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Betti numbers and Euler characteristics

Esben Bistrup Halvorsen

January 28, 2003

This paper constitutes a course project related to the course

Homological Algebra (Math253), which I took in the spring

semester of 2002 at the University of California, Berkeley.

The presentation is based on a discussion paper by my ad-

visor at the University of Copenhagen, Hans–Bjørn Foxby,

who has guided me through the entire process.

Introduction

The purpose of this paper is to introduce and investigate the concepts of Betti

numbers and Euler characteristics, which arise in algebraic topology as invariants

describing topological spaces. In this presentation we will ignore the origin of the

concepts, however, and instead discuss them completely within the language of

homological algebra.

Betti numbers and Euler characteristics are useful tools in describing modules

and complexes. Knowing these provides enough information to calculate the ﬂat

and projective dimensions of a complex (see corollary 4.4), and to determine

whether or not a ﬁnitely generated module is free (see theorem 5.9). In this

paper we shall also obtain results that can be formulated completely without Betti

numbers and Euler characteristics, although not easily proved without them. The

most remarkable of these states that a suﬃcient condition for a ﬁnitely generated

module to have trivial annihilator is that the annihilator consists only of zero-

divisors (see theorem 5.7).

In our presentation, we will limit ourselves to consider modules over a local

and noetherian ring; on the other hand, we will allow ourselves to designate Betti

numbers and Euler characteristics not only to modules, as is traditionally the case,

but also to complexes. This complies with the dogma “It’s better to work with

complexes than modules” from Hans–Bjørn Foxby’s notes [FoxbyHHA]. Towards

the end of the discussion, however, we will return to modules and see how the

Euler characteristics behaves.

1

Rings and modules 2

1 Rings and modules

Throughout this section, Rdenotes a (nontrivial, unitary, commutative) local

ring. The unique maximal ideal of Ris denoted by mand the quotient ﬁeld

R/mis denoted by k. All modules are, unless otherwise stated, assumed to be

R-modules.

In this section we obtain a variety of results about modules over local rings.

These will be needed later to prove general facts about Betti numbers and Euler

characteristics.

1.1 Free modules

Suppose Mis a module. The quotient M/mMhas a natural structure of a k-

vector space, the scalar multiplication being deﬁned by [r] [m]M= [rm]M.

Furthermore, any homomorphism ϕ:M→Nof modules induces a linear map

M/mM→N/mNof k-vector spaces given by [m]M7→ [ϕ(m)] N.

Deﬁnition 1.1. If Mis a module, Mdenotes the k-vector space M/mM, and

βR

0(M) denotes the dimension of M. If ϕ:M→Nis a homomorphism of

modules, ϕdenotes the induced k-linear map M→Ngiven by ϕ([m]M) =

[ϕ(m)] N.

For the next theorem, notice (see for example [Lang] proposition XVI.2.7) that

if Iis an ideal of Rand Mis a module, then M/IM ∼

=(R/I)⊗RM; in particular,

M∼

=k⊗RM. Furthermore, it is easy to see that under this isomorphism, any

homomorphism M/I M →N/IN induced by a homomorphism ϕ:M→N(that

is, on the form [m]IM 7→ [ϕ(m)]IN ) corresponds to the homomorphism (R/I)⊗Rϕ;

in particular, the k-linear map ϕ:M→Ncorresponds to k⊗Rϕ. In other words,

deﬁnition 1.1 simply gives new symbols to the functor k⊗R−.

Theorem 1.2. If Fis a ﬁnitely generated free module, then F∼

=Rnfor n=

βR

0(F).

Proof: We already know that F∼

=Rnfor some n∈N(see for instance [Lang]

corollary III.4.3), so it only remains to check that Rnis a k-vector space of

dimension n. This follows from the isomorphisms Rn∼

=k⊗RRn∼

=(k⊗RR)n∼

=

kn.

Deﬁnition 1.3. When Fis a ﬁnitely generated free module, the rank of Fis

the number rankRF=βR

0(F).

Normally one would deﬁne the rank of a ﬁnitely generated free module Fto

be the n∈N0such that F∼

=Rn, but for this to be well-deﬁned, one has to

show that Rn∼

=Rmimplies n=m. This is exactly what theorem 1.2 does: if

Rn∼

=Rm, then n=βR

0(Rn) = βR

0(Rm) = m.

Rings and modules 3

Lemma 1.4. Let 0→K→F→G→0be an exact sequence of modules,

and suppose that Fand Gare ﬁnitely generated free modules. Then Kis ﬁnitely

generated and free, and rankRF= rankRK+ rankRG.

Proof: Since Gis free and thereby projective, the sequence splits, hence there

is an isomorphism F∼

=K⊕G; in particular, Kmust be ﬁnitely generated.

According to [Lang] theorem X.4.4 a ﬁnitely generated projective module over a

local ring is free, hence for the freeness of Kit suﬃces to show that Kis projective.

This, however, follows from the fact (see for example [H&S] proposition I.4.5) that

a direct sum of modules is projective if and only if each summand is projective.

We now have an isomorphism F∼

=K⊕Gof free and ﬁnitely generated

modules, so it follows by the remarks preceding the lemma that rankRF=

rankRK+ rankRG.

Theorem 1.5. Let Fbe a ﬁnitely generated free module and (f1, . . . , fn)a set of

generators for F. Then (f1, . . . , fn)is a basis if and only if n= rankR(F).

Proof: Let (e1, . . . , en) be the standard basis for Rn, and consider the homo-

morphism ϕ:Rn→Fdeﬁned on the basis by ϕ(ei) = fi. We then have the

exact sequence

0−→ ker ϕ−→ Rnϕ

−→ F−→ 0,

and from the previous lemma it follows that kerϕis a free module with

n= rankR(F) + rankR(ker ϕ).

Now, if (f1, . . . , fn) is a basis, then kerϕ= 0, and the above equation yields n=

rankR(F). Conversely, if n= rankR(F), then the equation implies rankR(ker ϕ) =

0, hence ker ϕ= 0.

A similar result is found in

Theorem 1.6. Let Mbe a ﬁnitely generated R-module and (m1, . . . , mn)a set

of generators for M. Then ([m1]M,...,[mn]M)is a basis for Mif and only if

n=βR

0(M).

Proof: Since m1, . . . , mngenerate M, [m1]M, . . . , [mn]Mgenerate Mwhich

is a k-vector space of dimension βR

0(M), and we know that a generating set in a

vector space is a basis if and only if the number of generators is the same as the

dimension.

A simple consequence of theorem 1.5 and theorem 1.6 is the following.

Corollary 1.7. Let Fbe a ﬁnitely generated free module. Then (f1, . . . , fn)is a

basis for Fif and only if ([f1]F, . . . , [fn]F)is a basis for F.

Rings and modules 4

If m1, . . . , mngenerate a module M, then [m1]M,...,[mn]Mgenerate the

vector space Mwhose dimension is βR

0(M). In particular we must have n≥

βR

0(M), that is, any generating set for Mcontains at least βR

0(M) elements. This

observation leads us to

Deﬁnition 1.8. A set (m1, . . . , mn) of generators for a module Mis said to be

minimal if n=βR

0(M).

