

32 34 19 45 28 38 23 41

33 31 46 20 37 27 42 24

48 18 35 29 44 22 39 25

49 15 62 4 53 11 58 8

64 2 51 13 60 6 55 9

1 63 14 52 5 59 10 56

16 50 3 61 12 54 7 57

+ 4 + 4

+

+ 3

1 0 1 1 1 0 1 1

0 2 0 1 0 2 0 1

1 0 1 1 1 0 1 1

1 1 0 1 1 1 0 1

h11h9 h10 h12

h13 h14 h4 h3

h15 h2 F2

Figure 5.14: Another construction of Benjamin Franklin’s 8 ×8 square F2.

algebraic dependencies of Hilbert basis elements is IHB(CMn), the toric

ideal of HB(CMn). Consequently, the monomials in the quotient ring

RCMn=k[x1, x2,··· , xr]/IHB(CMn)are in one-to-one correspondence

with magic squares.

Example 5.4.1. For example, in the case of 3 ×3 magic squares,

there are 5 Hilbert basis elements (see Figure 5.7) and hence there are

148

5 variables x1, x2, x3, x4, x5which gets mapped by ϕas follows:

x17→ 



102

210

021

7→ t11t13

2t21

2t22t32

2t33

x27→ 



201

012

120

7→ t11

2t13t22 t23

2t31t32

2

x37→ 



021

210

102

7→ t12

2t13t21

2t22t31 t33

2

x47→ 



120

012

201

7→ t11t12

2t22t23

2t31

2t33

x57→ 



111

111

7→ t11t12 t13t21t22 t23t31t32 t33

We use the Software CoCoA [16] to compute the toric ideal

IHB (CM3)= (x1x4−x2

5, x2x3−x1x4).

Algorithms to compute toric ideals are provided in Appendix A.

Thus, the monomials in the ring

RCM3=Q[x1, x2, x3, x4, x5]

(x1x4−x2

5, x2x3−x1x4)

are in one-to-one correspondence with the 3 ×3 magic squares.

5.5 Hilbert Functions.

Deﬁnition 5.5.1. A module over a ring R(or R-module) is a set M

and a mapping µ:R×M→Msuch that, if we write af for µ(a, f ),

where a∈Rand f∈M, the following axioms are satisﬁed.

1. Mis an abelian group under addition.

2. For all a∈Rand all f, g ∈M,a(f+g) = af +ag.

149

3. For all a, b ∈Rand all f∈M,(a+b)f) = af +bf.

4. For all a, b ∈Rand all f∈M,(ab)f) = a(bf).

5. If 1is the multiplicative identity in R,1f=ffor all f∈M.

Example 5.5.1. 1. An ideal Iof Ris an R-module. Consequently,

Ritself is an R-module.

2. If Ris a ﬁeld kthen a R-module is a kvector space.

3. The set of all m×1 column vectors in Rmis a R-module with

component wise addition and scalar multiplication, that is, let

a1, a2, . . . , am, b1, b2, . . . , bm, c ∈R, then







a1

a2

.

am







+





b1

b2

.

bm







=





a1+b1

a2+b2

.

am+bm







, c 





a1

a2

.

am







=





ca1

ca2

.

cam







.

Let M, N be R-modules. A mapping f:M→Nis an R-module

homomorphism if

f(x+y) = f(x) + f(y)

f(ax) = af(x)

for all a∈Rand all x, y ∈M.

Asubmodule M′of Mis a subgroup of Mwhich is closed under

multiplication by elements of R. The abelian group M/M′inherits a

R-module structure from Mdeﬁned by a(x+M′) = ax +M′. The

R-module M/M′is called a quotient module of M.

Example 5.5.2. If f:M→Nis a R-module homomorphism, the

kernel of fis a submodule of M; the image of f(denoted by Im(f)) is

a submodule of N; the cokernel of f,N/Im(f), is a quotient module

of N.

Agraded ring is a ring Rtogether with a family (Rn)n≥0of sub-

groups of the additive subgroup of Rsuch that R=∞

n=0 Rnand

RmRn⊆Rm+nfor all m, n ≥0. If Ris a graded ring, a graded R-

module is an R-module Mtogether with a family (Mn)n≥0of subgroups

150

of Msuch that M=∞

n=0 Mnand RmMn⊆Mm+nfor all m, n ≥0.

Let xi∈Mbe such that every element of a R-module Mcan be writ-

ten as a ﬁnite linear combination of xiwith coeﬃcients in R, then the

xiare said to be a set of generators of M. A R-module is said to be

ﬁnitely generated if it has a ﬁnite set of generators.

Let RCMn(s) be the set of all homogeneous polynomials of degree s

in the ring RCMn. Then RCMn(s) is a k-vector space, and RCMn(0) = k.

The dimension dimk(RCMn(s)) of RCMn(s) is precisely the number of

monomials of degree sin RCMn. Since R=k[x1, x2, ..., xr] is a graded

Noetherian ring, and RCMnis a ﬁnitely generated graded R-module,

RCMncan be decomposed into a direct sum of its graded components

RCMn=RCMn(s). The function H(RCMn, s) = dimk(RCMn(s)) is

the Hilbert function of RCMnand the Hilbert-Poincar´e series of RCMn

is the formal power series

HRCMn(t) = ∞



s=0

H(RCMn, s)ts.

In other words, the Hilbert-Poincare series is the generating function

of the Hilbert function. See Appendix A for a discussion on generating

functions.

If the variables xiof a polynomial ring k[x1, x2, . . . , xr] are as-

signed nonnegative weights wi, then the weighted degree of a monomial

xα1

1···xαr

ris r

i=1 αi·wi. If we take the weight of the variable xito

be the magic sum of the corresponding Hilbert basis element hi, then

dimk(RCMn(s)) is exactly the number of magic squares of magic sum s.

Lemma 5.5.1. Let Mn(s)denote the number of n×nmagic squares

with magic sum s. Let the weight of a variable xiin the ring R=

k[x1, x2, ..., xr]be the magic sum of the corresponding element of the

Hilbert basis hi. With this grading of degrees on the monomials of R,

the number of distinct magic squares of magic sum s,Mn(s), is given

by the value of the Hilbert function H(RCMn, s).

Example 5.5.3. For example, in the case of 3 ×3 magic squares, be-

cause all the elements of the Hilbert basis have sum 3, all the variables

are assigned degree 3, and

M3(s) = H(RCM3, s).

151

A sequence of R-modules and R-homomorphisms

··· −→ Mi−1

fi

−→ Mi

fi+1

−→ Mi+1 −→ ·· ·

is said to be exact at Miif Im(fi) = Ker(fi+1).

Example 5.5.4. 1. The sequence 0 →M′f

→Mis exact if and only

if fis injective.

2. The sequence Mg

→M′′ →0 is exact if and only if gis surjective.

3. The sequence 0 →M′f

→Mg

→M′′ →0 is exact if and only if

fis injective, gis surjective, and ginduces an isomorphism of

Coker(f) = M/f (M′) onto M′′. A sequence of this type is called

ashort exact sequence.

Let Cbe a class of R-modules and let Hbe a function on Cwith

values in Z. The function His called additive if for each short exact

sequence

0−→ M′f

−→ Mg

−→ M′′ −→ 0

in which all the terms belong to C, we have

H(M′)−H(M) + H(M′′) = 0.

Proposition 5.5.1 (proposition 2.11, [7]).Let 0→M0→M1→

··· → Mn→0be an exact sequence of R-modules in which all the

modules Miand the kernels of all the homomorphisms belong to C.

Then for any additive function Hon Cwe have

n



i=0

(−1)iH(Mi) = 0.

Proof. The proof follows because every exact sequence can be split

into short exact sequences: if Ni= Im(fi) = Ker(fi+1), we have short

exact sequences 0 →Ni→Mi→Ni+1 →0 for each i.

Theorem 5.5.1 (Hilbert-Serre Theorem).Let kbe a ﬁeld, R:= k[x1, x2, ..., xr],

and let x1, x2, ..., xrbe homogeneous of degrees di>0. Let Mbe a

ﬁnitely generated R-module. Let Hbe an additive function, then the

152

Hilbert Poincar´e series of M(with respect to H), HM(t)is a rational

function of the form:

HM(t) = p(t)

Πr

i=1(1 −tdi),

where p(t)∈Z[t].

Proof.

Since R:= k[x1, x2, ..., xr] is a graded Noetherian ring, we can write

R=∞

n=0 Rnsuch that RmRn⊆Rm+nfor all m, n ≥0. Let M=

Mn, where Mnare the graded components of M, then Mnis ﬁnitely

generated as a R0-module. The proof of the theorem is by induction

on r, the number of generators of Rover R0. Start with r= 0; this

means that Rn= 0 for all n > 0, so that R=R0, and Mis a ﬁnitely-

generated R0module, hence Mn= 0 for all large n. Thus HM(t) is a

polynomial in this case. Now suppose r > 0 and the theorem true for

r−1. For any R-module homomorphism ϕof Minto N, we have an

an exact sequence,

0→ker(ϕ)→Mϕ

→N→coker(ϕ)→0,

where ker(ϕ)→Mis the inclusion map and N→coker(ϕ) = N/im(ϕ)

is the natural homomorphism onto the quotient module. Multiplication

by xris an R-module homomorphism of Mninto Mn+dr, hence it gives

an exact sequence, say

0→Kn→Mn

xr

→Mn+dr→Ln+dr→0.(5.3)

Let K=nKn,L=nLn. These are both ﬁnitely generated R-

modules and both are annihilated by xr, hence they are R0[x1, . . . , xr−1]-

modules. Applying Hto 5.3 we have

H(Kn)−H(Mn) + H(Mn+dr)−H(Ln+dr) = 0;

multiplying by tn+drand summing with respect to nwe get

(1 −tdr)H(M, t) = H(L, t)−tdrH(K, t) + g(t),

where g(t) is a polynomial. Applying the inductive hypothesis the

result now follows. 

153

By invoking the Hilbert-Serre theorem, we conclude that the Hilbert-

Poincar´e series for magic squares is a rational function of the form

HRCMn(t) = p(t)/Πr

i=1(1 −tdegxi), where p(t) belongs to Z[t]. We use

the Software CoCoA [16] to compute Hilbert-Poincar´e series. Algo-

rithms to compute this series are discussed in Appendix A. We also

refer the reader to [1], [7], [10], or [33] for information about the Hilbert-

Poincar´e series.

Example 5.5.5. 1. In the case of 4 ×4 magic squares, the Hilbert-

Poincar´e series is given by

∞

s=0 M4(s)ts=t8+4t7+18t6+36t5+50t4+36t3+18t2+4t+1

(1−t)4(1−t2)4=

1 + 8t+ 48t2+ 200t3+ 675t4+ 1904t5+ 4736t6+ 10608t7+ 21925t8+. . .

Observe that the number of magic squares of magic sum is 0,1,2,3,4, . . .

is 1,8,48,200,675, . . . respectively.

2. Let F8(s) denote the number of 8×8 Franklin squares with magic

sum s, then the Hilbert-Poincar´e series is given by

∞

s=0 F8(s)ts=

{(t36 −t34 + 28 t32 + 33 t30 + 233 t28 + 390 t26 + 947 t24 + 1327 t22 + 1991 t20

+1878 t18 + 1991 t16 + 1327 t14 + 947 t12 + 390 t10 + 233 t8+ 33 t6+ 28 t4

−t2+ 1)}/{(t2−1)7(t6−1)3(t2+ 1)6}

= 1 + 34 t4+ 64 t6+ 483 t8+ 1152 t10 + 4228 t12 + 9792 t14 + 25957 t16 +···

5.6 Ehrhart Polynomials.

A polytope Pis called rational if each vertex of Phas rational coordi-

nates. The dilation of a polytope Pby an integer sis deﬁned to be the

polytope sP={sα :α∈ P} (see Figure 5.15 for an example).

