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THESIQ

Will/ll]?lllli/Iflllllillljlli/l

This is to certify that the

thesis entitled

Maximal Commutative Subalgebras
of n by n Matrices Over a Field

presented by

Young Kwon Song

has been accepted towards fulfillment
of the requirements for

Ph.D. _degree in Mathematics

WW

Major professor

April 18, 1996

MS U is an Affirmative Action/Equal Opportunity Institution

 

 

LIERARY
Michigan State
University

 

 

 

 

 

 

PLACE IN RETURN BOX to roman this checkout from your rocord.
To AVOID FINES return on or bdoro data duo.

    

DATE DUE DATE DUE DATE DUE

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

L__I::J

MAXIMAL COMMUTATIVE SUBALGEBRAS
OF n BY n MATRICES OVER A FIELD

By

Young Kwon Song

A DISSERTATION

Submitted to
Michigan State University
in partial fulfillment of the requirements
for the Degree of

DOCTOR OF PHILOSOPHY

Department of Mathematics

1996

ABSTRACT

MAXIMAL COMMUTATIVE SUBALGEBRAS OF 11 BY n MATRICES OVER A
FIELD

BY

Young Kwon Song

The existence of RC Courter’s counterexample to M. Gerstenhaber’s conjecture
suggests some interesting questions about the isomorphism classes of local algebras
in the ring of 14 by 14 matrices. It was conjectured for a long time that Courter’s

example is unique up to isomorphism.

In Chapter 2, we will show that the class of maximal, local, commutative algebras
which are isomorphic to 8 K N 2 has only one isomorphism class. Next, we will show
the class of pairs (R, V) which are (a, T)-isomorphic to (3 IX N 2, 82 EB N) has only one

isomorphism class.

In Chapter 3, we will construct a new algebra S which is maximal, local, commuta-
tive, index of Jacobson radical 3, and dimension 13. We will use 8 to show the (B, N )-
construction depends on the field It. The algebra S is not a (B, N)-construction if k
is the real numbers and is a (B, N )-construction if k is an algebraically closed field.
Finally, we will answer the above conjecture by showing the algebra S is not isomor-

phic to the Courter’s algebra.

DEDICATION

To my parents, brother, sisters, loving wife, and pretty daughter,
Christina.

iii

ACKNOWLEDGEMENTS

I would like to give heartfelt thanks to my advisor, Professor William C. Brown for
his helpful guidance, encouragement, patience, and understanding. I would also like

to thank Professors Wei-Eihn Kuan, Christel Rotthaus, Charles Seebeck, and Bernd
Ulrich for their helpful advice.

iv

TABLE OF CONTENTS

1. Notation and History
2. Uniqueness of Algebras in fig and 901
2.1 Algebras in 9
2.2 Classification of Isomorphism Classes in fig and QC;
3. Nonuniqueness of Algebras in Q
3.1 Construction of New Algebra 8 in O
3.2 The Algebra 8
APPENDIX

BIBLIOGRAPHY

13

42

42

50

65

84

LIST OF TABLES

1. Multiplications of 6,’s.

2. Multiplications of Ai’s.

vi

49

57

Mn(k)
CT" (5)

(R, J, k)

(a, r) : (G, H) —> (G’,H’)
H0mG(H,H)

pg : H —+ H

(p : G —* HomG(H, H)

MX

LIST OF SYMBOLS

a field

the natural number

set of m x n matrices over It

Mnxn(k)

k-algebra

k-vector space dimension of R

the set of maximal, commutative, k-subalgebras of Tn
the centralizer of S in Tn

the Schur algebra of size 4

the Courter’s algebra

local k-algebra with maximal ideal J

and residue class field It

category whose objects are ordered pairs (G, H), where G
is a finite dimensional, local, commutative, k-algebra and H
is a finitely generated, faithful G-module

a morphism of X

the set of G-module homomorphisms from H to H
the G—module homomorphism given by ug(h) = hg
k-algebra homomorphism given by

My) = #9 for g E G

full subcategory of X whose objects are (C, H) E X

for which H has a small endomorphism ring G

vii

ubnhrhh

LIST OF SYMBOLS (continued)

[R]

{(R, V)]

(0) 1v J
L(a1, . . . , an)
Soc(R)

MBUV)

M lxn(k)

H0mk(V, V)

the canonical basis of k”

matrix representation

right R—module V

index of Jacobson radical of R

idealization of M

{(R, J, k) E M14(k)ldimkR = 13, 2(1) = 3}

the category of all finitely generated, faithful,
B—modules of dimension 4

{(R, J, k) E DIR -—"-’ B x N2 for some N E MB(4)}
{(R, V) E MX|(R,V) go”) (8 o< N2, 82 x N)
for some N E MB(4)}

the k-algebra isomorphism class containing R
the (a, r)—isomorphism class containing (R, V)
{v E Vle = 0}

the linear span of 011, . . . ,0;n

AnnRU)

the minimal number of generator of B—module N

the k-algebra defined on Equation (104)

O1

mNNUIU‘CflCfl

10
10
11
18
48

Chapter 1

Notation and History

In this thesis, I: will denote an arbitrary field. We will let N denote the natural
numbers, i.e. N = {1,2, 3, . . ..} If m,n E N, then men(k) will denote the set of all

m x n matrices with entries in k.

If m = n, then we will abbreviate menUc) by T". We will assume n 2 2
throughout this thesis. An associative ring R will be called a k-algebra if R is a
k-vector space and a(rr’) = (ar)r’ = r(ar’) for all a E k and r, r’ E R. In this
thesis, all k-algebras will be assumed to contain a (multiplicative) identity 1 7e 0. In
particular, if R is a k-algebra, then dimk(R) Z 1. Tn is an example of k-algebra.
A k-subspace R0 of a k-algebra R will be called a k-subalgebra of R if R0 is closed
under multiplication from R and R0 contains the identity of R. We will assume all

k-algebra homomorphisms take the identity to identity.

Let R be a commutative, k-subalgebra of Tn. Thus, my = yr for all 2:,y E R.
R is called a maximal, commutative, k-subalgebra of Tn if R satisfies the following
property : If R’ is a commutative, k-subalgebra of Tu and R _C_ R’, then R = R’. Thus,
a maximal, commutative, k-subalgebra of Tn is a maximal element with respect to
inclusion in the set of all maximal, commutative, k-subalgebras of Tu. We will let

Mn(k) denote the set of all maximal, commutative, k-subalgebras of Tn.

1

Thus, if CTn(S) = {A 6 Tu I As = 3A, for all s E S} is the centralizer of a
set S in T n, then a commutative, k—subalgebra R of Tn is maximal if and only if

Cum) = R.

Maximal, commutative, k-subalgebras of Tn come in many different shapes and

sizes. Here are a few examples.

Example 1: Let p,q E N such that Ip— q |g 1. Set n =p+q. Let

_ 31,, Z
(1) R_{(Oq)<p qu)eTn|xek,ZeMpxq(k)}.

In Equation (1), 1,, denotes the identity matrix of size p by p and qup denotes
the zero matrix of size q by p. Then, R is a commutative, k-subalgebra of Tu and it

is easy to check that CTn(R) = R. Thus, R 6 Mn(k). We will call R a Schur algebra

of size n.

Throughout this thesis, we will denote the Schur algebra of size 4 by B. Thus,

|$,a,b,c,d€ 1:

COOH
OOHO
can:
HOQ..c-

Cl

Example 2: Let R = k[D], where D 6 Tu is a nonderogatory matrix. Since D
is nonderogatory, the characteristic polynomial CD of D and the minimal polynomial
mg of D are the same. Consequently, dimk(k[D]) = deg(mp) = deg(cD) = n.
Suppose D’ E CTn(R). Then, DD’ = DD and by [7: Theorem 2], n = dimk(k[D]) g
idimk(k[D, 17]) g n. Thus, k[D,D’] = k[D] and in particular, D’ e k[D] = R. We
conclude R e M,(k). [:1

Example 3: Let

0.1 O 0
a2 a1 O .

(3) R= : : o. : ETnlaiek,z=1,2,...,n
an 0 a1

Then, R is clearly a commutative k-subalgebra of Tn, and CTn(R) = R. Thus,
R e Mn(k). C]

Example 4 (R.C. Courter): Let C = 1:114 @J g T14, where J is the set of all

matrices of the following form:

{ 02x2 02x10 02x2 \

$11 0
0 $11
$12 0
0 $12

0
(4) $021 $21 010x 10 010X2

 

$22 0
0 $22
211 212
221 222
3111 3112 2n 212 221 222 0 0 0 0 $11 $12

K 02x2

3121 3122 0 0 0 0 211 212 221 222 $21 $22 /

 

 

 

 

 

In Equation (4), xij,yij, and Zij E k. In [6], R.C.Courter showed CE M14(k).
Notice dim;c (C) = 13. We will call C Courter’s algebra. [:1

Recall that a commutative ring R is called a local ring if R has precisely one
maximal ideal. If R is a local ring, then the Jacobson radical J (R) of R is the unique
maximal of R. R/ J (R) is called the residue class field of R. We shall use the notation
(R, J, k) to indicate that R is a local ring with maximal ideal J and residue class field

It. For example, the Schur algebras discussed in Example 2 are all local rings with

maximal ideal J = {( 0”” Z

| Z G Mpquc) and residue class field k.
02X? 0qu

If M is a module over a commutative ring R, then M is called a faithful, R—module
if AnnR(M) = (0). Here, AnnR(M) = {r E R I rM = (0)} is the annihilator of M.

For example, k4 = M1x4(k) is a faithful, B-module via right multiplication.

Let X denote the category whose objects are ordered pairs (G, H), where G is
a finite dimensional, local, commutative, k-algebra and H is a finitely generated,
faithful, G-module. If (G, H), (G’, H’) are two objects in X, then a morphism from
(G, H) to (G’,H’) is an ordered pair (0,1'), where a : G —> G’ is a k-algebra
homomorphism, 7' : H —-+ H’ is a k-vector space homomorphism and T(hg) =
T(h)a(g) for all h E H and g E G. We will use the notation (0,7) : (G, H) ——r
(G’, H’) to indicate the morphism (0,?) from (G, H) to (G’, H’). We call a morphism
(0,7) : (G, H) ——r (G’, H’) an isomorphism if a is a k-algebra isomorphism and 'r
is a k-vector space isomorphism. In this case we will use the notation (G, H) EMT)
(G’, H’). The reader can easily check that (a, T) is an isomorphism if and only if (a, 7')

is an isomorphism in the category X.

Let (G, H) E X. We denote the set of G-module homomorphisms from H to
H by H omG(H, H ) Since G is a commutative ring, each 9 E G determines a G-
module endomorphism pg of H given by pg(h) = hg for h E H. We then have
a map cp : G —2 HomG(H, H) given by <p(g) = ,ug. Note that cp is a k-algebra
homomorphism. The map (,0 is called the regular representation of G given by H.
We say the G-module has a small endomorphism ring if HomG(H, H) “=“ G via the
regular representation. Let MX denote the full subcategory of X whose objects are
those (G, H) E X for which the G—module H has a small endomorphism ring. If
(G, H) E X, then (G, H) E MX if and only if HomG(H, H) g G via the regular

representation. Consider the following examples.

Example 5: Let G be a local, commutative, k-algebra with dimk(G) = n.

Then, G is a finitely generated, faithful, G-module. Notice that H 0mG(G, G) ’5 G
via the regular representation. Thus, G has a small endomorphism ring G, and

(G, G) e MK. [I]

Example 6: If R is a commutative, k-subalgebra of Tn, then V = k" = Mlxn(k)
is a finitely generated, faithful, (right) R—module via the usual matrix multiplication.
In particular, if V = k” and R = C, Courter’s algebra described in Example 4, then
(C,V) E X. Since CE M14(k), it follows from [4, Proposition 1] that (C,V) E MX.