Minimal generating sets do exist. In fact we have

Proposition 1.9. Any set of generators for a ﬁnitely generated module Mcan

be reduced to a minimal set of generators.

Proof: A generating set for Minduces a generating set for Mwhich can be

reduced to a basis for Mwith βR

0(M) elements. Thus we can ﬁnd a subset of the

generators for Mthat induce a basis for M. This subset must contain βR

0(M)

elements, and by Nakayama’s lemma (as stated in [Lang] lemma X.4.3) it will

still be generating.

We will now prove a theorem which states that any homomorphism ϕ:F→G

between ﬁnitely generated free modules splits up into two parts, one being an

isomorphism, and the other being “minimal” in the sense that the induced map ϕ

is zero. This will become very useful later when we construct a similar “splitting”

of free resolutions. First we need a lemma:

Lemma 1.10. Suppose Fand Gare ﬁnitely generated free modules and ϕ:F→

Gis a homomorphism such that ϕ(F)*mG. Then there exist bases (f1, . . . , fn)

and (g1, . . . , gm)for Fand Grespectively, such that ϕ(f1) = g1and ϕ(Rf2+· · ·+

Rfn)⊆Rg2+· · · +Rgm.

Proof: Consider the induced linear map ϕ:F→Gof k-vector spaces. Let

n= rankR(F) and m= rankR(G), and pick an element e1∈Fwith g1=

ϕ(e1)/∈mG. Notice that we must have e1/∈mFas well, such that we now have

[e1]F6= 0 and [g1]G6= 0. Now extend [e1]Fto a basis ([e1]F, . . . , [en]F)

for Fand extend [g1]Gto a basis ([g1]G, . . . , [gm]G) for G. By corollary 1.7,

(e1, . . . , en) is a basis for F, and (g1, . . . , gm) is a basis for G.

We will now transform the basis (e1, . . . , en) into a basis (f1, . . . , fn) with the

desired property. Let f1=e1, and choose for i= 2, . . . , n elements ri1, . . . , rim ∈

Rsuch that ϕ(ei) = ri1g1+· · · +rimgm. Let fi=ei−ri1e1. Then (f1, . . . , fn) is

a generating set for Fsince e1=f1and ei=ri1e1+fifor i= 2, . . . , n, hence it

is a basis by theorem 1.5. Furthermore, ϕ(f1) = g1, and for i= 2, . . . , n,

ϕ(fi) = ri2g2+···+rim gm,

so ϕ(Rf2+· · · +Rfn)⊆Rg2+· · · +Rgmas desired.

Complexes 5

Theorem 1.11. Suppose Fand Gare ﬁnitely generated free modules, and that

ϕ:F→Gis a homomorphism. There then exist direct sum decompositions

F=K⊕Land G=M⊕Ninto ﬁnitely generated free submodules, such that

ϕ(L)⊆mN, and such that ϕinduces an isomorphism ϕ|K:K∼

=

−→ M.

Proof: If ϕ(F)⊆mGwe are done. If not, we can apply lemma 1.10 to obtain

bases (f1, . . . , fn) and (g1, . . . , gm) for Fand Grespectively with ϕ(f1) = g1

and ϕ(Rf2+· · · +Rfn)⊆Rg2+· · · +Rgm. Let F0=Rf2+· · · +Rfnand

G0=Rg2+· · · +Rgm, and notice that F=Rf1⊕F0and G=Rg1⊕G0and that

ϕmaps Rf1isomorphically onto Rg1. Thus, if ϕ(F0)⊆mG0, we are done. If not,

lemma 1.10 applies to F0,G0and the induced homomorphism ϕ|F0:F0→G0, and

we alter the bases (f2, . . . , fn) and (g2, . . . , gm) for F0and G0(but omit changing

the symbols) such that we get ϕ(f2) = g2and ϕ(Rf3+· · ·+Rfn)⊆Rg3+· · ·+Rgm.

Now we let F00 =Rf3+· · · +Rfnand G00 =Rg3+· · · +Rgm, and notice that

F= (Rf1+Rf2)⊕F0and G= (Rg1+Rg2)⊕G00 and that ϕmaps Rf1+Rf2

isomorphically onto Rg1+Rg2. Therefore, if ϕ(F00 )⊆mG00, we are done. If not,

we repeat the process. Eventually the process will terminate, say at the i’th step,

at which point we have obtained bases (f1, . . . , fn) and (g1, . . . , gm) for Fand G

respectively, such that ϕ(f1) = g1, . . . , ϕ(fi) = gi, and ϕ(Rfi+1 +· · · +Rfn)⊆

m(Rgi+1+· · ·+Rgm). (Notice that it is possible that i=nsuch that ϕis simply an

isomorphism in itself.) We can now let K=Rf1+· · ·+Rfi,L=Rfi+1 +···+Rfn,

M=Rg1+· · · +Rgi, and N=Rgi+1 +· · · +Rgm, and the theorem is proved.

2 Complexes

In this section we introduce notation, terminology and constructions related to

complexes and the morphism between them. The presentation is based exclu-

sively on [FoxbyHHA] sections 0, 1, 2, 4 and 7.

2.1 Categories of complexes and the homology functor

Throughout this subsection, Rdenotes a (nontrivial, unitary) ring. All mod-

ules are, unless otherwise stated, assumed to be left R-modules.

We now move our focus from modules to the more general concept of complexes.1

Recall that a complex Xis a family (Xn)n∈of modules together with a family

(∂X

n)n∈of homomorphisms ∂X

n:Xn→Xn−1, which are called the diﬀerentials

of X, such that ∂X

n∂X

n+1 = 0 for all n∈Z:

X=· · · −→ Xn+1

∂X

n+1

−→ Xn

∂X

n

−→ Xn−1−→ · · · .

1Notation and deﬁnitions are from [FoxbyHHA] sections 1 and 2.

Complexes 6

The complex Xis said to be bounded at the left if Xn= 0 for suﬃciently large n.

In that case, when presenting Xas above, we omit the superﬂuous zero modules

and simply write

X= 0 −→ Xs−→ Xs−1−→ · · · .

Similarly, Xis said to be bounded at the right if Xn= 0 for suﬃciently small n,

and we then write

X=· · · −→ Xt+1 −→ Xt−→ 0.

A complex is said to be bounded if it is bounded at the left and at the right. In

the sequel, modules will always be thought of as complexes concentrated in degree

0, that is, complexes Xwith Xn= 0 for all n6= 0.

When Xand Yare complexes, a morphism α:X→Yis a family α= (αn)n∈

of homomorphisms αn:Xn→Ynmaking the following diagram commutative.

· · · //Xn+1

∂X

n+1 //

αn+1

²²

Xn

∂X

n//

αn

²²

Xn−1//

αn−1

²²

· · ·

· · · //Yn+1

∂Y

n+1 //Yn

∂Y

n//Yn−1//· · ·

All complexes together with all morphisms of complexes form a category which

is denoted by C(R). We introduce the following full subcategories of C(R) by

specifying their objects.

