EFFECTIVE WEDDERBURN AND APPLICATIONS

A thesis presented to the faculty of San Francisco State University In partial fulfilment of A *5 The Requirements for 3 ^ The Degree

Master of Arts * f(p[p ^ Mathematics

by

Justin Fong

San Francisco, California

December 2017 Copyright by Justin Fong 2017 CERTIFICATION OF APPROVAL

I certify that I have read EFFECTIVE AND APPLI­

CATIONS by Justin Fong and that in my opinion this work meets the criteria for approving a thesis submitted in partial fulfillment of the requirements for the degree: Master of Arts in Mathematics at San Fran­ cisco State University.

3 - Joseph Gubeladze Professor of Mathematics

Professor of Mathematics

Dustin Ross Assistant Professor of Mathematics EFFECTIVE WEDDERBURN AND APPLICATIONS

Justin Fong San Francisco State University 2017

We propose a new algorithmic proof of the classical Wedderburn’s theorem in the representation theory of finite abelian groups. Namely, the algebra isomorphisms, known to exist from the classical theory, are explicitly found. This leads to a new succinct resolution of the well-known isomorphism problem for group rings and an alternative description of the automorphisms of these algebras. Employing ideas from algebraic geometry, the global variety of all representations of fixed dimension for a finite group is introduced and its geometric properties are explored. The last step involves an extensive computer assisted analysis of the resulting high dimen­ sional polynomial ideals.

I certify that the Abstract is a correct representation of the content of this thesis.

Chair, Thesis Committee ACKNOWLEDGMENTS

I would like to thank Dr. Joseph Gubeladze for allowing me to start working on this project during my senior year as an undergraduate and allowing it to grow to become my Master’s thesis. His guidance and patience was very helpful for the past few years. I would also like to thank my committee members Dr. Serkan Hosten and Dr. Dustin Ross for their comments and corrections while writing this thesis. Their help was much appreciated.

v TABLE OF CONTENTS

1 Introduction...... 1

1.1 Preliminaries ...... 3

1.1.1 Ring Theory Preliminaries...... 4

1.2 Group Rings...... 9

1.2.1 Generalities...... 9

1.2.2 Maschke-Wedderburn...... 13

2 Effective Proof Wedderburn’s Theorem for Commutative Group Rings . . . 16

2.1 Vandermonde M atrix...... 16

2.2 Effective Wedderburn...... 19

3 Finite Modular Group Rings...... 29

3.1 The Isomorphism Problem ...... 31

4 Automorphisms of G roups...... 38

4.1 Generators and Relations of Algebras...... 39

4.2 The p-group C a se...... 44

4.3 The General Finite Abelian Group Case ...... 50

5 Modules Over the Group Ring /c[G ]...... 55

6 The Variety of A^Gj-modules...... 65

6.1 Dimension Theory of Varieties...... 67

vi 6.2 Irreducible Components of Varieties...... 70

6.3 Representation Varieties...... 71

6.3.1 The Free-Algebra Representation Variety...... 80

6.3.2 Computer Assisted Computations...... 84

Bibliography ...... 86

vii LIST OF TABLES

6.1 Underlying field k has characteristic z e r o ...... 85

6.2 Underlying field k has positive characteristic...... 85

viii 1

Chapter 1

Introduction

Algebraic objects known as group rings forms a blend of group theory and ring the­ ory. They were first conceived in the middle of the 19th century, where they were mentioned in the same paper that gave the first definition of an abstract group, due to Arthur Cayley. It was not until later that group rings were recognized to be the objects that established the connection between group representation theory and the structure theory of algebras, making them vital objects in representation theory where it is known that there is a one-to-one correspondence between group representations and modules over the group ring of these groups. In the 1940s group rings started to become a serious topic of study in their own right, beginning with the isomorphism problem of group rings, and since then they have found applica­ tions to many branches of algebra such as field theory, algebraic number theory, algebraic topology and so on [10]. Our main motivation for this paper is studying the well known classification theorem of Artin-Wedderburn from representation the­ 2

ory in the context of group rings, which are natural objects to apply this theorem.

As mentioned before, group rings have many applications to different areas and in our work we demonstrate this phenomenon by drawing ideas from field theory, al­ gebraic geometry, finite dimensional algebras, and representation theory to fuel our investigations.

This thesis work evolves around Wedderburn’s theorem for finite abelian groups and closely related representation theory topics. Let be a finite abelian group.

The main results are:

1. A new effective proof of the Wedderburn isomorphism k[G] = where

k is an algebraically closed field of characteristic 0 (chapter 2). This leads

to explicit isomorphisms between the two /c-algebras. Only the existence was

known from the Wedderburn theory. The recovery of these isomorphisms relies

on an iterated use of the Chinese Remainder theorem and the method of proof

is therefore algorithmic in nature.

2. In the modular case, i.e., when char(/c) = > 0 and is a p-group, we obtain

a new presentation of k[G] as a monomial quotient of a polynomial ring over k.

A well known isomorphism theorem (3.4) from the 1960s says that the modular

case is exactly opposite to the characteristic 0 case [5]: different groups have

non-isomorphic group algebras. Using the new monomial presentations, we

derive a very quick alternative proof of the same result by invoking a general

result from [8] on monomial quotients. 3

3. In Section 4, we explicitly describe all fc-algebra automorphisms of the group

ring k[G]. The main result (Theorem 4.7) is an amalgam of the two comple­

mentary (i.e., characteristic 0 and modular) cases, which need to be considered

separately.

4. In Section 5, in the modular case, we interpret the representations of G in

terms of nilpotent matrices (Theorem 5.5). We may reduce these nilpotent

matrices to upper triangular form, however in the general multivariable case

this assumption cannot be made (Theorem 5.6).

5. In Section 6, we develop a moduli space approach to the representation theory

of G. Namely, for every natural number d, we introduce the global variety

of d-dimensional G-representations over k. After reviewing relevant algebro-

geometric concepts, we conclude with a computer aided analysis of these rep­

resentation varieties (dimension, number of irreducible components). Com­

putations are carried out for small examples of G, with use of the software

programs Macaulay2 [7] and Singular [4]. The results are summarized in

Tables 6.3.2 and 6.3.2.

1.1 Preliminaries

We introduce and develop the main definitions and theorems that are central to this paper. Throughout R will denote a commutative ring containing the multiplicative 4

identity 1 and G a finite abelian group unless otherwise stated.

The order of a group G is denoted by |G|. A cyclic group of order n is denoted as Zn (the integers modulo n). A p-group is a group of prime power order pn. The following is a useful classification of finite abelian groups.

Theorem 1.1. (Structure Theorem of Finite Abelian Groups) LetG be a nontrivial finite abelian group of order n> 1. Then

G — Z “1 © ZF2 a2 © • • • © Z "t “i

such that pi < P2< • • • < Ptand n = pTpV where each pi are (not necessarily distinct) primes and the ai ’s are positive integers.

1.1.1 Ring Theory Preliminaries

The characteristic of a field F, denoted char(F), is the smallest positive integer p such that p ■ lp = 0. A ring R is called a graded ring if it is the direct sum of additive subgroups: R = Ro© R\© R^© • • • such that RtR3 C Ri+j, where

RiRj = {xy | x e Ri and y e Rj}.The following tells us how we can break a quotient ring into a direct product.

Theorem 1.2. (Chinese Remainder Theorem) Let Ii, / 2, • • •, be ideals in a ring

R, which are pairwise comaximal (i.e, R + Ij = R for i j ). Then the map 5

/ : R — > R/Ii x R/I2 x • • • x R/Ik

ri—* (r + /1; r + ..., r + is a surjective ring homomorphism with ker(f) = D H • • • fl = • • •

Therefore, R/(RR • • ■ R) = R/Rx R/R x • • • x

The Frobenius map F : R — > R is defined by F(r) = rp.

Lemma 1.3. R be a ring of characteristic p. The Frobenius map is a ring homomorphism. The nth composition of the Forbenius map, ) = is also a ring homomorphism.

Proof. Let F : R — > R be defined by F(r) = rp. This map is a ring homomorphism.

Let a, b G Rthen F(ab) — ( ab)p = apbp = fand F(a+b) = ( = =

F(a) + F(b), the latter follows from the fact that the binomial coefficients of the middle terms are multiples of p. It follows that the nth composition of F is a ring homomorphism as well. □

Lemma 1.4. If two k-algebras A and B are isomorphic then the k-algebras k ® k A and k ®k B are isomorphic as k-algebras.

Proof. Let f : A —> Bbe a /c-algebra isomorphism. Define the mapping

ip : k A k®kB 6

E ,(k ® at i—> ® i = l i= l From the way it is defined

n (f( c • ' 2 2 ® ai) = v(D cc<)® ai) = ® ^ (a*) = c-^ V i=i / \j=i / t=i

Multiplication on the tensors is preserved as seen below

n m \ / n m ' f "22Ci ® ( E dj ® j = v? ( ® M j ) i= 1 j = l / \ i = 1 j = l / n m

= E m ) ® M M j ) i= 1 j = l n m dj ® 0(6j »=i j=i n = v? Ci ® ai J ip dj ® bj

From the way it is defined is clearly surjective. To prove tp is injective we show that ker ip = {0}. Let 537=1 ci® a% ^ ker tp and

n \ n

p ( E j C{® ai) =E , Ci® ^ ) = k i = 1 / i = l 7

But because 0 is injective each 0(aj) are distinct, implying that each C; ® 0(a,-) are distinct, so Ci®0(aj) — 0 for i = 1,..., n(at least one can be zero while q = 0).

Therefore ker<£> = {0}, hence we have proven k ®k A = k ®*. B as ^-algebras. □

We review some basic facts concerning cyclotomic polynomials, which will be important in describing further isomorphisms of the type of quotient rings seen in

Proposition 1.7.

Definition 1 .1 . The nth cyclotomic polynomial is defined to be the polynomial whose roots are the primitive nth roots of unity (over Q), i.e.,

= n <*-o l

In order to express the polynomial xn —1 in terms of the cyclotomic polynomial we need the following.

Proposition 1.5. Let R(n) = {£, ( 2, ..., £n} be the set of nth roots of unity and let P(d) = | gcd(6, d)= 1 where 1 < b < be the set of primitive dth roots of unity. Then

R(n) = and P{d\) D P(d2) = 0 whenever d\ ^ d2. d\n 8

Proof. We prove the second statement first. For distinct divisors di and of n, let

£ G P(d\) fl P(c?2) then

(-di _ 2 — Q&2 ^ 1-^2 _ 2

But di — c?2 < d,2 . contradicting the fact that £ is a dfl primitive root of unity.

Therefore P(di)nP(c?2) = 0- Now to prove the equality of the two sets let Q G for some d \ n. Then

Q = ( & ' d = r '* =! = * & £ B(n).