Let i(P, s) denote the number of integer points inside the polytope

sP. If α∈Qm, let den αbe the least positive integer qsuch that

qα ∈Zm.

154

(1,2) (2,2)

(2,1)(1,1)

(2,2) (4,2)

(2,4) (4,4)

Figure 5.15: Dilation of a polytope.

Theorem 5.6.1. Let Pbe a rational convex polytope of dimension

din Rmwith vertex set V. Set F(P, t) = 1 + n≥1i(P, s)ts. Then

F(P, t)is a rational function, which can be written with denominator

α∈V(1 −tden α).

The proof of Theorem 5.6.1 involves Combinatorics and is not in the

scope of this book. We refer the reader to [33] for a proof. To extract

explicit formulas from the generating function we need to deﬁne the

concept of quasi-polynomials.

Deﬁnition 5.6.1. A function f:N7→ Cis a quasi-polynomial if there

exists an integer N > 0and polynomials f0, f1, ..., fdsuch that

f(n) = fi(n)if n ≡i(modN).

The integer Nis called a quasi-period of f.

For example, the formula for the number of 4 ×4 magic squares of

magic sum sis a quasi-polynomial with quasi-period 2. We now state

some properties of quasi-polynomials.

Proposition 5.6.1. The following conditions on a function f:N7→ C

and integer N > 0are equivalent:

1. fis a quasi-polynomial of quasi-period N.

2. n≥0f(n)xn=P(x)

Q(x),

where P(x)and Q(x)∈C[x], every zero αof Q(x)satisﬁes αN=

1(provided P(x)/Q(x)has been reduced to lowest terms) and deg

P < deg Q.

155

3. For all n≥0,

f(n) = k

i=1Pi(n)γn

i

where each Piis a polynomial function of nand each γisatisﬁes

γN

i= 1.The degree of Pi(n)is one less than the multiplicity of the

root γ−1

iin Q(x)provided P(x)/Q(x)has been reduced to lowest

terms.

A proof of Theorem 5.6.1 is given in [33] and is not discussed here be-

cause of its combinatorial nature. Theorem 5.6.1 together with Propo-

sition 5.6.1 imply that i(P, s) is a quasi-polynomial and is generally

called the Ehrhart quasi-polynomial of P. A polytope is called an in-

tegral polytope when all its vertices have integral coordinates. i(P, s) is

a polynomial if Pis an integral polytope (see [33]).

Verify that F(P, t) is the same as HRCMn(t) in Section 5.5. Recall

that the coeﬃcient of tsis the number of magic squares of magic sum

s. This information along with Proposition 5.6.1 enable us to recover

the Hilbert functions M4(s) and F8(s) from their respective Hilbert-

Poincar´e series by interpolation.

Example 5.6.1. 1.

M4(s) =











1

480 s7+7

240 s6+89

480 s5+11

16 s4+779

480 s3+593

240 s2+1051

480 s+13

16 ,

when sis odd,

1

480 s7+7

240 s6+89

480 s5+11

16 s4+49

30 s3+38

15 s2+71

30 s+ 1,

when sis even.

156

2.

F8(s) =











23

627056640 s9+23

17418240 s8+167

6531840 s7+5

15552 s6+2419

933120 s5+1013

77760 s4+701

22680 s3

−359

10206 s2−177967

816480 s+241

17496

if s ≡2 (mod 12) and s̸= 2,

23

627056640 s9+23

17418240 s8+167

6531840 s7+5

15552 s6+581

186624 s5+1823

77760 s4+6127

45360 s3

+10741

20412 s2+113443

102060 s+3211

2187

if s≡4 (mod 12),

23

627056640 s9+23

17418240 s8+167

6531840 s7+5

15552 s6+2419

933120 s5+1013

77760 s4+701

22680 s3

−5

378 s2−3967

10080 s−13

8

if s≡6 (mod 12),

23

627056640 s9+23

17418240 s8+167

6531840 s7+5

15552 s6+581

186624 s5+1823

77760 s4+6127

45360 s3

+11189

20412 s2+167203

102060 s+5771

2187

if s≡8 (mod 12),

23

627056640 s9+23

17418240 s8+167

6531840 s7+5

15552 s6+2419

933120 s5+1013

77760 s4+701

22680 s3

−583

10206 s2−608047

816480 s−20239

17496

if s≡10 (mod 12),

23

627056640 s9+23

17418240 s8+167

6531840 s7+5

15552 s6+581

186624 s5+1823

77760 s4+6127

45360 s3

+431

756 s2+1843

1260 s+ 1

if s≡0 (mod 12),

0

otherwise.

Summary.

To conclude the method to construct and enumerate nonnegative inte-

ger solutions of a linear system of equations Ax =bis as follows:

1. If bis the 0-vector, then

(a) Compute the Hilbert basis H={h1, . . . , hr}of the cone

Ax = 0. The Hilbert basis enables us to construct solutions.

(b) Associate variable yito a Hilbert basis element hi, and com-

pute the toric ideal Iof the Hilbert basis.

157

(c) Compute the Hilbert Poincare series of the ring k[y1, . . . , yr]/I

to enumerate the integer solutions.

(d) Interpolate using the coeﬃcients of the series to get formulas

for the number of nonnegative solutions.

2. If bis not the 0-vector, then introduce a new variable sand solve

the system Ax −bs = 0 using the steps in 1. Set s= 1 in the

solutions of Ax −bs = 0 to get the solutions of Ax =b.

Exercises.

1. Prove Pick’s Theorem: Let Abe the area of a simply closed lat-

tice polygon. Let Bdenote the number of lattice points on the

Polygon edges and Ithe number of points in the interior of the

polygon, then A=I+ 1/2B−1.

2. A labeling of a graph Gis an assignment of a nonnegative integer

to each edge of G. A magic labeling of magic sum rof Gis a

labeling such that for each vertex vof Gthe sum of the labels of

all edges incident to vis the magic sum r(loops are counted as

incident only once). Graphs with a magic labeling are also called

magic graphs. Let Gbe the complete graph on 3 vertices.

(a) Use the methods in this chapter to construct and enumerate

magic labelings of a graph G.

(b) Prove that the perfect matchings of Gare the minimal Hilbert

basis elements of the cone of magic labelings of Gof magic

sum 1. Count the number of perfect matchings of G.

3. Show that the number of 3 ×3 magic squares

M3(s) = 





2

9s2+2

3s+ 1 if 3 divides s,

0 otherwise.

158

Chapter 6

Miscellaneous Topics in

Applied Algebra.

If I saw further than other men, it was because I stood on the shoulders

of giants - Isaac Newton.

In this chapter, we look at some miscellaneous applications of the

concepts developed in this book. In the following sections, we count

and generate orthogonal Latin squares, prove the Chinese Remainder

Theorem, encrypt and decrypt messages, and generate error correcting

codes.

6.1 Counting Orthogonal Latin squares.

In 1781 Euler proposed the problem of seating 36 oﬃcers of six diﬀerent

ranks from six diﬀerent regiments in an array such that each row and

each column contains one oﬃcer of each rank and one oﬃcer from each

regiment. In this section, we relate this problem to Latin squares.

Deﬁnition 6.1.1. A Latin square of order nis an n×narray in which

each one of nsymbols occurs once in each row and once in each column.

We denote the nsymbols as 0,1, . . . , n −1.

Theorem 6.1.1. For each n≥2the n×narray deﬁned by

L(i, j) = i+jmod n

is a Latin square.

159

Proof. Suppose the symbols in positions (i, j ) and (i, j′) are the

same. Then

i+j=L(i, j) = L(i, j ′) = i+j′.

Since Zmcontains an element −i, we add −ito both sides of the above

equation to get j=j′. Hence each symbol occurs at most once in row

i. Consequently, since there are nsymbols and ncolumns, each symbol

occurs exactly once. A similar argument holds for columns. Thus Lis

a Latin square.

Example 6.1.1. By Theorem 6.1.1

L=

012345

123450

234501

345012

450123

501234

is a Latin square of order 6.

Theorem 6.1.1 shows that there is always at least one Latin square

of any given order.

A pair of Latin squares L1and L2of the same order are orthogonal

if for each pair of symbols (k, k′), there is just one position (i, j) for

which

L1(i, j) = kand L2(i, j ) = k′.

Thus, Euler’s problem of seating 36 oﬃcers is equivalent to ﬁnding

two orthogonal Latin squares L1and L2of order 6, such that L1is

the Latin square with the ranks as symbols, and the symbols of L2are

the regiments. Consequently, when the two squares are superimposed,

the cell (i, j) contains an oﬃcer of rank iand from regiment j, thereby

solving the arrangement problem. Euler correctly conjectured there

was no solution to this problem and Gaston Tarry proved this in 1901.

We will show that pairs of orthogonal squares with orders that are

powers of a prime number always exist. Before that we provide an

upper limit to the number of orthogonal squares possible for any order.

Theorem 6.1.2. There cannot exist a set of more than q−1mutually

orthogonal Latin squares of order q.

160

Proof. Suppose there exists a set of mmutually orthogonal Latin

squares of order q. By renaming the symbols we can transform each

square to the standard form such that the initial row is occupied by

the symbols 0,1, . . . , q −1 in order. Thus in each square, the cell (0, j)

contains the symbol j, where 0 ≤j≤q−1. The standardized squares

are mutually orthogonal. Since the cell (0,0) contains the symbol 0,

the symbol in the cell (1,0) must be diﬀerent from 0 for each of the m

standardized squares. When two diﬀerent squares are superimposed,

the pair of symbols (j, j) occurs in the cell (0, j). Hence the symbols

in the cell (1,0) of these two squares must be diﬀerent. Thus the cells

(1,0) of the mstandardized orthogonal Latin squares are occupied by

diﬀerent nonzero symbols. Since there are only q−1 nonzero symbols,

m≤q−1.

By Corollary 3.4.11, we know that for each positive prime pand

positive integer r, the splitting ﬁeld of xpr−xis a ﬁeld of order q=pr.

Denote this ﬁeld by Fqand its elements by αi.

Theorem 6.1.3. Let q=prsuch that pis a prime number. Take a

q×qsquare Lt, and in the cell (i, j )of this square, put the integer u

given by

αu=αtαi+αj,(6.1)

where αtis a nonzero element of Fq.Ltdeﬁnes a Latin square.

Furthermore, when t̸=t′, the Latin squares Ltand Lt′are orthogonal.

There are q−1mutually orthogonal Latin squares of order q.

Proof. To prove that Ltis Latin square, we need to show that the

symbols 0,1, . . . , n −1 occur in each row and column exactly once. In

the row ithe symbol uoccurs in the column jgiven by

αj=αu−αtαi.

In the column jthe symbol uoccurs in the row igiven by

αi=αu−αj

αt

.

Thus Ltis a Latin square. Consequently, we get q−1 Latin squares

from Formula 6.1 corresponding to the nonzero values of αt. Let Lt

and Lt′,t̸=t′, be two of these Latin squares. When superimposed the

161

symbol uof the ﬁrst square occurs together with the symbol u′of the

second square in the cell (i, j) if and only if

αu=αtαi+αj,

αu′=αt′αi+αj.

Solving these two equations we get

αi=αu−αu′

αt−αt′

, αj=αtαu′−αt′αu

αt−αt′

.

Thus Ltand Lt′are mutually orthogonal Latin squares.

Since there cannot exist a set of more than q−1 mutually orthogonal

Latin squares of order qby Theorem 6.1.2, we have exactly q−1 Latin

squares when qis a power of a prime number.

Example 6.1.2. By Exercise 3.4.8, the four elements of the ﬁeld F4

are

α0= 0, α1= 1, α2=x, α3=x2=x+ 1.