C(R): complexes that are bounded at the left

C(R): complexes that are bounded at the right

C(R): complexes that are bounded

C0(R): complexes that are concentrated in degree 0 (that is, modules)

Cf(R): complexes of ﬁnitely generated modules

Cl(R): complexes of modules of ﬁnite length

CI(R): complexes of injective modules

CF(R): complexes of ﬂat modules

CP(R): complexes of projective modules

CfP(R): complexes of ﬁnitely generated projective modules

CL(R): complexes of ﬁnitely generated free modules

We will freely use any combination of the subscripts {@,A,¤,0}and superscripts

{f,l,I,F,P,fP,L}be setting C?

#(R) = C?(R)∩C#(R), such that for example CL(R)

denotes the category of complexes which are bounded to the right and consist of

ﬁnitely generated free modules.

An important functor in the category C(R) is the homology functor H : C(R)→

C(R) which takes a complex Xto the complex H(X) deﬁned by H(X)n=

Hn(X) = ker ∂X

n/im ∂X

n+1 and ∂H(X)

n= 0 for all n∈Z, and a morphism α:X→Y

to the morphism H(α): H(X)→H(Y) given by

H(α)n([x]im ∂X

n+1 ) = Hn(α)([x]im ∂X

n+1 ) = [αn(x)]im ∂Y

n+1

Complexes 7

for all n∈Zand x∈ker ∂X

n. H(X) is called the homology complex of X, and

Hn(X) is called the homology module in degree n.

The functor H entails a new series of subcategories of C(R). Letting ?repre-

sent any of the superscripts {f,l,I,F,P,fP,L}, we deﬁne the subcategory C(?)(R)

by

X∈ C(?)(R)def

⇐⇒ H(X)∈ C?(R).

Letting # represent any of the subscripts {@,A,¤,0}, we likewise deﬁne the

subcategory C(#)(R) by

X∈ C(#)(R)def

⇐⇒ H(X)∈ C#(R).

As before, the subscripts and superscripts can be combined freely by setting

C(?)

#(R) = C(?)(R)∩ C(#)(R). The names for the objects in these subcategories

are obtained by adding the words “homology” and “homologically” to the names

for the objects in the corresponding subcategories C?

#(R), such that for instance

C(f)

( )(R) is the category of complexes which are homologically bounded to the

right and have ﬁnitely generated homology modules.

The homology functor H entails another concept. Recall that a morphism

α:X→Yof R-complexes is said to be an isomorphism if there exists a morphism

α−1:Y→Xsuch that α−1α= 1Xand αα−1= 1Y. In this case we say that X

is isomorphic to Y, and we write α:X∼

=

−→ Yor simply X∼

=Y.

Deﬁnition 2.1. Suppose α:X→Yis a morphism of complexes. We say that

αis a homology isomorphism, and we write α:X'

−→ Y, if H(α): H(X)→H(Y)

is an isomorphism; in other words,

X'

−→

αYdef

⇐⇒ H(X)∼

=

−→

H(α)H(Y).

The relation X'

−→ Yof homology isomorphism is reﬂexive and transitive,

but not necessarily symmetric. To help this we introduce

Deﬁnition 2.2. Two complexes Xand Yare said to be equivalent, and we

write X'Y, if there exist a third complex Zand homology isomorphisms

X'

−→ Z'

←− Y.

One can show2(and this also shows that 'is an equivalence relation) that

X'Ywhenever there exist n∈Nand n+ 1 complexes V(0), . . . , V (n+ 1) such

that X=V(0), Y=V(n), and we are in one of the following situations.

2This is actually the deﬁnition (or can easily be derived from the deﬁnition) of equivalence

as it is given in (1.28) in [FoxbyHHA].

Complexes 8

X=V(0) '

−→ V(1) '

←− · · · '

−→ V(n−1) '

←− V(n) = Y,

X=V(0) '

−→ V(1) '

←− · · · '

←− V(n−1) '

−→ V(n) = Y,

X=V(0) '

←− V(1) '

−→ · · · '

−→ V(n−1) '

←− V(n) = Y,

X=V(0) '

←− V(1) '

−→ · · · '

←− V(n−1) '

−→ V(n) = Y.

Isomorphic complexes are equivalent, and equivalent complexes have isomor-

phic homologies. Thus we have the implications

X∼

=Y⇒X'Y⇒H(X)∼

=H(Y).

A ﬁnal deﬁnition having to do with the homology functor is found in

Deﬁnition 2.3. Let X∈ C (R). The supremum of Xis given by supX=

sup{n∈Z|Hn(X)6= 0}and the inﬁmum of Xis given by inf X= inf{n∈Z|

Hn(X)6= 0}.

Here we follow the usual conventions that sup ∅=−∞ and inf ∅=∞, so

sup X=−∞ and inf X=∞whenever Xis homologically trivial, while −∞ ≤

inf X≤sup X≤ −∞ otherwise. Notice also that we obviously have X∈

C( )(R)⇐⇒ sup X < ∞, and X∈ C( )(R)⇐⇒ inf X > −∞.

2.2 Tensor products and dot products

Throughout this subsection, Rand Sdenote (nontrivial, unitary) rings, and

Qdenotes a (nontrivial, unitary) commutative ring over which Rand Sare

algebras. All modules are, unless otherwise stated, assumed to be left R-

modules.

Recall3that for an R◦-module (that is, a right R-module) Mand an R-module

N, the tensor product M⊗RNof Mand Nis a Q-module; in particular it is

an R-module if Ris commutative. If in addition Mhas an S-module structure

such that Mis an (S, R)-bimodule, then M⊗RNhas the structure of an (S, Q)-

bimodule. Similarly, if Nhas an additional S◦-module structure such that Nis

a (R, S)-bimodule, then M⊗RNhas the structure of a (Q, S)-bimodule.

The tensor product generalizes to complexes4: For X∈ C(R◦) and Y∈ C(R)

the tensor product of Xand Yis the Q-complex X⊗RYwhose n’th module is

the Q-module

(X⊗RY)n=M

`∈

X`⊗RYn−`,

and whose n’th diﬀerential ∂X⊗RY

n: (X⊗RY)n→(X⊗RY)n−1is deﬁned on

generators x`⊗yn−`∈X`⊗RYn−`⊆(X⊗RY)n,`∈Z, by

∂X⊗RY

n(x`⊗yn−`) = ∂X

`(x`)⊗yn−`+ (−1)`x`⊗∂Y

n−`(yn−`),

3See [FoxbyHHA] section 0.

4See [FoxbyHHA] section 4.

Complexes 9

which is an element of X`−1⊗RYn−`+X`⊗RYn−`−1⊆(X⊗RY)n−1.

From the tensor product we derive the dot product5: Given X∈ C(R◦) and

Y∈ C( )(R), the dot product X·RYis deﬁned as the equivalence class represented

by G⊗R◦Xfor any G∈ CF(R) with G'Y; this is well-deﬁned since such

complexes Gexist (as shown in [FoxbyHHA] theorem 2.6(P) which is reproduced

in theorem 3.2 of this paper), and since any G, G0∈ CF(R) satisfying G'Y'G0

must have G⊗R◦X'G0⊗R◦X. If instead we are given X∈ C( )(R◦) and

Y∈ C(R), the dot product X·RYis deﬁned as the equivalence class represented

by Y⊗R◦Ffor any F∈ CF(R◦) with F'X; this is also well-deﬁned. The

two deﬁnitions do not contradict: whenever they both apply, they yield the same

equivalence class.