D

In Example 6, we begin to see the connection between algebras in MnUc) and
objects in MX. Let R be a k—subalgebra of Tu and set V = k". Then, V is
a finitely generated, faithful, R-module with scalar multiplication given by vr for
v E V,r 6 R. Set ’H = H omk(V, V). Let u : Tn —> ’H be the representation given
by u(A)('u) = 12A. Notice that p is an anti k—algebra isomorphism of Tn onto ’H. For
each i = 1,2,...,n, set 5,- = (0,...,1,...,0) E V. We will calI§_= {€1,...,5n} the
canonical basis of V. Let I‘ : ’H ——» Tn denote the matrix representation of elements
of H via g. Thus, if f E ’H and f(5.-) = zy=10ij5jai = l,...,n, then I‘(f) = (a,,-).
1" is an anti k-algebra isomorphism of ‘H onto Tn. The reader can easily check that
I‘p = 1T", the identity map of Tn. We have now constructed the following sequence

of k-algebras and anti k-algebra isomorphisms.

(5) T, i) H L Tn with 1m = 1T".

For any commutative, k-subalgebra R Q Tn, u(CTn(R)) = CH(u(R)). Here,
CH(,u(R)) is the centralizer in 'H of [J(R). Likewise, I‘(Cu(p(R))) = CTn(P/J(R)) =
CTn(R). Now, suppose R E Mn(k). Then, R = CTn(R) and Cn(u(R)) = HomR(V, V).

Since R is commutative, p : R —> MR) is a homomorphism. Hence, the map

(6) R = CT..(R) A #(R) = u(CT,.(R)) = Cu(u(R)) = H 077mm V)

given in Equation (6) is a k-algebra isomorphism. This map is just the regular
representation of R afforded by V. Therefore, VR (the right R-module V) has a small
endomorphism ring. Thus, if R is local and R E Mn(k), then (R, V) E MX.

Conversely, suppose R is a commutative, k-subalgebra of Tn such that VR has a
small endomorphism ring. Then, CH(u(R)) = u(R) and CTn(R) = I‘(Gn(u(R))) =
I‘u(R) = R. Thus, R E Mn(k). Hence, if (R, V) E MX, then R is a local and
R E Mn(k). In summary, if R is a local, k-subalgebra of Tn, then R E Mn(k) if and
only if (R, V) E MX.

Isomorphism classes in the category MX correspond to isomorphism classes of
local, k-algebras in Mn(k). Local algebras in Mp(k) with p S n are the fundamental
building blocks of algebras in general in Mn(k). To see this, let
R E Mn(k). Since dimk(R) < 00, R is an artinian ring. It follows from [9:
Theorem 3, p205] that R = 69.2134, a finite direct sum of artinian, local rings R4,
2' = 1,. . . ,6. Since R contains the identity, V = VR = EBf=1VR,-. Set V, = VR;,i =
1,. . . ,6. Then, V = $f=1V.-. and each V,- is a finitely generated, faithful, Ri-module.
Notice R 9-5 HomR(V, V) g Hf’j=1H0mR(V,-,Vj) = f=1HomR,-(V,-,V,-). It follows
that R,- E HomR,(V,-,V,-). Hence, R,- E Mn,(k), where n,- = dimk(V,-), z' = 1,...,€.
Thus, R,- E Mn(k) can be decomposed into local, maximal, commutative subalgebras
of smaller dimensions. Thus, it suffices to study maximal, commutative subalgebras
which are local to understand the structure of maximal, commutative subalgebras in
general. We will use the notation (R, J (R), k) E Mn(k) to denote a local, commu-
tative, k-algebra R E Mn(k) which has J (R) as its Jacobson radical and k as its

residue class field. If R is clear, then we will use J instead of J (R)

In [7], M.Gerstenhaber conjectured that dimk(R) Z n for any R E Mn(k). In [6],

RC. Courter constructed an algebra C E M1406) which is’local, dimk(C) = 13, and

i(J(C)) = 3. Here, i(J(C)) is the index of nilpotency of the ideal J (C) Courter’s
counterexample to Gerstenhaber’s conjecture is minimal with respect to both n and
i(= i(J(R))). In [8], T.J. Laffey showed that dimk(R) Z n for R E Mn(k) if n S 13.
Thus, n = 14 is the smallest integer for which dimk(R) can be less than n. In [6],
RC. Courter showed that 11(J (R)) S 2 implies that dimk(R) 2 n for any R 6 Mn(k).
Thus, 2' = 3 is the smallest index of nilpotency for which dimk(R) can be less than
n for R E Mn(k). The existence of Courter’s example in M14(k) suggests some
interesting questions about the isomorphism classes of local algebras in M14(k). For
example, one could ask if Courter’s example is unique up to isomorphism. To be
more specific, is (C, V) unique up to (a, T)-isomorphism in MX? It turns out the
Courter’s example is not unique and we will construct another example in this thesis.
Let B be a commutative ring and M a right B-module. The direct sum B EB M of the
B-modules B and M can be given the structure of a commutative ring by defining

multiplication in the following way.
(7) (b1,m1)(b2,m2) = (b1b2,m2b1+ m1b2),bi E B,m,~ E M,i = 1, 2.

The commutative ring thus defined is called the idealization of M and will be denoted

byBxM.

Suppose R E Mn(k). We say R is a (B, N )-construction if R is k-isomorphic to
B o< N f for some (B, N) E X and 8 E N. Here, N 5 denotes the direct sum of 3 copies
of B-module N. The B-module Bf EBN is a B x N ’-module with'scalar multiplication
defined as follows.

I
(8) (b1, . . . ,bg,n)(b,n1, . . . ,n()=(b1b, . . . , bgb,nb + 2mm).

t=l

It is easy to check that B‘ EB N is a finitely generated, faithful, B x N ‘-module. In
[3: Theorem 2], WC. Brown and F.W. Call showed that the B x N ‘-module B‘ 69 N

has a small endomorphism ring. Thus, (B (X N‘, B‘ 69 N) E MX for all (B, N) E X.
We call (G, H) E X a Cl-construction if (G, H) Em” (B x N‘,B’ 69 N) for some
(B,N) E X and K E N. In [3], W.C.Brown and F.W.Call showed that Courter’s
algebra C is a (B,N)-construction. In [4], W.C.Brown proved that (C,V) is a Cl-
construction. In fact, (C ,V) ’="(a,,) (B D<N2, 82 619 N), where B is the Schur algebra of

size 4, N = k4, and V = k”.

Let Q = {(R, J, k) E M14(k) | dimk(R) = 13,z'(J) = 3}. Example 4 shows that
C E Q. We are interested in how many algebras in Q are (B, N )-constructions. To be
more specific, we are interested in how many algebras in Q are (B,N)-constructions,
where B is the Schur algebra of size 4 defined in Equation (2). To this end, let
M 13(4) denote the class of all faithful, B-modules of vector space dimension 4. Let
98: {(R, J, k) E Q | R E BKN2 for some N E MB(4) }. We have noted that C6 08.
Let 901 = {(R, V) 6 MXI (R, V) EMT) (B KN2,32 63 N) for some N E MB(4) }.
We have noted that (C,k14) E 901.

In Chapter 2, we will prove that if R, R’ 6 9,, then R E R’ 9-1 C as k-algebras.
Thus, QB has only one k-algebra isomorphism class [C]. If (R, V), (R’ ,V) E {201,
then we will show that (R, V) gum) (R’, V) Em“) (C, k”). Thus, QGI has only one

(a, T)-isomorphism class [(C,k14)].

In Chapter 3, we will study the following question: If (R, J, k) E Q, is R a (B, N )-
construction and is (R, k“) a Cl-construction? If k = R, the real numbers, then we
will construct a k-algebra (S, J, k) E Q that is not a (B, N )-construction. Further-
more, we will prove that (S, k“) is not a (II-construction in Chapter 3. Finally, we
will prove that S is not k-algebra isomorphic to C. and conclude that Q has at least

two k-algebra isomorphism classes [8] and [C].

Chapter 2

Uniqueness of Algebras in Q); and {201

2.1 Algebras in $2

In this section, we will prove two important theorems about the algebras in set 9.
Let R E 9. By replacing R with a suitable k-algebra isomorphic copy, we can assume
the elements in J have a particularly simple form. Let R1 and R2 be k-algebras in
Tn. If R1 = P'leP for some P E GL(n, k), then we say R1 and R2 are conjugate.

Lemma 2.1: Let R1 6 Mn(k) and let R2 be a commutative, k-subalgebm of Tn.

If R1 and R2 are conjugate, then R2 6 Mn(k).

Proof: Since R2 is commutative, it is enough to show that CTn(R2) E R2. Let
r E CTn(R2). Since R2 is conjugate to R1, R2 = P’lRlP for some P E GL(n,k).
Hence, r(P'1r1P) E (P‘lrlP)'r for all n E R1. Therefore, (PrP‘1)r1 = r1(PrP‘1).
Thus, PM"1 E CTn(R1) = R1. Hence, r E P‘IRIP = R2. Cl

Lemma 2.2: Let R1 6 Mn(k) and let R2 be a commutative, k-subalgebm of T".

If R2 = R? (transpose of R1), then R2 6 Mn(k).

Proof: Again it is enough to show that CT" (R2) E R2. Let r 6 CT” (R2). since
’ R2 = {, 'rr'ir = rfr for all n E R1. Then, 1"er = rTrl and hence TT 6 C7,,(R1) = R1.

' Thus, r 6 R3" = R2. [3

10

Theorem 2.3: Let (R, J, k) E 52. Then, there exists (R1, J(Rl), k) E (2 such that
R and R1 are conjugate and each element 7‘ E J (R1) is a matrix of the following form
02 0 O
(9) A 010 o .
C B 02

Here, 0,, denotes the zero matrix of size n by n, A E M10x2(k),B E M2x10(k), and
C 6 T2.

Proof: Let V = k”, V1 = (0) :V J = {v E V | uJ = (0)},V2 = (0) 1v J2,
p = dimk(V1),q = dimk(V2/V1), and E = 14 — p — q. Since i(J) = 3, VJ2 (_2 V1 g V2.
Suppose p = 0. Then, V1 = (0) and consequently, VJ2 = (0). Since V is a faithful,
R-module, J2 = (0). This is impossible since i(J) = 3. Thus, p Z 1. Suppose q = 0.
Then, VJ g V2 = V1 and again VJ2 = (0). This is a contradiction. Hence, q _>_ 1. In
[8: p 203], T .J . Laffey showed that

>qw+l)

(10) dimk(R) _ 1 +123 + 1 +123.

Let

dp+0
1+p€

 

14(p + K) + (p2 -1)(€2 — 1)

1 E:
+ +1) 1+p€

OD fwmflh=

Since dimk(R) = 13, f (8, p, q) S 13. An easy computation shows (3 = p = 2 and q = 10
are the only positive integers satisfying the inequality in (10). Since dimk(V1) =
p = 2, V = L(al,a2), i.e. V is a linear span of two linearly independent vectors
011,02. Similarly, V2 = L(al, . . . ,(112) and V = L(oz1, . . . ,014) for some k—basis A =
{(11, . . . ,oz14} of V. Then, for any a E V,a = ofilaix, for some x,- E k,i = 1,. . . , 14.
Ifr E J, then air = 0 for i = 1,2,a,-r E V1 for i = 1,...,12, and air E V2 for
i = 13, 14. Thus, each 1' E J has the following matrix representation with respect to

the basis A.

02 O O
(12) FA(R) = ( AU) 010 0 ) -
C(r) B(r) 02

11
Here, A(7’) E M10x2(k), B(7‘) E M2x10(k), and C(T) 6 T2.

Let R1 = I‘A(R). The matrix representation of each 7‘ E R with respect to
g is r and with respect to A is I‘A(R). Hence, there exists P E GL(14,k) such
that P‘er = I‘A(r) for all r E R. Thus, P'lRP = R1. Since R g R; as k-
algebras, dimk(R1) = 13 and i(J(R1)) = 3. By Lemma 2.1, R1 E M14(k). Thus,
(R1, J (R1), k) E Q and the elements in J (R1) have the form given in (9). E]

For an algebra R E (2, Theorem 2.3 has the following. interpretation. Any algebra
conjugate to R lies in the isomorphism class [R] of R. Hence, in studying [R], we
can assume the elements in J (R) are described as in Equation (9). We will use those

ideas to study the socle of an algebra R in 9.