The three mutually orthogonal Latin squares L1, L2, L3are:

[L1] [L2] [L3]

αu=α1αi+αjαu=α2αi+αjαu=α3αi+αj

0123

1032

2301

3210

,

0 1 2 3

2 3 0 1

3 2 1 0

1 0 3 2

,

0123

3210

1032

2301

.

Corollary 6.1.4. Let pbe a prime number. Let tbe a non-zero element

of Zp. Then the rule

Lt(i, j) = ti +jsuch that i, j ∈Zp

deﬁnes a Latin square. Furthermore, when t̸=t′, the Latin squares

Ltand Lt′are orthogonal. There are p−1mutually orthogonal Latin

squares of order p.

Proof. When q=p,Fp=Zp, therefore αu=ti +jin Theorem

6.1.3.

162

Example 6.1.3. When p= 3 the two mutually orthogonal squares are

L1=

012

120

201

, L2=

0 1 2

2 0 1

1 2 0

.

Is it possible to construct orthogonal pairs of Latin squares when q

is not a prime power? We already said that there are no such pairs for

order 6. Bose, Parker, and Shrikande succeeded in constructing a pair

of orthogonal Latin squares for n= 10. Whether there are more such

pairs for order 10 and higher is an open problem in combinatorics. See

[12] for an in-depth study of Latin squares.

6.2 Chinese Remainder Theorem.

The Chinese Remainder Theorem is a famous result in number theory

that was known to Chinese mathematicians in the ﬁrst century A.D.

The Chinese Remainder Theorem, supposedly, helped bandits divide

their gold coins in ancient China. Let us consider an example.

A band of 17 bandits steal a certain quantity of gold coins. When

they try to evenly distribute the coins amongst themselves, they end

up with 3 left over. A ﬁght breaks out over the remaining coins and

one pirate is killed. The 16 bandits left alive attempt to once again

divide the coins up between themselves. However, this time, there are

10 coins left over. Being the greedy bandits they are, another ﬁght

ensues, and another pirate is killed. Figuring that the third time is a

charm, the 15 remaining bandits try once again to evenly distribute the

coins. This time, they are successful. What is the minimum amount

of coins they could have stolen?

To solve this problem, denote the number of gold coins by x. Then

a solution to the bandit’s problem is a solution of the system of con-

gruence equations

x≡3( mod 17)

x≡10( mod 16)

x≡0( mod 15) (6.2)

We solve such systems of congruence equations using the Chinese

Remainder Theorem.

163

Theorem 6.2.1 (Chinese Remainder Theorem).Let m1, m2, . . . , mrbe

pairwise relatively prime positive integers and let m=m1m2m3···mr.

Let a1, . . . , arbe integers. Consider the system of congruence equations

x≡a1(mod m1)

x≡a2(mod m2)

.

x≡ar(mod mr).

Let Mk=m/mkand let Mkdenote the inverse of Mkmodulo mk, then

x=a1M1M1+a2M2M2+···+arMrMr

is a unique solution modulo m.

Proof. If j̸=k, then mkdivides Mj. Therefore,

ajMjMj≡0 mod mk,when j̸=k.

Consequently,

x≡akMkMk≡ak·1 = akmod mk.

Hence xis a solution of the system of congruence equations. If zis

any other solution of the system, then for each i= 1,2, . . . , r,

z≡ai( mod mi) and x≡ai( mod mi).

By transitivity z≡x(mod mi). Thus midivides z−xfor each iand

hence m1m2···mrdivides z−x. Hence z≡x(mod m1m2···mr).

Conversely, if z≡x(mod m1m2···mr), then m1m2·· ·mrdivides

z−x. Consequently, since m1, m2,··· , mrare relatively prime num-

bers, midivides z−xfor each i. Hence z≡x(mod mi) for each i.

Consequently, x≡ai(mod mi) implies z≡ai(mod mi), for each i, by

transitivity. Therefore zis a solution of the given system.

Example 6.2.1. We return to the bandits problem.

x≡3(mod 17)

x≡10(mod 16)

x≡0(mod 15) (6.3)

164

Here,

a1= 3, a2= 10, a3= 0, m1= 17, m2= 16, m3= 15, m = 4080,

and

M1= 16 ×15 = 240, M2= 17 ×15 = 255, M3= 17 ×16 = 272.

We need to ﬁnd the inverse of M1mod m1. Now M1= 240 ≡2

mod 17. Since the gcd(2,17) = 1, we use the Euclid’s algorithm to

write

17 −8×2 = 1.

Reducing this equation modulo 17, we see that the inverse of 2 mod 17

is −8≡9 mod 17. This implies that the inverse of 240 mod 17 is 9,

that is, M1= 9. Similarly, we show that M2= 15 and M3= 8.

By the Chinese Remainder Theorem, we get

x=a1M1M1+a2M2M2+a3M3M3

= 3 ×240 ×9 + 10 ×255 ×15 ×+0 ×272 ×8

= 44730 ≡3930 mod m.

So the minimum number of gold coins stolen by the bandits is 3930.

We illustrate an alternate method of multiplying numbers using the

Chinese remainder Theorem. Every computer has a limit on the size of

integers called the word size. Computer arithmetic with integers larger

than the word size requires time consuming multiprecision techniques.

In such scenarios, the alternate method of addition and multiplication

using the Chinese Remainder Theorem is quite eﬃcient.

Suppose we want to ﬁnd the product of the numbers t1, t2, . . . , tn.

Let m1, . . . , mrbe pairwise relatively prime positive integers. We

choose m1, . . . , mrsuch that the product of these numbers is larger

than the result we want to derive so that the solution is unique and

the method is well deﬁned. The method proceeds as follows.

1. Represent each integer tkas an element of Zm1×Zm2×·· ·×Zmr

by reducing tkmodulo mifor each i.

2. Represent the product as an element of Zm1×Zm2× ·· · × Zmr

thereby making the product the solution to a system of congru-

ence equations.

165

3. Use the Chinese Remainder Theorem to solve the system.

We illustrate this procedure with an example.

Example 6.2.2. In this example, we multiply the numbers 219 and

172 using Chinese Remainder Theorem. We begin by choosing sev-

eral numbers that are pairwise relatively prime, and are such that the

product of all these numbers are larger than the product of 219 and

172. For this example, we chose 4,7,11,15,13. Next we reduce the two

numbers and their product modulus each prime:

219 ≡3 mod 4

219 ≡2 mod 7

219 ≡10 mod 11

219 ≡11 mod 13

219 ≡9 mod 15

172 ≡0 mod 4

172 ≡4 mod 7

172 ≡7 mod 11

172 ≡3 mod 13

172 ≡7 mod 15

219 ×172 ≡0 mod 4

219 ×172 ≡8≡1 mod 7

219 ×172 ≡70 ≡4 mod 11

219 ×172 ≡33 ≡7 mod 13

219 ×172 ≡63 ≡3 mod 15

In other words, the integer 219 = (3,2,10,11,9) and 172 = (0,4,7,3,7)

in Z4×Z7×Z11 ×Z13 ×Z15. Moreover, 219 ×172 is a solution of the

system

x≡0 mod 4

x≡1 mod 7

x≡4 mod 11

x≡7 mod 13

x≡3 mod 15 (6.4)

We use the Chinese Remainder Theorem to solve this system of

congruences and get x= 37668 as the solution. We know that 219 ×

172 <4×7×11 ×13 = 60060. Also no two numbers between 0

and 60060 can be congruent modulo 60060. Therefore, we must have

219 ×172 = 37668.

The procedure to add large numbers using Chinese Remainder The-

orem is very similar to multiplication and is explored in the exercises.

6.3 Cryptology

Codes have been used since ancient times by friends, merchants, and

armies to transmit secret messages. For example, in Julius Caesar’s

166

coding system, each letter is shifted three letters forward in the alpha-

bet and the last three letters are send to the ﬁrst three letters.

Message: A B C D E F G H I J K L M N O P

Code: D E F G H I J K L M N O P Q R S

Message: Q R S T U V W X Y Z

Code: T U V W X Y Z A B C

The steps to implement Caesar’s code are as follows.

1. Replace each alphabet by an integer from 0 to 25:

A B C D E F G H I J K L M N O

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

P Q R S T U V W X Y Z

15 16 17 18 19 20 21 22 23 24 25

2. The Caesar’s encryption is a function ffrom the set of numbers

representing the alphabets of the message to the set of integers

{0,1,2, . . . , 25}, such that, f(p) = p+ 3 mod 26.

Example 6.3.1. In Caesar’s code, the message YOU ARE IN XANADU

is coded as follows.

Y O U A R E I N X A N A D U

p: 24 14 20 0 17 4 8 13 23 0 13 0 3 20

(p+ 3) mod 26 : 1 17 23 3 20 7 11 17 0 3 16 3 6 23

B R X D U H Y R A D Q D G W

Thus, the message YOU ARE IN XANADU becomes BRX DUH YR

ADQDGW.

To decrypt the message, we use the inverse function f−1(y) = y−3

mod 26.

167

Example 6.3.2. Decrypt the message Z HO F RP H.

Code: Z H O F R P H

p: 25 7 14 5 17 15 7

(p−3) mod 26 : 22 4 11 2 14 12 4

Message: W E L C O M E

In the generalized Caesar’s code, ais an integer which is relatively

prime to 26, and the message is encrypted using the function f(p) =

ap +bmod 26, where bis any integer. The choice of aensures that f

has an inverse.

Example 6.3.3. When f(p) = 7p+3 mod 26, the message WELCOME

is coded as

Message: W E L C O M E

p:22 4 11 2 14 12 4

7p+ 3 mod 26 : 1 5 2 17 23 9 5

Code: B F C R X J F

Caesar’s code is easy to break, and is not useful when high security

is desired. In recent times, the coding system developed by R. Rivest,

A. Shamir, and L. Adleman, called the RSA system, is popularly used.

Its security depends on the diﬃculty of factoring large integers. We

describe this coding system now.

Algorithm 6.3.1 (The RSA Algorithm).

1. Let Mbe the message to be encrypted. Choose two large primes

pand q. Let n=pq and t= (p−1)(q−1). Choose a lock

Lsuch that gcd(L, t) = 1. We also require gcd(M, p) = 1 and

gcd(M, q) = 1 for the algorithm to work. But, since pand qare

very large, this follows automatically.

2. Encrypt the message Mto get the code Cas follows:

C=MLmod n.

3. Determine the key Kwhich is the inverse of Lmod t.

4. Decrypt Cto get Mas follows:

M=CKmod n.

168

Example 6.3.4. Encode HOWDY using the RSA method with p= 3,

q= 11, and L= 3.

Like before we associate integers from the set {0,1, . . . , 25}to the

alphabets of the message:

message: H O W D Y

M: 7 14 22 3 24

Here n=pq = 3times11 = 33. Note that gcd(L, (p−1)(q−1)) = 1.

Hence Lis a valid lock. Compute MLmod n:

73≡13 mod 33

143≡5 mod 33

223≡22 mod 33

33≡27 mod 33

243≡30 mod 33

Consequently, the encrypted code Cis

C: 13 05 22 27 30.

Example 6.3.5. The following message was encoded using the RSA

method with p= 3, q= 11, and L= 3.

18 5 5 27 3 5 1

We now decode the message. Here t= (p−1)(q−1) = 20. The key

Kis the inverse of Lmod t. Since gcd(3,20) = 1, we use the Euclid’s

algorithm to write 1 = 7 ×3−20. Consequently, the inverse of 3 mod

20 is 7, that is, K= 7. Compute CKmod n:

187≡6 mod 33

57≡14 mod 33

277≡3 mod 33

37≡9 mod 33

17≡1 mod 33

C: 18 5 5 27 3 5 1

M: 6 14 14 3 9 14 1

G O O D J O B

Thus the message was GOOD JOB.

169

We could use more than 1 letter blocks to make encryption more

secure. This is explored in the next example.

