7 Representation theory, modules, extensions, derivations, and complements
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Background Material
49
In this book, mainly in Chapter 7, we shall attempt to cover only those
parts of computational representation theory that have applications to the
analysis of the structure of finite groups. This does not require that the
reader have a very deep knowledge of representation theory, but it does
involve certain topics that are not always covered by the most elementary
books on the subject, such as representations over finite fields rather than
over so it is not easy to find references at the appropriate level. Two
possibilities are the books by Isaacs [Isa76] and Rotman [Rot02], which we
shall use here for references to theoretical results.
2.7.1 The terminology of representation theory
Let us briefly review the basic definitions and results from the representation
theory of finite groups. Let K be a commutative ring with 1, and let G be a
finite group. The group ring KG of G over K is defined to be the ring of
formal sums
with the obvious addition and multiplication inherited from that of G. In
fact KG is an associative algebra with 1 and thus it is a ring with 1 and a
module over K. It is also known as the group algebra of G over K.
Let M be a right (unital) KG-module. We shall write m.x (m∈M, x∈KG)
to represent the module product in M, but when x∈K and we are thinking
of M primarily as a K-module, then we may write xm rather than m.x.
Since K is commutative, this does not cause any problems. From the module
axioms, and the fact that (m. g) .g -1=m for m∈M, g∈G, we see that
multiplication by a group element g∈G defines an automorphism of M as a
K-module. So we have an associated action ϕ:G→AutK(M), and we shall
sometimes use the group action notation mg as an alternative to m.g.
Conversely, if M is a K-module, then any action ϕ:G→AutK(M) can be used
to make M into an KG-module.
We shall always assume that M is finitely generated and free as a Kmodule, and so, after fixing on a free basis of M, we can identify M with Kd
for some d. Then, using the same free basis of M, AutK(M) can be identified
with the group GL(d, K) of invertible d×d matrices over K. So the action
homomorphism ϕ is ϕ:G→GL(d, K), which is the standard definition of a
representation of G of degree d over K.
According to basic results from representation theory, two KG-modules
are isomorphic if and only if the associated representations ϕ1, ϕ2 are
equivalent, which means that they have the same degree and there exists
␣∈GL(d, K) with ␣.ϕ2(g)=ϕ1(g).␣ for all g∈G.
When K is a field, a KG-module and its associated representation is
called simple or irreducible if it has no proper nonzero KG-submodules. (A
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slightly different definition of irreducibility is normally used when
for example.)
Since we are dealing only with finite-dimensional modules, any such
module M has a composition series, (that is, an ascending series of submodules
with simple factor modules), and the Jordan-Hölder theorem (Thereom 8.18
of [Rot02]) asserts that any two such series have isomorphic factors, counting
multiplicity, and so we can refer to the composition factors of M.
For any KG-module M, we can define the K-algebra EndKG(M) of
endomorphisms of M (= KG-homomorphisms from M to M). This is also
known as the centralizing algebra of M and its associated representation. It
contains the scalar automorphisms, which form a subalgebra isomorphic
to K. When K is a field and M is a simple KG-module, then Schur’s lemma
(Theorem 8.52 [Rot02] or Lemma 1.5 of [Isa76]) says that EndKG(M) is division
ring. This can be noncommutative in general, but in this book we shall be
particularly concerned with the case when K is a finite field, in which case
a well-known theorem of Wedderburn (Theorem 8.23 of [Rot02]) tells us
that EndKG(M) is a field, and it can be regarded as an extension field of K.
When K is a field, the KG-module M is called absolutely irreducible if it
is irreducible and remains irreducible when regarded as an LG-module for
any extension field L of K. By Theorem 9.2 of [Isa76], M is absolutely
irreducible if and only if EndKG (M) consists of scalars only.
So, if K is a finite field and L=EndKG (M), then we can use the action of L
on M to make M into an LG-module with dimL(M)|L:K|=dimK(M), and M is
absolutely irreducible as an LG-module. In particular, there is a finite
extension L of K for which all irreducible LG-modules are absolutely
irreducible, and such an L is called a splitting field for G.