Let R be a commutative, k-algebra with Jacobson radical J and dimk(R) < 00.

The socle of R, Soc(R), is the annihilator of J. Thus, Soc(R) = AnnR(J) = {r E R |

rJ = (0)}. The following Lemma is obvious from the definition.

Lemma 2.4: Let R and R1 be finite dimensional, commutative, k-algebras. If

R E R1 as k-algebras, then Soc(R) E Soc(Rl). El

Theorem 2.5: Suppose (R, J, k) E (2. Then, dimk(Soc(R)) = 4. Furthermore,
R is conjugate to an (R1, J (R1), k) E 9 such that each element of Soc(Rl) has the

following form.

02 O 0
(13) 7": 0 010 0 .

C(T) 0 02

Proof: Using Theorem 2.3, we may assume that each r E J has the form in

Equation (9). Let V = k”. Since i(J) = 3, we have the following strict containments.

(14) (O)<(0).VJJ<VJ<V

12

Let W1 = (0) :w J, W2 = VJ, u = dimk(W1), A = dimk(W2/W1), and
u = dimk(V/W2). Then, u+ /\+l/ = 14. Let

- 02 O O
(15) E,,-= 0 010 0 ,1Si,jS2.
Eu‘ 0 02
In (15), Eij is the i, j-th matrix unit of T2. Equation (9) implies Bar = 0 = r3,-
for all r e J(R) and 1 g 7;, j g 2. Thus, 177,-,- e R. Clearly, 173,-,- e Soc(R) for

l S i,j S 2 and hence, dimk(Soc(R)) 2 4.

Since 81 = 51313311 and 52 = 51417312, 5.- E W2 for i = 1,2. Since elJ = €2J = (0),
e,- E W1,i = 1,2. Thus, u 2 2. We had seen from Theorem 2.3 that dimk((0) iv
J) = 2. Since W1 (_2 (0) :V J,u S 2. Therefore, u = 2. The strict inclusions in
Equation (14) imply that A 2 1,12 2 1. Since {51,52} is a k-vector space basis of
W1, we can extend it to bases of W2 and V. Let {61, . . . ,BA,51,52} be a basis of W2
and A = {rylp..,*y,,,51,...,fi,\,el,52} be a basis of V. If r E J, then air = 0 for
i = 1,. . . , ll. Thus, we have the following matrix representation of r E J with respect
to the basis A.

( 0,, A(r) C(r) )
(16) FAQ") = 0 0), 8(1) .
0 0 02

Let R1 = I‘A(r). Since R1 = P‘IRP for some P E GL(14, k), R1 g R as k-algebras.
Thus, by Lemma 2.1, R1 E 9. Let R2 = RIT, the transpose of R2. Then, each element

r E J(Rg) is of the form

0,, 0 0
(17) T‘ = ( A1(T) 03 0 ).

01 (r) B1 (r) 02

Here, 'A1(r) = A(r)T, Bl(r) = B(r)T, and 01(1') = C(r)T. Since R1 is commutative,
R1 g R2 as k-algebras. By Lemma 2.2, (R2, J(R2), k) E Q.

13

Let r E SOC(R2). Horn (17), e,- E (0) Iv J(R2) fori = 1,...,u. Thus,
dimk((0) Iv J(R2)) 2 u. Since (R2,J(R2),k) E Q, the proof of Theorem 2.3 im-
plies that dimk((0) Iv J(R2)) = 2. Thus, V S 2.

Let r E Soc(R). Then, 5,1" = 0 for i = 1,2,B,-r = 0 for i = 1,...,A and 7,1" E W1
for i = 1, . . . , 11. Thus, we have the following matrix representation of r E Soc(R)

with respect to the basis A

DOC)

0
(18) I‘A(r) = ( 0A
0

By Lemma 2.4, 4 S dimk(Soc(R)) = dimk(Soc(R1)) S 21/ S 4.. Therefore,

dimk(Soc(R)) = 4 and each element r E J has the form in Equation (13). C]

Putting Theorem 2.3 and 2.5 together, we can always assume that a specific
representative R of an isomorphism class [R] has the following form. Every element

r E J (R) can be written in the form

02 0 O
(19) 1': A 010 0 .

CBO;

Furthermore, the socle of R is the set of all matrices of the form

02 O 0
(20) SOC(R) = {( 0 010 0 ) [C E T2} .
C O 02

2.2 Classification of isomorphism classes in $23 and
S201

The reader will recall that 93: {(R, J, k) E Q | R E Bt><N2 for some N E MB(4)},
where B is the Schur algebra defined in Equation (2). We had noticed that the

14

Courter’s example (C,J(C),k) E {23. In [3: Example 5], Brown and Call showed
that CE B I><(k4)2. This leads to a natural question about the role 1:4 is playing
in this example. One can ask whether other finitely generated, faithful, B-modules
N E M B(4) give algebras B x N 2 which determine other isomorphism classes in 93?
In this section, we will show QB has only one isomorphism class [C]. Thus, varing N

in M B (4) yields no new isomorphism classes in Q.

The reader will also recall that $201 = {(R, V) E MXI (R, V) EMT) (BMNZ, B2 69
N) for some N E M B(4) }. In this section, we will show that the set {201 has only

one (o, r)-isomorphism class [(C, 1:14)].

The questions above make sense because M B (4) has at least two isomorphism
classes. To see this, we first need a B—module presentation of k4. We will denote the

i,j-th matrix unit of T, by E,,~. Notice that Eij E B ifi = 1, 2,j = 3, 4.

Lemma 2.6: Let

_ E23 E24 E13 E14 0 0
(21) A‘(—E13 -E14 0 0 323 E2.)EMM(B)'

Then, B2/CS(A) E MB(4).

Proof: Obviously, B2 / CS (A) is a finitely generated, B—module. Since dimk(B2) =

10 and dimk(CS(A)) = 6, dimk(Bz/CS(A)) = 4. Suppose r E Ann3(BZ/CS(A)).

Then, r I4 ,r 0 E CS(A). Thus, r , O E CS(A) which implies
0 I4 0 1‘

that for some x,,y,- E B, 1 S i,j S 6

T = $1E23 + $2E24 + $3E13 + $4E14
(2 ) 0 = —$1E13 - $2E14 + $5E23 + $6E24
2
0 = y1E23 + y2E24 + y3E13 + 314E”

7‘ = "ylEia — y2Ei4 + 315523 + 3165324

15

Since J(B)2 = (0), we can assume x,,yj E k = k1,, for 1 S i,j S 6. The sec-
ond and third equations in (22) imply x1,x2,x5,x6,y1,y2,y3,y4 are all zero. Thus,
r = x3E13+x4E14 = y5E23+y6E24. Therefore, 1' = 0. Hence, AnnB(Bz/GS(A)) = (0)

and B2 / GS (A) is a faithful, B-module. C]
Lemma 2.7: Let A be the matrix in Equation {21). Then B2 / CS (A) is B-module
isomorphic to k4.

Proof: Let f : 82 ——> k4 be the map defined by f ( :5 ) = 5225 + 813/. Here,
81 = (1,0,0, 0) and 52 = (0,1,0,0). Then, f is a surjective, B—module homomor-
phism. If ( I: ) E ker f, then 2 = alI4 + a2E13 + a3E14 + a4E23 + a5E24 and
w = b114 + b2E13 + b3E14 + b4E23 + b5E24 for some a,,b,- E k,i = 1,. ..,5. Since
(C(17) ) =52z+51w= 0,a1 = b, = 0,b2=—a4, and b3: —a5.

Thus,

(23)

Hence, ( 1: ) E CS(A). It is easy to check that CS(A) Q kerf. Therefore, CS(A) =
kerf. Hence, 82/05 (A) E k4 as B-modules. D

We can now construct a faithful, B-module of dimension 4 which is not isomorphic

to k4 as B-modules.

Theorem 2.8: Let

(24) C: ( E13 E14 E23 E24 0 0

Mx 8.
E24 E23 0 0 E13 E14)€ 26‘)

Then, Bz/CS(C) E M 8(4) and Bz/CS(C) is not B-module isomorphic to 19“.

16

Proof: Obviously, Bz/GS(C) is a finitely generated, B—module. Since
dimk(Bz) = 10 and dimk(CS(C)) = 6, dimk(Bz/CS(C)) = 4. Suppose
r e AnnB(B2/CS(C)). Then, ( 8 ) ,( 0

r ) E CS(C’) which implies that for some
xiiyj 68,132,]. S 6

7" = $1E13 + 11321314 + $3E23 + $4E24

(2 ) 0 = $1E24 + $2E23 + $5E13 + $6E14
5

0 = ylEl3 + 2125714 + y3E23 + 314E211

7’ = y1E24 + y2E23 + y5E13 + y6E14

Since J(B)2 = (0), we can assume x,,yj E k = 1:14 for 1 S i,j S 6. The sec-
ond and third equations in (25) imply x1,x2,x5,x6,y1,y2,y3,y4 are all zero. Thus,
r = x3E23+x4E24 = y5E13+y5E14. Therefore, r = 0. Hence, AnnB(B2/CS(G)) = (0)
and B2/GS(C') E MB(4).

Suppose BQ/CS(C) is B—module isomorphic to 16‘. Then, there exists a B—module

isomorphism g : Bz/CS(C) ——» 16‘. Let ,61 = (8) = (g) + C'S'(C) E

0
I4

523,9(51) = 812:1 + 621:1 and 9(fl2) = 613:2 + 62312 for some why.- 6 B, i = 1,2.

Bz/CS(G). and B2 = ( > . Then, B2/C'S(C’) = BIB+figB. Since k4 = 518 +

Notice that x1 or 3;; is unit. To see this, suppose x1,y1 E J(B). Then, gwl) =

51x1 + 52y} E k4J (B). The inclusions
(26) k4 = 9(fllll3 + 903le E k4J(B) + 9(fi2)J(3) 9 k4

imply k4 = k4J(B) + g(fl2)J(B). By Nakayama’s Lemma, k4 = g(flg)J(B). This
implies B is isomorphic to k4 as B-modules and hence dim,c (B): 4. Since dimk(B)= 5,

' this is impossible. Hence, x1 or yl is unit in B. Similarly, $2 or yg is unit.

Let A be the matrix given in Equation (21) and let f be the B—module homomor-

17

phism given in the proof of Lemma 2.7. If ( j} ) E CS(C), then

f ( :1: : 3:3 ) = 510512 + a32w) + 62(y1z + yzw)
= (515151 + €2yllz + (51132 + 52312)“)
(27) = 9(fiilz +glfl2l'w
= 9(512 + B210)
= 9(0) = 0-
Thus,
3/1 92 Z _ 3112 + 29210 _
(28) ($1 x2)(w)_(xlz+x2w)€keTf—CS(A)'

Now, there are two cases to consider.

Case 1: Suppose x1 is a unit. Since ( E13 ) e CS(C), ( yl 3’2 ) ( E13 )

E CS(C') by the Equation (28). Hence,

311 312 E13 _ E23 E24 E13
(., ME.) -aI(-E.)+a2(—E.)+as( )

(29)

for some a, E k,1 S i S 6 (See the comments after Equation (22)). Thus,

ylEia + y2E24 = aiE23 + a2E24 + a3E13 + 04E”
(30)
$1E13 + $2E24 = “015313 — 02E14 + 05E23 + (165724-

Let x1 = t114 + 31 with t1 E k and 31 E J(B). The first equation in (30) then
implies a1 = a4 = 0. The second equation in (30) then implies t1 = 0. Thus,

x1 E J (B) Since we are assuming x1 is a unit, this is impossible.

18

Case 2: Suppose yl is a unit. Since ( £3023

v
m
0
£2
£3
A
ES
as
v
A
as
V

E C'S’(G) by the Equation (28). Hence,

311 92 E23 _ E23 E24 E13
(,1,2_)(0)..b.(_,,m)+b.(_m)+b.(0)

(31)

for some b,- E k, 1 S i S 6. Thus,

311E23 = blE23 + 52321 + 535313 + b4E14
(32)
$1E23 = -blE13 - b2E14 + bsE23 + b6E24-

The second equation in (32) implies bl = 0 and the first equation in (32) implies
y; E J (B) This is impossible. We conclude there is no B-module isomorphism g

between Bz/CS(C’) and k“. C]

Thus, M B(4) has at least two isomorphism classes [Bz/CS(A)] and [B2/CS(C)]

But as we will see, the idealizations of these modules are k-algebra isomorphic.