Next, let £9 G P(n) where 1 < q < n.Since R(n) is also a finite cyclic group and we know that the order of every element of R(n) divides its order, each element belongs to a cyclic subgroup of P(n), i.e., (£9) < R(n). Let d = nj gcd(n, q) and

[Cq)d = (C9)”/ gcd(n,

Also (£9)fc 7^ 1 for k < dsince |£9| = 0.cd")? ■ Therefore C € P{d) and we have proven the equality. □

Lemma 1.6. xn - 1 = Ildin^W-

The proof of this lemma follows from observing that the roots of 1 are the nth roots of unity and then applying the previous proposition. 9

1.2 Group Rings

1.2.1 Generalities

Definition 1.2. Agroup ring (or group algebra) of an arbitrary group G and a ring

R is the set R[G]consisting of formal sums agg, where 6 and ag G

Addition in R[G] is defined as:

^ ag9 + ^ bgg = ^ + bg)g. geG geG geG

Multiplication is defined to be:

E agd)(£2 bhh) = g€.G h(zG gh£G htzG

where c:k = J2gh=kagbh- Together, as one can check, these operations make R[G] into a ring.

When the ring R is a field F the group ring also has the structure of an A1-vector space, and the group elements of G forms a basis. Thus the dimension of F[G] is

|G|. The group ring also has the additional structure of an algebra, thus is the reason for the alternative term group algebra when emphasizing this structure on

F[G], 10

The next three statements will be important for our proof of Theorem 1.13. For cyclic groups, we have an explicit description of their group rings.

Proposition 1 . 7 . The group ring of a cyclic group over a ring R is isomorphic to a truncated polynomial ring, i.e., R[Zn\ R[X]/(xn — 1).

Proof. We define the map (p between R[X] and R[Zn] as

4>: R[X] — ► R[Zn\

f (x)/(e)> where f(x) = J/r/o akxk and e € Zn is some generator, is well-defined since xk (->• ek and extends by linearity as above, i.e., each element in n] is unique since

1, e,..., en_1 is a basis for this group ring. This also implies that is surjective. d> a ring homomorphism follows from usual polynomial addition and multiplication.

Next we claim ker = (xn — 1). Let f(x) G ( — 1) =>• f(x) = ( — l)s(x) for some s(x)G R[X]. Then

Euclidean algorithm: f{x) = ( x n— 1 )q(x) + r(x), where deg r(x) or r(x) = 0.

Then cp(f(x)) = /(e) = 0 => r(e) = 0 b0 + b\e -I + 6n_ = 0

bo = bi = ••• = 6n_i = 0, because 1, e,..., en~1 are linearly independent.

Therefore f(x) G (xn — 1) and our claim follows. Since we know

Im

When the underlying group and ring of a group ring are direct products the next two lemmas tells us how they can be simplified.

Lemma 1 .8. Let R and S be rings and G be arbitrary group. Then ( x S)[G] =

-R[G]xS,[G']. In general, for a collection rings we have (R\X- ■ =

R^G) x ---x R k[G\.

Proof. Define the map (R x S)[G] — > R[G] x S[G] by

( E TT1(r,s)gg, ^ 7r2(r,s)g^ , g£G g€G g€G where (r, s)9 € R Sx and wi : R x S —> R, 7r2 : i? x S' —>• S'are projection maps, i.e., 7ra(r, s)g= rg, 7r2(r, s)g = sg. This map is well-defined because the elements of

G form a (possibly infinite) basis for their corresponding group rings and so each element in the above mapping is unique. It is straightforward to verify that this map is an isomorphism. The general case can be proven inductively. □

Lemma 1.9. Let R be a ring and G\ and G2 be arbitrary groups. Then f?[Gi©G2 = i?[Gi][G2]. More generally, we have R[G\ © • • • © Gr] = R[G\ © • • • © Gr_i][Gr].

Proof. We define ip: R[Gi© G2] — » #[Gx][G2] by

5 ^ aM (g,h)^J2{J2 a(g,h)9^ h (g,h)(zGi®G2 /1GG2 g€-G\ where a{gM£ R. The map ip is well-defined due to the elements of G acting as the basis of each group ring and so each element is uniquely expressed. Since the 12

group ring elements closely correspond to polynomials in two variables proving that ip is a ring homomorphism follows easily from addition and multiplication of those elements. It is also apparent that ipis bijective. □

When dealing with group rings over arbitrary fields we can always extend the field of scalars to algebraically closed fields using tensor products.

Lemma 1.10. For a group ring k[G]one has k k = as k-algebras.

Proof. Observe that each element of k k k[G] is of the form

m ^ ® (augi H----- + ^ni9n) ? i=l where |G| = n. But this can be rewritten as

m m Y^ C-i <8>ipligi + • • • + anign) = <8> (flli^l) + ‘ ' • + Cj ® i=l i= 1 m n

= y Y ^ a^ ® i-1 j—i n = Y X j ® S j .7=1 where Xj Ek. Now, define the map

® 9j ► 5 ] j=i i=i

Checking that this is a Ac-algebra morphism is almost no different than that in the previous lemma. It is obviously surjective and to show it is injective let X o;=1Aj® = iA gj (z ker

(p ( Xi® 9j I = Y 1 Xigi = °' 0=1 / j=i Then each Xj = 0, because the group elements gj are linearly independent, so each

A j 0 g 1 = 0 0 g :j = 0 and we have ker = {0}. Therefore cp is a A;-isomorphism. □

Thus, whenever k[G] = k[H] over a field k of characteristic p, then k 0*. k[G] = k 0 fc k[H] = » k[G]= k[H}.

1.2.2 Maschke-Wedderbur n

An /^-module M is called semisimple if every submodule N of is a direct sum­ mand, that is, there is a submodule Pof Msuch that M © A ring R is called semisimple if it is semisimple as a left module over itself.

Theorem 1 .1 1 . (Maschke’s Theorem) LetG be a finite group and k be afield. Then the group ring k[G]is semisimple if and only if the characteristic of k (including char(/c) = 0)does not divide the order of G.

We now introduce the main theorem mentioned in the title of this paper. 14

Theorem 1 .1 2 . (Artin-Wedderburn’s Theorem) Let semisimple k-algebra

(not necessarily commutative). Then R = nj=i M i,(A ), where ) is a matrix ring of Hi x rq matrices with entries from Du a finite dimensional division algebra over k.

The positive integers r, rq, and division algebras Di are uniquely determined by R up to isomorphism. Each matrix ring ) is called a Wedderburn com­ ponent while the direct product above is called the Wedderburn decomposition of

R. The next statement is a special case of Wedderburn’s theorem in the setting of commutative group rings.

Theorem 1.13. (Wedderburn’s Theorem for finite group rings) Let G be a finite abelian group and k an algebraically closed field of characteristic 0 (pa prime) such that p does not divide |G| when p > 0. Then = k ^ as k-algebras.

This result easily follows from both Maschke’s and Wedderburn’s theorem above since char(fc) { |G'| we have that k[G] is semisimple and by Wedderburn’s the group ring k[G is] a finite direct product of matrix rings over Di. However, each division

A;-algebra Dl is a finite field extension of k and hence must be algebraically closed.

But k is already algebraically closed and we must have that Dt = k for each i. Also, since k[G] is commutative it must be the direct product of 1 by 1 matrix rings over the field k, thus we have k[G] = kn where = |G|. Although the proof of this statement was a simple consequence of Wedderburn’s theorem we present a more constructive proof in chapter 2, which is one of the main results of this paper. 15

We make a side note that the underlying field must be algebraically closed in order for the statement of Theorem 1.13 to hold. Otherwise, we get a coarser decomposition for our group ring as demonstrated in the following example.

Example 1 .1 . Let Q be the field of rationals and Z3 be a cyclic group of order three. We compute Q[Z3]. It is known that Q[Zy = Q[X]/(:r3 — 1), however the polynomial x3 — 1 over Q only factors as 1 = (x l)(x2 + + 1), so

Q [X ]/(x3 — 1) = Q[X]/ [x — 1) x Q [X ]/( x2+x + ). It is clear that Q [X]/(x — 1) = Q.

Since the polynomial x2 + x + 1 is irreducible over Q we have by field extensions

Q[X]/{x2 + x + 1) = Q(C), where ( = js primjtive 3rd root of unity.

Therefore Q[Z3] = Q x Q(C) ^ Q x Q x Q .

There is in fact a more general result of Theorem 1.13 due to both to S. Perlis and G. Walker referenced in [10].

Theorem 1.14. Let G be a finite abelian group of order n and F be a field with characteristic not dividing n. Then

d\n where (d is the primitive dth-root of unity, ad = > andnd = £ G 1} |. 16

Chapter 2

Effective Proof Wedderburn’s Theorem for

Commutative Group Rings

2.1 Vandermonde Matrix

The following provides a clean description of the inverse of a Vandermonde matrix whose entries are the nth roots of unity. It is in fact the matrix of the discrete

Fourier transform [3, pg. 276], which happens to be one of the explicit maps we will recover in our proof of Wedderburn stated in part (1) of Theorem 2.2 below. This matrix’s use will become apparent in the proof of that the aformentioned theorem. 17

Proposition 2.1.

1 1 1 1

i C c2 • • cn— 1 V = 1 c2 c4 £2(n-l)

C"-1 c2(n_1) .. c(n-1)2/

be the vandermonde matrix of nth roots of unity (. Then i inverse matrix is

V -1 = -Y H, where V ;/ is the conjugate transpose of V . i - 1 is £-i+1 II for 1 < i < n. Then the conjugate transpose of V is:

1 1 i i \

1 ^n-1 £2(n-l) £(n-l)2 1 \ H = 1 to ^2(n—2) £(n-2)(n-l)

V1 CC2 ■•• C” - 1 )

Now let ^ ^ be the kth entry of the ith row of V and let 1^n fc+1^ be the kth entry of the j thcolumn of V H, where 1 <

matrix product V V ^ is,

k= 1

If i = j) then £(fc-1)(*-i)+(.?-i)(n- fc+1) = i)—(t— 1)—ft(j—1)+0—i) _ 1) _ ^

If i^ then ^(A:_1)(i_1)+t/- i)(«-fc+i) = ^(fc-i)(»-i)-(fc-i)(j-i) —

Since 1 < i — j < n,then is just another nth root of unity. So using the finite geometric series formula,

V (■<*-»«-» = 1 - ^ = x~ x = o i-~is i _ (i-j i _ ri-j k= 1 ^

Therefore we have

fc=i I 0 , ± j

It follows that V V " = nl V(-^V” ) = I. Because V is invertible we conclude that V - 1 = ±VH. □ 19

2.2 Effective Wedderburn

Let G be a finite abelian group and k an algebraically closed field of characteristic p

(possibly zero) not dividing |G| = nin2 • ■ • nt, where each n, is a prime power. By the structure theorem of finite abelian groups we have the cyclic group decomposition

G = Z ni © Z n2 © • • • © Znt. The elements of the group ring Zn] are of the form

71—1 m = E die1, e e Zn a generator, »=o while elements of k [ZTO © Zn] are of the form

71—1 771—1 ) = E E dij(x\ yJ), xG Z n, y G Z m are generators. 7 = 0 j = 0

Let Cm, Cn ^ ^ be primitive mth and nth roots of unity respectively and let

71—1 771—1 9(eV) = E (E “«<®)c 7 = 0 j=0 be the evaluation of a polynomial g(s,t) € k[s,t], where a = 0,1,..., n — 1 and

/5 = 0,l,...,m — 1 . More generally, elements of k[Zni ffi Zn2 © • • • © Znt] are of the form ni — 1 nt —1

i 1==0 it —0 and evaluation of a multivariate polynomial at the roots , • • • > C 's 20

n\ — 1 nt — 1

ii=0 it—0 We now present our proof of Theorem 1.13, but only demonstrate it for the direct product of two cyclic groups as the general case becomes rather complex, however it follows the same pattern.