2.7.2 Semidirect products, complements, derivations,
and first cohomology groups
This and the following subsection contain a very brief description of the
first and second cohomology groups of groups acting on modules, insofar as
they are relevant to the (computational) study of group extensions. For a
more complete treatment, the reader may consult Chapter 10 of [Rot02],
particularly Sections 10.2 and 10.3. But, in common with the majority of
published material on this topic, the account in [Rot02] is in terms of left
modules, whereas ours uses right modules, so there will be some differences,
such as in the definitions of cocycles.
We defined the notion of a (split) extension of one group by another in
Subsection 2.54. Let G and M be groups, and suppose that we are given a
homomorphism ϕ:G→Aut(M). We define the semidirect product of M by G
using ϕ to be the set G×M endowed with the multiplication (g, m)(h, n)=
(gh, mhn), for g, h∈G, m, n∈M, where, as usual, mh is an abbreviation for
mϕ(h) The standard notation for a semidirect product is
or
.
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Background Material
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The semidirect product is an extension of M by G, using the maps
defined by
It is a split
extension, with complement {(g, 1M)|g∈G}.
Conversely, if the group E has a normal subgroup M with a complement
G then any e∈E can be written uniquely as e=gm for g∈G, m∈M, and
gmhn=ghmhn, so we have:
PROPOSITION 2.70 Any split extension E of M by G is isomorphic to the
semidirect product
where the action ϕ of G on M is defined by the
conjugation action of a complement of M in E on M.
In general, different complements could give rise to different actions ϕ.
However, if M is abelian, then the actions coming from different
complements are the same. We shall assume for the remainder of this
subsection that M is abelian and use additive notation for M.
We shall also assume that M is a K-module for some commutative ring
K with 1. This is no loss of generality, because any abelian group can be
regarded as a -module just by defining n.m=nm for n∈ , m∈M. In the
case when M is an elementary abelian p-group, we can take K to be the
field .
As we saw in Subsection 2.7.1, an action ϕ:G→AutK(M) of G on the Kmodule M corresponds to endowing M with the structure of a KG-module,
and so we can talk about the semidirect product
of the
, using additive
KG-module M with G. The multiplication rule in
notation in M, is (g, m)(h, n)=(gh, mh+n).
A general left transversal of the subgroup
isomorphic to M in
has the form T={(g(g)) | g∈G}, for a map :G→M.
Then T is a complement of
in
if and only if (g, (g)) (h,
(h))=(gh,(gh)) for all g, h∈G or, equivalently,
(†)
If M is a KG-module, then a map :G→M is called a derivation or a crossed
homomorphism or a 1-cocycle if (†) holds. Notice that by putting h=1G in
(†), we see that (1G)=0M for any derivation .
We denote the set of such derivations by Z1(G, M). By using the obvious
pointwise addition and scalar multiplication, we can make Z1(G, M) into a
K-module. We have proved:
PROPOSITION 2.71 If M is a KG-module, then the set T defined above is
a complement of in
if and only if ∈Z1(G, M).
Notice that for a fixed m∈M, {(g, 0M)(1,m)=(g, m–mg)|g∈G} is a complement
of
in
, and so
is a derivation. Such a map is called a
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principal derivation or 1-coboundary. The set of all principal derivations is
denoted by B1(G, M) and forms a K-submodule of Z1(G, M).
DEFINITION 2.72 The first cohomology group H1 (G, M) is the quotient
K-module Z1(G, M)/B1(G, M).
From the discussion above, it follows that H 1 (G, M) is in one-one
in
correspondence with the set of conjugacy classes of complements of
The following result tells us that derivations are uniquely determined
by their action on a generating set of a group.
PROPOSITION 2.73 Suppose that
extends to a derivation :G→M, then:
(i)
(ii) Let g∈G with
and let : X → M be a map. If
for all x∈X.
where each εi=±1. Then
where
PROOF By (†), we have
(ii) is proved by repeated use of (†).
when εi=1 or –1; respectively.
which proves (i).
In general, given any map :X→M, we can use (i) and (ii) of the above proposition
to extend to a derivation FX→M, where FX is the free group on X.