To classify the isomorphism classes in the sets fig and 901, we need Theorem 2.9.

We will denote the minimal number of generators of B-module N by u3(N).
Theorem 2.9: Let N E MB(4). Then, u3(N) = 2.

Proof : Since dimk(N) = 4,1 S uB(N) S 4. Suppose u3(N) = 1. Then, N = (1B
for somea E N. Letf : B—r Nbeamap defined byf(b) = ab for b E B. Then,f isa
B-module epimorphism. If b E kerf, then ab = 0. Thus, b E Ann3(a) = Ann3(aB).
Since N is a faithful, B-module, Ann3(aB) = (0). Therefore, b = 0 and hence f is
a B-module isomorphism. Thus, 5 = dimk(B) = dimk(aB) = 4. This is impossible.

Hence, 2 S uB(N) S 4.

Suppose uB(N) = 4. By Nakayama’s Lemma, u3(N) = dimk(N/NJ(B)). There-
fore, dimk(N J (B)) = 0. Thus, N J(B) = (0). Since N is a faithful, B-module, we

19

conclude J (B) = (0). This is impossible.

Suppose uB(N) = 3. Then, N = alB+agB + 0138 for some 01,-,i = 1, 2,3. After
relabeling the a,’s if need be, we can assume (11,02, 03 satisfy precisely one of the
following four conditions :

Case 1: a,J(B) = (0) for i = 1,2,3.

Case 2: a,J(B) = (0) for i = 1,2 and a3J(B) 75 (0).

Case 3: oz,J(B) = (0) and a,J(B) # (0) for i = 2, 3.

Case 4: a,J(B) 75 (0) for i = 1, 2, 3.

We will show all four cases lead to a contradiction.

Case 1: Suppose a,J(B) = (0) for all i = 1, 2, 3. Then, NJ(B) = (0). Since N is
a faithful, B-module, J (B) = (0). This is impossible.

Case 2: Suppose a,J(B) = (0) for alli = 1, 2 and 03.](3) 79 (0). Suppose aab = 0
for some b E B. If b is a unit, then 013 = 0. This is impossible. Thus, b E J (B)
Hence, b E Ann3(N). Since N is a faithful, B-module, we conclude b = 0. Thus,
Ann3(ag) = (0) and hence B E a3B§ N as B-modules. Since dimk(B) = 5, this is

impossible.

Case 3: Suppose alJ(B) = (0) and a,J(B) 51$ (0) fori = 2,3. Since

I4 0 0
B1 = O , B2 = I4 ,,B3 = 0 is a free B—module basis of B3, the map
0 0 I4

cp : B3 -—> N defined by MEL, ,bi) = L, a,b,-,b,- E B,i = 1, 2, 3 is a well defined
B-module epimorphism. Thus, B3/kercp E N as B-modules. Since dimk(Ba) = 15

and dimk(N) = 4, dimk(kergo) = 11. Hence, kercp has the following form.

3i

11
(33) kergo=2(y,-)B, x,,y,-,z,-EB,i=1,...,11.
i=1

Zi

20

x
Furthermore, if ( y ) E kercp, then x, y, z are not units in B. For example, suppose
z
x
x is a unit in B. Since 31 E kercp, a1 = (—1/x)(a2y + 0132). Thus, u3(N) < 3
2

which is impossible.

Since J (B)2 = (0), kercp can be written in the following form.

171'
(34) kerr = 23:11: [ y.- ) .

21'

Here, x,,y,-,z,- E J(B),i = 1,...,11. Since alJ(B) = (0)1(fi1 + kergp)J(B) = (0) in

J(B)
B3/kercp. Thus, ( 0 ). Since a,J(B) 79 (0) for i = 2,3, 1 S dimk(Ann3(a,-)) <
0

4 for i = 2, 3. Therefore, we have the following six subcases to consider.
Subcase 1: dimk(Ann3(a,-)) = 1 for i = 2, 3
Subcase 2: dimk(Ann5(a2)) = 2 and dimk(Ann3(a3)) = 1
Subcase 3: dimk(Ann3(a,~)) = 2 for i = 2, 3
Subcase 4: dimk(Ann3(ag)) = 3 and dimk(AnnB(a3)) = 1
Subcase 5: dimk(Ann3(a2)) = 3 and dimk(AnnB(a3)) = 2

Subcase 6: dimk(Ann3(a,~)) = 3 for i = 2, 3

We will show all six subcases lead to a contradiction.

Subcase 1: Suppose dimk(Ann3(a,-)) = 1 for i = 2, 3. Let Ann3(a,-) = k5,, s, E

0 0 J(B)
J(B),i = 2, 3. Then, 32 , 0 E kergo. Since a1_J(B) = (0), 0 g
0 S3 0

21

E13 E14 E23 E24 0
0 , 0 , 0 , 0 2 S2 1 i
0 0 0 0 0 33
x1 x2 $3 $4 $5
311 i 312 , 313 . 94 . 95
zl 22 23 24 25
be a basis of kercp. Since dimk(J(B)) = 4 and x,- E J(B) for i = 1,...,5,x,- E
L(E13, E14, E23, E24) for Z = 1, . . . , 5. Thus,
E13 E14 E23 E24
61: O ,62= 0 ,63= 0 ,64= O ,
0 0 0 0
0 O 0 O
(36) 65 = 32 , 66 = 0 a 67 = y]. i 68 = y2 :
0 83 21 22
O 0 0
59 = ya i 510 = 314 ,511 = 315
23 Z4 Z5
is a basis of kergo. Therefore, kercp can be written in the following form

‘ J o o 5 o
(37) ker<p=(O)$k(sz)®k(0)$zk(yi).
0 0 S3 i=1 2;

Since dimk(J (B) = 4, {32,111, . . . , 115} is a linearly dependent set. Thus, there exist

kercp. Let

00

(35)

d,c1,...,c5 E I: not all zero such that ds2+cly1+---+c5y5 = 0. Ifc, = 0 for

all i = 1,. . . ,5, then d 76 0 and d32 = 0. This implies S2 = 0. This is impossible
0

0
since 32 is a basis vector of kergo. Hence, some c,- is not zero. We can assume

c5 75 0. Thus, y5 E L(sz,y1,...,y4). We can repeat this proof on 32,y1,...,y4

and assume in E L(82,y1,v2,y3)- Hence. we may assume 114,115 6 Hem/1,112,313).

22

Therefore, 31., = dsg + clyl + c2y2 + c3y3 for some d,c1,c2,c3 E k. If d65 + c167 +
c263 + c369 — 610 = 0, then {155,157, 68, 69, 610} is linearly dependent which is impossible.

0
Thus, (165 + c167 + c268 + c369 — 610 = ( 0 ) with z E 0 in J(B). If 2 = ts3 for

Z
some t E k, then (165 ‘1‘ C167 ‘1' C268 + C369 '- 610 — LOG = 0 and {65,66,67, 63,69,610}

0
is linearly dependent which is impossible. Thus, ( 0 ) E kercp\k66. Therefore,
2

dimk(Ann3(a3)) Z 2. This is a contradiction.

Subcase 2: Suppose dimk(Ann3(a2)) = 2 and dimk(Ann3(a3)) = 1. Then,

Ann3(a2) = 1:31 + [CS2 and Ann3(a3) = k33 for some 3,- E J(B),i = 1, 2, 3. Let
E13 E14 E23 E24 0 0
0 , 0 , O , O a 51 a 82 9
O 0 0 0 0 O
0 $1 _ $2 $3 $4
0 . yl , 312 , vs . 114
$3 , 21 Z2 23 24

be a basis of hemp. Since dimk(J(B)) = 4 and x,- E J(B) for i = 1,...,4,x,~ E

(38)

L(E13, E14, E23, E24) for 1:: 1, . . . ,4. Thus,

E13 E14 E2
61 = 0 ,62 = 0 , (53 = 0
0 0 O
0 0 0 0
(39) 55: 51 :66: S2 ’57: 0 ’58: 311 ,
0 0 S3 21
0 0 0
59 = 92 ,510 = y3 ,511 = 314
Z2 Z3 Z4

is a basis of kercp Since dimk(J(B)) = 4, {33,21, . . . , 24} is a linearly dependent set.

23

Thus, there exist d, c1, . . . ,c., E It not all zero such that (183 + c121 + - ~- + 6424 = 0.
If c,- = 0 for all i = 1,. . . ,4, then at E 0 and (133 = 0. This implies 33 = 0. This is
0

impossible since ( 0 ) is a basis vector of kergo. Hence, some c,- is not zero. We
33

can assume a, E 0. Thus, 24 = dsa + c121 + 0222 + c323 for some (1,61,62,03 E k. If
d67 + c168 + c269 + c3610 — 611 = 0, then (67,68,159, 610,611} is linearly dependent which

0
is impossible. Thus, d67 + 6168 + c269 + c3610 — 611 = ( y ) with y E 0 in J(B). If
0

y = t181+t282 for some t1, t2 6 k, then d57+6163+c269+636m—611—t165—t265 = 0 and
0

{65, 66,67,153, 69, 610,611} is linearly dependent which is impossible. Thus, y E
O

kercp\k65 + 1:66. Therefore, dimk(Ann3(ag)) 2 3. This is a contradiction.

Subcase 3: Suppose dimk(Ann3(a,-)) = 2 for i = 2,3- Then, Ann5(a2) =

ksl + 1932 and Ann3(a3) = [mg + £38., for some 8,- E J(B),i = 1, 2, 3, 4. Let

be a basis of kergo. Since dimk(J(B)) = 4 and x,- E J(B) for i = 1,2,3, :17,- E

L(E13, E14, E23, E24) for i = 1, 2, 3. Thus,

24

is a basis of kercp. Since dimk(J(B)) = 4, {31, sg,y1,y2,y3} is a linearly dependent
set. Thus, there exist d1,d2, c1, c2,c3 E It not all zero such that (1131 + d2S2 + clyl +
c2112 + 033/3 = 0. If c,- = 0 for all i = 1,2, 3, then dlsl + d282 = 0. Since s1,32 are
linearly independent vectors in J (B), d1 = d2 = 0. This is impossible. Thus, c,- E 0
for some 1 S i S 3. We can assume c3 E 0. Hence, ya = (1131 + d282 + clyl + C2y2 for

some (11, (12, C1, C2 E k.1fd165+d266 +C169 +C2610 “-611 = 0, then {(55, (56, 69, (510, 611} 18

0
linearly dependent which is impossible. Thus, (1165 + d266 + c169 + c2610 — 611 = ( O )
z

with z E 0 in J(B). Ifz = t333+t4s4 for some t3, t4 E k, then d165+d265+c169+c2610—
611 — t367 - 13468 = 0. This is a contradiction since the vectors in Equation (41) form

0
a basis of kercp. Thus, ( 0 ) E ker<p\k67 + [$68. Therefore, dimk(AnnB(a3)) Z 3

Z

and this is a contradiction.

Subcase 4: Suppose dimk(Anng(a2)) = 3 and dimk(Ann3(a3)) = 1. Then,

Ann5(02) = ksl + ksz + 198;; and Ann3(013) = 1:34 for some 3,- E J(B),i = 1,2, 3, 4.