Theorem 2.2. Let G = Zni © Zn2 © • • • © Znt. The explicit k-algebra isomorphisms of k[G] = k^G are obtained as follows:

1.The k-algebra isomorphism ofk[ Zn] = kn, for a positive is given by

the map

2. The k-algebra isomorphism of k[Zm © Z„] = kmn, where m = n2 and n — n\,

is an iteration of the previous one and is given by

h(i—>

3. Continuing to reiterate the previous maps the isomorphism of

k[Zni© Zn2 © • • • © Znt] = kC\ is

: h(x i, ...,xt) 21

4- The k-algebra isomorphism of k[ Znin2...nt] = © Z„2 © • • • © is given

by the composition y?~V.

Proof. (1) We may assume that k = C since the field of complex numbers is alge­ braically closed and has characteristic zero. We first prove the existence of (p. Propo­ sition 1.7 implies that C[Zn] = C[X]/ (xn— 1). The polynomial 1 splits over C by the fundamental theorem of algebra. More specifically, we have — 1 = f|d|n by Lemma 1.6 and together with Proposition 1.5, — C ) where ( is an nth root of unity. For each maximal ideal C C[X] it can be shown that they are all pairwise comaximal, i.e., (x — + (x — (f) = C[X] for all i ^ j.

Choose 1 and -1 in C[X] such that (x - Q) + (~l)(x has an inverse in the ideal (x — (f) + (x — Q).It follows (x — Q) + (x — Q) = C[X] for all i^ j.Now, by the Chinese Remainder theorem

C[x\i(x" - 1) = q x ] / (n>-c*)) =nc[v]/(*-c). / ' i=0 ' i=0

Next, observe that C [X]/(x — Qt) = C for each = 0,..., n — 1 . This follows from defining the map

C[X] — > c

n —1 f(x) =Y2akxk I—> f(Cn), k=0 which is a surjective ring homomorphism whose kernel is the ideal ( — C C[X]. 22

The result follows by applying the first isomorphism theorem. Therefore we have

n—1 7i—l c [ z n] s c[x}/{xn - 1) = n c w /(;r - c )= nc=c x ■ ■ ■x c = c ”- i—0 2=0

By keeping track of the explicit maps involved in the chain of isomorphisms

above we obtain the map

p : C[Zn] — ► Cn

fie)' * (/(!)> /(Cn). •••./(C "-1))-

It is apparent that this is a C-algebra isomorphism, due to it being a composition

of other C-algebra isomorphisms.

(2) Using Lemmas 1.8 and 1.9 we obtain

C[Zm©Zn] “ C[ZTO][Zn] “ (C x • • • x C)[Z„] = C[Zn] x • • • x C[Zn] s Cn x • • • x C"y = Cmn, ' V ' V------v------' ------m m m

which proves the existence of our isomorphic map.

To obtain our map we carefully keep track of the explicit maps used in the chain

of isomorphisms above as follows:

C[Zm © Zn] — > C[Zm][Zn]

n—1 m — 1 Ti—1 h(x, i y)—¥ i= 0 j = 0 2=0 23

C[Zm][Zn] > (C x • • • x C)[Zn]

n—1 n—1 J2fi{y)xr 1 > 1))®* 2=0 2=0

(C x • • • x C)[Zn] — > C[Zn] x • • • x C[Zn]

n—1 n—1 n—1 E (/*«■ *«"■)- • • • mc~')V /-(c1)*') 2=0 2=0 2=0

C[Z„] x • • • x C[Zre] — > Cn x • • • x C"y = Cmn V v------' '------^------" m m (g(x , i 3(i, c r 1)) >-> (»(i,i) S(cr\i) 9(i.cr,),---,9(C‘ 1, a _1)), where yk i-* Cn between C[Zn] and Cn for all 0 < h < n — 1. We thus have obtained the map 24

Cmn /»(*, y) — > (y(i, i),..., y ( c r i ) , ..., i, c r 1),...,

(3) The isomorphism : C[Zni © Zn2 © • • • © ZnJ i—> ClG' in the general case is obtained in a similar process through an iteration of the previous map with each cyclic group component Zni added and stops at the t-th step.

(4) Using the maps

C[Zmn] by composing

C[Zm © Zn] — > C[Zmn]

To describe the inverse ip~l let us observe the matrix representation (ip) of

The matrix M(ip) turns out to be the square Vandermonde matrix consisting of the mnih roots of unity £ and is of size mn x mn. Using Proposition 2.1 we calculate the inverse of M (p) as 25

1 1 1 \

y £rrm—1 ^2(mn-l) ^(mn-1)2

= — £mn—2 ^2(mn-2) ^(mn- 2)(mn-1) mn

^mn-1 V1 C

The matrix representation of 4> with respect to the basis {(V , yJ) | 1 < < m, 1 < j < n} of C[Zm © Zn] and the standard basis of Cmn is presented as the block matrix

/ \ Mn Mi 2 A /l mn

M21 A/22 M 2 m n M(0) =

\ AI pin \ A/mn2 • • • A/mnmn J where each block is an m x n minor of the form: 26

i-l > (»-1)C j- 1 ) \ an Sn

ri—1 /•i—l/-(n—1)0—1) 'sm S>n

>2(i 1) /*2(i 1) — 1 Mij — Sm Sm Sn Sm

There are mn of these minors that partitions the ran x mn matrix M(

To describe the isomorphic map p~l

C[Zmn] we take the matrices of both cp_1 and (j) then multiply them, The isomorphism

C[Zni...n(] = C[Zni © Zn2 © • • • © Znt] is done similarly by iterating the previous steps. □

We illustrate the above method with a simple example.

Example 2.1. Consider the groups Z4 and Z2 © Z 2. By Wedderburn’s theorem we know that C[Z4] = C4 = C[Z2 © Z 2] and hence the isomorphisms between them exits. We have the isomorphisms

ip : C[Z4] — ► C4

f(x) i—> (/(!),/(-!),/(*),/(-*))»

where f(x) = ao + a%x + a2ar + a%x and

$ : C[Z2 © Z 2] — > C4

ao(l, 1) + 1) + a2( 1,x) + (23(x,x) i—>

(ag + ai + a2 + a3, ao + ai — a2 — a3, ao — c l\ + a2 — a3, ao — ai — a2 + a3).

Taking the matrix representations of both ip and (f> with respect to the standard basis of C4 28

1111 'l 1 1 ^

1 i—1 11-1-1 , M(0) = 1-11-1 1-11-1

1 — i — 1 i 1 - 1 - 1 1

we obtain the isomorphic map C[Z2 ® Z2] — > C[Z4] by first inverting M() to get

^1 0 0 0 ^

0 ^ 0 ^ M ( ^ V ) = 0 0 10

(1-0 , 0 ^ 0 2 /

This matrix represents the map

Chapter 3

Finite Modular Group Rings

When the field k has characteristic dividing the order of the group G we notice that Wedderburn’s theorem for finite group rings breaks down, since the group ring

/c[G] is no longer semisimple by Maschke’s theorem. In this setting k[G] is called a modular group ring. However, k[G]can be decomposed into a much coarser direct product of objects, namely the truncated polynomial rings. For the rest of this section let k be any field of characteristic 0.

Lemma 3.1 . The group ring k\Lvn] of a c is isomorphic to the truncated polynomial ring {ao + ajy 3-----+ 1 | a* € as k-algebras.

Proof. We have the isomorphism k[Zpn] = k [ X ] / — 1). Since k is of characteristic p > 0 we have xpn — 1 = (x — l)p" and so 30

By a change of variable 1,

*[*]/((* - l)p“) = k[Y + !]/(/”) = k[Y\l(y”") and this is the truncated polynomial ring. □

Theorem 3.2. LetP be a finite abelian p-group Then k[P] is isomor­ phic to the multivariable truncated polynomial ring i,..., X m]/(x^ 1, as k-algebras.

Proof. By the structure theorem of abelian groups we have k[P] = j Zph\. We induct on m, the number of cyclic group summands. When 1 this is Lemma

3.1. Suppose that for m — 1,

m — 1 ^ ^ Z pe{ = k[Xi,... ,Xm-i]/(x\ , ...,x^l_1 ). i= 1

Then

m 771—1

k ^ Zpei Zpei J [Zp em ]

i—1 2 — 1

“ (k[Xlt . . . , X m_ l ] / ( z f , . . . , x C i 1 )) [V » ]

- ( V i , ..., Xm-il/Crf,..., a C T 1)) [*m] 31

where the second to last isomorphism follows from the mapping gl xlm for all

<7* G Zpem . im

Theorem 3.3. Let G be a finitely generated abelian group |G| = such that gcd(p, n) = 1 . Then the group ring k[G] is isomorphic to the n-fold product of a truncated multivariate polynomial ring.

Proof. Let Hbe a subgroup of order n and P the sylow p-group of order pr such that G — ©H P .Using all of our prior knowledge of the isomorphisms of group rings we have

k[G\ = k[H® P] = k[H][P] = ( ]“ (k[P})n “ An

where A = k[Xi,..., X m\/{x\ 1,..., ) such that e\-4------1- em = r, and k[P] = A by Lemma 3.2 . □

3.1 The Isomorphism Problem

Recall that when working over fields whose characteristic didnot divide the order of the group, we always hadthe fact that: for any two groups G and H of equal order, k[G] = k[H], even when G H^ .This was of course due to Wedderburn’s theorem. However, this is no longer true when char(/c) = p | ]G’ |:

Example 3.1. The non-isomorphic p-groups Zp2 and ZP®ZP do not have isomorphic group rings over k. 32

Proof. We know that k[Zp2] = k[X]/(xp2)and © Zp] are the truncated polynomial rings in one variable and two variables respectively. The

/c-algebra k[X]/(xp2) has the nilpotent element x such that = 0, but xn 7^ 0 for all positive integers n < p2. We claim that k[U,V]/(up,vp) does not possess any nilpotent elements that share this same property. So, suppose that it does.

Then there exists a nilpotent element a Ek[U, V]/(up, vp) such a p2 = 0 and a :" ^ 0 for any n < p2. Notice that every nilpotent element in this fc-algebra is a linear combination of non-zero powers of u and v with no non-zero constant coefficients, since a field element is never nilpotent. That is a = Ciullvn H + cnulrivh\ where

.... cn E k and each power of u and v ranges between 1 and — 1. However,

ap = (< Z\Unv-71 -| + cnulnvin)p = (c\Unv^)p -\ + {cnulnv^n)p = 0

because up = 0 = vp.This contradicts our assumption that an 0 for all n < p2.

Therefore k[Zp2]¥k[Zp ® Zp\. □

Example 3.2. The groups Zp© Zp© Zp2 and Zp2 0 Zp2 do not have isomorphic group rings over k.

We should mention that the method used in the previous example will not help us prove the non-isomorphism above, due to the fact that there are nilpotent elements with smaller powers than p2 in both of these /c-algebras. We therefore must search for a different kind of invariant of the truncated polynomial rings, that of the prime 33

power nilpotent subspaces, i.e,

Vt := ker(F*) ={a e A| F^a) = api = 0},

where A is a truncated polynomial ring and F : A — > A is the Forbenius map.