2.7.3 Extensions of modules and the second
cohomology group
Let E be any extension of an abelian group M (regarded as subgroup of E)
by a group G. So we have an epimorphism ρ : E → G with kernel M. For
g∈G, choose ∈E with ρ( )=g and, for m∈M, define
Since M is
abelian, this definition is independent of the choice of g, and it defines an
action of G on M. In general, this action makes M into a G-module, but if
M happens to be a module over a commutative ring K with 1, and the
conjugation actions of g∈G define K-automorphisms of M, then M becomes
a KG-module. In particular, this is true with
in the case when M is
an elementary abelian p-group.
DEFINITION 2.74 Let G be a group and M a KG-module for some
commutative ring K. We define a KG-module extension of M by G to be a
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Background Material
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group extension E of M by G in which the given KG-module M is the same
as the KG-module defined as above by conjugation within E.
Given E as above, the elements { |g∈G} form a transversal of M in G. For
g, h∈G, we have
for some function :G×G→M, where the
associative law in E implies that, for all g, h, κ∈G,
A function :G×G→M satisfying this identity is called a 2-cocycle, and the
additive group of such functions forms a K-module and is denoted by Z2(G,
M).
Conversely, it is straightforward to check that, for any ∈Z2(G, M), the
group E={(g, m) g∈G, m∈M} with multiplication defined by
is a KG-module extension of M by G that defines the 2-cocycle when we
choose =(g, 0).
A general transversal of M in E has the form =(g, (g)) fora function
:G→M, and it can be checked that this transversal defines the 2-cocycle
+cx, where c is defined by c(g, h)=(gh)–(g)h-(h). A 2-cocycle of the form
c for a function :G→M is called a 2-coboundary, and the additive group of
such functions is a K-module and is denoted by B2(G, M).
Two KG-module extensions E1 and E2 of a KG-module M by G are said
to be equivalent if there is an isomorphism from E1 to E2 that maps M1 to
M2 and induces the identity map on both M and on G. From the above
discussion, it is not difficult to show that the extensions corresponding to
the 2-cocycles 1 and 2 are equivalent if and only if 1-2 ∈ B2 (G, M) and,
in particular, an extension E splits if and only if its corresponding 2-cocycle
∈ B2(G, M).
DEFINITION 2.75 The second cohomology group H2(G, M) is the quotient
K-module Z2(G, M)/B2(G, M).
So H2(G, M) is in one-one correspondence with the equivalence classes of
KG-module extensions of M by G.
Checking directly whether a given 2-cocycle is a 2-coboundary can be
difficult, but there is an alternative approach to deciding whether a KGmodule extension E splits, which we shall now describe.
Suppose that we have a finite presentation
of G, and let us identify
G with the group F/N defined by the presentation, where F is the free
group on X and
. For each x∈X, choose an element
with
Then there is a unique homomorphism θ:F→E with
for
all x∈X and, since ρθ(x)=G x, ρθ is the natural map from F to G. Hence
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Handbook of Computational Group Theory
ker(ρθ)=N. It follows that ker(θ)≤N in any case and, by Theorem 2.52,
ker(θ)=N if and only if θ induces
in which case
is an
isomorphism and
is a complement of M in E.
Furthermore, ker(θ)=N if and only if θ(w)=1E for all w∈R, so we have proved
the following lemma.
LEMMA 2.76 If
are chosen with
and θ:F→E is defined by
then the elements generate a complement of M in E if and only if
θ(w)=1E for all w∈R.
The elements θ(w) for w∈R always lie in ker(ρ)=M. Let
be another choice of the inverse images of x under ρ, where :X→M is a
map, and let θ:F → E be the associated homomorphism with
. If we use (i) and (ii) of Proposition 2.73 to extend to
: F → M, then a simple calculation shows that θ(w)=θ(w)(w) for all w∈R.
Since the elements in these equations all lie in M, we can switch to additive
notation and write them as θ(w)=θ(w)+(w). So we have the following
result.
PROPOSITION 2.77 With the above notation, the elements
generate a complement of M in E if and only if (w)= -θ(w) for all w∈R.
We shall use this result later, in Section 7.6, to help us determine
computationally whether an extension splits.