Example 6.3.6. Encrypt the message STOP using RSA with p= 43,

q= 59, and lock L= 13. Use two letter blocks.

Note that gcd(L, (p−1)(q−1)) = 1, so that Lis a valid lock. Here

n= 43 ×59 = 2537. Hence

C=M13 mod 2537.

The integer representation of STOP is 18,19,14,15. Since we are

using two letter blocks, STOP is represented as 1819,1415. Conse-

quently, STOP is encrypted as 2081 2182, since

181913 mod 2537 = 2081,141513 mod 2537 = 2182.

To prove the RSA Algorithm, we have to look at a theorem known

as Fermat’s Little Theorem.

Theorem 6.3.1 (Fermat’s Little Theorem).If pis a prime and ais

an integer not divisible by p, then

ap−1≡1mod p.

Furthermore, for every integer a,

ap≡amod p.

Proof. If pis a prime and ais an integer not divisible by p, then p

does not divide ka for any ksuch that 0 < k < p. Therefore, each of the

numbers 1,2a, ···,(p−1)amust be congruent to one of 1,2,3, . . . , p−1.

If ra ≡sa mod p, then since gcd(a, p)=1, we get that r≡smod p.

This is not possible because no two of the numbers 1,2, . . . p −1 are

congruent modulo p. Therefore, in some order, a, 2a, . . . , (p−1)aare

congruent to 1,2,3, . . . , p −1, that is,

a·2a·3a···(p−1)a≡1·2· ··(p−1) mod p.

Hence

ap−1·1·2···(p−1) ≡1·2· ··(p−1) mod p.

Since pdoes not divide 1 ·2··· (p−1), we get ap−1≡1 mod p. To

prove that for every integer a,ap≡amod p, ﬁrst consider the case

170

when pdivides a. Then, pdivides ap−a. Hence ap≡amod p.

Now if pdoes not divide a, then by Fermat’s little Theorem ap−1≡

1 mod p. Multiply the congruence equation on both sides by ato get

ap≡amod p.

Finally, we prove the RSA algorithm.

Proof of the RSA algorithm:

Since gcd(L, (p−1)(q−1)) = 1, the inverse Kof Lmod (p−1)(q−1)

exists and

LK ≡1 mod (p−1)(q−1).

Therefore for some integer t,LK = 1 + t(p−1)(q−1).Now

CK= (ML)K=MLK =M1+t(p−1)(q−1) mod n.

Assume gcd(M, p) = 1 and gcd(M , q) = 1, then by Fermat’s Little

Theorem: Mp−1≡1 mod p,

Mq−1≡1 mod q

Ck≡M1+t(p−1)(q−1) =M·(Mp−1)t(q−1) ≡M·1≡Mmod p.

Ck≡M1+t(p−1)(q−1) =M·(Mq−1)t(p−1) ≡M·1≡Mmod q.

Since gcd(p, q) = 1, we get CK≡Mmod pq by the Chinese remainder

Theorem.

6.4 Algebraic codes.

When a message is transmitted over a long distance there may be some

interference, and the message may not be received exactly as it is sent.

In such cases, we need to be able to detect and, if possible, correct

errors. In this section, we discuss these issues for messages represented

in the binary alphabet {0,1}.

Let B(n) denote the Cartesian product Z2×Z2×Z2× · ·· × Z2of

ncopies of Z2. Verify that with coordinate-wise addition B(n) is an

additive group of order 2n. In this section, the elements of B(n) will be

written as strings of 0’s and 1’s of length n. When B(n) is listed such

that the successor of an n-tuple diﬀers from it in only one position,

then B(n) is called a Gray code of order n. The following algorithm

generates a Gray code of order n.

171

Algorithm 6.4.1 (Gray Code Algorithm).1. The Gray code of or-

der 1 is 0

1

2. Suppose n > 1and the Gray code of order n−1is already con-

structed. To construct the Gray code of order n, we ﬁrst list the

(n−1)-tuples of 0s and 1s in the order of the Gray code of order

n−1, and attach a 0at the beginning of each (n−1)-tuple. We

then list the (n−1)-tuples in the order which is reverse of that

given by the Gray code of n−1, and attach a 1at the beginning.

Example 6.4.1. Gray code of order 2 is

00

01

11

10

and the Gray code of order 3 is

000

001

011

010

110

111

101

100

We refer the reader to [13] for the connection of Gray codes to unit

cubes and other details.

A code C∈B(n) is linear if whenever aand bare in C, then

a+b∈C. Equivalently, a (n, k)binary linear code Cis a subgroup

of B(n) of order 2k. The elements of Care called codewords. Only

codewords are transmitted, but any element of B(n) can be a received

word.

Example 6.4.2. C={0000,1111}is a (4,1) code since Cis a sub-

group of order 21of the group B(4) = Z2×Z2×Z2×Z2.

172

Deﬁnition 6.4.1. The Hamming weight of an element uof B(n)is

the number of nonzero coordinates in u, and it is denoted W t(u).

Example 6.4.3. For the codeword u= 010110, W t(u) = 3, and for

the codeword v= 110110, W t(v) = 4.

Deﬁnition 6.4.2. Let u, v ∈B(n). The Hamming distance between

uand v, denoted d(u, v), is the number of coordinates in which uand

vdiﬀer.

Example 6.4.4. For the codewords u= 010110 and v= 110110, the

Hamming distance d(u, v) = 1.

Lemma 6.4.1. If u, v, w ∈B(n), then d(u, v) = W t(u−v), and

d(u, v)≤d(u, w) + d(w, v).

Proof. A coordinate of u−vis nonzero if and only if uand vdiﬀer in

that coordinate. So the number of nonzero coordinates in u−v, namely

W t(u−v), is the same as the number of coordinates in which uand

vdiﬀer. Therefore d(u, v) = W t(u−v). We prove d(u, v)≤d(u, w) +

d(w, v) by proving W t(u−v)≤W t(u−w) + W t(w−v). For this

purpose, suppose that the i-th coordinate of u−v,ui−vi, is nonzero,

and the i-th coordinate of u−w,ui−wi, is zero. Consequently, since

ui=wi,wi−vi, the i-th component of w−vis ui−vi, which is nonzero

by our assumption. Thus (ui−wi)+(wi−vi) is nonzero whenever ui−vi

is nonzero. Therefore W t(u−v)≤W t(u−w) + W t(w−v).

If a codeword uis transmitted and the word wis received, then the

number of errors in the transmission is the Hamming distance d(u, w).

Assuming there are only few transmission errors, a received word is

decoded as the codeword that is nearest to it in Hamming distance

and this process is called nearest-neighbor decoding. A linear code is

said to correct t-errors if every codeword that is transmitted with tor

fewer errors is correctly decoded by nearest-neighbor decoding.

Theorem 6.4.1. A linear code corrects terrors if and only if the Ham-

ming distance between any two codewords is at least 2t+ 1.

Proof. Assume that the distance between any two codewords is at

least 2t+ 1. If the codeword uis transmitted with tor fewer errors

and received as w, then d(u, w)≤t. If vis any other codeword, then

d(u, v)≥2t+ 1 by hypothesis. Therefore by Lemma 6.4.1

2t+ 1 ≤d(u, v)≤d(u, w) + d(w, v)≤t+d(w, v).

173

Subtracting tfrom both sides of 2t+ 1 ≤t+d(w, v), we get d(w, v)≥

t+ 1. Since d(u, w)≤t,uis the closest codeword to w, so the nearest-

neighbor decoding correctly decodes was u. Hence the code corrects

t-errors. The proof of the converse is Exercise 9.

A linear code is said to detect t-errors if it detects that a received

word with at least one and not more than terrors is not a codeword.

Theorem 6.4.2. A linear code detects terrors if and only if the Ham-

ming distance between any two codewords is at least t+ 1.

Proof. Assume that the distance between any two codewords is at

least t+ 1. If the codeword uis transmitted with at least one, but not

more than terrors, and received as w, then

0< d(u, w)≤t, and hence d(u, w)< t + 1.

So wcannot be a codeword. Therefore the code detects terrors. The

proof of the converse is Exercise 10.

Corollary 6.4.3. A linear code detects 2terrors and corrects terrors

if and only if the Hamming weight of every nonzero codeword is at least

2t+ 1.

Proof. Let wbe a nonzero codeword. Since W t(w) = W t(w−0) =

d(w, 0), the minimum hamming distance between any two codewords

is the minimum Hamming weight of all the nonzero codewords. The

proof then follows by Theorems 6.4.1 and 6.4.2.

Ak×nstandard generator matrix is a k×nmatrix Gwith entries

in Z2of the form







100··· 0 0 a11 ··· a1n−k

010··· 0 0 a21 ··· a2n−k

.

..

.

..

.

..

.

..

.

..

.

..

.

000··· 1 0 a(k−1)1 ··· a(k−1)n−k

000··· 0 1 ak1· ·· akn−k







= [Ik|A]

where Ikis the k×kidentity matrix and Ais a k×(n−k) matrix.

Example 6.4.5. The 3 ×6 matrix

G=



100011

010101

001110



is a generator matrix.

174

Theorem 6.4.4. If Gis a k×nstandard generator matrix, then

{uG|u∈B(k)}is a (n, k)code.

Proof. Deﬁne a function f:B(k)→B(n) by f(u) = uG. Since

f(u+v) = (u+v)G=uG +vG =f(u) + f(v),

fis a homomorphism of groups. Verify that the ﬁrst k-coordinates

of uand uG are the same. Therefore fis injective. Consequently

Im fis isomorphic to B(k) and hence has order 2k. Therefore Im

f={uG|u∈B(k)}is a (n, k) code.

Example 6.4.6. Suppose we want to code the message “Hello World”,

then we choose B(3) because this group is suﬃcient to represent all the

letters in our message.

Symbols Message words

Blank space

H

E

L

0

W

R

D

000

001

011

010

110

111

101

100

We use the matrix Gin Example 6.4.5 to generate a (6,3) code.

For example, Let u= 011, then

uG =011



100011

010101

001110

=0 1 1 0 1 1 .

175

Operating with Gon all the message words in B(3), we get

Message words Codewords

000

001

011

010

110

111

101

100

000000

001110

011011

010101

110110

111000

101101

100011

Since all the code words have Hamming weight at least 3, this code

can correct single errors. The message “Hello World” will be coded as

001110 H

011011 E

010101 L

110110 0

000000

111000 W

110110 O

101101 R

010101 L

100011 D

For (n, k) codes with large k, brute force method of searching for

the nearest neighbor is impractical. So we develop more systematic

decoding techniques. We now look at a decoding technique based on

the cosets of the code C. We form a coset decoding table. Its rows

are the cosets of C, with Citself as the ﬁrst row. A coset leader of a

coset is an element of the smallest weight in the coset. Each row of

the decoding table is of the form e+C, where eis the coset leader.

The coset leader is always listed ﬁrst in the row. The decoding rule is:

decode a received word was the codeword at the top of the column in

which wappears.

Example 6.4.7. Consider the (6,3) code from Example 6.4.6:

C={000000,001110,011011,010101,110110,111000,101101,100011}.

176

Then the coset decoding table of Cis

000000 001110 011011 010101 110110 111000 101101 100011

100000 101110 111011 110101 010110 011000 001101 000011

010000 011110 001011 000101 100110 101000 111101 110011

001000 000110 010011 011101 111110 110000 100101 101011

000100 001010 011111 010001 110010 111100 101001 100111

000010 001100 011001 010111 110100 111010 101111 100001

000001 001111 011010 010100 110111 111001 101100 100010

101010 100100 110001 111111 011100 010010 000111 001001

The received words 011110 (third row) is decoded as 001110, the

word 101000 (again third row) is decoded as 111000, whereas the word

111111 (eighth row) is decoded as 010101 using the decoding rule.

We prove in the next theorem that a coset decoding is the nearest

neighbor decoding.