Proof. Let k[X, Y]/(xp2, yp2) = A = k[Zp2 © Zp2] and = 6 = 0} be a subspace of A. The basis elements of V\are of the form

{ xayb \ p < a < p2 — 1, 0 < b < p2 — 1} U {x°yd 0 < c < 1, 1}, and therefore has dimension dimfc \\ = (p2 — p)p2 + p(p2 — p) = p4 — p2.

As for £;[Zp©Zp©Zp2] let W\ be thep-nilpotent subspace of Y, Z} / ( , , zp2).

The basis elements of W\ are of the form

{xaiyblzCl | 1 < a\< p —1, 0 < b 1, 0 < ci < 1}

[J {xa2yb2z°2| ai = 0, 1 < b2 < p — I, 1 < c2 < 1}.

Therefore its dimension is dim/. W\ = (p — 1 )pp2 + 1) 1) = 1.

Both V\and W\have different dimensions, implying that /c[Zp© Zp© Zp2] ^ /c[Zp2©

Zp2], □

The two groups in example 3.2 are both isomorphic to a finite abelian p-group P of order p4. In fact, by the structure theorem, there are five different isomorphism 34

types that P is isomorphic to, with each type corresponding to one of the five dif­ ferent partitions of 4. We list them as follows,

Z p4

Zp2 © Zp2

Z p © Zp3

Z p © Z p© Zp2

Z p © Z p© Z © Z

None of these five isomorphism types are isomorphic to each other, but what about their /c-algebras? Example 3.2 already disproved two of them, so lets check the remaining p-nilpotent subspaces’ dimensions.

Vi C k[Zp4\ : dimfc Ei = p4 — p3

Ei C k[Zp2© Zp2] : dimfc Ei = p4 — p2

Ei C k[Zp© Zp3~j: dim/,, Ei =p 4 — p2 35

Vi C k,[Zp © Zp© Zp2] : dimfc V1 = p4 - p

V\ C A:[ZP © Zp © Zp ® Zp] : dim*, V\ = p4 — 1

All of the dimensions are different except for those of fc[Zp2©Zp2] and fc[Zp©Zp3].

In this situation we determine the dimensions of the p2-nilpotent subspaces. They are dim*, V2 = pA — 1 fork[ Zp2 © Zp2] and dim*, = p4 —p for k[Lp © Zp3]. Therefore k[Zp2 © Zp2] ^ /c[Zp © Zp3]. We now conclude that none of the A'-algebras of the isomorphism types of order p4 are isomorphic to each other.

In general, for any finite abelian p-group G of order pn, there are p(n) (the number of partitions of n) many isomorphism types of G, none of which are isomorphic to each other. Their corresponding truncated polynomial rings are not isomorphic to each other as well, which follows from a special case of the well-know isomorphism problem of modular group rings.

Theorem 3.4. LetG and H be finite abelian p-groups. Ifk[G] = ] as k-algebras, then G = H.

Remark. This statement was proven in [5] with the assumption that A; is a finite field of p elements. The details of the proof involves calculating the dimensions of the kernel of powers of the Frobenius map F on both group rings, which forms a unique sequence of numbers associated to each group ring that is invariant to it. This proof still holds in the truncated polynomial setting we have developed up to now 36

and we briefly sketch it as follows: Let k[G] ,..., xm]/(xf , x^ m) and k[H] = ,..., k[yi y n\/ {y\ 1,..., y£n) where m = n. Assume that ai < 02 < • • • < and b\ < b2< • • • < bm.Then the dimension of the kernel of FK (the iteration of the Frobenius map) on k[G] is calculated to be dim (ker = pd, where d is the number of times at repeats in ai < 02 < • • • < an.This gives a unique sequence

(c?i,..., ds), of numbers di = logp(dim(ker Fh)), which are invariant to k[G}. Hence we must have G = H.

More generally, one can consider for any field k the monomial fc-algebra k[x 1,..., xm}/(Mi,..., Ms)where each M% is a monomial in the variables xi,..., xrn and each are of degree at least 2. This algebra becomes a truncated polyno­ mial ring whenever each M% = In [8] it was proven that for two collections of variables {rri,..., xm} and {yi,..., yn} such that 1,..., xm]/(Mi,..., = k[yi, • • •, Vnl/iNx,...,Nt)siS A:-algebras and {x1;..., xm} n (Mi,..., = 0,

{jt/i,..., yn} D (iVx,..., iVt) = 0, then m = and there exists a bijection be­ tween {xi,..., xm} and {y\,..., yn} that sends the monomial ideal (Mi,..., Ms) to (N\, ..., Nt).It turns out that we can apply this more general isomorphism problem to the modular group rings above by first expressing those group rings k [G] and k[H] in the hypothesis of Theorem 3.4 as truncated polynomial rings. This then solves the isomorphism problem for modular group rings, giving an alternative proof in addition to the original one stated above.

We state necessary and sufficient conditions for the group rings of two finitely 37

generated abelian groups to be isomorphic to each other. Let G\ and be finite abelian groups of equal order. By the structure theorem of abelian groups consider their decomposition G\ = Hi © P] and G2 = Ho © P2, where Pi, P2 are p-groups whose order is divisible by p and H i, H2 are the parts whose order is coprime to p.

Theorem 3.5. We have k[G\] = k[G2] as k-algebras if and only if Pi = P2 and

\Hi \ = \H2\.

Proof. Suppose that k[Gi] = k[G2}. Then by first breaking of the parts of the groups that is coprime to p we have

= * k[Pi© Hi] “ © H2]

= 4> ( fc [ A ] ) |Hl1 = (k[P2]) W

=$> k[Pi] = k[P2], by Theorem 3.4

= * Pi = P2

=>\Pi\ = \P2\=pr

We now have |Gh| = |G2| =*>• prm = prn => where |Pi| = \H2\ = , and p is coprime to both m and n. Now, suppose we have Pi = P2 and \Hi\ = \H2\.

It follows

k[Gi] “ A:[Pi 0 Hi] “ (fc[Pi])|Hl1 = ( 21 S and the statement has been proven. □ 38

Chapter 4

Automorphisms of Groups

In this section, we give a description of the automorphisms of the finite group rings k[G\ in the three scenarios involving the characteristic of and the order of G we have seen earlier:

1. char(fc) | |G|

2. char(fc) { |(7|

3. char(fc) | \H\and char (A:) \\K\, where = ®

When wanting to emphasize that the automorphisms are compatible with the

/c-algebra structure of our group rings we add the subscript k in the notation of the automorphism group of k[G] as follows: Autfc(/c[G]). Before describing these automorphisms we first review some known facts about the generators and relations of algebras. 39

4.1 Generators and Relations of Algebras

Let k be an arbitrary field and consider a fc-algebra

A := k[xi, ..., xn] , ..., xn) = 0,..., (.Xi,..., 0, where {xi,..., xn} a generating set of A and /i(xi,..., xn) = 0,..., i,..., ) =

0 are the polynomial relations that the generators must satisfy. We point out that this construction of an algebra with generators and relations is equivalent to the quotient of a ring, where the ring modulo by one of its ideals corresponds to the relations that the generators of the ring must satisfy, i.e. ,

k[x1}..., xn]/(F1: ...,Fr)=k[xi, ..., ®»], /i(xi, • • •, 0,....., /r(xi,..., = 0

where Fj = fi(xi,..., xn) is fromk[x \,..., xn] and i 6 r}. Because is a field A is also compatible with a vector space structure.

Proposition 4.1. Let A = k[x\,..., x^/ipc.^1... be a k-algebra. The set of monomials {x^1 • • • x c*| < dj, i = 1,..., n) forms a basis of A.

Proof. It is immediate that this set spans A. Now suppose ax<\ '' ‘ x

k[xi,..., £n]}. The expression ^ ax)1 ■ ■ ■ x))1 = 0 is equivalent to this sum belonging to the ideal (x)*1,..., x^n), i.e.,

aXl• ' ■ Xn = d\+ • • • + 9nXdn

We can gather together all the monomials containing an x f in the left hand side of

(4.1), however this term will not equal qxf4 because q < and there are no Xj’s of power di in the left hand side of (4.1). Therefore ^ • • • «n ^ ( > unless all a =0, which proves linear independence. □

Now let a =

f : A — > A

The above map is completely determined by the generators of because

/ being a /c-algebra homomorphism means that these generators can be multiplied together under / and thereby obtaining the basis (x)1 • • • | c, < dt} of A, which can then be linearly extended to any element of A. However, because the polynomial

Proposition 4.2. Let * a = / ( a q ) for1 < < The endomorphism is well-

defined if and only if one has fi(oti,..., an) = 0,..., i,..., an) = 0.

Proof. Suppose that / is well-defined. Each of the expressions i,..., xn) = 0

are linear because they are the relations of the ring ,..., xn] so applying / to

both sides yeilds,

fj{a lt . . . ,On) = f j ( f ( x i), . . . , fM ) = f(fj{x 1, . . . , Xn)) = /(0 ) = 0.

Each Oj = f(xi) in the r many relations of A will be the same due to / being a

well-defined map.

On the other hand suppose Oi,..., ansatisfies all the relations of A. Consider

the commutative polynomial algebra k[x\,.., over k. Then by the universal

property of A;-algebras there is a unique map 7r :k\x \,..., — > A, which is

the projective homomorphism that sends xt to its coset image (one of the gen­

erating elements of ^4). Recall that A = k[xj,, xn] /{F\,... ,Fr). Clearly ker

7r = (F\...., Fr)and the kernel of the map 0 ; k\xi,..., xn] — > A contains ker n.

So ker ip factors uniquely through the projection n, i.e., there is a unique A:-algebra

endomorphism / such that ip — f o tt. □

k[xi,... ,x n\ A f A 42

We are interested in this general discussion of algebra generators and relations because it will aid us in describing the automorphisms of the group ring of a p-group over a field of characteristic p. As a demonstration consider A)[Zp0l © • • • ® Zpo„] with char (A:) = p, which comes with the relations g\ 1 = 1,..., g^‘" = 1 . However, describing the elements of this group ring that the generators gt must map to and showing that they satisfy the above relations can be rather difficult, since it is not so clear how one goes about finding the right group ring element so that raising it to the appropriate power equals the group identity 1. To make things manageable recall that

k[Zpai© ••• © Zpan] = k[Ult.. ■, Un]/ {xC ,..., x C ) :=

where U\ = pi — 1,..., un — gn— 1 and 1 = 0,..., i/tf n = 0. Re-expressing our original group ring in this form with these slightly new set of relations makes it easier to check whatever automorphisms we define on the generators than before, due to the fact that generators when raised to the appropriate power now equals zero.

We begin our investigation of the automorphisms of h[G] by first consider the case when the characteristic of k is relatively prime to \G\. It will be easier to describe the automorphisms of fc[G] by focusing on its Wedderburn decomposition kn instead of having to deal with linear combinations of group elements. Let Autj be the group of h-algebra automorphisms of kn where char(/c) { |G'| = n. 43

Theorem 4.3. The only automorphisms of kn are the ones that permute the basis elements ofk n.

Proof. Define / : kn — > kn on the standard basis e i,... ,en of by = e0 and

/(a ) = a for all a G k.This map is well-defined since it is also a linear transformation and can extend uniquely. Verifying that / is a A;-algebra homomorphism is routine so we omit it. Now, to prove that there are no other types of automorphisms suppose that € Autfc(/c"). One only needs to consider how (p acts on the standard basis e i ,..., e n.For any one of the basis elements e* we have =

((e;))2, which implies that

(p is an algebra isomorphism it must also respect products so,

0 = (f)(0) = 0 ^ ) = 0(ei)<^(ej) = (]T es)(J^ et).