25

{(ilfi‘lfialli‘Hills?)
(§)’(§l(§£l(§§)l§§)}

be a basis of kercp. Since dimk(J(B)) = 4 and x,- E J(B),x,- E L(E13, E14, E23, E24)

Let

(42)

for i = 1, 2, 3. Thus,

E13 E14 E23 E24
61: 0 ’62: 0 :63: :64: 0 a
0 O 0
0 0 0
(43) (55 = 31 , 65 = 82 ,67 = 0
0 0
0 0 0 -
59 = 311 510 = 312 1511 = 93
Z1 Z2 Z3

is a basis of kercp. Since dimk(J(B)) = 4, {S1, 32, 33,311, y2, y;} is a linearly dependent
set. Thus, there exist d1,d2,d3,c1,02,c3 E It not all zero such that dlsl + d282 +
dgsg + clyl + c2312 + c3y3 = 0. If c,- = 0 for all i = 1, 2, 3, then dlsl + (1232 + d333 = 0
Since 31,32,33 are linearly independent vectors in J (B),d1 = d2 = d3 = 0. This
is impossible. Thus, 0,- E 0 for some i. We can assume c;, E 0. Hence, 3);; =
d181+d232+d333+01y1+C2y2 for some d1, d2, d3, c1, C2 E k. If d165+d265+d367+0169+
c2610 — 611 = 0, then {65,156, 67, 69, 610,611} is linearly dependent which is impossible.

0
Thus, d165 + d266 + d367 + alt, + c2610 —- 611 = ( O ) with z E 0 in J(B). If 2 = ta,
2

for some t E k, then d165, + (1265 + (1367 + c169 + c2610 — 611 — tbs = 0. This is impossible

26

0
since {65,65,67,68,6g,610,611} is linearly independent. Thus, ( 0 ) E kercp\k63.
2

Therefore, dimk(AnnB(a3)) Z 2 and this is a contradiction.

Subcase 5: Suppose dimk(Ann3(ag)) = 3 and dimk(Ann3(a3)) = 2. Then,
Ann3(ag) = ksl + [€32 + ks3 and Ann3(a3) = ks4 + [935 for some 3.- E J(B),
i =1,2,3,4,5. Let

{(23%illilfiiiéllél
(§)’(§)’(§)’(§i)’(§§)}

be a basis of hemp. Since J(B) = L(E13,E14,E23,E24) and $1,132 E J(B),a:1,z2 E

[4313, E14, E23, E24)— ThUS,
E24
0 ,
0
0
0

(44)

E13 E14 E23,
{.1=(.),.=(.),.=(.a.=)
0 0 0
O 0
(45) .()6()6()6 (. )
0 0 0
' 0 0 0
59=(0),510=(y1),511=(92)}

is a basis of kercp. Since dimk(J(B)) = 4, {31, 32,33,y1,y2} is a linearly dependent
set. Thus, there exist (11, d2, d3,c1, c2 E It not all zero such that (1131 + 61232 + (1333 +
clyl +c2y2— — 0. If cl = c2: 0, then dlsl +d232 +d333- — 0. Since 31, 32, 33 are linearly
independent vectors in J (B),d1 = d2 = d3 = O. This is impossible. Thus, 6,- 75 O
for some i. We can assume c2 75 0. Hence, y2 = dlsl + (1282 + d333, + clyl for some

d1,d2,d3,C1 E k. If (£165 + (1265 + (1367 + 01610 — 611 = 0, then {65,65,67,610,511} lS

27

0
linearly dependent which is impossible. Thus, d165 + (1266 + (1367 + c1610 — 611 = ( 0 )

z
with z 9:5 0 in J(B). If 2 = t4s4+t5s5 for some t4, is E k, then d165+d266+d367+c1610-
611 — t468 — t569 = 0. This is again impossible since {65, 66, 67, 68, 69, 610, 611} is linearly

0
independent. Thus, ( O ) E ker<p\k68+k69. Therefore, dimk(Ann3(ag)) 2 3 which
2

is a contradiction.

Subcase 6: Suppose dimk(Ann3(a,-)) = 3 for i = 2, 3. Note that

(46) dimk(AnnB(ag)) + dimkk(Ann3(oz3)) = dimk(Ann3(ag) + Ann3(ag))
+dimk(Ann3(az) fl Ann3(a3)).

Since dimk(AnnB(a2) + Ann3(a3)) S dimk(J(B)) = 4, Equation (46) implies
dimk(AnnB(ag) fl Ann3(a3)) Z 2 Thus, there is O 34$ b E Ann5(ag) fl Ann5(a3).
This is a contradiction. We have now shown any of the subcases in Case 3 lead to a

contradiction. Hence, Case 3 is impossible.

Case 4: Suppose a,J(B) 75 (0) for i = 1,2, 3. Let n,- = dimk(Ann3(a,-)). By

relabeling the (123 if need be, there are ten subcases to consider.
Subcase 1: Suppose n,- = 1 for i = 1, 2, 3
Subcase 2: Suppose n1 = 2, 722 = 71.3 = ,1
Subcase 3: Suppose n1 = n2 = 2, 113 = 1
Subcase 4: Suppose n,- = 2 for i = 1, 2, 3
Subcase 5: Suppose in = 3, in = 713 = 1
Subcase 6: Suppose m = 3, 72.2 = 2, 713 = 1
Subcase 7: Suppose m = 3, n2 = 723 = 2

Subcase 8: Suppose m = 72.2 = 3, 713 = 1

28
Subcase 9: Suppose n1 = n2 = 3, 72.3 = 2
Subcase 10: Suppose n,- = 3 for i = 1, 2, 3.

A proof similar to that given in Case 3 will show that Subcase 1 through Subcase 9
are impossible. The reader can consult the Appendix for the details. Subcase 10 is
also impossible. To see this, let V be a vector space and suppose W,,i = 1, 2, 3 are
subspaces of V. Suppose dimk(V) = n. Then, we have the following equation which

can be found in [2 : Cor.2.15, p13].

( ) dimk(W1 0 W2 0 W3) = n — f=1(n — dimk(W,-)) + {(n — dimk(W1+ W2))
47
+(n — dimk((W1 0 W2) + W3))}.

Suppose V = B and W,- = Ann3(a,-),i = 1,2, 3. Then, Equation (47) implies
dimk(WlflW2r1W3) = 9—dimk(W1+W2)—dimk((WlflW2)+W3). Since dimk(W1+
W2) 3 4 and dimk((W1 0 W2) + W3) 3 4, we have dimk(W1 0 W2 0 W3) 2 1. Thus,
there exists 0 aé b E W1 0 W2 0 W3. Since W,- = Ann3(a,-),i = 1,2, 3, a,b = 0 for
i = 1, 2, 3. Thus, b E Ann3(N) = (0) which is a contradiction.

Therefore, all four cases are impossible. Hence, we conclude MB(N) = 2. D
We can now show that there is only one isomorphism class in 93.

Theorem 2.10: Let A be the matrix in Equation (21) and let Q = 82/05 (A)
Then, 8x622 g leN2 as k-algebras for any N E MB(4).

Proof: Let N E M 8(4). Then, u3(N) = 2 by Theorem 2.9. Thus, N = alB+agB

for some a, E N,i = 1,2. Since {71 = ( g ),’72 = ( IO )} is afree B-module basis
4

of 82, the map ’l/J : 32 —r N (defined by ¢(Zf=l'y,-b.-) = f=1aib,-,b,- E B,i = 1,2)

is a well defined surjective, B-module homomorphism. Hence, Bz/kerz/J '5 N as

B—modules. Since dimk(B2) = 10 and dimk(N) = 4, dimk(keri,0) = 6. Thus, ken!)

29

has the following form.

6
(48) ken/J = Z: ( :1 ) 8.

Here, why,- E B,i = 1, . . . ,6. Furthermore, if ( 1: ) E ken/2, then 2 and w are not
units in B. For example, if z is a unit, then alz + agw = 0 implies that N = (128.
Thus, MB(N) = 1 and this is impossible. Therefore, kerw has the following form.

6 1131'
Here, 13,-,3},- E J(B),i = 1,...,6. To exhibit an isomorphism between thQ2 and
BK N 2, we need to choose a good basis of herd). We may assume ken/J has the

following form.

(50) ker¢=k( E" )eak( Em )EBk( Em“ )eak( Eu” )eak( 0 )eak( 0 )
91 3/2 113 314 205 316

Here, y,- E J(B),1 S i S 6 and the ordered pairs (i,j), (p, q), (m, n), and (u, v) are
just (1,3), (1,4), (2,3), (2,4) in some order. To see this, we proceed as follows. Since
dimk(B) = 4, {2:1, . . . ,xe} in Equation (49) is a linearly dependent set. Thus, by

555

replacing the ( )’s, i = 1,. . . ,6 by suitable linear combination if need be, we

i
may assume $5 = $5 = 0. It now follows that {$1,12, $3, $4} is a linearly independent
set. For, if ($1,132,233, $4} is a linearly dependent, then by the same argument, we

may assume $4 = O in (49). Then, kerw has the following form.

..WWW)..(;)-(;)..(;).

Since dimk(J(B)) = 4, {y1,. . . ,ye} is a linearly dependent set. Thus, by the same

argument above, we may assume y1 = y2 = (0). Therefore, we have

(52) ker¢=k<ag)eak($02)eak(::)eak(z)eak(:5)eak(y06).

30

Now, let W1 = kxl + 163:2 and let W2 = 19314 + 16315 + kys. Then,
dimk(W1) = 2, dimk(W2) = 3, and dimk(W1 + W2) 3 4. Notice that

(53) dimk(W1 + W2) + dimk(W1 0 W2) = dlmk(W1) + dzmk(W2).
Therefore, dimk(W1 0 W2) 2 1. This implies that there exists 0 74 b E J (B) such that

b E WlflW2. Thus, (8),(?) E ken/1. Since 10(8) = mb and

b
This is impossible. Thus, {2:1, 2:2, 3:3, 1:4} is a linearly independent set in J (8) Since

11)( 0 ) = 02b, Gib = 02b = 0. Since N = alB+agB, b E AnnB(N) = (O).

dimk(J(B)) = 4, we may assume {$1,$2,$3,$4} = {E13, E14, E23, E24}. Hence, kerw

can be written as Equation (50).

We next show we can rearrange the six basis vectors given in (50) so that

(54) ker¢=k(zs)®k(26)$k(Eab)€Bk(E°d)Gak(0)$k(0).
21 22 Z3 Z4 25 26

Here, zl,...,26 E J(B),(a,b),(c,d) are distinct ordered pairs in {(1,3),(1,4),
(2,3), (2,4)}. To see this, we proceed as follows. In Equation (50), 315,316 E J (B)
ThUS, .115 = a1E13 + 02E14 + 031323 + 04E24 and ya = blEl3 + b21314 + b35323 + b41324
for some a,-,b,- E 16,1 3 i S 4. Since 3);, 74 O,a,- # 0 for some i. We can assume

a1 aé 0. By replacing ( 0 ) by ( 0

316 316

bl = 0. Hence, ya = b2E14 + b3E23 + b4E24. Since ya 76 0, some b,- 9é 0 for i = 2, 3, 4.

We can assume b2 75 0. By replacing (yO) by (0) —a2b§1(y0) 76 O, we
5 6

315

) — blal‘l ( yo ) if need be, we can assume
5

can assume y5 = 01E13 + a3E23 + a4E24. Thus, y5 = 01E13 + a3E23 + a4E24 and
ya = b2E14 + b3E23 + b4E24 with al 7e 0 and b2 91$ 0. The ordered pair (1,3) is one of
four ordered pairs appearing in (50). We can assume (1, 3) = (i, j). Since a1 75 0, we

can write ken/z as follows.

-55556:)55665562">56r>5<z>5<z>-

31

Here, 317 = alyl + a3y3 + am. The ordered pair (1,4) is one of three ordered pairs
appearing in (55). We can assume (1, 4) = (p, q). Since b2 75 O, we can write ken!) as

follows.

<555-5-4“)55(“)55(Em)5k(’3w>540>540)-
y7 ya 313 .714 95 ye

Here, ya = b2y2 + b3y3 + b4y4. Setting 21 = y7,22 = y8,23 = y3,z4 = y4,z5 =
315,26 = ya, (a, b) = (m,n), and (c,d) = (u,v), we have (54). Note that, 21 75 0.
For, if not, then ( 38 ) , ( Z ) E ken/J. This implies 315 E Ann3(N) = (0) which
is impossible. By the same argument, 22 3:9 0. Furthermore, {z1, 22, 25, 26} is linearly

independent. For, if not, then there exist t, E k,i = 1, 2, 3, 4, not all zero such that

tlzl + t222 + t3Z5 + t425 = 0. Thus,

t125+t225 _ Z5 25 O 0
(57) ( 0 )—t1(zl)+t2(z2)+t3(z5)+t4(z6)Eke'l‘w.

