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arXiv:0707.1539v1 [math.AC] 11 Jul 2007
Conjugacy classes and invariant subrings
of R-automorphisms of R[x]1 2
Jebrel M. Habeb and Mowaffaq Hajja
Yarmouk University
Irbid – Jordan
jhabeb@yu.edu.jo , mhajja@yu.edu.jo
and
William J. Heinzer
Purdue University
West Lafayette, IN 47907 – USA
heinzer@math.purdue.edu
1 Introduction and terminology
All rings are assumed to be commutative with an identity element. The
group of units of a ring Ris denoted by U(R), and the set of nilpotent
elements of Rby N(R). It is well known that N(R) is an ideal of Rcalled
the nilradical of R. If N(R) = {0}, then Ris said to be reduced.
Let R[x] be the polynomial ring in one indeterminate xover a ring R.
An endomorphism σof R[x] is called an R-endomorphism if σ(r) = rfor
all rin R. Clearly, an R-endomorphism σof R[x] is completely determined
by σ(x). A theorem of Gilmer [6, Theorem 3] asserts the following:
Fact 1.1 An R-endomorphism σis an R-automorphism if and only if
σ(x) = a+ux +x2f(x),(1)
where a∈R,u∈ U(R), and f(x)∈ N (R[x]). In other words, an element
y∈R[x]is such that R[y] = R[x]if and only if yis of the form given by
the right hand side of (1).
1This work is supported by a research grant from Yarmouk University
22000 AMS Classification Number: 13A50
1
It is well known that a polynomial f(x)∈R[x] is nilpotent if and only
if all the coefficients of fare nilpotent elements in R(see [1, Exercise 2(ii),
page 11]). Thus N(R[x]) = N(R)R[x]. Since the sum of a unit and a
nilpotent element is a unit, an equivalent formulation of Fact 1.1 is:
Fact 1.2 An R-endomorphism σof R[x]is an R-automorphism if and
only if
σ(x) = b+vx +g(x),(2)
where b∈R,v∈ U(R)and g(x)∈ N (R[x]).
Remark 1.3 Let Rbe an integral domain with field of fractions Kand
let Hbe a group of R-automorphism of R[x]. Each h∈Hextends in a
canonical way to an automorphism of the field K(x). Thus we may regard
Has a group of automorphism of the field K(x). The fixed field K(x)Hof
Hacting on K(x) contains the fixed ring R[x]Hof Hacting on R[x]. If His
infinite, then the fixed field K(x)His K. Therefore if His an infinite group
of R-automorphisms of R[x], where Ris an integral domain, then Ris the
ring of invariants of Hacting on R[x], i.e., R[x]H=R. Assume the group
His finite, say |H|=n, and let L=K(x)H. Then K(x)/L is a Galois field
extension with [K(x) : L] = nand {1,x,...,xn−1}is a vector space basis
for K(x) over L. Moreover, Lis the field of fractions of R[x]H[2, Corollary,
page 324]. For each h∈H, we have h(x) = uhx+ah, where uh∈ U(R) and
ah∈R. Let f=Qh∈H(uhx+ah) denote the norm of xwith respect to H.
Notice that
f=uxn+bn−1xn−1+···+b1x+b0,
where u∈ U(R) and each bi∈R. It follows that xsatisfies a monic
polynomial of degree nwith coefficients in R[f]. Therefore K(f) = Land
{1,x,...,xn−1}is a free module basis for R[x] as an R[f]-module, and as
Samuel observes in [9], we must have R[x]H=R[f].
For an integer n≥2, let Zndenote the residue class ring Z/nZ. Notice
that any automorphism of the polynomial ring Zn[x] maps 1 to 1 and thus
maps every element of Znto itself and is therefore a Zn-automorphism of
2
Zn[x]. . If pis a prime integer, and Gis the group of automorphism of Zp[x],
then the result of Samuel in Remark 1.3 implies that the ring of invariants
Zp[x]G=Zp[(xp−x)p−1]
Let Hbe a finite group of R-automorphisms of the polynomial ring
R[x]. Example 1.4 illustrates that computing generators over Rfor the
ring of invariants R[x]His more subtle in the case where Rhas nonzero
nilpotent elements. With R=Z4, there exists a group H=hαiof order 2
of R-automorphisms of R[x] such that R[x]Hproperly contains the subring
R[xα(x), x+α(x)] generated over Rby the norm and trace of xwith respect
to H.
Example 1.4 Let αdenote the automorphism of Z4[x] defined by α(x) =
−xand let H=hαidenote the cyclic group generated by α. Then Hhas
order 2 and Z4[x+α(x), xα(x)] = Z4[x2] is properly contained in Z4[x]H;
for it is clear that Z4[2x, x2]⊆Z4[x]H. Indeed, Z4[x]H=Z4[2x, x2], as we
discuss in more detail below in Section 3.
Notation 1.5 In [4], the group of R-automorphisms of R[x] is denoted by
G(R), and the subgroup of G(R) consisting of R-automorphisms in which
f(x), as given in (1), is zero by B(R). We retain the notation G(R); how-
ever, to highlight the dependence of B(R) on x, we denote it by Bx(R),
or rather by Bx(R). This dependence is illustrated in Example 2.1, where
it is shown that if R[x] = R[y], then Bx(R) need not coincide with By(R).
Of course it is true that Bx(R) and By(R) are isomorphic via the obvious
map that sends the R-automorphism sdefined by s(x) = a+ux to the
R-automorphism s′defined by s′(y) = a+uy. Thus, up to isomorphism,
one may denote Bx(R) by B(R). Identifying the element of B(R) defined
by x7→ ux +awith the element (u, a)∈ U (R)×R, we see that B(R) is the
semidirect product of the multiplicative group U(R) by the additive group
Rdefined by the multiplication
(u, a)·(v, b) = (uv, va +b).
At this point, it is convenient to prove the following simple theorem that
will be used later in the proof of Theorem 4.6.
3
Theorem 1.6 Let R×Sdenote the direct product of the rings Rand S.
Then the group B(R×S)is isomorphic to the direct product B(R)×B(S)
of the groups B(R)and B(S).
Proof. It is easy to check that the mapping from R[x]×S[x] to (R×S)[x]
defined by
Xrixi,Xsixi7→ X(ri, si)xi
is a ring isomorphism and that the mapping from Bx(R)×Bx(S) to Bx(R×S)
defined by
(x7→ ux +a, x 7→ vx +b)7→ (x7→ (u, v)x+ (a, b))
is a group isomorphism.