S t

But the right hand side above need not be equal to zero because there may be nonzero components in corresponding indices of both sums, due to the fact that the tuples are of length n. This however cannot happen and therefore must permute the standard basis elements. □

Thus we have that Autfc(A;n) = Sn (the symmetry group on n elements) in the 44

simplest case.

4.2 The p-group Case

With Proposition 4.1 we know that the A;-algebra A is a graded ring, i.e., A

/c© A ® A © - • ‘©A/ such that A A C A +j, where and d = (• is the maximum degree of all monomials. Each A is an additive subgroup of all homogeneous polynomials of degree i.Before describing the automorphisms of the algebra A it will be convenient to first examine a special case where all the nilpotent power relations of A are the same. This will lead to a clearer picture of the general case.

Proposition 4.4. Consider the k-algebra A := ,..., 0,..., = 0.

The map f : A — > A is a k-algebra automorphism if and only if f defined on the generators x* are of the form f ( xi) — anx i H \-ainxn where pi is the higher degree part and the coefficients from the linear part defines an invertible n x n matrix

(dis­ proof. Suppose / € Autfc(.4 ). Since A is a graded ring we have for each generating element Xj G A, f(xf) = an + H + apr-_x where € Since char(fc) = p we have

f(xi)p = (q;i + 02 + • • • + a pr-i)p — © © ' ' ‘ © l — 0 45

where a0 =0 and each a1’ = 0 because all monomials in has pr as its power relation. Therefore the expression f(x i)pr 0 is satisfied. Now let ® ©

A2 © • • • © Apr_1. Notice that /(x ,)d for d > 2 belongs to © Ad+i © • • • © part of A and no lower degree monomials appear since each term is multiplied after distribution. Therefore we focus only on the linear part of each /(x ,) by taking A/1 for the ideal I = A2© • • • © so that the © A\ part of A is relevant for our purpose, i.e., A/1 = k© A\. Then / : A/1 — > A/I is defined by f(xi) = anX\-\ \-dinx n . But since / £ Autfc(A) we know that / -1 : — >• exists and / / _1 = id implies that the n x nmatrix (a^) with respect to x\, , xn is invertible.

Now suppose that / defined on the generators is of the form stated in the hypothesis. We know that / : A — >■ A is an endomorphism. Let us define an automorphism 'ip : A — )■ A by ip{xt) = + • • • + such that the n x matrix (6^) is the inverse of (o,jy). Now let =— > A and we wish to show that 9 is an automorphism, which will then imply that / is one too. If we define 9 on each xt then

9(xi) = ipf (.Xi) = ip{anXi 4 b ainxn + ipi)

= an'ippx 1) H- b ain'tp(xn) + -ip(tpi),

where pi £ a2©• • •ffiapr _ 1. Now ip(

the linear terms in 9. The composition above corresponds to matrix multiplication, so the linear part an'iffxf) + • • • + ainip(xn) from above is the multiplication of the n x n matrices (a^) and (btj). But this is just the identity matrix, so for each of the n generators 9(xi ) =Xi + with ^ G 0:2 © • • • © a y -i- To complete our proof that 9 is an automorphism it is enough to show that it is surjective due to A being a finite dimensional vector space. First we subtract from 9(xt) the images of all degree two monomials which appear in 9{xi) each multiplied by the appropriate coefficients, i.e.,

9{xi) - i [crfix^x ) )] for some C\, ,cnG k. This eliminates all degree two monomials in 9{xi) and at the same time introduces new monomials of higher degree. The next step is to subtract from (4.2) all the images of the degree three monomials that are present in the same way we did with the degree two monomials. This gets rid of all the degree three monomials and we continue on to the next highest degree and repeat this procedure recursively. Because pr — 1 is the highest degree, new monomial terms will eventually stop appearing in our subtractions. The final step is to eliminate any remaining terms and we are then left with Xi G lm(0). This implies that the monomial basis of A is contained in lm(0) so we have lm(0) = A. therefore 9 is surjective and hence bijective. □

Theorem 4.5. Let A := k[x\,..., xn],x\ 1 = 0,..., x^ n = 0 where the exponents d i,...,d n are arbitrary. Then G f Autk(A) if and only if f{xi) an + where 47

cq is the linear combination of those generators such that their power relation exponents dj are restricted by dj < di and the coefficients of these generators defines an invertible matrix. The expression (fii is a linear combination of all monomials

XT' ‘' x°n tf and oniy if for at least one xt in these monomials we have pdt < ctpdi.

Proof. Let / € Autfc(A). For each generator xt of A we need f(x i)pdl = 0. So assume f(xi) = aiH + ad where a0 = 0 and cq- £ Aj from

Then

f(Xiy di = (a, + a2 + • • • + adf ' a f + a f + • • • + o f = 0

==> 0% = (aij1xjl H F aijmxjrn)p 1 — H------^ ( * = 0-

But the above holds if and only if for all s such that pds 0. Therefore

Q\ = aijixjiH + a ijmx jm where di > dJm,..., dn . As for the higher degree part av2 1 + • • • + apd 1 = 0, we know a 2 + ■ ■ ■ + ad is a linear combination of monomials x°\ ’ " xn"- So we must have (xcf ■ ■ • xcjy dl = xcfp 1 ■ ■ ■ x^ pdl = 0. Knowing that there are no mixed products of x i,..., xnas relations of A if we assume that none of these C\pdi, ..., cnpdi are greater than or equal to pd\ then x\lP 1 •.. x%ipdl A 0. a contradiction. Therefore the equality is equal to zero if and only if there is at least one exponent c3pdl >pdi. We have now shown that f{xf) = ayiXj1 H 1-

Due to d i,... ,dn being arbitrary let us assume the ordering • • • < dim 48

and

di= • • • = dh

dh+i '

dir+i = • • • =

Ignoring the higher power terms of each f we can interpret / on the generators x \ ,...,x n as the n x nmatrix (a^) = M (f). The shape of this matrix will be almost lower triangular because in our ordering d^ is the smallest exponent and the linear part of f { xh)is made up of only those generators with power relation pdii. The next higher exponent is d{2 and the linear part of f ( x l2) is made up of not only those generators with pdi2 as their power relation but also includes all the linear terms of /(x^). This pattern continues on up to the maximum exponent where the linear part of f ( x im) includes all generators x 1:. .., xn. In particular, all those generators that share the same power relations * will form m-many ik x ik minors in M (f) along its diagonal. Now because we know / € Aut/C(T) its inverse

/ -1 must also exists and / / -1 = id. Interpreting these automorphisms as matrices,

= I, notice that the corresponding i\ x , x minors of the identity matrix / are themselves identity matrices that make up a block diagonal.

It follows from this that the i\ x ii,... ,im x imminors of M(f) are invertible. 49

Now suppose f(xi) is of the form as described in the hypothesis and let (a,j) denote the invertible ikx ik matrix of the linear part. Assuming dim we define each of the m-many automorphisms ., € Autfc(A) as follows: For each generator xt with dik exactly being the exponent in its power relation let

'

such that the coefficients of each xt defines an invertible x ik matrix ( ) =

(ay)-1 and for any generator Xj such that dj / dik let = x3. Now let

4/ = • • • ipimf be the composition of our maps, which is an endomorphism of A.

To prove that / is an automorphism it is enough to show that T is an automorphism.

We already know that because of the graded structure of A all higher degree terms are mapped to other higher degree terms under T, so we only need to focus on the linear part of TlA;). For Xi(1 < i

V(xi) =V>*i

= ijjil • • • + • • • + ajjXjj + ifii)

= a^i^xi) H bOi1V’ii(^ii) +^ii •

= Xi + ipi, ■ • ■

where V'u ‘ ‘ £ A2 © • • • © A<*. We already know how to deduce that each

Xi € Im(T) since T on these generators is the same situation encountered in the 50

proof of Proposition 4.4. Now for generators Xj (i\ 1 < j < with power relation exponent dj2 take

^{xj) =V>ii

— V’ii ■ • • 1pim(a lx l 4------+ + Gii+l^ii+l + ‘ ‘ ’ + + (fj)

= xi + xj +ifil

To show that each Xj G Im(T) we perform T (x-j) — T(xj) = Xj + (higher degrees) and as before we know how to eliminate the higher degree terms of this expression.

We continue in this fashion until we reach the maximum exponent dlm, where by then all generators x\,..., xnare be contained in Im(T). Therefore 4/ = • • • b’lm/ is surjective and hence bijective, implying also that / is bijective. □

4.3 The General Finite Abelian Group Case

Next we will describe all the automorphisms of the group ring k,[G] for a general finite abelian group where |Gj = pmn and p \ n. Let A = ,..., xm]/ (xf1,..., x ^ ) and

An be the n-fold product. Also let a j =

Lemma 4.6. The only idempotent elements of A are 0 and 1. 51

Proof. We first show the statement is true for A = k[x]/(xpdl). There are two cases to consider: p= 2 or p^7 2. We deal withp = 2 first. Let a0+aiX-\ b be an idempotent element of A, then

(a0 + aix H b avd1_1xpdl~1)2 = + a\x2 b

shows us that no odd powers of x appears on the righthand side above. Therefore the only idempotents can be ao, which is either 0 or 1 since is a field. Now when p ^ 2 we have

(d 0 ~b CL\X + • • • + Clpd1_iXP 1)2 = Ofo "b CL\X + • • • + Opd1_i XP 1

= b Co + C\X H b = a0 + a\X H +

where c, = a0Oi + aia^i + a2ai_2 + • • • + First we have q = a0. Assuming that a0 is not the trivial idempotent observe that

a0ai -|- — &i

a0(a0ai + aiao) = a^ai

a0ai -b O1U0 = aoai

ax = 0. 52

By induction on i assume that ai = a2 = ■ ■ ■ = apd1_2 = 0. Then

a0Vi_i aiapdi-2 + a2dpdi-3 + • • • + upd1_1ao = Qpdi-i

a o V i-i i—i ®o = i—i

CLpdi_i — 0.

Therefore a0 is the only idempotent element of A, which is either 0 or 1. As for the general case we can induct on the number of variables as follows: Let

A = k[xu .. .,x m]/(x{d\.. = A[xm]/(x^m)

where K := k[xi,..., (.Xj 1,..., xpmZl 1). Since we know that the only idem- potents of K are 0 and 1 these must also be the only idempotents of A as well. □

Theorem 4.7. Let F : An — » An be the map that permutes the standard basis elements of An . Then there exists an €f Autfc(A) such that

F : An —> An

(ax,..., an) i—> (/(ai), • • •, f(an)) is a k-algebra automorphism, where each a in the ith position and each f{af) is in the n(i)th position for some tt€ Sn. 53

Proof. Observe that unlike in the kn case: F(0,..., a*,..., 0) 7^ OjF(0,..., 1,..., 0) for (0,..., dj,..., 0) G An because a* may not be a scalar from A and F must be a fc-algebra homomorphism. In order to determine where F sends (0,..., a*,..., 0) we do the following,

(0,..., a*,..., 0) = ei(0,...,Oi,...,0).