In particular, from the case
and
for all x∈X,
Proposition 2.71 yields:
THEOREM 2.78 If M is a KG-module with
then a map :X→M
extends to a derivation G→M if and only if (w)=0M for all w∈R, where
:F→M is defined by (i) and (ii) of Proposition 2.73.
This last result is the analogue of Theorem 2.52 for derivations. It can of
course be proved directly from the definition of derivations, without involving
semidirect products. It will be used later in Section 7.6 to help us to compute
Z1(G, M).
2.7.4 The actions of automorphisms on cohomology
groups
Two KG-module extensions E1 and E2 of M by G can be isomorphic as groups
without being equivalent as extensions. This remains true even if we
restrict our attention to isomorphisms that map M to M.
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Background Material
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Example 2.8
Let
with G acting trivially on M. It is
not difficult to check that there are eight equivalence classes of extensions
of M by G (in fact H2(G, M)≅C2×C2×C2), in which each of x2, y2 and [x, y] can
be equal to 1M or t. However, there are only four isomorphism classes of
groups E that arise, namely C2×C2×C2, C4×C2, D8, and Q8.
Suppose that ␣:E1→E2 is a group isomorphism mapping M to M, and inducing
µ ∈ AutK(M). Then, since E1/M E2/M G, ␣ also induces v∈Aut(G). We
have
for all m∈M, g∈G, because both expressions result from
applying α to -1m , where ∈E1 maps onto g∈G.
DEFINITION 2.79 Let M be a KG-module. If µ∈AutK(M) and v∈Aut(G) are
isomorphisms satisfying
for all m∈M and g∈G, then (v, µ) is
called a compatible pair.
This expression was introduced by Robinson [Rob81]. It was first used in
computational group theory by M.J.Smith in Section 4.2 of [Smi94], in
connection with the computation of the automorphism groups of solvable
groups; we shall return to that theme in Section 8.9.
The set Comp(G, M) of compatible pairs forms a group under composition,
and is a subgroup of Aut(G)ìAutK(M). If Z2(G, M) and (v, à) is a compatible
pair, then we can define (ν,µ) by the rule
for all g, h∈G. It is straightforward to check that (ν,µ)∈Z2(G, M). Indeed, the
mapping
defines an automorphism of Z2(G, M) that fixes B2(G,
M) setwise, and so it induces an automorphism of H2(G, M).
So Comp(G, M) induces a group of automorphisms of H2(G, M). From
the above discussion, it can be shown that the isomorphism classes of
KG-module extensions of M by G (where we are restricting attention to
isomorphisms that fix M) correspond to the orbits of Comp(G, M) on
H2(G, M).
Exercises
1.
2.
If G=Cn and
, then show that the smallest splitting field for G
containing L is Fqr where r is minimal with qr≡1 (mod n).
Let M be an KG-module, where G is finite with |G|=n.
(i)
(ii)
Let ∈Z1(G, M). By considering
n∈B1 (G,M).
Let ∈Z2(G, M). By considering
that n∈B2(G, M).
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show that
for h, k∈G, show
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(iii) Deduce that if the map
of M is an automorphism of M,
then H1(G, M)=H2(G, M)=0. This holds, for example, when M is finite
with |M| coprime to n.
3.
4.
5.
Calculate Comp(G, M) and its action on H2(G, M) in Example 2.8.
Show that Comp(G, M) induces a naturally defined action on H1(G, M).
Let M be a KG-module defined via ϕ:G→AutK(M).
(i) If ϕ is a faithful action, then show that Comp
(ii) If ϕ is trivial, then show that Comp(G, M)=Aut(G)×AutK(M).
6.
.
If M is a KG-module defined via ϕ:G→AutK(M), and ν∈Aut(G), then we
v
can define a KG-module M via the action
(i)
Verify that
defines an action of Aut(G) on the set of
isomorphism classes of KG-modules.
(ii) Check that (ν,µ)∈Comp(G, M) exactly when µ is a KG-module
isomorphism from M to Mv. Hence, for ν∈Aut(G), there exists (ν,
µ)∈Comp(G, M) if and only if
2.8 Field theory
For a detailed treatment of the material in this section, see any book on
abstract algebra.
2.8.1 Field extensions and splitting fields
If F is a subfield of K, then K is said to be a (field) extension of F. The degree
[K:F] of the extension is defined to be the dimension of K as a vector space
over F.