Remark 1.7 If the ring Rhas nonzero nilpotent elements, then the group
G=G(R) of automorphisms of R[x] is infinite, and properly contains the
subgroup Bx(R) defined in (1.5); however, Dowlen proves that the ring of
invariants R[x]Gis equal to the ring of invariants R[x]Bx[4, Theorem 1.2].
In the case where Ris a finite ring, for example R=Zn, the group Bx(R)
is finite and R[x] is integral over its invariant subring R[x]G=R[x]Bx.
Dowlen in [5] determines generators for the ring of invariants of Zn[x] with
respect to the group G(Zn) of automorphisms of Zn[x]. One of our goals is
to determine generators for the ring of invariants of Zn[x] with respect to
various subgroups Hof G(Zn).
In the hope that it may be useful in the task of describing rings of invari-
ants of R[x], we prove in Section 4 that every element of Bx(Zn) is equivalent
to an element having a certain simple representation. We determine con-
ditions for two elements of Bx(Zn) to be conjugate and give a formula for
the number of conjugacy classes of this group. In Section 2, we state and
prove results that hold in general, and raise several open problems concern-
ing Bx(R) and G(R). We examine in detail in Section 3 the structure of the
automorphism group G(Z4) of Z4[x] and enumerate the invariant subrings
of Z4[x] with respect to subgroups of G(Z4). In particular, for R=Z4, we
4
establish the existence of subrings of R[x] that are rings of invariants of sub-
groups of G(R), but are not rings of invariants of subgroups of Bx(R). We
prove, however, that each of these invariant subrings of R[x] is the ring of
invariants of a subgroup of Bz(R) for some z∈R[x] such that R[z] = R[x].
2 General results and open problems
Let Abe the ring of polynomials in one indeterminate over the ring R, and
let G=G(R) be the group of R-automorphisms of A. For x∈Asuch
that A=R[x] and for σ∈G, we say that σis x-basic if σ(x) is of the
form σ(x) = a+ux for some a∈Rand u∈ U(R). The group of x-basic
elements of Gis denoted by Bx=Bx(R). As mentioned in (1.5), Bxand
Byare isomorphic if yis such that R[x] = R[y]. However, they need not be
equal as is illustrated in Example 2.1.
Example 2.1 Let n=pk, where p≥3 is an odd prime and where k≥2.
Let R=Znand let σ∈G(R) be defined by σ(x) = 2x. Then σis x-
basic. However, if we let y=x+pk−1x2, then it follows from Fact 1.1 that
R[y] = R[x], and it is easy to see that σis not y-basic. Indeed:
σ(y) = 2x+pk−1(2x)2= 2y+ 2pk−1x2= 2y+ 2pk−1(y−pk−1x2)2
= 2y+ 2pk−1y2.
Two elements αand βin G(R) = Gare G-conjugate if α=g−1βg for
some g∈G(R). They are Bx-conjugate if gcan be chosen to belong to
Bx(R). They are B-conjugate if gcan be chosen to belong to By(R) for
some ysuch that R[x] = R[y]. The corresponding conjugacy classes are
denoted by [α]G, [α]Bx, and [α]B, respectively.
If α∈Bx(R), then we clearly have [α]Bx⊆[α]B⊆[α]G.Theorem 2.2
demonstrates that if Ris not reduced and α∈Bxis defined by α(x) = x+1,
then [α]Bx([α]G.
Theorem 2.2 Let Rbe a ring that is not reduced and let α∈Bxbe defined
by α(x) = x+ 1. Then [α]Gis not contained in Bx. Therefore Bx(R)is not
normal in G(R).
5
Proof. Since Ris not reduced, there exists a nonzero r∈Rsuch that r2= 0.
If [k(k−1)
2]r= 0 for 3 consecutive natural numbers k, say
n(n−1)
2r=(n+ 1)n
2r=(n+ 2)(n+ 1)
2r= 0,
then by subtracting we have nr = (n+1)r= 0. But this implies that r= 0.
Therefore there exists an integer n≥3 such that [ n(n−1)
2]r6= 0. For such r
and n, define σin G(R) by
σ(x) = x+rx2+rxn.
We prove that σ−1ασ /∈Bx(R). For suppose that
σ−1ασ(x) = ux +c, (3)
where u∈ U(R) and c∈R. From (3), it follows that
α(x+rx2+rxn) = u(x+rx2+rxn) + c
x+ 1 + r(x+ 1)2+r(x+ 1)n=u(x+rx2+rxn) + c.
Equating the coefficients of xand of x2, we see that
1 + 2r+nr =uand r+rn(n−1)
2=ur.
Hence
r+ 2r2+nr2=r+rn(n−1)
2,
and therefore [n(n−1)
2]r= 0, contradicting the choice of n.
Remark 2.3 Let α, β ∈Bx(R) be defined by
α(x) = ux +a, β(x) = vx +b, (4)
where u, v ∈ U(R) and a, b ∈R. It is easy to see that if αand βare Bx-
conjugate, then u=v. For if σ∈Bx(R) is defined by σ(x) = wx +c, then
6
σ−1(x) = w−1x−w−1c, and
σ−1ασ(x) = σ−1α(wx +c)
=σ−1(ws(x) + c)
=σ−1(wux +wa +c)
=wuσ−1(x) + wa +c
=wuw−1x−wuw−1c+wc +c
=ux −uc +wc +c.
Therefore if αand βin Bx(R) as in (4) are Bx-conjugate, then u=v.
We demonstrate in Example 2.4 that for α, β ∈Bx(R) as in (4), it may
happen that αand βare G-conjugate and u6=v.
Example 2.4 Let n=p2, where p≥3 is an odd prime. Let R=Zn, let
α∈Bx(R) be defined by α(x) = x+ 1, and let σ∈G(R) be defined by
σ(x) = x+px2, where we use elements in Zto represent their equivalence
classes in Zn. Notice that σ−1(x) = x−px2. We have:
σ−1ασ(x) = σ−1α(x+px2)
=σ−1(x+ 1 + p(x+ 1)2)
= 1 + σ−1(x) + pσ−1(x2+ 2x+ 1)
= 1 + (x−px2) + p[(x−px2)2+ 2(x−px2) + 1]
= 1 + p+ (1 + 2p)x.
Therefore β∈Bx(R), where β(x) = (1 + 2p)x+ 1 + pis G-conjugate to α
and 1 + 2p6= 1 mod p2.