Theorem 6.4.5. Let Cbe an (n, k)code. The decoding for Cusing

its coset decoding table is nearest neighbor decoding.

Proof. If w∈B(n), then w=e+v, where eis a coset leader and v

is a codeword at the top of the column containing w. Coset decoding

decodes was v. Therefore, we must show that vis nearest to w. If

u∈Cis any other codeword, then w−uis an element of w+C. But

w+C=e+C, because e=w−v∈w+C. By construction, the coset

leader ehas the smallest weight in its coset, so W t(w−u)≥W t(e).

Therefore, by Lemma 6.4.1

d(w, u) = W t(w−u)≥W t(e) = W t(w−v) = d(w, v).

Thus vis the nearest codeword to w.

Again when nis large, the coset decoding tables are diﬃcult to

construct. So we discuss other methods. For an (n, k) code with k×n

standard generator matrix G= [Ik|A], the parity-check matrix of the

code is the n×(n−k) matrix H=A

In−k.

Example 6.4.8. For the standard generator matrix Gin Example

177

6.4.5, the parity matrix

H=





011

101

110

100

010

001







Theorem 6.4.6. Let Cbe an (n, k)code with standard generator ma-

trix Gand parity-check matrix H. Then an element win B(n)is a

codeword if and only if wH = 0.

Proof. Deﬁne a function f:B(n)→B(n−k) by f(w) = wH.

Verify that fis a homomorphism. Let Kbe the kernel of f. Note that

w∈Kif and only if wH = 0. We can prove the theorem if we show

that K=C. By the deﬁnition of the generator matrix, every element

of Cis of the form uG for some u∈B(k). But (uG)H=u(GH)=0

because GH is the zero matrix by Exercise 11. Therefore C⊆K. Since

Cis a group of order 2k, it suﬃces to show that order of Kis also 2k

to conclude that C=K.fis surjective because if v=v1v2·· ·vn−k∈

B(n−k), then v=f(u), where u= 000 · ··0v1v2vn−k∈B(n). Applying

the First Isomorphism Theorem we get B(n−k)∼

=B(n)/K. By

Lagrange’s Theorem 4.4.1

2n=|B(n)|=|K||B(n) : K|=|K||B(n)/K|

=|K||B(n−k)|=|K|2n−k.

Dividing the ﬁrst and last terms of this equation by 2n−kwe get |K|=

2k.

Corollary 6.4.7. Let Cbe a linear code with parity-check matrix H

and let u, v ∈B(n). Then uand vare in the same coset of Cif and

only if uH =vH.

Proof. By Theorem 6.4.6 u−v∈Cif and only if (u−v)H= 0 if

and only if uH =vH.

If w∈B(n), then wH is called the syndrome of w. We now describe

a procedure for decoding called syndrome decoding.

Algorithm 6.4.2 (Syndrome Decoding).1. If wis a received word,

compute the syndrome wH of w.

178

2. Find the coset leader ewith the same syndrome (that is eH =

wH).

3. Decode was w−e.

Example 6.4.9. The Syndrome table for a (6,3) code is given below.

Syndrome 000 011 101 110 100 010 001 111

Coset Leader 000000 100000 010000 001000 000100 000010 000001 101010

For the received word w= 010111, the syndrome wH = 010 corre-

sponds to coset e= 000010. Therefore wis decoded as the codeword

w−e= 010101. So instead of the entire coset table, we need only the

coset leaders in the syndrome decoding technique.

For correcting only single errors the parity check matrix decoding,

which we describe next, is the best method because there is no need to

compute cosets or ﬁnd coset leaders.

Algorithm 6.4.3 (Parity check matrix decoding).1. If wis the re-

ceived word, compute its syndrome wH.

2. If wH = 0, decode was w.

3. If wH ̸= 0, and wH is the ith row of H, then decode was w−ei,

where eiis a vector such that the i-th entry of eiis 1and all other

entries of eiare zero.

4. If wH ̸= 0 and wH is not a row of H, do not decode and request

a re-transmission.

Example 6.4.10. Consider the (6,3) code with the Parity matrix H

in Example 6.4.8. The syndrome of the received word w= 011111 is

wH =0 1 1 1 1 1 





011

101

110

100

010

001







= 100,

which is the fourth row of H. Therefore the wis decoded as w−

(000100) = 011011.

179

For the received word v= 101010, the syndrome of vis

vH =101010





011

101

110

100

010

001







= 111.

Since vH is not a row of H,vis not decoded and a re-transmission

is requested.

The next theorem proves that the Parity check matrix decoding

corrects single error.

Theorem 6.4.8. Let Cbe a linear code with parity-check matrix H.

If every row of His nonzero and no two are the same, then the parity

check decoding corrects all single errors.

Proof. By Corollary 6.4.3, to prove that the code corrects one error,

we need to show that the minimum weight of the codewords wmin ≥3.

Suppose Ccontains a codeword uwith wt(u) = 1. Then uhas just one

bit equal to 1, suppose it is in the position i. Since uH is the i-th row of

H, the condition uH = 0 implies the i-th row of Hconsists entirely of

zeroes. This contradicts our assumption. Hence Ccontains no words

of weight 1. Suppose Ccontains a codeword vwith W t(v) = 2, then v

has a 1 in the positions iand jonly. Let hi, hjdenote the i-th and j-th

row of H. Then vH =hi+hj. The condition vH = 0 implies hi=hj

which contradicts the hypothesis. Hence Ccontains no words of weight

less than or equal to 2. When a codeword uis transmitted with exactly

one error in coordinate iand received as w, then w−u=ei. Hence

ei=w−u∈w+C, so eimust be the coset leader for w. Therefore w

is correctly decoded as w−ei=u.

Let a word aof length nbe denoted by a0a1,··· an−1. A code Cis

said to be cyclic if it is a linear code and if

a0a1. . . an−1implies an−1a0a1. . . an−2∈C.

Cyclic codes are popular because it is possible to implement these codes

using simple devices known as shift registers. Moreover, cyclic codes

can be constructed and investigated by means of rings and polynomials.

180

The word ˆa=an−1a0a1. . . an−2is the ﬁrst cyclic shift of the word

a. If Cis a cyclic code then the words obtained by performing any

number of cyclic shifts on aare also in C.

The key to the algebraic treatment of cyclic codes is the correspon-

dence between the words and polynomials which is given in the next

theorem.

Theorem 6.4.9. The function f:Z2[x]/(xn−1) →B(n)given by

f(a0+a1x+··· +an−1xn−1) = a0a1·· ·an−1is an isomorphism as

additive groups.

The proof of Theorem 6.4.9 is left as an exercise.

In this correspondence, the ﬁrst cyclic shift f(1)(x) of a polynomial

f(x) = a0+a1x+· ·· +an−1xn−1is

f(1)(x) = an−1+a0x+···+an−2xn−1

=x(ao+a1x+· ·· +an−1xn−1)−an−1(xn−1)

=xf(x)−an−1(xn−1).

Thus f(1)(x)≡xf(x) mod (xn−1). Let R(n) denote the ring

Z2[x]/ < (xn−1) >, then this fact leads to the following theorem.

Theorem 6.4.10. A code Cin B(n)is cyclic if and only if it corre-

sponds to an ideal ICin R(n).

Proof. Since ICcorresponds to a linear code, if a(x), b(x)∈IC, then

a(x) + b(x)∈IC. Since xia(x) represent successive cyclic shifts of a(x),

xia(x)∈IC. Any polynomial p(x)∈R(n) is the sum of the number of

powers of xi. Since ICis linear, p(x)a(x)∈IC. Hence ICis an ideal by

Proposition 1.3.1.

Conversely, if ICis an ideal, then by deﬁnition, if a(x), b(x)∈IC,

then a(x) + b(x)∈IC. Hence ICrepresents a linear code. Moreover,

since ICis an ideal, xa(x)∈IC, which implies Cis a cyclic code.

Observe that if f(x)∈R(n), then deg f(x)< n, by deﬁnition.

Example 6.4.11. Let f(x) = 1 + x+x2∈Z2[x]/ < (x3−1) >,

then a cyclic code corresponding to the ideal < f (x)>is generated as

181

described below.

p(x)p(x)f(x) mod(x3−1) Word

0 0 000

1 1 + x+x2111

x1 + x+x2111

1 + x0 000

x21 + x+x2111

x2+ 1 0 000

x2+x0 000

x2+x+ 1 1 + x+x2111

The ideal <1 + x+x2>has only two elements {0,1 + x+x2}in

R(3) = Z2[x]/ < (x3−1) >, and the corresponding code

C={000,111}.

Theorem 6.4.11. Let Cbe a cyclic code and let ICbe its corresponding

ideal in R(n). Then there is a polynomial f(x)∈R(n)such that IC=<

f(x)>.

Proof. If Cis the trivial code, then ICcontains only the zero polyno-

mial, hence IC=<0>. If not, then ICcontains a non-zero polynomial

f(x) of least degree. Suppose g(x) is any element of IC, then by the

Division Algorithm, we have

g(x) = q(x)f(x) + r(x)

where either degree of r(x) is less than degree of f(x) or r(x) = 0.

Because both f(x) and g(x) are in IC, and since ICis an ideal, it

follows that

q(x)f(x)−g(x) = r(x)∈IC.

Consequently, r(x) = 0, since f(x) is a polynomial of least degree in

IC. Recall that the zero polynomial has no degree. Thus

IC=< f(x)> .

In general, a cyclic code Cgenerated by < f(x)>will have many

generators, but only one of them will have the least degree (Exercise

14). We shall refer to the unique polynomial as the canonical generator

of C.

182

Theorem 6.4.12. The canonical generator f(x)of a cyclic code Cin

B(n)is a divisor of xn−1in Z2[x].

Proof. Using the division algorithm for Z2[x], we get

xn−1 = f(x)h(x) + r(x),

such that either r(x) = 0 or the degree of r(x) is less than f(x).

Consequently, since xn−1 = 0, r(x) = f(x)h(x) in Z2[x]/ < (xn−1) >.

Thus r(x)∈< f(x)>which contradicts the fact that f(x) has the least

degree in Cunless r(x) = 0. Therefore xn−1 = f(x)h(x) in Z2[x],

that is f(x) divides xn−1.

Example 6.4.12. The generator 1 + x+x2of the code Cin Example

6.4.11 is a canonical generator because

x3−1 = (1 + x)(1 + x+x2).

Theorem 6.4.13. Let Cbe a cyclic code and let IC=< f(x)>, where

f(x)is a canonical generator of C. Let xn−1 = f(x)h(x), where

h=h0+h1x+···hkxk, and let

HT=





hkhk−1hk−2··· h00 0 ·· · 0

0hkhk−1··· h1h00·· · 0

0 0 hk·· · h2h1h0··· 0

.

..

.

..

.

..

.

..

.

..

.

0 0 0 ··· hkhk−1hk−2··· h0







Then His a parity check matrix for C.

Proof. Let p(x) = fx)g(x) be any element of IC, where

g(x) = g0+g1x+···+gn−1xn−1.

Multiplying both sides by f(x) we get

p(x) = g0f(x) + g1xf(x) + ·· · +gn−1xn−1f(x).

Let pbe the word in Ccorresponding to p(x). Then

p=g0f+g1f(1) +· ·· +gn−1f(n−1),(6.5)

where f(i)denotes the i-th cyclic shift of the word corresponding to f.

183

If His a parity check matrix, then pH = 0 for every p∈C. Conse-

quently, by Equation 6.5, it is suﬃcient to prove that f(i)H= 0 for 0 ≤

i≤n−1. Equating the coeﬃcients of the equation xn−1 = f(x)h(x),

we get

f0h1+f1h0= 0 (coeﬃcient of x)

f0h2+f1h1+f2h0= 0 (coeﬃcient of x2)

.

fn−k−1hk+fn−khk−1= 0 (coeﬃcient of xn−1)

Also since the coeﬃcients of 1 and xnare both 1, we get

f0h0+fn−khk= 0.