Applying F to both sides gives

F(0,..., ..., 0) = 0,..., a*,..., 0)

F (0, . • . , Uj, • . • , 0) ^?r(i)(^l5 • • * > ^0 • * •

F(0,..., at,.. .,0) = (0,..., &»(»),..., 0).

Now consider the map / : A — >■ A defined by /(a ,) = bn^y We claim that this map is an automorphism. Since a*, bn^)G A, both of these elements can be expressed in terms of the monomial basis, so on the generators xl the map / must send each of these to other generating elements while also respecting f(xi)di = 0. However, we have already seen that in order to satisfy these relations ) can only be of the form described in Theorem 4.8. It follows that / G Autfc(A). Therefore f{af) = 54

and for any element (ai,..., an) €E A11,

F(ai,..., an) = F(ai,..., 0) + • • • + F(0, ..., an)

= (^tt(I)) • • • , 0) + • • ' + (0, . . . ,

= (/(ai),...,/(an)) with each /(a*) in their respective n(i)th position. It is immediate from above that

F is induced by / , therefore F is a A;-algebra automorphism of An. To show that there are no other automorphisms of An consider the idempotent elements of An.

The idempotent e € Anis an n-tuple consisting of only l ’s and 0’s because those are the only idempotents of A by the previous lemma. However, just like with Aut in Theorem 4.1, this is the same situation where idempotents are used to prove that

F can only permute the standard basis elements. Therefore no other elements of

Autfc(An) exists other than F. □ 55

Chapter 5

Modules Over the Group Ring k[G\

We will now study finitely generated modules over the finite group ring k[G] and see how the structure of these modules depend on the isomorphism type of k[G], that of the truncated polynomial rings. As will be seen shortly, each of these module structures correspond to certain nilpotent matrices and vice versa. For a ring R let M(i?) denote the set of all finitely generated A-modules M along with their R- module homomorphisms, i.e., M (R)is the category of finitely generated modules.

The following lemmas about modules and their vector space dimensions are known facts that will be useful in our discussion.

Lemma 5.1. Let R and S be rings. If M E M x S), then there are E M(l?) and M2 G M(S') such that M = Mi x M2 as an (R x S)-module.

Proof. Let Mi = (1, 0)M and M2 = (0 ,1)M be R and S submodules of M respec­ tively. Define the isomorphisms We must show bijectivity between these sets, i.e., = and 'tpcp = id. First consider

, m2) = ( ( 1 , 0)(mi + m2), (0, l)(m i + 777,2))

= ((1,0)777! + (1, 0)7712, (0, 1)777! + (0, l)m 2),

where mi G Mi and m2 G M2, which implies mi = (1,0)m'i and m2 = (0, l)m '2 for some m'i, m'2 G M .In the first coordinate above we have

(1, 0)mi + (1, 0)m2 = (1, 0)(1, 0)m'1 + (1, 0)(0, l)m 2 = (1, 0)m/1 + (0, 0) m2 = ma. 0

As for the second coordinate the same argument applies and we get

(0, l)mi + (l,0)m2 = m2. Therefore (f)ip(mi,m2) = (mi,m2), i.e., (j>ijj = id. For ip

we have

rjxf>{m) = (1,0 )m + (1, 0)m = ((1, 0) + (1, 0))m = (1,1= m

It follows ipcj) — id and we have proven M = x M2 as (R x 5')-modules. □

Lemma 5 .2. Let A be a finitely generated k-algebra that is also a finite dimensional vector space over k. If M G M(„4), then M is also a finite dimensional vector space over k.

Proof. Let G Mbe generators of M. Then for any m G M, m = a\m\ + • • • + apmp for some ai,... ,ap G A. However, A is also a vector space over k, say of dimension d. so let bi,... ,bd £ be a fc-basis. Therefore for any a £ A we have a = A161 + • — b \dfid, for some Ai,..., G Expanding the coefficients of m above we have

m = airai H + apmp

= (An^i + • • • + Xidbd)mi + • • • + (Xdibi + • • • + A

= An(&imi) + • • • + Aid(6

where each bimj G M. This shows M = span{6imi,..., 6ymp}. Since every span­ ning set can be reduced to a basis the remaining elements will be a fc-basis for M, hence dim*, M <00. □ 58

The following statement is a known fact from linear algebra that will allow us to simplify the nilpotent matrices. Its proof can be found in [1].

Proposition 5.3. For a nilpotent operator f on a vector space V with f n = 0, the null spaces of the powers of f forms an increasing chain of length at most n:

{0} C null(/) C null(/2) C • • • C null(/n_1) C null(/n) =

In fact, there exists a basis such that a nilpotent matrix is of (strictly) upper trian­ gular form.

We now state and prove our results of this section.

Theorem 5.4. Let k be any field and A :=0 k-algebra. Then the set of all modules from M (/l) of fixed k-vectorspace dimension d is in natural bijective correspondence with the set of all dx dnilpotent matrices of power n.

Proof. For a module M € M(A) let d = dim^ M be its vector space dimension over k (by Lemma 5.2). Let mi,... , m^ be a fc-basis of M and define a fc-linear map

M —> M by m xm for all m e M .The basis mi,..., m^ under this map gives rise to an d x dnilpotent matrix T such that = 0 due to the relation xn = 0 of

A.

Conversely, let Tbe a d x dnilpotent matrix such that Tn = 0 and let the d-dimensional /c-vector space, be the underlying set. Clearly is an abelian 59

group with respect to addition. For any v M' define Ws action on M' as

(ao T dix + • • • + an-\Xn • v = {clqI + a\T + • • ■ + an_iTn 1)(n)

= a0v + a±Tv H h an_iTn_1u,

where I is the identity matrix and a0 + a .\X + an-ix n~l £ A. This action

is well defined since each element of A is unique. We must check that this action

satisfies the rest of the axioms defining a module. Let and v,u £ For

summation of scalars

/ n— 1 \ n—1 n—1 n— 1 (a + /3)v — | + bi)xl J v = Y l^ai + bi)Tlv = + (3v. \ 2=0 / 2=0 2=0 2=0

For multiplication of scalars

n—1 (a(3)v = c{Tlv, where q = asbt £ k 2 = 0 s+t=i

and

n—1 n—1 n—1 'n —1 n—1 a(/3v) = Y a^ fa = Y aiTi^ v) = Y aiTi Y biTiv = Y CiTiv’ 2=0 2=0 2=0 , 2=0 2=0

where the last equality follows from the linearity and nilpotency of Tl and rearrang- 60

ing terms. Therefore (a/3)v = a(/3v). For summation of module elements

n—1 n—1 71—1 n—1 a(u + v) = a,iTl(u + v) = ai(Tlu + ) = a,iTlu + = av. i=0 1=0 i=0 i=0

Finally, la= afor all a E A is trivial and therefore is an ^-module. □

We denote the module to matrix and matrix to module correspondence in The­

orem 5.4 by MA Tand T —> M respectively. The above theorem gives us the

correspondences

f Modules M E M(A) 1 x nilpotent matrices 1 of dim/c M = d J \ of power n J

Note that because of Proposition 5.3 we may assume without loss of generality

that the nilpotent matrices T above are upper triangular. We use this simplification

in the hypothesis of the following theorem.

Theorem 5.5. LetA := k[x],xn = 0 for any field k.

1.For any M E M(T) of fixed k-vector space dimension d, we have the corre­

spondence A M T ^M' such that M as A-modules.

'Ip 2. For anyx d dupper triangular matrix T, we have the correspondence T —

M -4 t' such that V = ST5'- 1 ,for some invertible matrix S. In other words 61

T and V are similar to each other.

Proof. (1) Let Mand M 'be A-modules from the correspondence M A T A M'.

Both are vector spaces of dimension d over k, so dim*; = dimfc Since M is finitely generated let {mi,..., md} be its generating set. Then M — Am\ + • • • +

Am.d. Each Am.i is a submodule of M,so define the map

Arri\ © • • • © Amd — Arrt\ © • • • ©

(aim,i,..., adm.d) i—> aim x H h adm.d.

This is easily seen to be an A-module isomorphism. Also, notice that each Arnl is isomorphic to A under the mapping am,, i-» a, implying that A© - • - ©A =

Similarly, M' is a finitely generated A-module with generating set {m^,..., m'e} and by the same argument we have M' = Ae. Now, notice that the /c-algebra A is also a vector space over k, so let dim*, A = n.Then dimfc M = dim^ Ad — nd and dimfc M = dimfc Ae = ne.But since dimfc M = dimfc M', we must have d = e. Therefore M = M' as A-modules.

(2) Let T and T' be upper triangular x matrices from the correspondence

T M A t'. The matrix T must be with respect to a /c-basis Vi,... ,vd of some vector space V over k, while the matrix T' is with respect to the ©basis mi,..., md of the module M, due to the correspondence M A T'. Since V and M are both 62

vector spaces of the same dimension d over we must have V M So, define a linear map kd —> kd on these bases by t>, M- m, for each = 1,..., It is immediate that this map is surjective and since the vector spaces we are dealing with are finite dimensional this map must also be bijective. Therefore the matrix of this map is an invertible d x dmatrix S', with respect to v\, .., and mi,..., m(i, such that T' = STS~1, i.e., T and T'are similar. □

The next theorem generalizes Theorem 5.5, however in this case we cannot as­ sume that all matrices are simultaneously of upper triangular form. This is because in general upper triangular matrices do not commute with each other, yet we require this condition since the A’-algebra A has /-commuting variables.

Theorem 5.6. Let A := k[x\,..., = 0,..., =0 be a k-algebra for any field k. The set of all modules from M(A) of fixed k-vector space dimension d are in natural bijective correspondence with the set of t-tuples x nilpotent matrices

(Xi,..., Tt) such that T f1 = 0,..., T fl = 0 and TfTj = for = 1,...,

Proof. Given an M G M(T) of dim*, M =define d /-many G-linear maps —> M bymG Xim for each 1 < i < t.Let mi,..., be a A;-basis of M and applying the ith map m G to this basis gives rise to the x matrix Tj such that TT = 0.

This correspondence will be denoted as MA (7\,... ,Tt).

Now, given a /-tuple of dxdmatrices (Ti,..., Tf) such that T f1 0,..., T f1 = 0, we derive a finitely generated T-module. Let = kd be the underlying abelian group. For any ve kddefine A’s action on kd by 63

(J 2 OiX? • • • X?)■v = Y J • • • where 0 <

(Ti,..., Tf)^ M '.This completes the bijection:

f Modules M G M(A) 1 f Tuples of x nilpotent 1 ^ \ of dim/; M = d J \ matrices (Ti,..., T f ) J

Theorem 5.7. Let A := k[x\ , ..., xt], x"1 = 0,..., rr”‘ = 0 any field k.

1. Given any M € M(A) of fixed k-vectorspace dimension d we have the corre­

spondence M A (Ti ,,Tt)AT such that AT as A-modules.

2. Given any t-tuple of xddnilpotent matrices (Ti,... ,Tt) we have the corre­

spondence (Ti,..., Tf) ^ M (Ti,..., ^ such that = ST.-S' 1 for all

i = l,...,i and S is some invertible matrix.

Proof. (1) The proof for this statement is identical to the one seen in Theorem 5.5.