Remark 2.5 Let α1and α2be elements of G(R) defined, as in (1), by
αi(x) = ci+uix+x2fi(x), i = 1,2,
where ci∈R,ui∈ U(R) and fi(x)∈ N (R[x]).
1. In Example 2.6 we demonstrate that it is possible to have u16=u2
and yet α1and α2are Bx-conjugate.
7
2. In Theorem 2.7 we prove that if α1and α2are in Bxand are G-
conjugate and if u1=u2, then α1and α2are Bx-conjugate.
Example 2.6 Let n=p2, where p≥3 is an odd prime. Let R=Znand
let α∈G(R) be defined by α(x) = x+px2, where we use elements in Z
to represent their equivalence classes in Zn. Let σ∈Bx(R) be defined by
σ(x) = 1 + x. Notice that σ−1(x) = −1 + x. We have:
σ−1ασ(x) = σ−1α(1 + x)
=σ−1(1 + x+px2)
= 1 + σ−1(x) + pσ−1(x2)
= 1 + (−1 + x) + p(−1 + x)2
=x+p(1 −2x+x2)
=p+ (1 −2p)x+px2.
Theorem 2.7 Let αand βbe elements of Bx(R)defined by
α(x) = a+ux , β(x) = b+ux,
where u∈ U(R), and a, b ∈R. If αand βare G-conjugate, then αand β
are Bx-conjugate.
Proof. Let σ∈Gand let y=σ(x). We have
β=σ−1ασ ⇐⇒ σβ(x) = ασ(x) = α(y).
Since
σβ(x) = σ(b+ux) = b+uσ(x) = b+uy,
we have
α(y) = b+uy ⇐⇒ β=σ−1ασ. (5)
The assumption that αand βare G-conjugate implies there exists σ∈G
with y=σ(x) and α(y) = b+uy. The element yis of the form
y=d+wx +f(x) = d+wx +x2g(x),
8
where w∈ U(R), d ∈Rand f(x) and g(x) are nilpotent element of R[x].
To prove that αand βare Bx-conjugate, it suffices to find Y=c+vx,
where v∈ U(R) and c∈R, such that α(Y) = b+uY . For, by (5), if τ∈Bx
is defined by τ(x) = Y, then β=τ−1ατ ⇐⇒ α(Y) = b+uY .
It follows from
b+uy =α(y) = α(wx +d+f(x)) = w(ux +a) + d+f(ux +a)
=u(wx +d+f(x)) + wa −ud −uf(x) + d+f(ux +a)
=uy +wa −ud −uf (x) + d+f(ux +a)
that
b= (1 −u)d+wa +f(ux +a)−uf (x).
Putting x= 0 and noting that f(0) = 0 (since f(x) = x2g(x)), we obtain
b= (1 −u)d+wa +f(a) = (1 −u)d+a(w+ag(a)).
Therefore
α(y) = uy +b
=uy + (1 −u)d+a(w+ag(a))
=uy + (1 −u)d+av,
where v=w+ag(a) is a unit since g(a) is nilpotent. Let Y=vx +d. Then
R[Y] = R[x], and
α(Y) = α(vx +d)
=v(ux +a) + d
=u(vx +d) + va −ud +d
=uY + (1 −u)d+av
=uY +b,
as desired.
It is clear that Bx(R) = G(R) if and only if Ris reduced. One wonders
whether there is a more quantitative version of this statement that relates
the relative size of N(R) in Rand that of Bx(R) in G(R). In particular, we
ask:
9
Questions 2.8 Which rings Rhave the property that
G(R) = [{Bz(R)|z∈R[x] and R[x] = R[z]}?
If Ris not reduced, do there always exist elements
ξ∈G(R)\[{Bz(R)|z∈R[x] and R[x] = R[z]}?
If there exist elements ξ∈G(R)\S{Bz(R)|z∈R[x] and R[x] = R[z]},
what properties distinguish such elements ξ?
Remark 2.9 If σ∈G(R) is defined by σ(x) = z, then σ−1Bzσ=Bx.
Therefore every element of Bzis G-conjugate to an element of Bx. Hence if
ξis in the center of G(R) and ξ∈Bz(R) for some z∈R[x] with R[x] = R[z],
then
ξ∈\{Bz(R)|z∈R[x] and R[x] = R[z]}.
In Section 3 we observe that for R=Z4, the center of G(R) is not contained
in Bx(R). We deduce that for R=Z4we have
G(R)6=[{Bz(R)|z∈R[x] and R[x] = R[z]}.
We examine in detail in Section 3 the structure of the automorphism
group G(Z4) of Z4[x] and enumerate the invariant subrings of Z4[x] with
respect to subgroups of G(Z4). We use the following results that hold in
more generality.
Lemma 2.10 Let Rbe a ring and let f∈R[x]be a monic polynomial with
deg f=d≥1. For each nonzero polynomial g(x)∈R[x]there exists an
integer n≥0and a unique representation for g(x)as
g(x) =
n
X
k=0
gkfk,(6)
where each gk∈R[x],gn6= 0, and for each kwith 0≤k≤n, either gk= 0
or deg gk< d.
10
Proof. If deg g < d, then the statement is clear with n= 0. We use induction
on deg gand assume for some integer m≥dthat every polynomial Gwith
deg G < m can be represented as in (6). Let g∈R[x] be a polynomial with
deg g=mand write m=dn +r, where nand rare integers and 0 ≤r < d.
Let cdenote the leading coefficient of g. Then G=g−cxrfn∈R[x] and
either G= 0 or deg G < m. If G= 0, then g=cxrfnhas the form given in
(6). If G6= 0, then by induction G, and hence also g=G+cxrfn, has the
desired form.
To prove uniqueness, notice that if g(x) = Pn
k=0 gkfk, where each gkis
either 0 or deg gk< d, then g(x) = 0 only if all the gkare 0. For if some
gk6= 0, let s= max{k|gk6= 0}and let cdenote the leading coefficient
of gs. Then cis the leading coefficient of Pn
k=0 gkfk, so this polynomial is
nonzero.
Theorem 2.11 Let Rbe a ring and let βbe the R-automorphism of R[x]
defined by β(x) = −x+ 1. Then β2= 1 and the ring of invariants of the
cyclic group hβiacting on R[x]is R[x]hβi=R[y], where y=x(−x+ 1).