Since the degree of h(x) is kand degree of f(x) is n−k, the coeﬃcients

hk+1, . . . , hn−1and fn−k+1,··· , fn−1are all zero. Hence the above n

equations can be written as

hkfn−k+j+hk−1fn−k+j+1 +· ·· +h0fn+j= 0,

where j= 0,1, . . . , n −1. For suitable values of j, these are precisely

the expressions which occur in the evaluation of f(i)H. Hence f(i)H=

0.

Thus to describe the cyclic codes of length nwe must ﬁnd the factors

of xn−1 in Z2[x].

Example 6.4.13. Consider cyclic codes of length 7. Recall that x8−x

is the product of all irreducible polynomials of degrees that divide 3

(see Exercise 31, Chapter 3). Therefore

x7−1 = (1 + x)(1 + x+x3)(1 + x2+x3).

The equation shows that there are just eight divisors of x7−1 in Z2[x]:

they are the trivial divisors 1 and x7−1 together with

1 + x, 1 + x+x3,1 + x2+x3,

(1 + x)(1 + x+x3),(1 + x)(1 + x2+x3),(1 + x+x3)(1 + x2+x3).

Each of these divisors generate a cyclic code and these are the only

cyclic codes of length 7.

184

If C=< f(x) = (1 + x+x3)>, then h(x) = (1 + x)(1 + x2+x3) =

1 + x+x2+x4. Hence

HT=



1110100

0111010

0011101

.

Let w= 1101000 be the codeword corresponding to f(x) = 1 + x+

x3, then

wH =1101000







100

110

111

011

101

010

001







=0000000.

Theorem 6.4.14. A cyclic code of length nand designed distance 2t+1

corrects terrors.

The proof of this theorem is not in the scope of this book. The

reader may refer to [31] for more about cyclic codes.

Exercises.

1. List all the mutually orthogonal Latin squares of orders 5,7,8,

and 9.

2. Let f1(x), f2(x), . . . fk(x)∈Z[x] be polynomials of the same de-

gree d. Let n1, n2, . . . , nkbe integers which are relatively prime

in pairs (i.e (ni, nj) = 1 for all i̸=j). Prove that there exists a

polynomial f(x)∈Z[x] of degree such that

f(x)≡f1(x)( mod n1)

f(x)≡f2(x)( mod n2)

.

f(x)≡fk(x)( mod nk)

3. Solve the system of congruence equations given below.

185

(a)

x≡2(mod 3)

x≡3(mod 5)

x≡2(mod 7)

(b)

x≡3(mod 4)

x≡6(mod 7)

x≡6(mod 11)

x≡1(mod 13)

4. Use the Chinese Remainder Theorem to add the numbers 219 and

172.

5. Bill Gates decided to donate some computers to M University. He

decided to divide the computers equally among the 5 important

departments. But there were 2 computers left. Then, he decided

to divide it equally among 6 departments. Again, there were 2

computers left. Next, he divided it equally among 7 departments.

Lo and behold, again, there were two computers left. Finally, he

decided to divide the computers among all the 11 departments.

And Vow! No computers were left. Find the number of computers

Bill Gates is planning to donate.

6. Decode the message

47 15 20 49 23 1

which was encoded using the RSA algorithm with the prime num-

bers p= 5, q= 13, and the lock L= 11.

7. Decode the message

349 447 202 349 107 591 536

which was encoded using the RSA algorithm with the prime num-

bers p= 23, q= 31, and the lock L= 233.

8. Decode the message

61 60 112 22 25 80 123

which was encoded using the RSA algorithm with the prime num-

bers p= 7, q= 23, and the lock L= 61.

186

9. Prove that if a code corrects terrors, then the Hamming distance

between any two codewords is at least 2t+ 1 (Hint: If u, v are

codewords and d(u, v)≤2t, construct a word wthat diﬀers from

uin exactly tcoordinates and from vin tor fewer coordinates).

10. Prove that if a code detects terrors, then the Hamming distance

between any two codewords is at least t+ 1.

11. If G= [Ik|A] is the standard generator matrix for a linear code

and H=A

In−kis its parity check matrix, then prove that GH

is the zero matrix.

12. Prove that the ideal <1 + x2>has four elements in Z2/(x3−1).

13. Prove that the function f:Z2[x]/(xn−1) →B(n) given by

f(a0+a1x+···+an−1xn−1) = a0a1···an−1is an isomorphism as

additive groups.

14. Show that the canonical generator of a cyclic code is unique.

15. What is the number of cyclic codes of length 15?

16. Describe the cyclic code of length 15 generated by the polynomial

1 + x+x2.

17. What is the number of cyclic codes of length 31?

187

188

Appendix A

I examined my own heart and discovered that I would not care to be

happy on condition of being an imbecile - Voltaire.

A.1 The Euclidean Algorithm.

Deﬁnition A.1.1. Let aand bbe integers, not both 0. The greatest

common divisor (gcd) of aand bis the largest integer dthat divides

both aand b. In other words, dis the gcd of aand bprovided that

1. ddivides aand ddivides b

2. if cdivides aand cdivides b, then c≤d.

The greatest common divisor of aand bis denoted by (a, b).

Theorem A.1.1. Let aand bbe integers, not both 0and let dbe the

greatest common divisor. Then there exist integers uand vsuch that

d=au +bv.

Proof. Let S={am +bn ∈Z:m, n ∈Z}.Sis nonempty

because a2+b2=aa +bb ∈S. Moreover, since both aand bare

not simultaneously zero, a2+b2>0. Therefore, Scontains positive

integers. Let dbe the smallest positive integer in S, then dis of the

form d=au +bv for some integers uand v. We will prove that dis the

gcd of aand b. Divide aby dto write a=dq +r, such that q, r ∈Z

and 0 ≤r < d. Consequently,

r=a−dq =a−(au +bv)q=a(1 −uq) + b(−vq).

Thus ris an integer combination of aand b, therefore r∈S. Con-

sequently, the condition 0 ≤r < d, and the fact that dis the smallest

189

positive integer in Simplies r= 0. Thus, ddivides a. A similar argu-

ment proves that ddivides b. Hence dis a common divisor of aand b.

Let cbe any other common divisor of aand b. Then a=cr and b=cs

for some integers rand s. Therefore

d=au +bv = (cr)u+ (cs)v=c(ru +sv).

Therefore cdivides d. Hence c≤ |d|. Since dis positive |d|=d. Hence

c≤d. Therefore dis the gcd of aand b.

Lemma A.1.1. If a, b, q, r ∈Zand a=bq +r, then (a, b) = (b, r).

Proof. If cis a common divisor of aand b, then a=cs and b=ct

for some s, t ∈Z. Consequently,

r=a−bq =cs −(ct)q=c(s−tq).

Hence cdivides r, which implies that cis also a common divisor of b

and r. Conversely, if eis a common divisor of band r, then b=ex and

r=ey for some x, y ∈Z. Then

a=bq +r= (ex)q+ey =e(xq +y).

Thus edivides a, so that eis a common divisor of aand b. Thus

the set Sof common divisors of aand bis the same as the set Tof

common divisors of band r. Hence the largest element in S, namely

(a, b), is the same as the largest element in T, namely (b, r).

Theorem A.1.2. [The Euclidean Algorithm] Let aand bbe positive

integers with a≥b. If bdivides a, then (a, b) = b. If bdoes not divide

a, then apply the division algorithm repeatedly as follows:

a=bq0+r0,0< r0< b

b=r0q1+r1,0≤r1< r0

r0=r1q2+r2,0≤r2< r1

r1=r2q3+r3,0≤r3< r2

r2=r3q4+r4,0≤r4< r3

.

The process ends when a remainder 0is obtained. This must occur

after a ﬁnite number of steps because the sequence ristrictly decreases.

That is, for some integer t

rt−2=rt−1qt+rt,0< rt< rt−1

rt−1=rtqt+1 + 0

190

The last nonzero remainder rtis the greatest common divisor of aand

b.

Proof. If bdivides a, then a=bq + 0, so that (a, b) = (b, 0) = b

by Lemma A.1.1. If ais not divisible by b, then apply Lemma A.1.1

repeatedly to each division to get

(a, b) = (b, r0) = (r0, r1) = ···= (rt−1, rt) = (rt,0) = rt.

Example A.1.1. In this example, we compute (312,272) using Eu-

clid’s Algorithm.

312 = 272 ×1 + 40 (A.1)

272 = 40 ×6 + 32 (A.2)

40 = 32 ×1 + 8 (A.3)

32 = 8 ×4+0

Thus (312,272) = 8. We use back substitution to write 8 as an

integer combination of 312 and 272 as follows.

8 = 40 −32 ×1 (by Equation A.3)

= 40 −32

= 40 −(272 −40 ×6) (by Equation A.2)

= 7 ×40 −272

= 7(312 −272) −272 (by Equation A.1)

= 7 ×312 −8×272

Thus, we write 8 = 7 ×312 −8×272.

The Euclidean algorithm carries over to k[x], where kis a ﬁeld.

Deﬁnition A.1.2. Let kbe a ﬁeld and f(x), g(x)∈k[x], not both zero.

The greatest common divisor (gcd) of f(x)and g(x)is the monic

polynomial d(x)of highest degree that divides both f(x)and g(x).

Example A.1.2. Consider the polynomials

f=x4−15x3+ 73x2−129x+ 70,

g= 2x3−9x2+ 13x−6.

191

Apply the Euclidean Algorithm:

f=g1

2x−21

4+77

4x2−231

4x+77

2

g=77

4x2−231

4x+77

28

77 x−12

77 + 0

Hence, the last non zero remainder is 77

4x2−231

4x+77

2. Since the

gcd of fand gis a monic polynomial, we multiply this remainder by

(4/77) to get:

(f, g) = 4

77 77

4x2−231

4x+77

2=x2−3x+ 2.

A.2 Polynomial irreducibility.

In this section, we list a few results (without proof) that help us deter-

mine irreducibility of a polynomial. The interested reader can refer to

[24] for proofs of the results presented in this section.

Theorem A.2.1 (The Remainder Theorem).Let kbe a ﬁeld, f(x)∈

k[x], and a∈k. The remainder when f(x)is divided by the polynomial

x−ais f(a).

Example A.2.1. Consider the polynomial f(x) = x3−8x2+x+ 42.

The remainder, when f(x) is divided by (x+2), is 0, but the remainder,

when f(x) is divided by x−2, is 20. Verify that f(−2) = 0 and

f(2) = 20.

Theorem A.2.2 (The Factor Theorem).Let kbe a ﬁeld, f(x)∈k[x],

and a∈k. Then ais a root of the polynomial f(x)if and only if x−a

is a factor of f(x)∈k[x].

Example A.2.2. x+ 2 is a factor of the polynomial f(x) = x3−8x2+

x+ 42. Hence −2 is a root of f(x).

Corollary A.2.3. Let kbe a ﬁeld and f(x)a nonzero polynomial of

degree nin k[x]. Then f(x)has at most nroots in k.

Corollary A.2.4. Let kbe a ﬁeld and f(x)∈k[x], with deg f(x)≥2.

1. If f(x)is irreducible in k[x], then f(x)has no roots in k.

192

2. If f(x)has degree 2or 3and has no roots in kthen f(x)is

irreducible in k[x].

Example A.2.3. To show that x3+x+ 1 is irreducible in Z5[x], you

need only verify that none of 0,1,2,3,4∈Z5is a root.

Theorem A.2.5 (Rational Root Test).Let f(x) = anxn+an−1xn−1+

···+a1x+a0be a polynomial with integer coeﬃcients. If r̸= 0 and the

rational number r/s (in lowest terms) is a root of f(x), then rdivides

a0and sdivides an.