2.3 Localization

Throughout this subsection, Rdenotes a (nontrivial, unitary) commutative

ring. All modules are, unless otherwise stated, assumed to be R-modules.

Let Ube a multiplicatively closed subset of the commutative ring R. Recall that

the localization U−1Mof Mat Uhas the structure of an R-module as well as

of an U−1R-module. Notice also that U−1M∼

=(U−1R)⊗RM(see for example

[Eisenbud] lemma 2.4) from which the preceding statement follows. It is easy

to see that under this isomorphism, any homomorphism U−1ϕ:U−1M→U−1N

induced by a homomorphism ϕ:M→Ncorresponds to the homomorphism

(U−1R)⊗Rϕ, so that the functor U−1(−) is actually the same as the functor

(U−1R)⊗R−. From this it follows that U−1(−) is a covariant additive (right

exact) functor C0(R)→ C0(U−1R), so by (1.51) in [FoxbyHHA] it must induce a

functor C(R)→ C(U−1R), which is likewise denoted by U−1(−). We can therefore

generalize the notion of localization to complexes: Given an R-complex X, the

localization of Xat Uis the U−1R-complex (and thereby R-complex)

U−1X=· · · −→ U−1Xn+1

U−1∂X

n+1

−→ U−1Xn

U−1∂X

n

−→ U−1Xn−1−→ · · · ,

where the n’th diﬀerential U−1∂X

nmaps xn/u ∈U−1Xnto ∂X

n(xn)/u ∈U−1Xn−1.

A complex morphism α:X→Yis mapped by the functor U−1(−) to the complex

morphism U−1α:U−1X→U−1Ygiven by (U−1α)n(xn/u) = αn(xn)/u for xn∈

Xn.

According to 2.4(2) in [FoxbyHA], U−1(−) is exact as a functor C0(R)→

C0(U−1R), hence it follows from (1.51) in [FoxbyHHA] that H(U−1X)∼

=U−1(H(X))

for all X∈ C(R), and that U−1(−) preserves homology isomorphisms and equiv-

alences.

As is the case with modules we write Xfor (R\p)−1Xwhenever pis a prime

ideal.

5See [FoxbyHHA] section 7.

Resolutions 10

3 Resolutions

This section is devoted to resolutions, not only of modules, but also of complexes.

The results obtained here will be fundamental to the following sections in which

we ﬁnally encounter Betti numbers and Euler characteristics.

3.1 Injective, projective, and ﬂat resolutions

Throughout this subsection, Rdenotes a (nontrivial, unitary) ring. All mod-

ules are, unless otherwise stated, assumed to be left R-modules.

Recall that a projective resolution of a module Mis a complex

P=· · · −→ Pn

∂P

n

−→ Pn−1−→ · · · −→ P1

∂P

1

−→ P0−→ 0

of projective modules together with a homomorphism α0:P0→M, such that

· · · −→ Pn

∂P

n

−→ Pn−1−→ · · · −→ P1

∂P

1

−→ P0

α0

−→ M−→ 0

is exact. Considering Mas a complex concentrated in 0, we can formulate this

as follows: A projective resolution of Mis a complex P∈ CP(R) with Pn= 0 for

n < 0 together with a homology isomorphism α:P'

−→ M.

P

α

²²

· · · //Pn

∂P

n//

²²

Pn−1//

²²

· · · //P1

²²

∂P

1//P0//

α0

²²

0

M· · · //0//0//· · · //0//M//0

This observation (together with similar observations about ﬂat, free, and injective

resolutions) leads to the following generalization.6

Deﬁnition 3.1. An injective resolution of a complex X∈ C( )(R) is a homology

isomorphism X'

−→ I∈ CI(R). A ﬂat resolution of a complex X∈ C( )(R) is a

homology isomorphism X'

←− F∈ CF(R); a projective resolution of a complex

X∈ C( )(R) is a homology isomorphism X'

←− P∈ CP(R); and a free resolution

by ﬁnite modules of a complex X∈ C(f)

( )(R) is a homology isomorphism X'

←−

L∈ CL(R).

Injective and projective (and thereby ﬂat) resolutions exist for modules. The

generalization of this is found in [FoxbyHHA] theorem 2.6 which we bring here

without proof.

6See (2.5) in [FoxbyHHA].

Resolutions 11

Theorem 3.2.

(I) Every X∈ C( )(R)has an injective resolution X'

−→ Isuch that In= 0

for all n > sup X.

(P) Every X∈ C( )(R)has a projective (and thereby ﬂat) resolution X'

←− P

such that Pn= 0 for all n < inf X.

(L) If Ris noetherian, then every X∈ C(f)

( )(R)has a free resolution X'

←− L

by ﬁnite modules such that Ln= 0 for all n < inf X.

The concepts of injective, projective, and ﬂat dimensions of modules are gen-

eralized in the following deﬁnition7(in which we follow the usual (and logical)

conventions about inﬁmum of sets containing ±∞).

Deﬁnition 3.3. If X∈ C( )(R), then the injective dimension of Xis given by

idRX= inf{sup{n∈Z|I−n6= 0} | X'I∈ CI(R)}; and if X∈ C( )(R), then

the projective dimension of Xis given by pdRX= inf{sup{n∈Z|Pn6= 0} | X'

P∈ CP(R)}, and the ﬂat dimension of Xis given by fdRX= inf{sup{n∈Z|

Fn6= 0}| X'F∈ CF(R)}.

3.2 Minimal free resolutions

Throughout this subsection, Rdenotes a (nontrivial, unitary, commutative)

local, noetherian ring with maximal ideal mand quotient ﬁeld k=R/m. All

modules are, unless otherwise stated, assumed to be R-modules.

When dealing with Betti numbers we will often be concerned with free resolutions

by ﬁnite modules; in fact we will consider the special kind of free resolutions

deﬁned below.

Deﬁnition 3.4. A complex Lis said to be minimal if im ∂L

n⊆mLn−1for all

n∈Z. A free resolution X'

←− L∈ CL(R) by ﬁnite modules of a complex Xis

called a minimal free resolution of Xif Lis a minimal complex.

Theorem 3.5. Suppose X'

←− L∈ CL(R)is a free resolution by ﬁnite modules of

a complex X. Then there exist subcomplexes Aand Bof Lwith ﬁnitely generated

modules, such that Ais exact, Bis minimal, and L=A⊕B.

Proof: By theorem 1.11 we can for each n∈Zﬁnd direct sum decom-

positions Ln=Fn⊕Gn=Mn⊕Nn, where Fn,Gn,Mn, and Nnare free

ﬁnitely generated submodules of Ln, such that ∂L

n(Gn)⊆mNn−1, and such that

∂L

ninduces an isomorphism ∂L

n|Fn:Fn

∼

=

−→ Mn−1. The properties ensure that

7See [FoxbyHHA] section 8.