Proof. Every polynomial f(x)∈R[x] has a unique representation as
f(x) =
n
X
k=0
(akx+bk)xk(x−1)k,
for some integer n≥0, where akand bkare in R, 1 ≤k≤n. Assume that
f(x) is fixed by β. Then f(x) = f(1 −x), and therefore
n
X
k=0
(akx+bk)xk(x−1)k=
n
X
k=0
(−akx+ak+bk)xk(x−1)k.
By uniqueness of representation, we conclude that ak= 0 for every k.
Remark 2.12 Let Rbe a ring and let
f=a0+a1x+···+anxn∈R[x]
be a polynomial. Consider the following assertions:
11
1. The surjective R-algebra homomorphism of R[x] onto R[f] defined by
mapping x7→ fis injective.
2. The subring R[f] of R[x] is R-isomorphic to R[x].
3. The R-algebra R[f] is a polynomial ring over R.
4. The annihilator in Rof the ideal I= (a1,...,an)Ris zero.
It is readily seen that the first 3 assertions are equivalent, and it is well
known that these are also equivalent to assertion 4 [6, Theorem 2]. In
considering condition 4, since R[f] = R[f−a0], one may assume that a0= 0.
With this assumption, if b∈Ris nonzero and bI = 0, then bx ∈R[x] is a
nonzero polynomial in the kernel of the R-algebra homomorphism defined
by x7→ f. On the other hand, if this R-algebra homomorphism is not
injective, let
bmxm+···+b1x+b0∈R[x]
be a nonzero polynomial of minimal degree such that
bmfm+···+b1f+b0= 0.
Then a0= 0 implies b0= 0. Therefore
(bmfm−1+···+b1)f= 0,
and bmfm−1+···+b1is a nonzero polynomial because of the minimal degree
assumption. Thus
f=anxn+···+a1x
is a zero-divisor in R[x]. A well known theorem of McCoy [7, page 290] im-
plies that the ideal Iof Rhas a nonzero annihilator (see also [8, Theorem 4,
pages 34-36], [6, page 330] and [1, Exercise 2(iii), page 11]).
Theorem 2.13 Let Rbe a ring and let f∈R[x]be such that R[f]is a
polynomial ring over R. If gis a nilpotent element of R[x], then the ring
R[f, g]is a polynomial ring over Rif and only if g∈R[f].
12
Proof. Let N=N(R) be the nilradical of R. The nilradical of the poly-
nomial ring R[x] is given by N(R[x]) = NR[x]. Let R/N =F. Then
R[x]/(NR[x]) = F[x].
We denote the image of r∈Runder the canonical map R→R/N by
r=r+N. This map extends to a map from R[x] to F[x]. We denote the
image of f∈R[x] under this map by f=f+N[x]. Thus if f=Prixi∈
R[x], then f=Prixi.
Assume that R[f, g] = R[y] for some yin R[x]. Passing to quotients
mod N R[x], we see that
F[y] = F[f , g] = F[f].
By Fact 1.2, we have
y=Uf +C,
where Uis a unit in Fand C∈F. Thus U=u+N,C=c+N, and
(u+N)(v+N) = 1 + Nfor some v∈R. Hence uv = 1 + νfor some ν∈N
and uv and therefore uis a unit in R. Hence
y=uf +c+H(x),
where H(x)∈NR[x]. Again, by Fact 1.2, R[y] = R[f], as desired.
3 Automorphisms of Z4[x]
In this section, we determine the structure of the group G(R) of R-automorphisms
of the polynomial ring R[x] in the case where R=Z4and we describe the
ring of invariants R[x]Hfor every subgroup Hof G(R). We also describe
the conjugacy classes in G(R).
Remark 3.1 Let Rbe a ring and let f∈ N (R[x]) be a nilpotent element
of the polynomial ring R[x]. We associate with fthe R-automorphisms αf
and βfin G(R) defined by
αf:x7→ x+f , βf:x7→ −x+ 1 + f. (7)
13
Notice that the correspondences f7→ αfand f7→ βfare both one-to-one.
Moreover, the set A:= {αf:f∈ N (R[x])}is a subgroup of G(R). Also
the constant term of αf(x) is a nilpotent element of R, while the constant
term of βf(x) is a unit of Rfor each f∈ N (R[x]). Therefore the sets A
and {βf:f∈ N (R[x])}are disjoint. In the special case where 2Ris a
maximal ideal of Rwith R/2R=Z2, so, in particular, in the case where
R=Z4, every automorphism3of R[x] is of the form αfor βffor some
f∈ N (R[x]). Indeed, in this case, G(R/2R) is a group of order 2, and Ais
the normal subgroup of G(R) that is the kernel of the canonical surjective
homomorphism of G(R) onto G(R/2R) while {βf:f∈ N (R[x])}is the
unique nonidentity coset of Ain G(R).
We start by describing the ring of invariants R[x]Hin the case where
R=Z4and His a cyclic subgroup of G(R). By Remark 3.1, Hhas the
form H=hαfior H=hβfifor some f∈ N (R[x]). The fixed rings of each
of these types is described in Theorems 3.2 and 3.4 below. Lemma 3.3 is
used in the proof of Theorem 3.4.
Theorem 3.2 Let R=Z4and let α∈G(R)be defined by
α(x) = x+f,
where f∈ N (R[x]). If f= 0, then the order of αis 1. If f6= 0, then the
order of αis 2 and the fixed ring of αis R[x2,2x]and is not a polynomial
ring.
Proof. It is clear that αis the identity element of G(R) if and only if f= 0.
Assume that f6= 0. Since N(R[x]) = 2R[x], we have 2f=f2= 0. It
follows that α2(x) = xand αhas order 2. Also f= 2g, where ghas the
form, for some integer m≥0,
g=
m
X
k=0
bkxk,where 2bm6= 0.
3In the case where R=Z4, every automorphism of R[x] is an R-automorphism.
14
Let S=R[x2,2x]. Clearly, S⊆R[x]hαi. To show that this inclusion is an
equality, assume that there exists an element h∈R[x]hαi\S. Subtracting
from han element in S, we obtain for some positive integer nan element
h′∈R[x]hαi\Sof the form
n
X
k=0
akx2k+1,where an= 1.
Since h′is α-fixed, it follows that
n
X
k=0
akx2k+1 =
n
X
k=0
akx2k(x+ 2g)
n−1
X
k=0
akx2k+1 = 2x2ng+
n−1
X
k=0
akx2k(x+ 2g).
Comparing the coefficients of the highest degree terms in this last equation,
we see that 2bmx2n+m= 0. This contradicts the assumption that 2bm6= 0.