If F≤K≤L, then we have [L:F]=[L:K][K:F]. This is proved by showing
that for bases [ui] 1≤i≤[K:F]] and [vj| 1≤j≤[L:K]] of K over F and L over K,
[uivj| 1≤i≤[K:F], 1≤j≤[L:K] ] is a basis of L over F.
An element α∈K is said to be algebraic over F if f(α)=0 for some
polynomial f∈F[x]. Otherwise α is transcendental over F. If [K:F] is finite,
then all elements of K satisfy polynomials of degree at most [K:F] over F,
and so are algebraic over F.
In any case, F(α) is defined to be the subfield of K generated by F and a;
that is, the intersection of all subfields of K that contain F and α. The
elements of F(α) are all quotients f(α)/g(α) with f, g∈F[x], g≠0.
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If α is algebraic over F, then the set of all f∈F[x] with f(α)=0 forms an
ideal, and hence a principal ideal (p) of F[x], where p can be chosen to be
monic. Then p is called the minimal polynomial of a over F. Since F has no
zero divisors, p must be irreducible over F.
If p is the minimal polynomial over F of α∈K, then (p) is the kernel of
the ring homomorphism :F[x]→K in which (␥)=␥ for all ␥∈F and (x)=α.
Hence im () F[x]/(p). Since p is irreducible, (p) is a maximal ideal of F[x]
and so F[x]/(p) is a field and
with [F(a):F]=deg(p).
On the other hand, if F is any field and p∈F[x] is an irreducible
polynomial, then K:=F[x]/(p) is a field, which can be thought of as an
extension of F of degree deg(p) by identifying F with its natural image in K.
If we define α to be the image x+(p) of x in K, then p(α)=0. Hence α is a root
of p in K, and p factorizes in K as (x–α)q(x) for some q∈K[x].
More generally, if f∈F[x] has an irreducible factor p of degree greater
than 1, then p factorizes nontrivially in the extension field K=F[x]/(p). By
repeating this construction, we can show by a straightforward induction
argument on the largest degree of an irreducible factor of f, that there is
an extension field K of F, with [K:F]≤ n!, in which f factorizes into deg(f)
linear factors.
If K has this property, and no proper subfield of K containing F has this
property, then K is called a splitting field of f over F.
It is an important result that if K and KЈ are two splitting fields of f
over F, then there is a field isomorphism from K to KЈ that fixes every
element of F. Here is a very brief outline of the proof. Use induction on
the minimum of [K:F] and [KЈ:F], let α∈K, αЈ∈KЈ be roots of the same
irreducible factor p of f with deg(p)>1, observe that F(α) F[x]/(p) F(αЈ),
and then apply the inductive hypothesis to K and KЈ regarded as splitting
fields of f over F[x]/(p).
The characteristic char(F) of a field F is defined to be the smallest integer
n>0 such that n1F=0F, or zero if there is no such integer n>0. So the familiar
fields , and all have characteristic zero. If char(F)>0 then, since F has
no zero divisors, char(F) must be a prime p.
It is easily shown that a polynomial f∈F[x] has repeated roots (that is,
repeated linear factors) if and only if gcd (f, fЈ)≠1, where fЈ is the derivative
of f. If F=K, then gcd(f, fЈ)=1 in F[x] if and only if gcd(f, fЈ)=1 in K[x], so we
can use this condition in F[x] to check whether f has repeated roots in its
splitting field.
If f is irreducible then gcd(f, fЈ)=1 if and only if fЈ≠0, which is certainly the
case when char(F)=0. If char(F)=p>0, then fЈ=0 if and only if f is a polynomial
in xp.
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2.8.2 Finite fields
The main result on finite fields is that all finite fields have prime power
order and that, for each positive prime power q=pn, there is, up to
isomorphism, a unique finite field of order q. We shall outline the proof of
this fact in this subsection. Although it is only defined up to isomorphism,
and can be constructed in different ways, it is customary to regard ‘the’
finite field of order q as a fixed object, and to denote it by .