Example A.2.4. Consider the polynomial f(x)=4x4−12x3+x2−

4x+ 3. By Theorem A.2.5, r/s is a root of f(x) if and only rdivides

3 and sdivides 4. Therefore r=±1,±3 and s=±1,±2,±4. So the

possible roots of f(x) are

1,−1,3,−3,1

2,−1

2,3

2,−3

2,1

4,−1

4,3

4,−3

4.

We substitute each of these values in f(x), and we ﬁnd that only

f(1/2) = 0 and f(3) = 0. So these are the only roots of f(x) in this

list. By the Factor Theorem A.2.2, (x−3) and (x−1/2) are factors of

f(x). Verify with long division that

f(x) = 2(x−1

2)(x−3)(2x2+x+ 1).

Theorem A.2.6 (Eisenstein’s Criterion).Let f(x) = anxn+an−1xn−1+

···+a1x+a0be a nonconstant polynomial with integer coeﬃcients. If

there is a prime psuch that pdivides each of a0, a1, . . . , an−1but pdoes

not divide anand p2does not divide a0, then f(x)is irreducible in Q[x].

Example A.2.5. 1. The polynomial x7+ 6x5−15x4+ 3x2−9x+ 12

is irreducible in Q[x] by Eisenstein’s criterion with p= 3.

2. The polynomial xn+ 5 is irreducible in Q[x] for each n≥1 by

Eisenstein’s criterion with p= 5. Thus there are irreducible poly-

nomials of every degree in Q[x].

Finally, we discuss irreducible polynomials in R[x] and C[x].

Theorem A.2.7. A polynomial f(x)is irreducible in R[x], if and only

if, f(x)is a ﬁrst-degree polynomial or

f(x) = ax2+bx +cwith b2−4ac < 0.

193

Theorem A.2.8. A polynomial is irreducible in C[x], if and only if,

it has degree 1.

A.3 Generating Functions.

Let h0, h1, . . . , hn, . . . be an inﬁnite sequence of numbers. Its generating

function is deﬁned to be the inﬁnite series

g(x) = h0+h1x+h2x2+· ·· +hnxn+···

Example A.3.1. 1. The generating function of the inﬁnite sequence

1,1,1, . . . , 1, . . .

is

g(x) = 1 + x+x2+···+xn+···

g(x) is a geometric series and hence

g(x) = 1

1−x,for |x|<1.

2. Similarly, the generating function of 1,−1,1,−1, . . . , (−1)n, . . . is

1

1 + x= 1 −x+x2−x3+· · · + (−1)nxn+. . .

3. The generating function of 1,1

1! ,1

2! , . . . 1

n!, . . . is

ex= 1 + 1

1!x+1

2!x2+·· · +1

n!xn+. . .

Proposition A.3.1. There are n+r−1

rr-combinations from a set with

nelements when repetition of elements is allowed.

Proof. Each rcombination of a set with nelements can be rep-

resented by a list of n−1 bars and rstars. The number of ways of

choosing rpositions to place rstars from the n+r−1 possible positions

is n+r−1

r=n+r−1

n−1.

194

Example A.3.2. How many ways are there to select ﬁve bills from a

cash box containing $1 bills, $2 bills, $5 bills, $10 bills, $20 bills, $50

bills, and $ 100 bills?

Imagine a cash box with 7 compartments. Selecting ﬁve bills corre-

$1$100 $50 $20 $10 $5 $2

sponds to placing 5 stars and 6 dividers between them. For example,

we choose one 50 dollar bill and 4 one dollar bills as shown below. Thus

** ***

the number of ways of selecting ﬁve bills is the same as the number

of selecting ﬁve positions to place ﬁve stars among the 11 possible po-

sitions. Thus there are 11

5ways to choose ﬁve bills from a cash box

with seven types of bills.

Example A.3.3. How many solutions does the equation

x1+x2+x3= 11

have, where x1, x2and x3are nonnegative integers?

A solution corresponds to choosing 11 items of 3 types with x1items

of the ﬁrst type, x2items of the second type, and x3items of the third

type. Hence the answer is

11 + 3 −1

11 =13

11=13

2= 78.

Example A.3.4. Example: How many solutions does the equation

x1+x2+x3= 11

have, where x1≥1, x2≥2, and x3≥3?

Like before, a solution corresponds to choosing 11 items of the 3

types, but now x1≥1, x2≥2, and x3≥3. So choose 1 item of the

195

ﬁrst type, 2 items of the second type, and 3 items of the third type.

Then the remaining 5 items can be chosen in

5+3−1

5=7

2= 21.

Consider the sequence h0, h1, h2. . . , hn, . . . where hnequals the

number of nonnegative integral solutions of

x1+x2+· ·· +xk=n.

Then by the above argument of sticks and stars, we have

hn=n+k−1

n,(n≥0).

Proposition A.3.2. The generating function of hnis

g(x) = ∞

n=0n+k−1

nxn.

Proof. We will ﬁrst show that

1

(1 −x)k=∞

n=0n+k−1

nxn.

Observe that

1

(1−x)k=1

1−x×1

1−x× · ·· × 1

1−x(kfactors)

= (1 + x+x2+··· )(1 + x+x2+···)·· ·

···(1 + x+x2+· ··)

=∞

x1=0 xx1∞

x2=0 xx2···∞

xk=0 xxk.

Now xx1xx2···xxk=xnprovided x1+x2+··· +xk=n.

Thus the coeﬃcient of xnequals the number of nonnegative integral

solutions of this equation, that is n+k−1

n. Consequently,

g(x) = 1

(1 −x)k=∞

n=0n+k−1

nxn.

196

Example A.3.5. Determine the number of ways of making ncents

with pennies, nickels, dimes, quarters, and half-dollar pieces.

Answer: The number hnequals the number of nonnegative integral

solutions of the equation

x1+ 5x2+ 10x3+ 25x4+ 50x5=n.

We create one factor for each type of coin, where the exponents are

the allowable numbers in the n-combinations for that type of coin.The

generating function is

g(x) = (1 + x+x2+· ··)(1 + x5+x10 +· ··)(1 + x10 +x20 +. . . )×

(1 + x25 +x50 +. . . )(1 + x50 +x100 +...)

=1

1−x

1

1−x5

1

1−x10

1

1−x25

1

1−x50

We can use Maple to expand this generating function using the

following command.

series((1/(1-x))*(1/(1-x^5))*(1/(1-x^10))*(1/(1-x^25))*(1/(1-x^50)),x=0,50);

A.4 Algorithms to compute Hilbert bases.

We describe an algorithm to compute the Hilbert basis of a cone CA=

{x:Ax= 0,x≥0}.

Let A be an m×nmatrix. We introduce 2n+mvariables t1, t2, ..tm,

x1, .., xn,y1, y2, .., ynand ﬁx any elimination monomial order such that

{t1, t2, ..tm}>{x1, .., xn}>{y1, y2, .., yn}.

Let IAdenote the kernel of the map

C[x1, . . . , xn, y1, . . . , yn]→C[t1, . . . , tm, t−1

1, . . . , t−1

m, y1, . . . , yn],

xj→yj

m



i=1

taij

i

and yj→yjfor each j= 1, . . . , n.

We can compute a Hilbert basis of CAas follows.

197

Algorithm A.4.1. 1. Compute the reduced Gr¨obner basis Gfor the

ideal IAwith respect to the monomial ordering given above.

2. The Hilbert basis of CAconsists of all vectors βsuch that xβ−yβ

appears in G.

Example A.4.1. Let

A=1−1

−2 2 

To handle computations with negative exponents we introduce a

new variable tand consider the lexicographic ordering

t > t1> t2> x1> x2> y1> y2.

Then the given map acts as follows

x1→y1t1

1t−2

2

x2→y2t−1

1t2

2

Set tt1t2−1 = 0 and the Kernel of the map is given by IA=

(x1−y1t3

1t2, x2−y2t3

2t, t1t2t−1).

We compute the Gr¨obner basis of IAwith respect to the above

ordering and get:

IA= (x1x2−y1y2, t1y1−t2

2x1, t1x2−t2

2y2, t3

2ty2−x2, t3

2tx1−y1, t1t2t−1)

Therefore, the Hilbert basis is {(1,1)}.

See [18] and [39] for more details about this algorithm. See [26] for

more eﬀective algorithms to compute the Hilbert basis.

A.5 Algorithms to compute toric ideals.

Computing toric ideals is the biggest challenge we face in applying the

methods we developed in Chapter 5. Many algorithms to compute toric

ideals exist and we present a few of them here.

Let A={a1, a2, ..., an}be a subset of Zd. The additive group gen-

erated by Ais a lattice, that is, the group is generated by linearly

198

independent vectors. The set of linearly independent vectors that gen-

erate the lattice is called a basis of the lattice. See [32] for more details

about lattices.

Consider the map

π:k[x]7→ k[t±1] (A.4)

xi7→ tai(A.5)

Recall that the kernel of πis the toric ideal of Adenoted by

IA. The most basic method to compute IAwould be the elimina-

tion method. Though this method is computationally expensive and

not recommended, it serves as a starting point. Note that every vector

u∈Zncan be written uniquely as u=u+−u−where u+and u−are

non-negative and have disjoint support.

Example A.5.1. For the given vector u= (−1,−1,1), u+= (0,0,1)

and u−= (1,1,0). Thus, ucan be written as u= (0,0,1) −(1,1,0).

We describe an algorithm to compute toric ideals given in [39].

Algorithm A.5.1.

1. Introduce n+d+ 1 variables t0, t1, .., td, x1, x2, ..., xn.

2. Consider any elimination order with {ti;i= 0, . . . , d}>{xj;j=

1, . . . , n}. Compute the reduced Gr¨obner basis Gfor the ideal

(t0t1t2...td−1, x1ta1−−ta1+, ...., xntan−−tan+).

3. G∩k[x] is the reduced Gr¨obner basis for IAwith respect to the

chosen elimination order.

If the lattice points aihave only non-negative coordinates, the vari-

able t0is unnecessary and we can use the ideal (xi−tai:i= 1, . . . , n)

in the second step of the Algorithm A.5.1.

To reduce the number of variables involved in the Gr¨obner basis

computations, it is better to use an algorithm that operates entirely

in k[x1, . . . , xn]. We now present such an algorithm for homogeneous

ideals. Observe that all the toric ideals we face in our computations in

Chapter 5 are homogeneous.

199

The saturation of an ideal Jdenoted by (J:f∞) is deﬁned to be

(J:f∞) = {g∈k[x] : frg∈Jfor some r∈N}.

Let ker(A)∈Zndenote the integer kernel of the d×nmatrix with

column vectors ai. With any subset Cof the lattice ker(A) we associate

a ideal of IA:

JC:= (Xu+−Xu−:u∈ C).

We now describe another algorithm to compute the toric ideal IA

from [39].

Algorithm A.5.2.

1. Find any lattice basis Lfor ker(A).

2. Let JL:= (Xu+−Xu−:u∈L).

3. Compute a Gr¨obner basis of (JL: (x1x2···xn)∞) which is also a

Gr¨obner basis of the toric ideal IA.

Example A.5.2. Let A={(1,1),(2,2),(3,3)}. Consider the matrix

whose columns are the vectors of A

1 2 3

1 2 3 .

Then kerA={[−2,1,0],[−3,0,1]}. We use the software Maple

to compute a lattice basis of kerA:{[−1,−1,1],[−2,1,0]}. Therefore

JL= (x3−x1x2, x2−x2

1) and

(JL: (x1x2x3)∞) = (x3−x1x2, x2−x2

1, x2

2−x1x3)

which is also IA(see Algorithm A.5.2). Note that many available com-

puter algebra packages including CoCoA [16] can compute saturation

of ideals.