(2) Each of the d x dmatricesT\,... ,Ttmust be with respect to a basis of an d-dimensional vector space V over k. so let v\,... ,Vd be the basis of each of the matrices Ti,..., Tt.By the correspondence M A (T(,..., each of the matrices 64

T[,..., T[ are with respect to the /c-basis mi,, rnr{ of M. Since V and M are both vector spaces of the same dimension d over we must have V =

Define a linear map kd —>• kd on these bases by Vi rrii for each 1 < and the matrix of this map will be an invertible d x matrix S, with respect to Vi,... ,va and mi,..., such that T' = _1 STiS for each = 1,..., □ 65

Chapter 6

The Variety of /c[G']-modules

In this final section we present our global variety consisting of finite group represen­ tations on k. Before introducing it we review a number of main concepts and results from algebraic geometry (in affine space) essential to the rest of this paper. Let An be the affine n-space over a field k. Up to isomorphism one can say A

Definition 6.1 . An affine algebraic variety is the common zero set of a system of polynomial equations from k[xi,... ,xn].More formally

V{I) = {(0l, ..., o € An I /(«!, . . . , On) 0 V/ e /}, where I C k[x\,...,xn] is an ideal.

A variety A is said to be reducible if there exists proper subvarieties Aj, A 2 C l such that X = X \U AT Otherwise, A is said to be irreducible.

Definition 6 .2. If A is a variety and A = Aj, where each Aj is irreducible 66

and Xj Xifor all % f j, each Xi is called an irreducible component of X.

The radical of an ideal I C k [ x \xn] ,, is defined to be

V7 = {aG I | are I for some positive integer r}.

An ideal I is called a radical ideal if I — \fl The coordinate ring of an affine variety

X in A” is the quotient ring k[X] = k[xi,.., where

i(x) = {f ek[Xl,...,xn]\f(x) = o}

is an ideal of k\x\, ..., xn\.

Theorem 6.1 . (Hilbert’s Nullstellensatz) Let k be an algebraically closed field. Then

Z(V(I)) = VIfor any ideal I C k[xi,... ,x n\, i.e., the maps T and V are bijective

inverses of each other:

j f Affine algebraic sefsl — > J Radical ideals 1 \ I C A " J \ I = '

Through the Nullstellensatz we know that every affine algebraic variety X =

V(I), with I a radical ideal, has a unique coordinate ring k[X] = i,..., xn\/I. 67

6.1 Dimension Theory of Varieties

We give the geometric and algebraic definition of dimension below, along with some useful facts that are used in computing the dimension of varieties.

Definition 6.3. Let X be an irreducible affine variety and let C C • • • C be a chain of irreducible subvarieties of X. The dimension of X is defined to be the supremum of the length of all chains in X i.e.,,

dim X = maxjn | Z qC Z \C • • ■ C —

For a general variety one can write X = V [J and its dimension is defined to be

dimX = max{dim ,..., dim

Definition 6.4. Let Rbe a ring and P a prime ideal in R. The height of P, denoted ht(P), is defined to be the supremum of the lengths of all chains of distinct prime ideals in P, i.e.,

ht(P) = max{n | P0 C Pi C • • • C

For a proper ideal I ^ Rits height is defined to be

ht(/) = min{ht(P) | prime F 7 /}. 68

The Krull- dimension of the ring R,denoted dim R, is the supremum of all heights of prime ideals in R, i.e.,

dimP = max{ht(P) | C prime ideal}.

Lemma 6.2. Let R be a commutative ring and C ideal. Then for a prime ideal P , IC P if and only if \fl C P.

Proposition 6.3. Let I be an ideal of a ring R. Then ht = ht(\/7).

Theorem 6.4. The dimension of an irreducible variety X is equal to the Krull- dimension of its coordinate ring k[X],

Given a field extension E/F a set {ai,..., an} C is algebraically independent over F if there does not exist a nonzero polynomial / G i,..., xn] such that f(ai,...,an) = 0. A set SC E is algebraically independent over F if every finite subset is algebraically independent. A transcendence base of E/F is a maximal subset B C Ethat is algebraically independent over F. The cardinality of the transcendence base B is called the transcendence degree of E/F and is denoted by trdegF(S).

Theorem 6.5. Let k be a field and A an integral domain that is also a finitely generated k-algebra. 69

1. The dimension of A is equal to the transcendence degree of its field of fractions

Frac(A) over k.

2. For any prime ideal C P A,we have ht(P) + dim = dim

By (1) above for an irreducible affine variety I C A" its dimension is equal to the transcendence degree over k of the field of fractions k(X) of the coordinate ring k[X], i.e., dimX = trdegfcA;(X). By (2) we have dim/c[X] = ht(P) + dim/c[X]/P for P C k[X] a prime ideal.

Theorem 6.6. Let R be a finitely generated k-algebra. If R is an integral domain with Krull- dimensionn, then all maximal ideals of R have height n and every chain of prime ideals from 0 to M has length n.

In general, statement (ii) of Theorem 6.5 does not always hold:

Theorem 6.7. Let R be a commutative ring and P a prime ideal of R. Then ht(P) + dimP/P < dim P. If R is a finitely generated k-algebra then equality holds.

The following extends statement (ii) of Theorem 6.5 for any ideal.

Theorem 6 .8. Let R be a finitely generated k-algebra and I an ideal of R. Then ht(/) + dim R/1 = dim R.

Using the above theorem, for any affine variety I C A" let T{X) C ,..., xn] be a radical ideal and

dimfc[xi,... ,x„] = dim(k[xi,..., xn]/l(X)) +ht(J(X)), 70

and thus, dim A = dim /c[X] = n— ht(Z(X)).

6.2 Irreducible Components of Varieties

One can explicitly find the irreducible components of a variety defined by an ideal via primary decomposition of the ideal’s radical.

Lemma 6.9. For an ideal I Ck[x\ ,..., xn] we have V(I) —

A prime ideal P in a ring R is said to be minimal if for every prime Q C R if

Q C P then Q = A P. prime ideal P is said to be minimal over an ideal I if it is minimal among all primes Q D I.

Lemma 6.10 . For any proper ideal I of a commutative ring R we have \fl = n r = i^ where each Pi is a minimal prime ideal over L

Now, given a variety defined by an ideal I C k[x\,..., xn] we can say that

m m m v(i) =v (V 7 ) = p n o = u y=u*‘' f6-1) i= 1 i= 1 i= 1 where each X t — V(Pi) is an irreducible subvariety. Throwing out any redundancies among these subvarieties so that X3 Xi results in X expressed in its irreducible components. It should be noticed that each irreducible component above corre­ sponds to a minimal prime ideal over /. 71

Therefore, if one wants to know the number of irreducible components of a variety

V(I) it is enough to determine the number of minimal prime ideals over I since by the Nullstellensatz irreducible varieties correspond to prime ideals. In our case, when determining the number of irreducible components of the variety of modules

X = V (I),where I = (/i,..., f m)is the ideal generated by the equations obtained from the matrix relations, the minimal primes of the finitely generated k-algebra A = k[x\,... ,xtd2]/Iare in one-to-one correspondence with the minimal primes over I in k [ x \ ,,xup\. Thus, it is not necessary to take the radical of our ideal I and use the expression in (6.1) when computing the number of irreducible components.

6.3 Representation Varieties

We introduce a new algebraic variety whose points correspond to finite group rep­ resentations on the vector space kd (i.e., fc[G]-module structures). This particular variety captures all the possible fixed dimensional representations of the group G as a single mathematical object, which allows for algebra-geometric analysis to be applied. We refer to this object as the variety of /e[G']-modules or the representation variety. After defining this variety we will study some of its basic geometric proper­ ties through computations with the help of mathematical softwares. First, we will need to define a bijective correspondence between the set of A,[G]-module structures from M(A:[G]) and points of a certain algebraic variety. 72

Theorem 6.11. The set of all k[G]-module structu on the fixed dimensional vector space kd is in bijective correspondence with the points of a variety in Atd2 determined by a system of matrices satisfying the relations of the group G.

Proof. Let M € M(/c[G]) be a module with dim*, = d and let M = Fix the standard basis e\, ..., of M. For a finite abelian group G with |G| take its decomposition

G = Zp«i © Zp«2 © • • • © Z pH, where

The group G has presentation:

G = (g \,..., gt| gp 1 = • • • = gT =L gigj = ^ ^ )-

Form each cyclic group component take a generator gi G Z'h. Define t-many k- "i linear maps M —>■ M by m 4 c/,m. The standard basis e\,, e(i under each of pai these maps gives rise to a d x dmatrix Tt such that Ti i —I (we have the identity matrix I because gfipai =1). We now have a collection of x matrices Ti,..., and substituting each with the corresponding generators ,..., of G we get the relations T f1 = ••• = Tf* = I and TfTj = TjT, for each i,j = 1 It is from these matrix relations where we obtain a system of equations that determines a variety. Each d x dmatrix Tj: has d2-many entries and since there are f-many of these matrices our polynomial equations that form the system will be in td2-many 73

variables. Thus, our variety described by these equations lie in the affine space .

There is a total of td2+(*) d2 equations that describe our variety, where td2 equations come from T f1 = • • • = Tf* = I and Q)d2 equations come from = T]Tl for each =

Conversely, there are as many /c[G]-module structures as there are solutions to the system of equations that arise from the tuple of x matrices (Ai,... satisfying A1 j = • • • = .4f‘ = I and AtAj = AjAi: i.e., given a variety V C Atd2 determined by (Ai,..., Ad) one can take a point and determine a fc[G]-module

M e M(fe[G]).

First, notice that for a point a € V

Oi (cki,..., ,..., cr^2_^2,..., G k , take d2-many coordinates from a and arrange them as entries of a x matrix At.

Since there are a total of td2 coordinates of a we can form t many x matrices

Ai, ..., At, hence we identify the point a with the matrix tuple [A\,..., At). Now let

M = kd, the d-dimensional vector space over k, be our underlying set and m, E M,

(p E k[G], Define the A"[G']-action on M by

(p.m= atgl1 ■ ■ ■ g?) ■ m = (J T ■ ■ ■ A m .

This action is well-defined because each element of the group ring is unique. It can 74

be checked that this ring-action satisfies all the required axioms of a module by poly­ nomial addition and multiplication (compare with the discussion preceding Theorem

5.5). Therefore M G M(fc[G]) and we have obtained a /c[G]-module structure from a point aG V. □

This establishes the following bijection:

f &[G]-modules ) J Points of variety 1 \ up to fc-linear isomorphism J \ determined by matrices Ti,... ,Tt j

We mention that the correspondence we defined between the set of modules from M(A) and the set of tuples of equal sized nilpotent matrices in Section 5 is a special class of these more general representation varieties. The solutions to the equations from the nilpotent matrices T and their conjugates T' = _1 forms this representation variety of .4-module structures.

Proposition 6.1 2 . The set of all A-module structures on a fixed dimensional vector space kd is in bijective correspondence with the points of a variety in Atd2 determined by the tuple (Ti,..., Tt)of equal sized nilpotent matrices.