Let K be a finite field. Then we must have char(K)=p>0, and the subset
forms a subfield of K, which is isomorphic
to the field Fp of integers modulo p. In particular, all fields of order p are
isomorphic to Fp.
The degree n:=[K:F] must be finite, and then q:=|K|=pn is a prime power,
so all finite fields have prime power order.
The multiplicative group K# of K\{0K} has order q-1, and hence αq–1=1K
for all α∈K#, and αq-α=0K for all α∈K. So K contains q distinct roots of the
polynomial xq-x∈F[x]. Clearly no proper subfield of K can have this property,
so K is a splitting field of xq–x over F. It now follows from the uniqueness of
splitting fields that all fields of order q are isomorphic.
To prove the existence of a field of order q for any prime power q=pn,
let K be the splitting field of xq–x over F:=FP. The derivative of xq–x is–1,
which is nonzero so, as we saw earlier, xq–x has q distinct roots in K. It is
easily checked that, if α, ß are roots of xq–x in K then so are α±ß, αß, and
α/ß if ß≠0K, so the set of these roots form a subfield of K which, by
minimality of the splitting field, must be equal to K itself. So |K|=q, as
required.
PROPOSITION 2.80 Any finite subgroup of the multiplicative group of a
field K is cyclic. In particular, the multiplicative group of a finite field is
cyclic.
The proof of this depends on the following result from group theory.
The exponent of a group G is defined to be the least common multiple of
the orders of its elements or, equivalently, the least n>0 such that gn=1
for all g∈G.
LEMMA 2.81 If G is a finite abelian group of exponent n, then G has an
element of order n.
PROOF Let
elements gi of order
as required.
for distinct primes pi. Then G must have
for each i. Since G is abelian, g1g2…gr has order n,
Note that the final step in the proof of the lemma is not necessarily true
for nonabelian groups G, and indeed the lemma itself is not true in general.
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To prove the proposition, let the finite subgroup H in question have
order n. If H had exponent m
by all elements of H and hence have n>m distinct roots, which is impossible.
So H has exponent n, and the result follows from the lemma.
An element a of multiplicative order q-1 in Fq is called a primitive element
of Fq. Clearly Fq=F(α) with F=Fp, and so the minimal polynomial f of α over
F must be of degree n, where q=pn. An irreducible polynomial of degree n
over of which the roots are primitive elements of Fq is called a primitive
polynomial.
(This meaning is distinct from and unconnected with the meaning of a
primitive polynomial over as one in which the greatest common divisors
of the coefficients is 1. This clash of meanings is unfortunate, but since the
concept of a greatest common divisors of field elements is trivial, there is
probably little danger of confusion.)
It is easily verified that the map x→xp defines an automorphism of Fq of
order n (it is called the Frobenius automorphism, and generates the
automorphism group of Fq, but we shall not need that fact). So, if f is a primitive
polynomial of degree n over F with root α∈Fq, then the n elements in the set
are all roots of f in Fq. Hence Fq is a splitting field of f.
If w is a primitive element of Fq then, for 0≤k
only if gcd(k, q-1)=1, and so the total number of primitive elements is Φ(q1), where Φ is the Euler Phi-function. Each primitive polynomial f has n
roots in Fq so there are a total of Φ(q-1)/n primitive polynomials.
2.8.3 Conway polynomials
Although Fq is unique up to isomorphism, it can, and often does, arise as
the splitting field of many different irreducible polynomials of degree n
over Fp. For computational purposes, it is useful to agree on a standard
primitive polynomial, so that different computer algebra systems can use
the same representation of the elements of Fq. Unfortunately, there appears
to be no natural mathematical way of choosing such a standard polynomial.
The standard that has been generally agreed upon is known as the
Conway polynomial for Fq. (This is an unfortunate choice of name, because
there is another meaning of Conway polynomial in knot theory!) They
were originally introduced by Richard Parker, who also computed many
examples. To define them, we first need to define an ordering on the set of
all polynomials of degree n over F=Fp, and it is here that an apparently
arbitrary choice had to be made.
We order Fp itself by 0<1<2<…
is mapped onto the word αn-αn-2…α1α0, and the resulting words are ordered
lexicographically using the above ordering of Fp.
© 2005 by Chapman & Hall/CRC Press