Therefore R[x]hαi=S. By Theorem 2.13, S=R[x2,2x] is not a polynomial
ring over R
Lemma 3.3 Let R=Z4and let θbe the automorphism of R[x]defined by
θ(x) = x+ 1.
Then
R[x]hθi=R[w2,2w],
where w=x(x+ 1). In particular, R[x]hθiis not a polynomial ring over R.
Proof. Let M=x(x+ 1)(x+ 2)(x+ 3) be the norm of xwith respect to θ.
Then θ(w) = (x+ 1)(x+ 2) = w+ 2(x+ 1).Therefore, both w2and 2ware
fixed by θ. It remains to show that every g∈R[x]hθibelongs to R[w2,2w].
By Lemma 2.10, the polynomial ghas a unique representation as
g(x) =
n
X
k=0
gkMk,
15
for some integer n≥0, where each gkis either 0 or a polynomial in R[x] of
degree less than 4. Since θfixes Mand does not increase degrees, it follows
from Lemma 2.10 that θ(g) = gif and only if θ(gk) = gkfor all k. Thus
it suffices to show that if h=ax +bx2+cx3∈R[x] is fixed by θ, then
h∈R[w2,2w]. We have
θ(h) = h⇐⇒ ax +bx2+cx3
=a(x+ 1) + b(x2+ 2x+ 1) + c(x3+ 3x2+ 3x+ 1)
⇐⇒ 3c= 3c+ 2b=a+b+c= 0
⇐⇒ c= 0, a =−b, 2b= 0
⇐⇒ h=b(x2−x),2b= 0
⇐⇒ h= 2d(x2−x) = 2d(x2+x),for some d∈R
⇐⇒ h= 2dw.
Therefore R[x]hθi=R[M, 2w] = R[w2,2w], since M=w2+2w. By Theorem
2.13, R[x]hθiis not a polynomial ring over R.
Theorem 3.4 Let R=Z4and let β∈G(R)be defined by
β(x) = −x+ 1 + f,
where f∈ N (R[x]). Let y=x(−x+ 1). The order of βis either 2 or 4,
and the following are equivalent:
1. The order of βis 2.
2. The element f∈R[y].
3. f= 2hfor some h∈R[y].
If the order of βis 2, then the fixed ring of βis R[y+xf], a polynomial
ring over Rgenerated by the element y+xf. If the order of βis 4, then
the fixed ring of βis R[y2,2y]and is not a polynomial ring.
Proof. It is clear that
β2(x) = x+g, (8)
16
where g=β(f)−f. Since β(N(R[x]) = N(R[x]), β(f)−f∈ N (R[x]). By
Theorem 3.2, the order of β2is 1 or 2. Since βis not the identity element
of G(R), the order of βis 2 or 4.
We consider first the case where the order of βis 2. Clearly,
order (β) = 2 ⇐⇒ β2(x) = x⇐⇒ β(f)−f= 0.(9)
We show that this happens if and only if f= 2hfor some h∈R[y].
It is easy to see that β(xk) = (−x+ 1)kif kis even and β(2xk) =
2(−x+ 1)kif kis odd. Thus letting β0be the automorphism x7→ −x+ 1,
we see that βand β0coincide on 2xkfor every kand therefore coincide on
every element of N(R[x]). Therefore
β(f)−f=β0(f)−f.
From this and (9) it follows that the order of βis 2 if and only if fbelongs
to R[x]hβ0i. By Theorem 2.11, R[x]hβ0i=R[y]. Therefore
order (β) = 2 ⇐⇒ f∈R[y].
However an element in R[y] that 2 multiplies to 0 must be of the form 2h
for some h∈R[y]. Therefore
order (β) = 2 ⇐⇒ f= 2hfor some h∈R[y].
Assume that the order of βis 2. Thus f= 2h∈2R[y]. Notice that
β(y) = y+ 2h, and therefore β(y2) = y2and β(2y) = 2y. Therefore
β(2yk) = 2ykfor all kand hence β(2g) = 2gfor all g∈R[y]. In particular,
this holds for f= 2h, and we have
β(2xh) = (−x+ 1 + 2h)(2h)
= 2xh + 2h.
Let z=x+ 2xh. By Fact 1.2, R[x] = R[z]. Also,
β(z) = (−x+ 1 + 2h) + (2xh + 2h)
=−z+ 1.
17
Therefore
R[x]hβi=R[z]hβi
=R[z(−z+ 1)] by Lemma 2.11
=R[(x+ 2xh)(−x−2xh + 1)]
=R[x(−x+ 1) + 2xh]
=R[y+ 2xh]
=R[y+xf],
as claimed. Note that R[y+xf] contains R[y2,2y] properly since R[y2,2y]
is not a polynomial ring over Rby Theorem 2.13.
Assume that the order of βis 4. Then the order of β2is 2. Since β2(x)
has the form given in (8), Theorem 3.2 implies that R[x]hβ2i=R[x2,2x].
This does not depend on f. Nor does the action of βon this ring, since
β(x2) = (−x+ 1)2, β(2x) = 2x+ 2.
Thus for the purpose of finding R[x]hβi, one may take f= 2x. Then β:
x7→ x+ 1. By Lemma 3.3, R[x]hβi=R[w2,2w], where w=x(x+ 1), Also
R[w2,2w] = R[y2,2y], so R[x]hβi=R[y2,2y]. This completes the proof of
Theorem 3.4.
Notation 3.5 To consider R[x]Hfor an arbitrary subgroup Hof G(R),
we use the following notation. Referring to (7), we observe that α0is the
identity of G(R) and β0is the automorphism βdefined by
β:= β0:x7→ −x+ 1,
that is considered in Theorem 2.11. The automorphism θ:x7→ x+ 1 of
Lemma 3.3 is β2xand its inverse is β2(−x+1). We also let ybe defined by
y=x(−x+ 1),(10)
as in Theorem 3.4. For h∈R[x], we denote β0(h) by h′. Thus h′is obtained
from hby replacing xby −x+ 1. Hence h′′ =hfor all h∈R[x] and
h=h′⇐⇒ h∈R[x]hβi=R[y].(11)
18
Also the map φ:R[x]→R[y] defined by φ(f) = f+f′is onto. In fact,
φxn+1(−x+ 1)n=xn+1(−x+ 1)n+xn(−x+ 1)n+1 =xn(−x+ 1)n
=yn.