Resolutions 12

Mn= im ∂L

n+1|Fn+1 ⊆im ∂L

n+1 ⊆Mn⊕mNnand ker ∂L

n= ker ∂L

n|Gn⊆Gn, so that

for all n∈Zwe have the following chain of inclusions:

Mn⊆im ∂L

n+1 ⊆ker ∂L

n⊆Gn.(1)

For every n∈Zwe now let An=Fn+Mnand Bn=Gn∩Nn. We claim

that Anand Bnare free and ﬁnitely generated, and that Lnis the direct sum of

Anand Bn. To see this, notice ﬁrst that since Fn∩Gn= 0 and Mn⊆Gn, we

must have Fn∩Mn= 0, and it follows that An=Fn⊕Mnis a direct sum; in

particular, Anis free and ﬁnitely generated.

Now, suppose that x∈An∩Bn. Let x=m+fbe the unique decomposition

of xinto f∈Fnand m∈Mn. Then f=x−m∈Bn+Mn⊆Gn, so we must

have f= 0 (since Fn∩Gn= 0). Consequently x=m, but since x∈Nn, it

follows that x= 0 (since Mn∩Nn= 0). This shows that An∩Bn= 0.

Finally, suppose x∈Lnis arbitrarily chosen. Let x=f+gbe the unique

decomposition of xinto f∈Fnand g∈Gn, and let g=m+nbe the unique

decomposition of ginto m∈Mnand n∈Nn. Then n=g−m∈Gn+Mn=Gn,

so n∈Bnand x= (f+m) + n∈An+Bn. This shows that An+Bn=Ln,

and we now have the (inner) direct sum Ln=An⊕Bn. In particular there is an

exact sequence 0 →Bn→Ln→An→0, and it follows from lemma 1.4 that Bn

is free and ﬁnitely generated.

The inclusion im ∂L

n⊆Gn−1from (1) implies that ∂L

nrestricts to a homomor-

phism ∂L

n|Bn:Bn→Gn−1. We already know that ∂L

nmaps Gnto mNn−1, so the

image of the restriction ∂L

n|Bnis actually contained in Gn−1∩mNn−1⊆Bn−1,

hence ∂L

nrestricts to a homomorphism Bn→Bn−1. Let us for each n∈Zdenote

by ∂B

nthe homomorphism Bn→Bn−1which is the restriction of ∂L

nto Bn. We

have then obtained a complex with free and ﬁnitely generated modules:

B=· · · −→ Bn+1

∂B

n+1

−→ Bn

∂B

n

−→ Bn−1−→ · · · .

The inclusion Mn⊆ker ∂L

nfrom (1) and the isomorphism ∂L

n|Fn:Fn

'

−→ Mn−1

implies that ∂L

n(An) = Mn−1⊆An−1, hence ∂L

nrestricts to a homomorphism

An→An−1which we denote by ∂A

n. We have then obtained a complex with free

and ﬁnitely generated modules:

A=· · · −→ An+1

∂A

n+1

−→ An

∂A

n

−→ An−1−→ · · · .

From the above it follows that Lis the direct sum A⊕Bof the complexes A

and B.

It is clear from the construction of Athat im ∂A

n+1 =Mn= ker ∂A

n, so A

is exact. To see that Bis minimal, suppose x∈Bnfor some n∈Z. Since

∂L

nmaps Gnto mNn−1we can ﬁnd r1, . . . , rt∈mand n1, . . . , nt∈Nn−1such

that ∂B

n(x) = Pt

i=1 rini. Now, for each ilet ni=ai+bibe the direct sum

Betti numbers 13

decomposition of niinto ai∈An−1and bi∈Bn−1. We then have ∂L

n(x) =

Pt

i=1 riai+Pt

i=1 ribi, and since ∂L

n(x)∈Bn−1it follows that Pt

i=1 riai= 0,

hence ∂L

n(x) = Pt

i=1 ribi∈mBn−1. This shows that Bis minimal.

The above theorem together with theorem 3.2(L) now yields

Corollary 3.6. Every X∈ C(f)

( )(R)has a minimal free resolution X'

←−

αB∈

CL(R)such that Bn= 0 for all n < inf X.

Proof: Since Ris assumed to be noetherian, we can by theorem 3.2(L) ﬁnd

a free resolution X'

←−

αL∈ CL(R) such that Ln= 0 for all n < inf X. Using

the above theorem, we write Las a direct sum L=A⊕B, where Ais exact

and Bis minimal free. We must of course have An=Bn= 0 for all n < inf X,

so in particular Bn∈ CL(R). Now, the homology isomorphism αinduces an

isomorphism H(X)∼

=

←− H(L). Since it is clear that the homology functor takes

direct sums to direct sums, H(L) = H(A)⊕H(B) = H(B), and the restriction

of αto Bmust still induce an isomorphism H(X)∼

=

←− H(B). Thus α|Bis a

homology isomorphism, hence X'

←−

α|B

Bis a minimal free resolution.

On the level of modules (or complexes concentrated in 0), corollary 3.6 sounds

as follows.

Corollary 3.7. Given any ﬁnitely generated module M, there exists an exact

sequence

· · · −→ Fn

∂F

n

−→ Fn−1−→ · · · −→ F1

∂F

1

−→ F0

α0

−→ M−→ 0,

such that the modules F0, F1, F2, . . . are ﬁnitely generated and free, and such that

∂F

n(Fn)⊆mFn−1for all n∈N.

4 Betti numbers

Throughout this section, Rdenotes a (nontrivial, unitary, commutative) local,

noetherian ring with maximal ideal mand quotient ﬁeld k=R/m. All modules

are, unless otherwise stated, assumed to R-modules.

We have ﬁnally reached the point where we are able to deﬁne the n’th Betti

number of a complex. Let X∈ C(f)

( )(R) and recall that X·Rkis the equivalence

class of complexes represented by the complex k⊗RF, which (since kis a complex

concentrated in 0) is given by

k⊗RF=· · · −→ k⊗RFn+1

k⊗R∂F

n+1

−→ k⊗RFn

k⊗R∂F

n

−→ k⊗RFn−1−→ · · · ,

Betti numbers 14

for any complex F∈ CF(R) with F'X. Since equivalent complexes have

isomorphic homologies, it makes sense to consider the n’th homology module

Hn(X·Rk) as the isomorphism class of modules represented by Hn(k⊗RF). The

ﬁeld kis a (k, R)-bimodule, so k⊗RFis also a k-complex (since its diﬀerentials are

k-linear maps), and Hn(k⊗RF) is a k-vector space; and if F0∈ CF(R) is another

complex with F0'X, then the R-isomorphism Hn(k⊗RF)∼

=Hn(k⊗RF0) is

also a k-isomorphism (since it is a bijection). It therefore makes sense to consider

Hn(X·Rk) as an isomorphism class of k-vector spaces.

Deﬁnition 4.1. Let X∈ C(f )

( )(R) and n∈Z. The n’th Betti number of Xis the

number βR

n(X) = dimk(Hn(X·Rk)).