From the computational point of view, computing (JL: (x1x2··· xn)∞)

is the most demanding step. The algorithms implemented in CoCoA

try to make this step eﬃcient [9]. For example, one way to compute

(JL: (x1x2···xn)∞), would be to eliminate tfrom the ideal H:=

JL+ (tx1x2···xn−1) but this destroys the homogeneity of the ideal.

200

It is well-known that computing with homogeneous ideals have many

advantages. Therefore, it is better to introduce a variable uwhose de-

gree is the sum of the degrees of the variables xi, i = 1, . . . , n. We then

compute the Gr¨obner basis of the ideal H:= JL+ (x1x2·· ·xn−u).

Then a Gr¨obner basis for (JL: (x1x2·· ·xn)∞) is obtained by simply

substituting u=x1x2···xnin the Gr¨obner basis of H.

Another trick to improve the eﬃciency of the computation of satu-

ration ideals is to use the fact

(JL: (x1x2···xn)∞) = ((. . . ((JL:x∞

1) : x∞

2). . . ) : x∞

n).

Therefore we can compute the saturations sequentially one variable

at a time. See [10] for other tricks. We refer the reader to [39] for

details and proofs of the concepts needed to develop these algorithms

and other algorithms.

A.6 Algorithms to compute Hilbert Poincar´e se-

ries.

In this section, we will describe a pivot-based algorithm to compute the

Hilbert Poincar´e series. Variations of this algorithm is implemented in

CoCoA [16].

Let kbe a ﬁeld and R:= k[x1, x2, ..., xr] be a graded Noetherian

ring. let x1, x2, ..., xrbe homogeneous of degrees k1, k2, .., kr(all >0).

Let Mbe a ﬁnitely generated R-module. Let Hbe an additive function

on the class of R-modules with values in Z. Then by the Hilbert-Serre

theorem, we have

HM(t) = p(t)

Πr

i=1(1 −tdegxi).

where p(t)∈Z[t].

Let Ibe an ideal of R, we will denote

HR/I (t) =

Πr

i=1(1 −tdegxi).

Observe that we only need to calculate the numerator since the

denominator is already known.

201

Let ybe a monomial of degree (d1, ..., dr) called the pivot. The

degree of the pivot is d=r

i=1 di. The ideal quotient (J:f) of an

ideal J⊂k[x1, . . . , xr] and f∈k[x1, . . . , xr] is

(J:f) = {g∈k[x] : fg ∈J}.

It is proved in [10] that

HR/I (t) = HR/(I,y )(t) + td(HR/(I:y))(t),

which implies

= +td . (A.6)

When Iis a homogeneous ideal,

HR/I (t) = HR/in(I)(t),

where in(I) denotes the ideal of initial terms of I(see Chapter 1).

The pivot yis usually chosen to be a monomial that divides a gen-

erator of Iso that the total degrees of (I, y) and (I:y) are lower than

the total degree of I. The computation proceeds inductively.

Example A.6.1. Let R=k[x1, x2, . . . , xn] be the polynomial ring.

Let R=d∈NRdwhere each Rdis minimally generated as a k-vector

by all the n+d−1

dmonomials of degree d. Therefore,

HR/(0)(t) = HR(t) = ∞



d=0

dimRdtd=∞



d=0 n+d−1

dtd= 1/(1 −t)n.

Therefore we get <0>= 1. We will use this information to compute

HR/(I)(t), where I= (x1, x2, . . . , xn).

Let J= (x2, . . . , xn). Then, (J:x1) = J. Therefore by Equation

A.6, we get

<(J, x1)>= (1 −tdegx1)< J > .

That is,

< x1, x2, . . . , xn>= (1 −tdegx1)< x2, . . . , xn> .

Now, choosing the pivot x2, x3, . . . , xnsubsequently we get

< x1, x2, . . . , xn>=

i=1,...,n

(1 −tdegxi)<0> .

202

Now since <0>= 1, we get < x1, x2, . . . , xn>=i=1,...,n(1 −

tdegxi).

Therefore HR/(x1,x2,...,xn)(t) = 1.

See [10] for more information about computing the Hilbert Poincare

series.

203

204

Bibliography

[1] Ahmed, M., De Loera, J., and Hemmecke, R., Polyhedral cones of

magic cubes and squares, New Directions in Computational Ge-

ometry, The Goodman-Pollack Festschrift volume, Aronov et al.,

eds., Springer-Verlag, (2003), 25–41.

[2] Ahmed, M., How many squares are there, Mr. Franklin?:

Constructing and Enumerating Franklin Squares, Amer. Math.

Monthly, Vol. 111, 2004, 394–410.

[3] , Magic graphs and the faces of the Birkhoﬀ polytope, An-

nals of Combinatorics, Volume 12, Number 3, October 2008, 241-

269

[4] , Algebraic combinatorics of magic squares, Ph.D. disserta-

tion, Univ. of California UC Davis (2004).

[5] Anand, H., Dumir, V.C., and Gupta, H., A combinatorial distri-

bution problem, Duke Math. J. 33, (1966), 757-769.

[6] Andrews, W. S., Magic Squares and Cubes, 2nd. ed., Dover, New

York, 1960.

[7] Atiyah, M.F., and Macdonald, I.G., Introduction to Commutative

Algebra, Addison-Wesley, Reading, MA, 1969.

[8] Beck, M., Cohen, M., Cuomo, J., and Gribelyuk, P., The number

of magic squares, cubes and hypercubes, Amer. Math. Monthly,

110, no.8, (2003), 707-717.

[9] Bigatti, A.M., La Scala, R., and Robbiano, L., Computing toric

ideals, J. Symbolic Computation, 27, (1999), 351-365.

205

[10] Bigatti, A.M, Computation of Hilbert-Poincar´e Series, J. Pure

Appl. Algebra, 119/3, (1997), 237–253.

[11] Biggs, N.L., Discrete Mathematics, Revised edition, Oxford Uni-

versity Press Inc., New York, 1985.

[12] Bose, R.C., Manvel, B., Introduction to Combinatorial Theory,

John Wiley and Sons, Inc., USA, 1984.

[13] Brualdi, A.R., Introductory Combinatorics, 4th ed., Pearson Pren-

tice Hall, Upper Saddle River, N.J., 2004.

[14] Brualdi, A. R. and Gibson, P., Convex polyhedra of doubly stochas-

tic matrices: I, II, III, Journal of combinatorial Theory, A22,

(1977), 467-477.

[15] Carlitz, L., Enumeration of symmetric arrays, Duke Math. J., 33,

(1966), 771-782.

[16] Capani, A., Niesi, G., and Robbiano, L., CoCoA, A System for Do-

ing Computations in Commutative Algebra, available via anony-

mous ftp from cocoa.dima.unige.it (2000).

[17] Cox, D., Little, J., and O’Shea, D., Ideals, varieties, and Algo-

rithms, Springer Verlag, Undergraduate Text, 2nd Edition, 1997.

[18] , Using Algebraic Geometry, Springer-Verlag, New York,

1998.

[19] Dummit, D. S. and Foote, R. M., Abstract Algebra, Prentice Hall,

New Jersey, 1991.

[20] Fraleigh, J.B., A First Course in Abstract Algebra, second edition,

Addison-Wesley Publishing Company, Inc., World student series

edition, 1976.

[21] Giles, F.R. and Pulleyblank, W.R., Total dual integrality and in-

teger polyhedra, Linear Algebra Appl., 25, (1979), 191-196.

[22] Gupta, H., Enumeration of symmetric matrices, Duke Math. J.,

35, (1968), 653-659.

206

[23] Halleck, E.Q., Magic squares subclasses as linear Diophantine sys-

tems, Ph.D. dissertation, Univ. of California San Diego, (2000),

187 pages.

[24] Hungerford, T. W., Abstract Algebra, An Introduction, Saunders

College Publishing, New York, 1990.

[25] Lang, S., Algebra, third edition, Addison Wesley Longman, Inc,

1993.

[26] Hemmecke, R., On the computation of Hilbert bases of cones,

in Proceedings of First International Congress of Mathemati-

cal Software, A. M. Cohen, X.S. Gao, and N. Takayama, eds.,

Beijing, (2002); software implementation 4ti2 is available from

http://www.4ti2.de.

[27] MacMahon, P.A., Combinatorial Analysis, Chelsea, 1960.

[28] Pasles, P. C., The lost squares of Dr. Franklin: Ben Franklin’s

missing squares and the secret of the magic circle, Amer. Math.

Monthly, 108, (2001), 489-511.

[29] , Franklin’s other 8-square, J. Recreational Math., 31,

(2003), 161-166.

[30] L. D. Patel, The secret of Franklin’s 8×8magic square,

J.Recreational Math., 23, (1991), 175-182.

[31] Pretzel, O., Error-Correcting Codes and Finite Fields, Oxford Uni-

versity Press Inc., New York, 1992.

[32] Schrijver, A., Theory of Linear and Integer Programming, Wiley-

Interscience, 1986.

[33] Stanley, R.P., Enumerative Combinatorics, Volume I, Cambridge,

1997.

[34] , Combinatorics and commutative algebra, Progress in

Mathematics, 41, Birkha¨user Boston, MA, 1983.

[35] ) , Linear Homogeneous Diophantine Equations and Magic

Labelings Of Graphs, Duke Mathematical Journal, Vol. 40,

September 1973, 607-632.

207

[36] , Magic Labelings of Graphs, Symmetric Magic Squares,

Systems of Parameters and Cohen-Macaulay Rings, Duke Mathe-

matical Journal, Vol. 43, No.3, September 1976, 511-531.

[37] Stewart, B. M., Magic graphs, Canad. J. Math., vol. 18, (1966),

1031-1059.

[38] , Supermagic complete graphs, Canad. J. Math., vol. 19,

(1967), 427-438.

[39] Sturmfels, B., Gr¨obner Bases and Convex Polytopes, University

Lecture Series, no. 8, American Mathematical Society, Providence,

1996.

[40] Wallis, D., Magic Graphs, Birkh¨auser Boston, 2001.

208

Index

Abelian group, 82

Buchberger’s Algorithm, 19

Characteristic of a ring, 69

Codewords, 172

Cone, 138

Coset of an ideal, 65

Cyclic code, 180

Cyclic group, 89

Ehrhart quasi-polynomial, 156

Elimination ideal, 33

Field, 10

Franklin square, 136

Galois group, 110

Generating function, 194

Gray code, 171

Group, 81

Gr´

’obner basis, 18

Hamming weight, 173

Hilbert basis, 141

Hilbert-Poincare series, 151

Homomorphism of groups, 83

Homomorphism of rings, 60

Ideal, 16

Ideal quotient, 202

Index of a subgroup, 95

Integral domain, 10

Isomorphism of groups, 83

Isomorphism of rings, 61

Latin squares, 159

Lattice, 198

Leading term, 11

Magic square, 135

Maximal ideal, 68

Minimal polynomial, 72

Module, 149

Noetherian rings, 17

Normal subgroup, 92

Orthogonal Latin squares, 160

Pointed Cone, 138

Polyhedron, 138

Polytope, 138

Principal ideal domain, 17

Quasi-polynomial, 155

Quotient Ring, 65

Radical of an ideal, 32

Rational polytope, 154

Reduced Gr´

’obner basis, 20

Resultant, 41

209

Ring, 9

S-polynomial, 18

Saturation of an ideal, 200

Separable polynomial, 75

Solvable group, 111

Splitting ﬁeld, 73

Standard generator matrix, 174

Subgroup, 85

Subring, 16

Sylvester matrix, 40

Toric ideal, 146

Variety, 27

210

Abstract Algebra for Polynomial Operations

Abstract and Figures

Recommended publications

Some classes of linear codes : observations about their structure, construction (non-) existence and...

Algebraic coding theory and information theory. DIMACS workshop algebraic coding theory and informat...

The structuring of design knowledge

Packings with block size five and index one: v≡2(mod4)