Suppose t1 = t2 = 0. Then, (57) implies t3 = t4 = 0. This is impossible. Thus, t1 # 0
or t2 aé 0 and hence t125+t226 75 0. Equation (57) implies t125+t2z5 E Ann3(N) = (0)
which is impossible. Therefore, {21, 22, 25, 26} is linearly independent. Thus, a basis
for [rent can be given as in (54) with {21, 22, 25, 26} is linearly independent.

Now, we are ready to define an isomorphism between 8x622 and B x (13’2 / kerzb)’.
For simplicity, we will denote cosets ( f: ) + OS (A) in Q by ( : )—. We will write
elements in thQ2 as orders triples (b, qhqg). Here, b E B,q1,q2 E Q. It is easy to

check that the following 13 elements form a k-vector space basis of 8x622.

32

:61 = (14:0: 0):/82 = (El3a03 0):fi3 = (E1430: 0)

54 = (E23,0,0)555 = (E241), (”.56 = (0, ( g ) ,0)

(58) fi7=(0,(104) ,0),fis=(0,(%3) ,0),,69=(0,(E024 )-,0)
I - 0 - E23 .—
510 =(0v0a(04) )afill=(0:0s(l4) )afil2=(0:0a( 0 ) )
E24 _-
513 —(0v02( 0 ) )

We will denote cosets ( Z ) + ken/1 in Bz/kerz/J by ( j: ) . It is easy to check that

the following 13 elements form a k-vector space basis of B x (15'2 / ken/2)?

61 = (I4: 0: 0): 62 = (_zl:010)a63 = (—22a0:0)

0

5 H W )5 Mm
510 =(0,0,(8)_),6u= (0,0, (104)),612=(0, o, (0 ))
613 =(o,o,(206)-).

Define a map a : 8x622 —-> B l>< (Hz/kerw)2 by

64 = (25,0,0),65= (26,0, 0), 66 = (0, (I46) ,0)

13 13
(60) 01254): 2236,, t.- e In = 1,. . . ,13.
i=1

i=1

Then, obviously, a is a k-vector space isomorphism. Notice that
[Bifij =0=6,-6_,- for 23i35, j=2,3,4,5,8,9,12,13
(61) flifij = O = 6‘65; for 6 _<_ Z,j S 13

fl? =0=¢§i2 for 2SiSI3.

I.

33
Furthermore, owl/6,) = 0(6) = 6,- = 616.- : 0(fil)a(fi,) for i = 1, . . . , 13. Thus, to

show 0 is a k-algebra isomorphism it remains to show the following.

0(51'56) = 0(fli)0(fl6), 0(5ifl7) = 0(fli)0(fl7)
0(fii1610) = 0(fii)0(fi10), 0(51'511) = 0(fli)0(511) for i: 2,3,4,5-

(62)

Notice that the third and fourth equations in (62) are actually the same as the first
and second equations in (62) but in the third slot. Thus, we will finish the proof by

verifying the first and second equations in (62).

563237) = 0(0, ( b5; )—,0) =50 ( ”233 )-.0) = (0, ( 0 )-,0)
0

(0, ( _,1 ) ,0) = <—zl.o,0><o, ( f3 ) ,0) = 5.5. = 5055(5)

E14 _ -21 -
0(fi3fl6) =0(0:( 0 ) ’0)=0(0’0,0)=(0a0a0)=(0i( 0 ) v0)
= (—22,0,O)(0, ( g ) ,0) = 6366 = 0(,83)0'(fi5)
0 - E24 - 26 _
50367) = 5(0. ( a.) ,0) =a(0.( 0 ) ,0) = (0.( 0) .0)
= (O, ( _022 ) ,0) = (—Zg,0,0)(0, ( [04 ) ,0) = (5367 = 0(fl3)0’(,37)
E23 - 25 -
0(fl4fi6) = 0(0: ( 0 ) :0) = (O: ( 0 ) :0)
= (25,050)(05 ( g ) ,0) = 5456 = 0(34)0(56)
0(6467) = 0(0,( E33 >10): 0(0,0,0) = (0,0,0) = (0, ( Z )_ ,0)

= (Z5,0,0)(0, ( [04 )- ,0) = (5467 = 0'(,84)0'(fi7)

34

(63)

26

E24 - _
arms/36> =a(0,( ) ,0)=<0,(0) ,0)

= (25,0,0)(0,( 0 )— ,0) = 6566 = 0(fl5)0(fl6)

Q

0(6567) =0(0,< £4 >10) =a(0,0,0) = (0,0,0) = (0, ( Z )-,0)
0

= (261 0:0)(0: ( I4 ) ,0) = 5557 = “500(57)-
Thus, 0(fi,~flj) = 0(fi,)0(fij), 1 S i, j S 13 and hence a is a k-algebra isomorphism.

Therefore, B [X Q2 g B x (Eff/hemp)?

Notice that BQ/kerz/J E N as B-modules. Let f : Bz/kerzb —> N be a B-
module isomorphism. Then, the map 0’ : l3 o< (13’2/lce1‘zb)2 ——» B x N 2 defined by
0’(b,n1,n2) = (b,f(n1), f(n2)) is a k-algebra isomorphism. Thus, 8x622 ’5 Bt><N2 as
k-algebras. C]

In [3], Brown and Call showed that C E Bx(k4)2. Thus, by Theorem 2.10,
C E’ 8 IX N 2 for any N E M 8(4). This implies there is only one isomorphism class [C]
in 93.

Recall (B,N) E X implies (BKN‘,B’€BN) E MX. To classify the isomorphism

classes in (201, we need the following lemma.

Lemma 2.11: Let (B,N), (B,M) E X. Suppose N E M as B-modules. Then,
(B [X N‘, B‘ 69 N) and (B x M‘,B’ 69 M) are (0,T)-isom0rphic.

Proof: Let f : N -—> M be a B-module isomorphism. Define amap 0‘ : BxN‘ —>
B D< M‘ by o(b,n1,...,ng) = (b,f(n1),...,f(m)) for b E B,n,- E N,i = 1,...,2. It

is easy to show 0 is a k-vector space isomorphism. If (b, m, . . . ,ng), (b’, n’l, . . . ,n}) E

35

B x N‘, then
(64)
0((b,n1,.. .,ng)(b’,n’1,...,n2)) = 0(bb’,n’1b + nlb’, . . . ,ngb + ngb’)
= (bb’,f(n§ b + n, b’), . . . ,f(n§b + ngb’))
= (bb’,f(n§)b +f(n1)b’,~.,f(n2)b+f(nz)b’)
= (b,f(n1),--5,f(ne))(b’,f(ni),~<,f(n2))

= 0(b,n1,.. .,ng)0(b’,n’1,...,n2).

Thus, a is a k-algebra isomorphism. If we define a map 1' : 3‘ EB N -—> B‘ 69 M

by r(b1,...,bg,n) = (b1,...,bg,f(n)). Then, T is a k-vector space isomorphism.

Obviously,
(65) T((b1, . . .,bg,n)(b,n1, . . .,Tlg» = T(b1, . . . ,b¢,n)0(b, n1, . . . ,ng).
Thus, (B x N’,B‘ 69 N) EMT) (B l>< M‘,B‘ 69 M). [I]

Now, we are ready to classify the isomorphism classes in QC).

Theorem 2.12: Let Q and N be as in Theorem 2.10. Then, (BxQQ, 8269Q) 9.7”)
(B x N2,B2 69 N).

Proof: Let

£1 = (14,090): {2 = (E13,0’0)) £3 = (E141010)
£4 = (E23: 0:0): 65 = (E241010)3 £6 = (031430)

E7 = (09 E13,0), £8 = (01E14)O)a £9 = (0,E23,0)
(65)

£10 =(0’E24)0), £11=(0303( 8 ) )3612 = (0:0: ( IO) )
4

513 = (0,0, ( L33 )3, 514 = (0:0, ( £84 )3

36

Then, {51, . . . ,{14} is a k-vector space basis of 82 EB Q and let

771 = (14:0:0)? 772 = (—Z11010)3 773 = (—22’ 0’ O)
774 = (Z5: 0: O), 775 = (‘36: 0: O): 776 = (091430)
777 = (0a —Zl: 0): 778 = (07 -Z2: 0)) 779 = (0: 25: O)

I - 0 -
7’10 = (0: 26:0): 7711 = (010’ 4 )1 7712 = (0:01 )
0 I4

1713 = (0.0, ( f; )3, m4= (00(3)).

Then, {171, . . . ,n14} is a k-vector space basis of 82 EB (BZ/kerzb), where 1b is a map in
Theorem 2.10. Let 7' : 32 EB Q -—+ B2 EB (132 /ker11)) be the map defined by

14 14
(68) T(Z tifi) = Ztinz’: ti 6 k, 1 S 2 S 14.
i=1 i=1

Then, T is a k-vector space isomorphism.

Let 0 be the k-algebra isomorphism in Theorem 2.10. Let b,-, n, E B for
i = 1,2,3,j = 1,2, 3,4, 5, 6. Then,

bi = 7214 + aiEl3 + biEM + CiE23 + diE24
(69)
”j = 33'14 + PjEis + qu14 + qu23 + vjE24

for some r,,a,,b,-,c,-,d,-,sj,pj,qj,uj,vj E k,i=1,2,3,j = 1,2,3, 4,5,6. Since ( E13 ),

0
(%4),(E2,),(E3,)ecm

n1 - = 8114 + P1E13 + 91E14 + 111323 + 7115324 - = 3114 + U1E23 + 111324 -
8214 + P2313 + 921314 + 71521323 + 7121324 8214 + P2E13 + (12314

37

Since E23 , E24 E CS (A), Equation (70) becomes
-Eis _E14

(71) 3214

(nl )— = ( 8114 + (U1 +p2)E23 + (’01 + Q2)E24 )— .

(72) n3 _ = 8314 + (U3 + p4)E23 + (113 + Q4)E24 —
n4 .9414 °

and

(73) n5 — = 3514 + (“5 +p6)E23 + (””5 + QG)E24 _
”6 8614 '

Notice that

r(b1,b2, ( Z; ) ) = T(b1,0,0)+T(0,bg,0)+1’(0,0,( Z; ) )
= (T114 — 0.121 - 0122 + C125 + d125, 0, 0)

(74) +(0, T214 — 0.221 — 0222 ‘1‘ C225 + d226, 0)
+(0, 0,( 8114 + (U1 +p2)z5 + ('01 + Q2)26 )-).
8214
Let
L1 = T114 — (1121 — 0122 + C125 + (1125
, L =rI—az—bz+cz+dz
(75) 2 2 4 2 l 2 2 2 5 2 6

L = 8114 '1“ (U1 +p2)25 + (’01 + Q2)25 -
3 S214 .