If Mis a module, we have already deﬁned βR

0(M) to be the dimension of the

k-vector space M. For the above deﬁnition to be consistent with the previous

one, we of course hope for the two deﬁnitions to agree whenever they both apply,

that is, whenever we consider ﬁnitely generated modules. This is fortunately the

case: If M∈ Cf

0(R), then by corollary 3.6 we can ﬁnd a minimal free (and thereby

ﬂat) resolution M'

←− F∈ CL(R) which by corollary 3.7 is on the form

· · · −→ Fn

∂F

n

−→ Fn−1−→ · · · −→ F1

∂F

1

−→ F0

α0

−→ M−→ 0,

where the modules F0, F1, F2, . . . are ﬁnitely generated and free, and where

∂F

n(Fn)⊆mFn−1for all n∈N. The latter of these properties ensures that

all the diﬀerentials in the complex k⊗RFare zero (for recall that for all n∈Z

there is an isomorphism k⊗RFn∼

=Fnunder which k⊗R∂F

ncorresponds to

∂F

n). Thus H0(k⊗RF) = k⊗RF0=F0, so the 0’th Betti number of Mis

dimkF0= rankRF0.

On the other hand, we have an exact sequence 0 →ker α0→F0→M→0, so

application of the right exact functor k⊗R−yields the exact sequence kerα0→

F0→M→0. Here, however, the map ker α0→F0must be zero, since it is

induced by the inclusion of kerα0= im ∂F

1⊆mF0into F0. Consequently the map

F0→Mis an isomorphism, and it follows that rankRF0=βR

0(M). This shows

that the 0’th Betti number of Mis exactly what we denoted by βR

0(M).

Proposition 4.2. Suppose that X∈ C(f )

( )(R)and that X'F∈ CF(R). Then

βR

n(X)≤βR

0(Fn)for all n∈Z, and if Fhas ﬁnitely generated modules, then

βR

n(X) = βR

0(Fn)for all n∈Zif and only if Fis minimal.

Proof: The Betti numbers of Xare by deﬁnition the dimensions as k-vector

spaces of the homologies of the complex

k⊗RF=· · · −→ k⊗RFn+1

k⊗R∂F

n+1

−→ k⊗RFn

k⊗R∂F

n

−→ k⊗RFn−1−→ · · · .

Betti numbers 15

Now, as observed several times before, the above complex is the same as the

complex

· · · −→ Fn+1

∂F

n+1

−→ Fn

∂F

n

−→ Fn−1−→ · · · .

The n’th Betti number βR

n(X) is therefore the dimension of

ker ∂F

n/im ∂F

n+1,(2)

which is a quotient of subspaces of Fnwhose dimension is βR

0(Fn). This immedi-

ately gives the inequality βR

n(X)≤βR

0(Fn) for all n∈Z.

In the case where Fhas ﬁnitely generated modules, the vector spaces involved

in (2) are ﬁnite dimensional, so the only way for the dimension of that quotient to

equal βR

0(Fn) is if ker ∂F

n=Fnand im ∂F

n+1 = 0. This happens for all n∈Zif and

only if ∂F

n= 0 for all n∈Z, which again is the case if and only if ∂F

n(Fn)⊆mFn−1

for all n∈Z, that is, if and only if Fis a minimal complex.

If X'

←− L∈ CL(R) is a minimal free resolution of X, then the above theorem

states that βR

n(X) = rankRLnfor all n∈Z. Together with corollary 3.6 we

therefore get

Corollary 4.3. Every X∈ C(f)

( )(R)has a minimal free resolution, and all min-

imal free resolutions X'

←− L∈ CL(R)must satisfy that rankRLn=βR

n(X)for

all n∈Z. In particular, βR

n(X)∈N0for all n∈Z, and βR

n(X) = 0 for all

n < inf X.

Proposition 4.2 and corollary 4.3 show that the Betti numbers of a complex

Xare the ranks of the modules in a minimal free resolution of X, and that these

ranks are minimal among corresponding ranks of all free resolutions of X. This

justiﬁes the terminology “minimal free resolution”.

Corollary 4.4. For all X∈ C(f)

( )(R),

fdRX= pdRX= sup{n∈Z|βR

n(X)6= 0}.

Proof: Let N= sup{n∈Z|βR

n(X)6= 0}. By corollary 4.3 we can ﬁnd a

complex L∈ CL(R) with X'Land rankRLn=βR

n(X) for all n∈Z. We then

have sup{n∈Z|Ln6= 0}=N, so Nbelongs to the set over which the inﬁmum

is taken in the deﬁnition of ﬂat dimension, hence fdRX≤N.

If F∈ CF(R) has X'F, then there exists a complex Zand homology

isomorphisms X'

−→ Z'

←− F. The complex Zmust necessarily belong to

C(f)

( )(R), and since the composition L'

−→ X'

−→ Zis a minimal free resolution

of Zwhile Z'

←− Fis a ﬂat resolution of Z, proposition 4.2 and corollary 4.3

yield

βR

n(X) = rankR(Ln) = βR

n(Z)≤βR

0(Fn)

Euler characteristics 16

for all n∈Z. Thus βR

n(X)6= 0 implies Fn6= 0, so N≤sup{n∈Z|Fn6= 0}.

Since this holds for all F∈ CF(R) with X'F,N≤fdRX. In conclusion,

N= fdRX.

Exactly the same arguments hold for the projective dimension, hence N=

pdRX.

For the next theorem, recall that since Ris (nontrivial, unitary, commutative)

local and noetherian, the localization Rat a prime ideal pis of course also (non-

trivial, unitary, commutative) local and noetherian. (See for example [Eisenbud]

corollary 2.3 for the part about being noetherian.)

Proposition 4.5. Suppose that X∈ C(f )

( )(R)and let pbe a prime ideal of R.

Then X∈ C(f )

( )(R), and βR

n(X)≤βR

n(X)for all n∈Z.

Proof: As mentioned in section 2.3, exactness of the localization functor on

modules implies that the localization functor on complexes preserves homology

isomorphisms and satisﬁes that H(X)∼

=(H(X)) . The latter of these properties

implies that X∈ C(f )

( )(R).

By corollary 4.3 we can ﬁnd a minimal free resolution X'

←− L∈ CL(R) such

that βR

n(X) = rankRLnfor all n∈Z. Since the localization functor preserves

homology isomorphisms, there is an induced homology isomorphism X'

←− L.

Now, as was also mentioned in section 2.3, the localization functor is additive.

In particular it preserves products, hence (Rt)∼

=(R)tfor all t∈N. Since

each Lnis free and ﬁnitely generated, this shows that (Ln) is free and ﬁnitely

generated as an R-module with rankR(Ln) = rankRLnfor all n∈Z.

We have now seen that X'

←− Lis a free resolution, and it follows from

proposition 4.2 that for all n∈Z,

βR

n(X)≤rankR(Ln) = rankRLn=βR

n(X).

5 Euler characteristics

Throughout this section, Rdenotes a (nontrivial, unitary, commutative) local,

noetherian ring with maximal ideal mand quotient ﬁeld k=R/m. All modules

are, unless otherwise stated, assumed to be R-modules.

Let X∈ C(f )

( )(R) be a complex with pdRX < ∞. Notice that we must have

X∈ C(f )

( )(R), since Xis equivalent to a projective module whose n’th module

is zero for n > pdRX. Now, according to corollary 4.3, βR

n(X) = 0 whenever

n < inf X, and according to corollary 4.4, βR

n(X) = 0 whenever n > pdRX.