Theorem 3.6 describes R[x]Hfor an arbitrary subgroup Hof G(R).
These fixed subrings are precisely those subrings fixed by cyclic subgroups
(as described in Theorems 3.2 and 3.4), together with the family of polyno-
mial rings R[y+xf], f ∈R[y], where y=x(−x+ 1). We let edenote the
identity element of the group G(R).
Theorem 3.6 Let R=Z4and let A={αf:f∈ N (R[x])}be the normal
subgroup of G(R)of index 2 defined in Remark 3.1. Let Hbe a subgroup of
G(R), and let y=x(−x+ 1).
(a) If His a subgroup of A, then R[x]His either R[x]or R[x2,2x]de-
pending on whether or not His trivial.
(b) If His not contained in Aand if H∩A 6=hei, then R[x]H=R[y2,2y].
(c) If His not contained in Aand if H∩ A =hei, then His cyclic
generated by an element βfof order 2 and R[x]H=R[y+xf], where
f∈ N (R[y]).
Proof. (a) Theorem 3.2 implies that if His a non-trivial subgroup of A,
then RH=R[x2,2x].
(b) Assume that His not contained in Aand that H0:= H∩ A is
nontrivial. By (a), R[x]H0=R[x2,2x].Therefore R[x]H⊆R[x2,2x]. Also,
every βh∈Hacts on R[x2,2x] as follows:
x27→ (−x+ 1)2,2x7→ 2(−x+ 1).
Therefore every βh∈Hfixes y2and 2yand hence
R[x]H⊇R[y2,2y].(12)
Letting α,β, and θbe the automorphisms on R[x] defined by
α:x7→ −x , β :x7→ −x+ 1 , θ =αβ :x7→ x+ 1,
19
we see that βrestricts to an automorphism of R[x2,2x], and
R[x]H=R[x2,2x]hβi
=R[x]hαihβi
⊆R[x]hαβi
=R[x]hθi
=R[y2,2y] (by Theorem 3.3).
From this and (12) it follows that R[x]H=R[y2,2y], as claimed.
(c) Assume that His not contained in Aand that H∩ A =hei. Since
[G;A] = 2, we have [H:H∩ A]≤2. Therefore His a cyclic group of
order 2, and H=hβgi, where β2
g= 1. By Theorem 3.4, g∈ N (R[y]) and
R[x]H=R[y+xg]. This completes the proof of Theorem 3.6.
Remark 3.7 Theorem 3.6 asserts that every ring of invariants of R[x]
with respect to a subgroup of G(R) is one of the following rings:
(i) R[x],(ii) R[x2,2x],(iii) R[y2,2y],or (iv) R[y+xg],
for some g∈ N (R[y]). The first three items are specific rings while item
(iv) describes an infinite family of polynomial subrings of R[x]. The ring
R[y2,2y] is the ring of invariants of R[x] with respect to G(R) and thus is
the unique smallest ring in the family. The rings that are fixed rings with
respect to subgroups of Bx(G) are the rings of the first three items and the
polynomial rings R[y] and R[x(x+ 1)]. Notice that R[y] corresponds to
g= 0 in (iv), and R[x(x+ 1)] corresponds to g= 2 or g= 2x. Letting
α2x:x7→ −x , β0:x7→ −x+ 1 , β2:x7→ −x−1, θ :x7→ x+ 1 ,
we easily see that
R[x] = R[x]hei
R[x2,2x] = R[x]hα2xi
R[y] = R[x]hβ0i
R[x(x+ 1)] = R[x]hβ2i
R[y2,2y] = R[x]hθi
20
The polynomial subrings R[y+xg] of item (iv) for g∈ N (R[y])\{0,2,2x}
are not rings of invariants of subgroups of Bx. However, each of these rings
is the fixed ring of an element σ∈Bz(R) for some zsuch that R[z] = R[x].
To see this, simply take z=x+xg and σ:z7→ −z+ 1. Then R[z] = R[x]
because g(and hence xg) is nilpotent,
R[x]hσi=R[z]hσi=R[z(−z+ 1)],
and
z(−z+ 1) = (x+xg)(−x+ 1 + xg) = x(−x+ 1) + xg =y+xg,
as desired.
The rings R[y+xg], g∈ N (R[y], are also pairwise different. To show
this, we associate with each subring Sof R[x], the subgroup
S∗={σ∈G(R)|σ(s) = sfor all s∈S}
of G(R) consisting of the automorphisms that restrict to the identity map on
S. For subrings S1and S2of R[x], it is clear that S∗
16=S∗
2implies S16=S2.
Therefore Theorem 3.8 implies that the rings of invariants enumerated in
Remark 3.7 are all distinct.
Theorem 3.8 For a subring Sof R[x], let S∗denote the subgroup of G(R)
whose elements fix S. Then
R[x]∗={e}
R[x2,2x]∗=A
R[y2,2y]∗=G(R)
R[y+xf]∗=hβfi.
for each f∈ N (R[y]).
Proof. The first equality is clear. The second equality follows because x2
and 2xare fixed by αffor all fand no βffixes 2xsince
βf(2x) = −2x+ 2 = 2x+ 2 6= 2x.
21
Since y2, and 2yare fixed by αfand βffor all f, the third equality is clear.
To establish the last equality, observe that
αg(y+xf) = y+xf ⇐⇒ αg(x(−x+ 1) + xf) = x(−x+ 1) + xf
⇐⇒ (x+g)(−x−g+ 1) + (x+g)f=x(−x+ 1) + xf
⇐⇒ x(−x+ 1) + g+xf =x(−x+ 1) + xf
⇐⇒ g= 0.
βg(y+xf) = y+xf ⇐⇒ βg(x(−x+ 1) + xf) = x(−x+ 1) + xf
⇐⇒ (−x+ 1 + g)(x+g) + (−x+ 1 + g)f=x(−x+ 1) + xf
⇐⇒ x(−x+ 1) + g−xf +f+gf =x(−x+ 1) + xf
⇐⇒ g+f= 0
⇐⇒ g=f.
This completes the proof of Theorem 3.8.
We now turn to the structure and conjugacy classes of G(R), again for
R=Z4.
Theorem 3.9 Let R=Z4and let αfand βfbe as defined in (7). Let
y=x(−x+ 1), and let
A={αf:f∈ N (R[x])},A0={αf:f∈ N (R[y])},C={α0=e, β0=β}.