38

Then, T(b1,b2, ( :1 ) ) = (L1, L2, L3). Notice that
2
(76)

0(b3, ( :3 )1 ( Z: )) = 0(b3,0,0) +0(0, ( Z: )-,0) +0(0,0, ( Z: ))

= (7.314 _ (1321 — b322 + 6325 "l' (1326: 0: 0)

+(0, ( 3314 + (“3 +P4)25 + (03 + Q4)26 )- ,0)
S414

+(0 0 8514 + (“'5 +p6)25 + (715 + gs)26 -)
a a 8614 -

Let
M1 = T314 — (1321 — 0322 '1' C325 ‘1' ((325
_ 3314 + (U3 + p4)z5 + (03 + (1“,);6 ‘
(77) M2 — ( 8414

M = 8514 + (“as + P6)25 + (’05 + (16)Zs _
3 8514

Then, 0(b3, ( :3 ) , ( n5 ) ) = (M1,M2,M3). Thus,

4 ”6

(78) T(b1,b2, ( Z; ) )0(b3, ( Z: ) , ( :1: ) )= (L1,L2,L3)(M1,M2,M3)
=(L1M19 L2M11L3M1+ M2111 + M3112),

Since y,- E J(B) for i = 5,6, 7, 8, we have

39

(79)
L1M1 = T1(T3I4 — (1321 — b322 + C325 + d326) — T3(a121 + b122 — C125 — (1126)
= r1r3I4 - (T103 + r3a1)zl — (rlb3 + r3b1)22 + (r1c3 + r301)z5 + (T1d3 + r3d1)26
L2M1 = T2(T3I4 — 0.321 — b322 + C325 + d326) —- T3(a221 + b222 — C225 — (1226)
= T2T3I4 — (T203 + T302)21 — (T2b3 + T3b2)22 + (T2C3 + T3C2)25 + (T2613 + T3d2)26
_ 31(T314 - 0321 — 19322 + 0325 + daze) + r3((u1 +P2)25 + (’01 + €12)26) —
L3M1 —
82(T3I4 — 0.321 — b322 + C325 + (1326)
S1T3I4 +(r3u1+ T3172 + 3103 + S203)25 + (T301 + T3Q2 + 81d3 + 82b3)26
32r31;
M L r1(s3I4 + (U3 + p4)25 + ('03 + q4)26) — 33(alzl + b122 — c125 — dlzs) _
2 1 84(T1I4 — 0.121 — b122 + C125 + d125)
713314 + (71713 + T1194 + 8301+ 840025 + (71113 + 7194 + 83611 + 84b1)25_
347114
M L T2(S5I4 + (115 + p5)z5 + (115 + q6)26) — 85(0221 + b222 — C225 — (1225) _-
3 2 86(T2I4 — G221 - b222 + C225 + (1226)
:( T2S5I4 + (T2115 + 72“; + 8502 + 860.2)25+ +(T2’U5 + T2q6 + S5d2 + 86b2)26 )—
36T21h

On the other hand,
(80)

n1 - n3 - n5 _ _ n1b3 _ n3b1 '— n5b2 _
(bl:b2, ( n2) )(b3a ( n4) ’(n6) )— (biba,b2b3, ( 71253) +( n4b1) +( 71652) )
Equation (71),(72),(73) imply that

(81)

blba = 7‘1(T314 + 03513 + b31314 + 031323 + (13324) + 73011513 + 51314 + 611323 + 611524)
= 1‘17‘3I4 + (T103 + T301)E13 + (T1b3 + T3b1)E14 + (T1C3 + T301)E23 + (1‘1d3 + T3d1)E24

5253 = 72(7314 + 03313 + 531314 + 631323 + 6131324) + T3(02313 + 521314 + 62323 + d2E24)
= 1'21'314 + (T203 + r3a2)E13 + (7‘2b3 + T3b2)E14 + (T263 + T362)E23 + (T2d3 + r3d2)E24

40

81(7‘314 + 03313 + b3E14 + 631323 + 613324) + r3((u1 + P2)E23 + (v1 + <12)Ez4) )—

S2(7‘3I4 + a3E13 + b3E14 + C3E23 + d3E24)
817‘314 + (8103 + T3111 + T3192 + 3203)E23 + (31d3 + T‘31)1 + 7'302 + 82b3)E24 )-
32T31h

84(7'114 + 01313 + 1711314 + 011323 + 611324)

331‘114 + (3361 + Tlua + T1114 + 3401)E23 + (33d1 + T103 + T194 + 8450324 )—
84T114

85(7‘214 + 02313 + 52514 + 62523 + d2E24) + 72((1‘5 + P6)E23 + (“Us + (16)Ez4) )-

86(7‘214 + 02E13 + b21314 + 62323 + (12324)

857'214 + (3502 + T2115 + T2136 + 8602)E23 + (85612 + T205 + 7206 + 36b2)E24 )—

n3b1 _- _ 33(7'114 + 01513 + b1E14 + C11323 + d1E24)+ T1((u3 +104)E23 + (2);, + Q4)E24) —
n4b1 —
( 861’214

Horn Equation (80), we have
(82)

7((b1,b2, ( 2: )-)(b3, ( ”'3 )1 ( ”5 )3) = T(b1b3,0,0) +T(0,b2b3,0)

nlbs — n3bl _
+r(o,o, ( ”2,3 ) )+T(o,o, ( "4,1 ) )
n5b2 _—
+T(0,0, ( nebz) ).

Since

(83)
1'(b1b3, 0,0) - = (r1r314 — (r103 + r3a1)zl — (r1b3 + r3b1)22 + (T103 + r3cl)zs + (r1d3 + r3d1)26,0,0)
T(0, b2b3,0) = (0, 7'21'3I4 — (T203 + T302)Z1 - (T2b3 + 7‘3b2)22 + (T263 + T362)25 + (1‘2d3 + T3d2)26, 0)

710,0, n1b3 _) = (0,0, 811314 + (8163 + mm + T3172 + 8203)25 + (sld3 + r3121 + r3q2 + 32(13):;6 ')
MI)" 827314

710,0, n3b1 -) ___ (0, O, 331‘114 + (3361 + r1113 + T1124 + 8401)25 + (33d1 + 1‘1'03 + 7104 + 84b1)zs -)

71451 847‘114
n5b2 _) ___ (0 0 351'214 + (8562 + rzus + 1‘ng + 36a2)25 + (85612 + T2115 + Tzqe + 86b2)26 -)
n6b2 , , 867214 '

710,0, (

41

Therefore, from Equation (78) to (83),

(84)

T((b1,b2,(,’;;) )(b3,(::j) ,(Zg) ))=T(b1,b2,(:;) )o(b3,(::) ,(Zg) ).
Thus,

(85) (B o< Q2, 132 ea Q) 210,) (B v< (Bz/kerw, 132 as (B2/ker¢)).

Since B2/ker2/z ’5 N as B-modules, (B [X (l3""'/lce1'¢)2,l3’2 EB (B2/ker¢)) EWJI)
(BKN2,B2$N) by Lemma 2.11. Therefore, (BxQ2,B2€BQ) $701.71) (BxN2,Bz®N),
where 01 = 0’0 and 7'1 = 7’7. [:1

We have now proven the following assertion. If N E M 8(4), then
(B x N2,l32 GB N) E1“) (8 IX (k4)2,82 EB 16“) $0,571) (C, k”). Thus, 901 has exactly

one (a, T)-isomorphism class [(C, k14)].

Chapter 3

Nonuniqueness of Algebras in 9

3.1 Construction of New Algebra 8 in Q

It has been conjectured for a long time that the set 9 = {(R, J, k) E
M14(k)| dimkR = 13 and i(J) = 3} has only one isomorphism class [C]. It turns
out the isomorphism class [C] is not unique. In this section, we will construct a new

k-algebra (8, J, k) E 52 such that [S] sé [C].

If (R, J, k) E Q, then by Theorem 2.3, we may assume every 1" E J has the form
in (9). From Theorem 2.5, we may assume every 1" E Soc(R) has the form in (13).

We can then write R = k[)\1, . . . , A8, E11, E12, E21, E22], where

02 0 0
(86) A1: Ai 010 0 , Z=1,....,8

Conversely, suppose R is a commutative, k-subalgebra of T14 of the form
R = k[)\1, . . . , A3, E11, E12, E21, E22], where dimkR = 13 and A1, . . . , A8 have the form
given in Equation (86). (We are not assuming R is maximal). Then, R is a local ring
with Jacobson radical given by J = (A1, . . . , A3, E11, E12, E21, E22) and residue class
field It. We will give a necessary and sufficient condition on the A,’s and B,’s which

will imply R E Q.
42

43

For a matrix A 6 menUc), we will let her A = {u E Mlxm(k)|uA = 0} and
NS(A) = {v E Mnx1(k)|Av = 0}.

Theorem 3.1: Let R = k[/\1,...,/\8,E11,E12,E21,E22] be a commutative,
k-subalgebra of T14. We assume dimkR = 13 and each A,- has the form given in
Equation (86). Suppose {)le ker(A,-) = (0) and (18:1 N S(B,-) = (0). If r E 07,,(R),

then 1' has the following form.

02 0 0
(87) 1‘ = P 010 O + a114, a E k.
( Z Q 02 )
X1 X2 X3
Proof: Let r = X4 X5 X6 6 CT,,(R). Here, X1,X9 6 T2 and X5 6 T10.
( X7 X8 X9 )

Then, TEL-J- = Eij r and

X1X2 X3 020 O 020 O X1X2 X3
(88) X4 X5 X6 A5 010 0 = Ai 010 0 X4 X5 X6

X7 X8 X9 W B; 02 W Bi 02 X7 X8 X9
for all i = 1,. . . ,8. Thus, we have the following equations.

(a) X2141 + X3W = 0 (8) AiX3 = 0
(89) (b) X38, = 0 (f) XgAi + XgW = WX1 + BiX4

(C) XSAi + XGW = AiXI (g) X9Bi = WX2 + BiXS

(d) XGB, = A,-X2 (h) WX3 + B,X6 = 0.

These equations hold for all i = 1, . . . ,8 and all W 6 T2. We also have the equations
obtained by replacing A,- and B,- in (a) through (h) with the zero matrix. Since
X3W = 0 for all W 6 T2, we have X3 = 0. Then, (a) implies X2A, = 0 for all
i = 1,. . . ,8. Thus, X2 6 “i=1 ker(A,~) = (0). Hence, X2 = 0. Equation (h) implies
13,-Xe = O for all i = 1,. . .,8. Thus, X5 6 (19:1 NS(B,) = (0). Hence, X5 = 0. Since
X 9W = WX1 for all W 6 T2, we have the following equations.

44

X9E11 = E11X1 X91312 = E12X1
X91321 = E21X1 X9E22 = E22X1-

(121 (I22 b2l (’22
Then, (90) implies an = an = bu = by and an = 0.21 = b12 = b21 = 0. Thus,

Let X1 = (all a”) and X9 = (bll b” ) Here, a,,-,b,-, e k, i,j = 1,2.

X1 = X9 = 0.1112. In (C), 1813 W = 0. Then, X5145 = A1X1 = A,(a1112) = 011.41.
Hence, (X5—a11I10)A,' = 0, for alli = 1, . . . ,8. Thus, X5—011110 E 018:1 k€T(Ai) = (0)

which implies X5 = anI 10. Therefore, 1' has the form in (86). [I]

Let R = k[)\1, . . . , A8, Eu, E12, [3321, E22] be as in Theorem 3.1. Theorem 3.1 implies

any 7‘ e CT,,(R) has the form given in (87). Notice that all matrices of the form
02 0 0
(91) O 010 O and a1”

are elements in CT,,(R). In the next theorem, we characterize those P’s and Q’s for

which 1' E CT,4(R).

Theorem 3.2: Let R = k[/\1,...,A8,E11,E12,E21,Ezg] be the k-subalgebra in

02 0 0
Theorem 3.1. Let r = P 010 0 + a1” 6 T14. Then, 1' E CT,,(R) if and
Z Q 02
(RolelT
only if (RODESV E NS(A). Here, Rom-Q is the i-th row of Q, Col,P is the i-th
I COl2P

column of P, and A E M32x4o(k) is the following matrix.

45

 

 

 

(COI1A1)T —R0’wlB1 0 O
(COZ2A1)T 0 0 —Row1B1
O "RO’LU2B1 (001114071 0
O 0 (0012241)T —ROIU2B1
(92) A:
((0011A8)T —R0’UJ1BS 0 0 \
(COI2A3)T 0 0 —R0wlBg
0 —R0’U)2B8 (COI1A3)T 0
_ \ O 0 (COI2A8)T —R0’I.U2Bg) ‘

 

Proof: Suppose r e CT,4(R). Then, for all i = 1, . . . , 8,
02 0 0 02 0 0 02 O O 02 0 0

(93) P 010 O A.- 010 0 = A,- 010 O P 010 O .

Z Q 02 W B,- 02 W B,- 02 Z Q 02

Therefore, QA, = B,P for i = 1,... ,8. Let

(i) a“)

011012 bu) . . . b6)

A,= ' 3‘ Bi=(b(lil) 5(1‘1)0),f0ri=1,...,8
.51. an 21

(94)

P11 P12 q g

P = : : = 11 ° " 110 .