Euler characteristics 17

Consequently βR

n(X) is nonzero only for the ﬁnitely many nthat lie between

inf Xand pdRX, so the sum Pn∈(−1)nβR

n(X) is well-deﬁned.

Deﬁnition 5.1. Let X∈ C(f )

( )(R) be a complex with pdRX < ∞. The Euler

characteristics of Xis the number χR(X) = Pn∈(−1)nβR

n(X).

By the above remarks, χR(X) = PpdRX

n=inf X(−1)nβR

n(X); in particular if Mis

a module, χR(M) = PpdRM

n=0 (−1)nβR

n(M).

Theorem 5.2. Let X∈ C(f)

( )(R)be a complex with pdRX < ∞, and suppose

that X'

←− L∈ CL(R)is a bounded free resolution by ﬁnite modules. Then

χR(X) = Pn∈rankRLn.

Proof: By theorem 3.5 we can ﬁnd subcomplexes Aand Bof Lwith free and

ﬁnitely generated modules, such that Ais exact, Bis minimal, and L=A⊕B. As

we saw in the proof of corollar 3.6, the homology isomorphism X'

←− Lrestricts

to a homology isomorphism X'

←− B, hence, by corollary 4.3, βR

n(X) = rankRBn

for all n∈Z.

Since Lis bounded, Aand Bmust also be bounded. In particular there exist

s, t ∈Zsuch that Ais an exact sequence on the form

0−→ At

∂A

t

−→ At−1−→ · · · −→ As+1

∂A

s+1

−→ As−→ 0.

For all n∈Zwe have an exact sequence 0 −→ ker ∂A

n−→ An

∂A

n

−→ im ∂A

n−→

0, so if im ∂A

nis free and ﬁnitely generated, then lemma 1.4 gives that im ∂A

n+1 =

ker ∂A

nis free and ﬁnitely generated, and that rankRAn= rankR(im ∂A

n+1) +

rankR(im ∂A

n). But im ∂A

s+1 is indeed free and ﬁnitely generated since it equals

As, hence it follows by induction that all the images are free and ﬁnitely generated,

and we see that we have a telescoping sum

X

n∈

(−1)nrankRAn=X

n∈

(−1)n(rankRim ∂A

n+1 + rankRim ∂A

n) = 0.

Thus

χR(X) = X

n∈

(−1)nrankRBn

=X

n∈

(−1)n(rankRAn+ rankRBn)

=X

n∈

(−1)nrankRLn

as desired.

We saw in proposition 4.5 that βR

n(X)≤βR

n(X) for all X∈ C(f )

( )(R) and

all prime ideals pof R. When considering Euler characteristics instead of Betti

numbers, we can do much better:

Euler characteristics 18

Proposition 5.3. Suppose that X∈ C(f )

( )(R)is a complex with pdRX < ∞, and

let pbe a prime ideal of R. Then pdRX < ∞and χR(X) = χR(X).

Proof: We know from proposition 4.5 that X∈ C(f )

( )(R) and βR

n(X)≤βR

n(X)

for all n∈Z, so it follows from corollary 4.4 that pdRX≤pdRX < ∞.

Now, by corollary 4.3 we can ﬁnd a minimal free resolution X'

←− L∈

CL(R) such that βR

n(X) = rankRLnfor all n∈Z. Since Xhas ﬁnite projective

dimension, it follows from corollary 4.4 that Lis bounded, that is, L∈ CL(R). As

we saw in proposition 4.5, the free resolution X'

←− Linduces a free resolution

X'

←− Lin which each (Ln) is a free R-module with rankR(Ln) = rankRLn.

As the localization Lof course also is bounded, it now follows from theorem 5.2

that

χR(X) = X

n∈

(−1)nrankRLn=X

n∈

(−1)nrankR(Ln) = χR(X).

We conclude this paper by investigating properties of the Euler characteristics

of modules. First a lemma:

Lemma 5.4. If pis an associated prime of R, then depth R= 0, and every

ﬁnitely generated R-module Pwith pdRP < ∞is free with rankRP=χR(P).

Proof: Suppose that p= AnnR(r0) is a prime for some nonzero element r0∈

R. In the localization R, the element r0/1 is nonzero since tr06= 0 for all

t∈R\p. For any element r/u ∈p, we have that (r/u)(r0/1) = rr0/u = 0

since r∈p= AnnR(r0), so all elements of pare zero-divisors in R. However,

all elements outside the maximal ideal pare invertible and therefore not zero-

divisors, so pis exactly the ideal Zd Rof zero-divisors of R. Another way of

saying this is that depth R= 0. According to [FoxbyHA] lemma 17.1(8) any

ﬁnitely generated module with ﬁnite projective dimension over a ring with depth

0 is free. Consequently Pis free, hence Pis its own minimal free resolution, so

βR

n(P) = 0 for all n6= 0, and it follows that rankRP=βR

0(P) = χR(P).

The lemma is particularly useful together with proposition 5.3: By [Lang]

corollary X.2.7 we can ﬁnd an associated prime pof R. Given a ﬁnitely generated

module Mwith ﬁnite projective dimension, Mwill be ﬁnitely generated with

ﬁnite projective dimension, and also free with rankRM=χR(M) = χR(M).

Thus ﬁnding the Euler characteristics of a module comes down to ﬁnding the

rank of a free module—a task which apparently is less complicated.

For the next theorem, notice that it is clear from the deﬁnition (and the

fact that the tensor product and the homology functor are additive) that Betti

numbers are additive, that is, that βR

n(X⊕Y) = βR

n(X)+βR

n(Y) for all complexes

X, Y ∈ C(f )

( )(R) and all n∈Z. This implies that whenever Xand Yare complexes

Euler characteristics 19

such that the Euler characteristics is deﬁned, then the Euler characteristics is

deﬁned for X⊕Y, and χR(X⊕Y) = χR(X) + χR(Y).

Theorem 5.5. Suppose 0→K→M→N→0is an exact sequence of ﬁnitely

generated modules, each having projective dimension <∞. Then χR(M) =

χR(K) + χR(N).

Proof: According to [Lang] corollary X.2.7 we can ﬁnd an associated prime

pof R, and by the preceding lemma, all ﬁnitely generated modules with ﬁnite

projective dimension over Rmust be free. Thus, by exactness of the localization

functor, we get an induced exact sequence 0 →K→M→N→0 of ﬁnitely

generated free R-modules. We therefore get a splitting M=K⊕N, and

it follows from proposition 5.3 and the additivity of Euler characteristics that

χR(M) = χR(M) = χR(K) + χR(N) = χR(K) + χR(N).

Notice how the above theorem resembles lemma 1.4. The theorem actually

also holds for complexes, but proving this requires a more constructional and

cumbersome approach than the one given above, so we leave the theorem as it is,

sacriﬁcing generality for simplicity.

Proposition 5.6. If M∈ Cf

0(R)is a ﬁnitely generated module with pdRM < ∞,

then 0≤χR(M)≤βR

0(M).