Then the groups Aand Care isomorphic to the additive groups Z2[x]and
Z2, respectively, and G(R)is the extension of Aby Cvia the multiplication
β−1αgβ=βg′,
where g′=β0(g). Also the center of G(R)is A0.
Proof. It is clear that αfand βfact on 2R[x] and that
αf(2h) = 2h , βf(2h) = β0(2h) = 2h′(13)
for all f, h ∈R[x]. Thus the actions of αfand βfon 2R[x] are independent
of f. It is also easy to verify that
αgαh=αg+h, βgβh=αg+h′, αgβh=βg+h, βgαh=βg+h′,(14)
22
and therefore
α−1
gαhαg=αh, β−1
gαhβg=αh′, α−1
gβhαg=βg′+h+g,
β−1
gβhβg=βg+h′+g′.(15)
The assertions in Theorem 3.9 follow immediately from (14) and (15).
Remark 3.10 To describe conjugacy classes in G(R), let
A={αf:f∈ N (R[x])},A0={αf:f∈ N (R[y])}
B={βf:f∈ N (R[x])},B0={βf:f∈ N (R[y])}.
We see that Gis the disjoint union of Aand B, that Ais a normal subgroup
of Gof index 2, and that A0is the center of G. We also see that each of
A0,A \ A0,B0,B \ B0is the union of conjugacy classes. In A0, a conjugacy
class has one element. In A \ A0, a conjugacy class has two elements αf
and αf′. In B, two elements βgand βhare conjugate if and only if g−hor
g−h′belongs to R[y]. In particular, B0is a conjugacy class. Identifying B
(as a set) with R[x], and B0with R[y], the conjugacy class B0corresponds
to R[y] and every other conjugacy class in Bcorresponds to the union of
two cosets in the group R[x]/R[y].
Since the center A0of G(R) is not contained in Bx(R), Remark 2.9
implies that
G(R)6=[{Bz(R)|z∈R[x] and R[x] = R[z]}.
Remark 3.11 In Theorem 2.7 we prove for a general ring Rthat certain
elements of Bx(R) that are conjugate as elements of G(R) are actually
conjugate as elements of Bx(R). For R=Z4, we show this holds without
any additional condition on the elements; that is, if two elements in Bx(R)
are G(R)-conjugate, then they are Bx(R)-conjugate. This is equivalent to
showing that
∀σ∈Bx(R),[σ]Bx= [σ]G∩Bx, (16)
where [σ]Bxand [σ]Gare the conjugacy classes in Bx(R) and in G(R) that
contain σ.
23
The 8 elements of Bx(R) are:
α0=e:x7→ x , α2:x7→ x+ 2 , α2x:x7→ −x , α2x+2 :x7→ −x+ 2,
β0:x7→ −x+ 1 , β2:x7→ −x−1, β2x:x7→ x+ 1 , β2x+2 :x7→ x−1.
We verify Condition 16 by observing that each of the pairs
(α2x, α2x+2),(β0, β2),(β2x, β2(x+2))
consists of Bx-conjugate elements. It is direct to check that
β−1
2xα2xβ2x=α2x+2 , α−1
2xβ0α2x=β2, α−1
2xβ2xα2x=β2x+2.
The question of which elements ξ∈G(R) do not belong to any Bz(R)
with R[z] = R[x] is answered for R=Z4as follows: These are precisely
those ξfor which [ξ]Gdoes not intersect Bx. These are
(i) All αfthat do not belong to Bx, i.e., all αfexcept
α0=e , α2:x7→ x+ 2 , α2x:x7→ −x , α2x+2 :x7→ −x+ 2.
(ii) All βfexcept
β0:x7→ −x+ 1 , β2:x7→ −x−1, βg, β2x+g, β2(x+2)+g,
where g∈ N (R[y].
4 Conjugacy classes in the group B(Zn)
Throughout this section, we fix a natural number n > 1. All numbers are
elements in Zand an element in Znis represented by one of its inverse images
under the natural map Z→Zn. In particular, if αis an automorphism4of
the polynomial ring Zn[x] that is x-basic, then α(x) has the form ux +a,
where u, a ∈Zand where uis a unit mod n. We denote the group of x-basic
Zn-automorphisms of Zn[x] by B(Zn).
4Recall that every automorphism of Zn[x] is a Zn-automorphism.
24
If a, b ∈Z, then (a, b) denotes the positive greatest common divisor of a
and b. Assume the elements αand βof B(Zn) are such that
α(x) = ux +a, β(x) = vx +b, (17)
where (u, n) = (v, n) = 1. We emphasize the trivial fact that
α=β⇐⇒ ux +a=vx +bin Zn[x]
⇐⇒ u≡v(mod n) and a≡b(mod n).
On the other hand, αand βare said to be equivalent, and we write α∼
=β, if
they are conjugate as elements in B(Zn). This happens if and only if there
exists X=wx +c, where (w, n) = 1, such that α(X) = vX +b. We have
α(X) = vX +b⇐⇒ w(ux +a) + c=v(wx +c) + bin Zn[x]
⇐⇒ wux +wa +c=vwx +vc +bin Zn[x]
⇐⇒ wu ≡vw (mod n) and wa +c≡vc +b(mod n)
⇐⇒ u≡v(mod n) and wa ≡(v−1)c+b(mod n).
We record the conclusion as:
Fact 4.1 If αand βare defined as in (17), then
α∼
=β⇐⇒ u≡v(mod n)and there exist w,cin Zsuch that (w, n) = 1
and such that wa ≡(v−1)c+b(mod n).
Our objective is to determine a canonical representation of each conju-
gacy class of the group B(Zn) in order to simplify the task of describing
rings of invariants of the polynomial ring Zn[x]. We use the following sim-
ple theorem that is an extremely special case of Dirichlet’s theorem on the
infinitude of primes in arithmetic progressions; see [3, pages 105–122].
Theorem 4.2 If a, b, n are positive integers such that (a, b) = 1, then the
sequence
a+kb :k= 0,1,2···
contains an element that is relatively prime with n.
25
Proof. Let rbe the product of all prime factors of nthat do not divide b.
Then (r, b) = 1. Let b′be an inverse of bmod rand let kbe a non-negative
integer such that k≡(1−a)b′(mod r). Then a+kb ≡1 (mod r). Therefore
(a+kb, r) = 1. By the definition of r, we conclude that (a+kb, n) = 1, as
desired.