Example 6.1. Let G = Z 2 ® Z 3 and M G M(fc[G]) with dim^ M = 2. From the matrix pair T±,T2, we obtain our system of equations from the relations T:T2 = TfiTi and Tf = /, Tf = I. 75

( \ 1a b e / Let T\ and T= dj V5 hJ

Carrying out the computations of matrix multiplication gives our system of equa­ tions:

( ( ( a 6^ e A e A a 10 T{T2 - t2= 0 dj dj vc V5 h \9 h> vc 1° °) bg-cf = 0 ce + dg — ag — ch = 0 af + bh — be — df = 0 cf -bg = 0

( a a T2 = I b) 61 d j vc dj V a2 + be — 1 = 0 ac + cd = 0 ab -f- bd — 0 be — d2 — 1 = 0

( A / A e / / e /e o' T2 = I \ 9 fc/ Vs 'V 'v \° V 76

e(e2 + fg) + g(ef +fh) - 1 = 0

/( e 2 + f d) + h(ef + fh) = 0 e(eg + gh) + g(fg + h2) = 0 f{e9 + 9^) + g + h2) — 1 = 0.

The 2 • 22 + (2) • 22 = 12 equations above generates an ideal I in k[xi, , x8].

This ideal determines the variety V(I) which lives in affine space A2 22 = A8.

Example 6.2. G = Z3 © Z 3 and M G M(fc[G]) with dim*; = 2. The matrices

f\ / a /Xb e f Ti = and T2 = \c d) \3 h)

satisfy T\T2 = T2T\ and Tf = I, Tf = I. The system of equations are:

T{T2 - TfT = 0 :

bg-cf = 0 ce + dg — ag — ch = 0 af + bh — be — df = 0 cf -bg = 0

Tf = I 77

qf T 2abc bed — 1 — 0 a2b + b2c + abd + bd2 — 0 a2c + be2 + acd + cd2 = 0 abc “I- 2 bedT 3 — 1 — 0

T3 = / :

e3 + 2 efg + fgh - 1 = 0 e2f + f 2g + efh + fh 2 = 0 e2g + fg2 + egh + = 0 efg + 2 fgh +h 1 =0 .

The 2 • 22 + (2) • 22 = 12 equations above generates the defining ideal I of the variety V(I) C A8.

Example 6.3. G = Z 2 © Z2 and M G M(/c[G]) with dim*. = 3.

a b c •i * o T\ = d e f ,T 2 = m n o

\g h i \P Q r) 78

Both matrices satisfy the algebraic relations T{T2 = T2T\ and Tf = I, T2 = I.

T{T2 - T2Ti = 0 :

—dk — gl + bm + cp = 0

—bj + ak — ek — hi + bn + cq = 0

—cj — f k + al — il + bo + cr = 0 dj — am + em — dn — go + fp = 0 dk — bm — ho + fq = 0 dl — cm — fn + eo — io+ fr = 0 gj + hm — ap + ip — dq — gr = 0 gk + hn — bp — eq + iq — hr = 0 gl + ho — cp — fq = 0

T? = I:

af “t- bd -I- eg — 1 — 0 ab be -j- ch — 0 ac + bf + ci = 0 ad -I- dc f g = 0 bd + e2 + fh — 1 = 0 79

cd + e f + f i = 0 ag + dh + gi = 0 bg -I- ehT hi = 0 eg + fh + i2 — 1 = 0

Ti = :I

j 2 + fcm + Ip 1 = 0 jk + kn + Iq = 0 jl + ko + Ir = 0 jm + mn + op = 0 fcm + n2 + oq — 1 = 0

Im, + no + or = 0 jp + m,q + pr 0 kp + nq + qr = 0

Ip + og + r2 — 1 = 0.

The 2 • 32 + (g) • 32 = 27 generating equations above determines an ideal / that determines the variety V(I) C A18. 80

6.3.1 The Free-Algebra Representation Variety

The variety of /c[G]-modules we have just defined can naturally be extended to a variety of noncommutative finite k-algebra representations on kd. Recall that a free k-algebra k{Xi,..., Xt) is a polynomial ring over k in non-commuting variables

Xi,.. . ,Xt.The free /c-algebra is also a (infinite dimensional) vector space over k and the set of monomials {x^ ’ " xinI • • • £ {!>••• A }} forms a natural basis.

Each element of the free algebra is therefore of the form

where G k. Multiplication in i,... ,Xt) is concatenation

(xh ■ • • xin)(xh• • • xjm)xinxn ■ ■ ■ xjm.

In this subsection let A be a noncommutative finitely generated A>algebra with generating set {a*,..., at} and k-vector space dimension d. The modules from the category M(.A) are assumed to be left M-modules.

Theorem 6.13. For a finitely generated k-algebra A we have A = k{X \,..., X t)/1, where I is the kernel of some surjective map from k(X \,..., to A.

Proof. Let a^,... ,atbe generators of A which implies A = k(ai,... ,at). Define the 81

map

<(>:k(Xu...,Xt) — >A

Xi i—» ^ for each i = 1 , ,t.This map is a A;-algebra homomorphism on the generators, so under 0 we can take products of the Xi’s and send the basis { of k(Xi,..., Xt) to the basis {a^ • • • ain}of By linearity

ol —• • • Uin = y ^ Xi(f)(xi1)■ ■ ■ 4>{xin) = cp • • • for some \ xh ''' xin £ k(Xi, • • •, Xt). The kernel ker is a two-sided ideal of k(X\,.... Xt), since every element commutes with 0. By the first isomorphism theorem for rings our claim follows. □

In commutative algebra Hilbert’s basis theorem implies that any ideal I in the polynomial ring k[Xi,.... Xt] is finitely generated, so Theorem 6.13 in this setting states that any finitely generated k-algebra A has a finite presentation, i.e., A = k[X-i,..., Xt]/I. However, Hilbert’s basis theorem does not hold in the free algebra setting. Consider the following counter example.

Example 6.4. In the free algebra k(X Y), let ({xy1} € N) := { JT fixV% I fi £ k(X, T )} be a left-sided ideal. This ideal is not finitely generated. 82

Proof. Let I = ({xy1} \ iG N). We check that is an ideal. For any /, g G I clearly

/ — g = Yi fixy1 —Y j9]xyJ £ P To prove that I absorbs products let G and for any g G k{X, Y) we get gf = g(Yi foxy*) = Y iid fi)xt^ € l -

Now suppose I = (/i,... ,/n) where each f t = Y K3xyl] is a finite sum. For any / G / we have / = gxfi H b gnfn for some G k(X, Y). However, each element in I is a finite sum with each term ending on the right with xy1 for arbitrary large igN. If fhas a term ending on the right with for i larger than any of the xytj in difi + 9nfn, due to non-commutativity no higher powers of xylj can be obtained in gxf\ + ■ • • + gnf n- Therefore this f f I and / cannot be a finitely generated left ideal. □

Therefore it is not sufficient to just take any ideal in the free algebra and form the quotient k{X \,..., Xt) to /I obtain a set of relations of a finitely generated /c-algebra.

Instead, we will consider only those fc-algebras A which are finitely , i.e., there exists an ideal / = (Fx,..., Fr), for some Fj,..., G k{Xx,..., such that A = k{Xx,.... X t)/1. This means that A can be presented by generators x\,... ,xt satisfying the relations Fx(xx,... ,xt) = 0 .

Theorem 6.14. For a finitely presented k-algebra A, the set of all A-module struc­ tures on the k-vector space kd is in bijective correspondence with the points of a variety in affine space Atd2 determined by a system of matrices satisfying the rela­ tions of A. Up to isomorphism the points of the variety corresponds to modules from

M (A). 83

Proof. Let M € M(A) be of dimfc M = d, which follows from Lemma 5.2. Take the generators aq, ..., xtG A satisfying the relations ,...,= 0,..., ,...,

= 0 of .4. Define t-many linear maps M —>• M by m 1-4 Xiiri, for each i = 1,...,

Let be a /c-basis of M and applying each of the t maps m i-)- Xim to this basis gives rise to d x dmatrices Ti,... ,Tt such that=

0,..., Fr(Ti, ..., Tt) = 0. From these relations one obtains rd2 equations in td2 variables that defines a variety in affine space 2.

Now, given a point a form a variety V C Atd'2 that is defined by a system of equations in fd2-many variables, we derive an .4-module structure. Since there are td2-many entries of a, for every d2-many of these coordinates arrange them in such a way that they become the entries of a d xmatrix Ti. As a result we obtain t-many matrices Ti,..., Tf, and hence the point a is associated with these matrices.

Let the k-vector space kd be the underlying abelian group. For and m G kd define A ’s action on kd by

a ■ m = E a

This action is well-defined because each element of the finitely generated /c-algebra is unique. It can be checked that this ring action satisfies all the required axioms of a module by polynomial addition and multiplication in noncommuting variables, 84

thus we have an .4-module structure on kd. We have shown the bijection

J *4-module structures 1 J Points of variety 1 \ on kdJ 4 * \ V C Atd2 J

Now, take a left module M C M(*4) with dim*, = which implies that M = kd as /c-vector spaces. We conclude the following correspondence:

( Left modules M € M(_4) 1 f Points of variety 1 \ of dimfc M = d J ^ l C Atd2 J '

6.3.2 Computer Assisted Computations

We present the computations of the number of irreducible components along with their dimensions of our representation varieties with the help of Macaulay2 and

Singular. The following tables organize our data in which all computations were done with the assumption that k is algebraically closed and has characteristic co­ prime to |(j|. The first two columns lists the group and vector space dimension, treated as input information which defines the variety, and the ?’s indicates that computation time to finish using both programs took too long and are thus un­ known. We investigate two cases regarding the characteristic of the underlying field k: char(fc) = 0 and char(ft) = p 0,> and see how this condition affects the properties of our varieties. 85

Group Vector space dimension Dimension # irreducible components and their dimensions Z2 2 2 3 ; { 0, 0, 2} Z2 3 4 4 ; {0,0, 4, 4} Z2 4 7 ? Z3 2 2 4; {0,0, 2, 2} Z2 © Z2 2 2 1 0 ; {0,0,0,0,2, 2,2, 2,2, 2} Z2 © Z2 3 6 ? Z2 © Z3 2 2 13; {0,0, 0,0, 2, 2, 2, 2, 2, 2, 2, 2,2; Z2 © Z3 3 6 ? Z3 © Z3 2 2 25 Z3 © Z3 3 ? ? Z2 © Z2 © TLi 2 2 36

Table 6.1: Underlying field k has characteristic zero

Group Vector space dimension Characteristic Dimension # irreducible compo­ nents

Z 2 2 2 2 1

Z 2 3 2 4 2

Z 2 4 2 7 ?

Z 3 2 3 2 1

Z 2 © Z 2 2 2 3 2

Z 2 ffi Z 2 3 2 6 ?

Z 2 © Z 3 2 2 2 6 3 2 5

Z 2 © Z 3 3 2 6 ? 3 6 ?

Z 3 © Z 3 2 3 3 2 ? Z 3 © Z 3 3 3 6

Z 2 © Z 2 © Z 2 2 2 4 2

Table 6.2: Underlying field k has positive characteristic 86

Recall that for the group ring k[Z„] we have the truncated polynomial represen­ tation k[x]/(xn — 1) where k has any characteristic. When k has positive charac­ teristic p dividing n we also have an alternate truncated polynomial representation k[y]/(yn) where y = x— 1. Both are /c-algebra representations of Zn\ for positive characteristic and one should expect that the representation varieties, obtained by substituting a matrix into the relations xn — 1 = 0 and yn = 0 respectively, will give isomorphic varieties. This is indeed the case. 87

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