Proof: Using (as usual) [Lang] corollary X.2.7 we can ﬁnd an associated prime

pof R, so by lemma 5.4, Mis a free R-module with rankRM=χR(M).

By proposition 5.3 we therefore conclude that χR(M) = χR(M) = βR

0(M). It

now follows from proposition 4.5 that 0 ≤χR(M)≤βR

0(M).

It is somewhat surprising that the alternating sum deﬁning the Euler charac-

teristics turns out to be nonnegative for modules. A natural question to ask next

is when the Euler characteristics of a module is zero. The question is answered

in

Theorem 5.7. If M∈ Cf

0(R)is a ﬁnitely generated module with pdRM < ∞,

then the following conditions are equivalent:

(i) χR(M)>0

(ii) SuppRM= Spec R

(iii) AnnRM= 0

(iv) AnnRM⊆Zd R

Proof: If M= 0 then all conditions are false, and the theorem holds. We can

therefore assume that M6= 0.

Euler characteristics 20

“(i) ⇒(ii)”. If pis a prime ideal not in SuppRM, then M= 0, and it follows

from proposition 5.3 that χR(M) = χR(M) = χR(0) = 0, which contradicts

condition (i).

“(ii) ⇒(iii)”. Suppose that AnnRM6= 0. There exists according to [Lang]

corollary X.2.7 a prime ideal associated with AnnRM, that is, we can ﬁnd a

nonzero element r0∈AnnRMsuch that AnnR(r0) = pis a prime ideal. Consider

now the element r0/1∈R, which must be nonzero since tr06= 0 for all t∈R\p.

For all m/u ∈Mwe have (r0/1)(m/u) = 0 since r0m= 0, so r0/1∈AnnRM,

hence AnnRM6= 0. However, since pis an associated prime (of AnnRM⊆R),

it follows from lemma 5.4 that Mis free, and the only free module with nontrivial

annihilator is the zeromodule, so M= 0, that is, p/∈SuppRMwhich contradicts

condition (ii).

“(iii) ⇒(iv)”. Obvious.

“(iv) ⇒(i)”. According to [Eisenbud] theorem 3.1, there are only ﬁnitely

many primes associated with R, and their union is exactly the set of zero-divisors

of R. From the assumption it therefore follows that AnnRMis contained in a

ﬁnite union of associated primes of R, hence by the prime avoidance theorem (see

for example [Eisenbud] lemma 3.3) AnnRMis contained in one of the associated

primes p. It follows (see for example [Eisenbud] corollary 2.7) that M6= 0.

On the other hand, since pis an associated prime, we have by lemma 5.4 that

Mis a free R-module with rankRM=χR(M) = χR(M). Consequently

χR(M)>0.

Notice that conditions (ii), (iii) and (iv) are formulated completely without

reference to Betti numbers or Euler characteristics; the fact that they are equiva-

lent makes sense (but would be hard to prove) in a world without Betti numbers

and Euler characteristics. A consequence of the theorem is given in the corollary

below, which can also be formulated without the mention of Betti numbers or

Euler characteristics, simply by replacing “χR(K)>0” by one of its equivalent

conditions.

Corollary 5.8. Suppose that Kis a ﬁnitely generated submodule of a free and

ﬁnitely generated module F, and that pdRK < ∞. Then either K= 0 or

χR(K)>0.

Proof: Suppose that the latter condition fails to hold. Then by the above

theorem, AnnRK*Zd R, so we can ﬁnd r∈Rsuch that rannihilates Kbut

is not a zero-divisor of R. Let (f1, . . . , fn) be a basis for F, and let kbe any

element of K. We can then ﬁnd r1, . . . , rn∈Rsuch that k=r1f1+· · · +rnfn,

and we must therefore have rr1f1+· · · +rrnfn=rk = 0. Since the fi’s formed

a basis, it follows that rr1=· · · =rrn= 0, but rwas not a zero-divisor, so we

conclude that r1=· · · =rn= 0, that is, k= 0. Consequently K= 0.

Theorem 5.7 answered the question of when the Euler characteristics χR(M)

of a module Mreaches its lower bound of 0. The last theorem of this paper

References 21

answers the question of when the Euler characteristics reaches its upper bound

of βR

0(M).

Theorem 5.9. If M∈ Cf

0(R)is a ﬁnitely generated module with pdRM < ∞,

then χR(M) = βR

0(M)if and only if Mis free.

Proof: If Mis free, then Mis its own minimal free resolution, and it follows

that βR

n(M) = 0 for all n6= 0, hence χR(M) = βR

0(M).

Suppose conversely that χR(M) = βR

0(M). By corollary 3.7 we can ﬁnd an

exact sequence

· · · −→ Fn

∂F

n

−→ Fn−1−→ · · · −→ F1

∂F

1

−→ F0

α0

−→ M−→ 0,

such that the modules F0, F1, F2, . . . are ﬁnitely generated and free, and such

that ∂F

n(Fn)⊆mFn−1for all n∈N. In particular, the module ker α0= im ∂F

1

is ﬁnitely generated and contained in mF0. Furthermore, the exact sequence

· · · → Fn→Fn−1→ · · · → F1→ker α0→0 is a minimal free resolution of ker α0,

so it follows from corollary 4.3 and corollary 4.4 that pdR(ker α0)≤pdRM < ∞.

We now have an exact sequence

0−→ ker α0−→ F0

α0

−→ M−→ 0,(3)

and it follows from theorem 5.5 that χR(F0) = χR(ker α0) + χR(M). But F0is

free, so χR(F0) = βR

0(F0) = rankRF0by the previous, and χR(M) = βR

0(M) by

assumption, so the equation now sounds

rankRF0=χR(ker α0) + βR

0(M) (4)

Application of the right exact functor k⊗R−to the exact sequence (3) yields

the exact sequence ker α0→F0→M→0 of k-vector spaces. Here, the ho-

momorphism ker α0→F0is induced by the inclusion of kerα0⊆mF0into F0

and therefore zero. Thus there is an isomorphism F0∼

=M, and it follows that

rankRF0=βR

0(M). Equation (4) therefore yields χR(ker α0) = 0, and since ker α0

is a ﬁnitely generated submodule of the free module F0and has pdR(ker α0)<∞,

it follows from corollary 5.8 that kerα0= 0. Thus F0∼

=M, so Mis free.

References

[Eisenbud] David Eisenbud: Commutative Algebra with a View Toward Alge-

braic Geometry, Springer 1995.

[FoxbyB&E] Hans–Bjørn Foxby: Bettital & Eulerkarakteristik, discussion paper

(in Danish) providing the basis for this course project, University

of Copenhagen 2002.

References 22

[FoxbyHA] Hans–Bjørn Foxby: Homologisk Algebra, lecture notes (in Danish)

for the graduate course Mat4ha, University of Copenhagen 1999.

[FoxbyHHA] Hans–Bjørn Foxby: Hyperhomological Algebra & Commutative

Rings, unpublished notes, University of Copenhagen 1998.

[H&S] P.J. Hilton & U. Stammbach: A Course in Homological Algebra,

Second Edition, Springer 1997.

[Lang] Serge Lang: Algebra, Third Edition, Addison–Wesley 1999.