Theorem 4.3 Let α, β ∈B(Zn)be given by
α(x) = ux +a , β(x) = ux +b
where (u, n) = 1. If (a, n) = (b, n), then αand βare equivalent. In partic-
ular, every element in B(Zn)is equivalent to one of the form x7→ ux +d
where nis divisible by d.
Proof. Let σ∈B(Zn) be defined by σ(x) = ux +d, where d= (a, n). It
suffices to show that αis equivalent to σ. Let a1=a/d and n1=n/d Since
(a1, n1) = 1, there exists tsuch that (a1+tn1, d) = 1. Also, (a1+tn1, n1) =
(a1, n1) = 1. Therefore (a1+tn1, n) = 1 and hence v:= a1+tn1is a unit
mod n. Also, vd =a+tn ≡a(mod n). Therefore σ(vx) = v(ux +d) =
u(vx) + vd =u(vx) + a, and σ∼
=α, as desired.
Theorem 4.4 Let α, β ∈B(Zn)be given by
α(x) = ux +a , β(x) = ux +b
where (u, n) = 1 and where nis divisible by both aand b. Then αand β
are equivalent if and only if (u−1, a) = (u−1, b).
Proof. If αand βare equivalent, then by Fact 4.1 there exist w, c in Zsuch
that (w, n) = 1 and wa ≡(u−1)c+b(mod n). Since nis divisible by a,
it follows that wa ≡(u−1)c+b(mod a) and b=ka −(u−1)cfor some
integer k. Thus bis divisible by (u−1, a). Hence (u−1, b) is divisible by
(u−1, a). By symmetry, we conclude that (u−1, a) = (u−1, b).
Conversely, assume that (u−1, a) = (u−1, b) = d, say. Let a1=
a/d, b1=b/d, r = (u−1)/d. Then (a1, r) = (b1, r) = 1. Therefore the
26
congruence aiξ≡b1(mod r) has a solution ξthat is necessarily relatively
prime with r. The sequence (ξ+kr :k= 0,1,2,···) consists of solutions
of the given congruence and it contains infinitely many primes. Therefore,
one of these solutions w, say, is a unit mod n. Thus there exists wsuch that
(w, n) = 1 and b1=wa1+cr. Multiplying by d, we have b=wa +c(u−1).
By Fact 4.1, αand βare equivalent.
We summarize in Corollary 4.5 the conclusions obtained in Theorems
4.3 and 4.4. In the statement of Corollary 4.5 we let
U={u∈ {1,2,···, n −1} | (u, n) = 1}.
Corollary 4.5 Let α, β ∈B(Zn)be given by
α(x) = ux +a , β(x) = vx +b
where uand vare in U. Then αand βare equivalent if and only if u=v
and (u−1, a, n) = (u−1, b, n), where (−,−,−)is the greatest common
divisor of the three numbers.
Consequently, every conjugacy class in B(Zn)has a unique representa-
tion of the form x7→ ux +a, where u∈U, and where both u−1and nare
divisible by a.
Theorems 4.6 and 4.7 yield the explicit formula given in Corollary 4.8
for the number of conjugacy classes in B(Zn).
Theorem 4.6 Let Ψ(n)denote the number of conjugacy classes in B(Zn).
Then Ψis multiplicative in the sense that Ψ(rs) = Ψ(r)Ψ(s)for all relatively
prime positive integers rand s.
Proof. If rand sare relatively prime, then the rings Zrs and Zr×Zs
are isomorphic by the Chinese remainder theorem. By Theorem 1.6, the
groups B(Zrs) and B(Zr)×B(Zs) are isomorphic. Denoting the number of
27
conjugacy classes of a group Hby µ(H) and using the fact that µ(H×K) =
µ(H)µ(K), we see that
Ψ(rs) = µ(B(Zrs )) = µ(B(Zr)×B(Zs))
=µ(B(Zr)) µ(B(Zs)) = Ψ(r) Ψ(s),
as desired.
Theorem 4.7 Let Ψ(n)denote the number of conjugacy classes in B(Zn),
and let pbe a prime. Then
Ψ(pe) = pe−1−1
p−1+pe.
Proof. According to Corollary 4.5, Ψ(pe) is the number of ordered pairs
(u, a), where
1≤u < pe,(u, pe) = 1 , a|pe, a|(u−1).
Let Sbe the set of pairs that satisfy these conditions, and let Sk, 0 ≤k≤pe
be those pairs (u, a) in Sfor which a=pk. Then S=∪e
k=0Sk. Also, it is
clear that if k≥1, then
(u, a)∈Sk⇐⇒ a=pkand u= 1 + rpkwhere r= 0,1,···, pe−k−1.
Thus card (Sk) = pe−kif k≥1. Also
(u, a)∈S0⇐⇒ a= 1 and uis a unit mod pein {1,2,· · · , pe}.
Thus card (S0) = φ(pe) = pe−pe−1. Therefore
Ψ(pe) = card (S) = card (S0) + card (S1) + ···+ card (Se)
=pe−pe−1+pe−1+pe−2+···+ 1
=pe−1−1
p−1+pe,
as desired.
28
Corollary 4.8 Let n=pe1
1···pek
kbe the factorization of nas a product of
distinct prime powers. The number of conjugacy classes in B(Zn)is
Ψ(n) = Ψ(pe1
1)···Ψ(pek
k),
where
Ψ(pei
i) = pei−1
i−1
pi−1+pei
i,
for each iwith 1≤i≤k.
Example 4.9 The group B(Z9) has order 54 and by Theorem 4.7, B(Z9)
has 3−1
3−1+ 32= 10 conjugacy classes. Representatives for these conjugacy
classes are
•x7→ x+ 9, the identity element.
•x7→ x+ 3, with x7→ x+ 6 as conjugate, so a conjugacy class with 2
elements.
•x7→ x+ 1, with x7→ x+u,u∈ {2,4,5,7,8}, as conjugates, so a
conjugacy class with 6 elements.
•x7→ 2x+ 1, a conjugacy class with 9 elements.
•x7→ 4x+ 3, with x7→ 4x+ 6 and x7→ 4x+ 9 as conjugates, so a
conjugacy class with 3 elements.
•x7→ 4x+ 1, a conjugacy class with 6 elements.
•x7→ 5x+ 1, a conjugacy class with 9 elements.
•x7→ 7x+ 3, a conjugacy class with 3 elements.
•x7→ 7x+ 1, a conjugacy class with 6 elements.
•x7→ 8x+ 1, a conjugacy class with 9 elements.
29
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[5] M. M. Dowlen, The fixed subring of some groups of ring automor-
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30