' ° Q ( 921 ° " (1210

P101 P102

Here, a111,, bmn, pm", q"m E 19. Since QA, = B,P for all i = 1,. . . ,8, we have

j: -—l glJaj1L-21‘01 bl?Pj1= 0, 1:1 quaJa-(B) —21-0=1b(13)pj2= O
(95)
21-0 1 qzja-(i) _2101 b91931: O 2101 (1210(1) :101b(i)pj2__ 0.

46

It is easy to check (95) is equivalent to

(Role)T
0,01113

(96) A (Row2Q)T
. 00l2P

=0.

Conversely, if P and Q satisfy Equation (96), then QA, = B,~P for all i = 1,. . . ,8.
Hence, by Equation (93), r E 07,,(R).

Theorem 3.3: Let R = k[/\1, . . . , A3, 1311,5312, 321, E22] be a commutative,
k-subalgebra of T14. We assume dimkR = 13 and each A,- has the form given in

(86). Then, the following two statements are equivalent.

(a) R E M14(k)

(b) §=1 ker(A,-) = (0), L1 NS(B,) = (O), and rank(A) = 32.

In Theorem 3.3, A is the 32 x 40 matrix given in (92).

Proof: (a) => (b) Let u = (111,. . . ,ulo) 6 {18:1 ker(A,-). Then,

02 0 0
(97) 0 010 0 6 Soc(R).
o (.., 02

0

Theorem 2.5 implies dim,c Soc(R) = 4. The elements 1.3),), i, j = 1, 2 are clearly in
Soc(R). Hence, Soc(R) = L(E11,E12,E21,E22). Thus, it = 0 and hence
13:1 Iced/4:) = (0).

Let U = ('01,...,v10)T E 0:; NS(Bi). Then,

02 0 0
(

(v0) 010 0 ESoc(R).
0 O 02

47

Since Soc(R) = L(E11,E12,E21,E22), v = 0. Therefore, 0le NS(B,) = (0).

Let
(R01111Bi)T
__ COI1A,‘ . __
(99) (2,-— (W213i? , i—1,...,8.
COl2A.‘

Since A,- E R = CT,,(R), oz,- 6 N S(A) by Theorem 3.2. Since A1, . . . , A8 are linearly
independent, (11,. .. ,as are linearly independent. Hence, dimkNS(A) Z 8. Let

w E NS(A). Since w E M40x1(k), we can write w as follows;

(RolelT
COllp
(R0102QlT
COZ2P

(100) w =

for some P E M10x2(k) and Q E M2x10(k). Let

02 O 0
(101) T: P 010 0 .

OQ02

Then, by Theorem 3.2, r E 07,,(R) = R. Thus, r = clAl + + CgAg for some
6,- E k, i = 1,. . .,8. Hence, w = 01011 + ---+ C8/\3. Therefore, dimkNS(A) S 8 and

hence dimkN S (A) = 8. We conclude rk(A) = 32.

(b) => (a) Since rank(A) = 32, dimkNS(A) = 8. Let 02,-, i = 1,...,8 be the
vectors defined by (99). Since dimkR = 13, A1, . . . , A8 are linearly independent over
1:. It easily follows that al, . . . ,as are linearly independent over It. Thus, {ab . . . , a8}
is a basis of N S (A) If r 6 07,,(R), then Theorem 3.1 implies r has the form given
in (87). Thus, by Theorem 3.2,

(RowinT
(102) (12311201215?
COZ2P

e NS(A).

48

This implies

02 O O
(103) P 010 0 E L()\1, . . . , A3).
0 Q 02
Therefore, r E R and hence CTI,(R) = R. We conclude R 6 M1409). E]
Thus, we can easily check whether a k—subalgebra R = k[)\1, . . . , A8,

E11, E12, E21, 322] of the type given in Theorem 3.3 is in 9.

Now, we will construct a new k-algebra (S, J, k) E Q with the following matrices.

Let

 

 

 

 

02 0 0
(104) 5,:(10, 010 o ), i=1,...,8.
0 Q1 02
Here,
I2 02 02 ) Q2 )
02 I2 02 2
P1: 02 1 P2: O2 1 P3“ I2 1 P4: 02 1
Q2 02 02 I2
02 02 02 j 02 f
02 O2 \ 02 \ 02
02 02 02 02
P5 = 02 1 P6 = Q2 1 P7 — Q2 , Pa = 02 ,
02 Q2 02 02
E11 E12 f E21 f E22

and
Q1 = (12 02 02 02 Eu), Q2 = (02 I2 02 02 En)
Q3 = (Q2 Q2 [2 02 E21), Q4 = (Q2 Q2 02 I2 E22)
Q5 = (E11 02 E21 02 02), Q6 = (E12 02 E22 02 02)
Q, = (02 E11 02 E21 02), (28 = (02 E12 02 E22 02).

49

Throughout the rest of this thesis, we will let S = k[61, . . . , 63, E11, E12, E21, E22] with
61, . . . , 68 given by (104). Notice that S is a k-subalgebra of T14. The multiplication
table for J (S ) is as follows:

Table 1: Multiplications of 635

 

 

 

 

 

 

 

 

 

51 62 (53 54 65 (55 67 68
(51 E11 + E22 0 O 0 E11 E12 0 O
62 0 E11 '1' E22 0 O O 0 E11 E12
(53 O 0 E11 '1' E22 0 E21 E22 0 O
64 O O 0 E11 + E22 0 0 E21 E22
55 Eu 0 in o o o o o
56 E12 0 E22 0 o o o o
6-, o 157,, 0 E21 0 o o o
53 0 E12 0 E22 0 O O O

 

 

 

 

 

 

 

 

 

 

 

We don’t include the multiplications for Eij’s since E,jJ(S) = (O) for all i, j = 1, 2.

Theorem 3.4: Let S = k[61, . . . ,63, E11, E12, E21, E22] be the k—subalgebra of TM

defined by the equations in (104). Then,
(a) S E M14013)
(b) (S, J, k) e 52

Proof: (a) It is easy to check that S is a local, commutative, k—subalgebra of T14
with dimkS = 13. Obviously, {P1, . . . , P3} is linearly independent. Furthermore,
8 kerm) = (0) and (1:. NS(Q.) = (0). Let

i=1

50

(0011P1)T —R0le1 0 0 q
(COI2P1)T 0 0 —R0’UI1Q1

0 —Row2Q1 (C'ollPl)T 0

O 0 (001213 1)T —Row2Q1

(105) A =

(COZ1P8)T —R0’le8 0 0
(0012P8)T o o —Roles

0 —R0’U)2Q8 (COl1P8)T 0

0 0 (COl2P8)T -R0’w2Q8 -

 

 

Then, A E M32x4o(k) and rank(A) = 32. Thus, by Theorem 3.3, 8 E M1406).

(b) We can easily check that dimkS = 13 and i(J) = 3. Thus, (8, J, k) E Q by

(a). C]

In Theorem 3.4, we constructed a new k-algebra (S, J, k) E Q. In the next section,

we will show [8] 75 [C]. Hence, 8 determines a new isomorphism class in 9.

3.2 The Algebra 5

In this section, we will prove the k-algebra S constructed in Theorem 3.4 is not
a (B, N )-construction if k = R and is a (B, N )-construction if k is an algebraically
closed field. We will prove that 8 is not k-algebra isomorphic to C. Therefore, we can
conclude that Q has at least two k—algebra isomorphism classes [8] and [C]. It also
follows that (S, k”) is not (a, T)-isomorphic to (C, k“). Furthermore, we will prove

(S, k“) is not a (II-construction.

Theorem 3.5: Suppose k = IR. Then, S is not a (B, N)-construction.

51

Proof: Suppose S is a (B, N )-construction. Then, by [3: Theorem 4], 8 contains

an ideal I which satisfies the following two properties.
(106)
(a) Ann5(I) = I

(b) O —» I ——+ S Li 8/ I —* 0 splits as k-algebras, i.e., there exists a k-algebra

homomorphism V6 = 15/].

Since 13,-I = 0 for 1,3' = 1,2, 1.53,,- e I, i, j = 1,2 by (a). Notice that a, e 1.
Otherwise, 6? = 0 by (a). Since 6? = E11 + E22, this is impossible. Thus, 61 ¢ I. Let
6 : S/I —» S be a splitting map. Then, 0(61+ I) = 61+ r, where r E I. Since 0 is a

k-algebra homomorphism, we have

6% + 261r = 6% + 2617' + r2
= (61 +1")2 = (0051+ 1))2
= 6((61 + 1)?) = 0(En + E22 + I)
= 9(0 + I) = 0.

(107)

Let r = ?=1t151 + Ziezi Sngjg, t,, 31-; 6 IR. Then, (107) implies

(108) (1 + 2t1 + 2t5)E,1+ 2t5E12 + (1 + 2m}?22 = 0.
Thus, t1 = —-,:;, t5 = t6 = 0. Hence, we have
1 2 -
(109) T = —§61 + t262 + t363 + t464 + t767 + t863 + Z Sngjg.
j,(=l

Since r E I, r2 = 0 by (a). Thus,

(1 +153 + t3 + t3 + 2t2t7)Eu + 2t2t3E12 + 20,1753,
.(110) _
+(fi + t3 + t3 + ti + 2t4ts)E22 = 0.

52

Therefore, we have the following four equations.

t2t8=0
t4t7=0
fi+t§+t§+ti+2t2t7=0
§+t§+t§+t§+2t4tg=0

(111)

We will show that there is no real solution of the equations given in (111). Since

t2t8 = 0, t2 = 0 or t8 = 0. Thus, we have the following two cases to consider.
Case 1: t2 = 0
Case 2: t3 = 0

We will show both cases lead to a contradiction.

Case 1: Suppose t2 = 0. Then, from the third equation in (111), we have

i + t3 + ti = 0. This is impossible since t3, t4 6 R.

Case 2: Suppose t8 = 0. Then, the fourth equation in (111) implies

% + t3 + t3 + t} = 0. This is again impossible since t2, t3, t4 6 R.

Thus, the equations in (111) have no real solutions. This implies that there is
no r E I such that 0(61 + I) = 61 + r. Thus, there is no splitting map of the exact

sequence given in (106). Therefore, 8 is not a (B, N )-construction. [I]

It was conjectured that every R E M1409) is a (B, N )-construction. Theorem 3.5
implies this conjecture depends on k. If k = R, then S is not a (B, N )-construction.
If k = C (complex numbers), then S is a (B, N )-construction. More generally, we

prove S E M14(k) is a (B, N )-construction if k is an algebraically closed field.

Theorem 3.6 Suppose k is an algebraically closed field. Then, S is a (B,N)-

construction.

53

Proof: Since k is an algebraically closed field, the polynomial f (1:) = :02 + 1 E k[:r]
has a root i. Set 021 = 61 — i152, (12 = 63 — 1'64, (13 = 65 — i157, and a4 = 66 —
i68. Then, c1165 = E11, (1166 = E12, (1265 = E21, (1265 = E22. Thus, the ideal
I generated by a1,qg,ag,a4 contains Emu for all m,n = 1,2. It is easy to check

I = L(a11a21a31a41E111E121E211E22)‘ Thus1 dzmk(I) = 8'

Let )6 E Ann5(I). Then, 6 = 28=1tn6n + 23nm=l smnEmn for some tmsmn E k.
Since an E I for all n = 1,2, 3,4, anfl = 0 for all n = 1, 2,3,4. From alfl = (123 = 0,

we have

( ) (t1 — it) + t5 — 11:7)E'11 + (16 — it3)E12 +(t1 — my?” = 0
112
(t3 — 2'14)E11 + (:5 — it7)E21 + (t3 - it4 +16 — its)E22 = 0.

Equation ( 112) implies

tl—it2+t5—it7=0
t3—it4+t5—’lt8=0

t —it =0
(113) 6 .8
t1-1t2=0
t3-it4=0
t5—lt7=0

Thus, we have t1 = itg, t3 = it4, t5 = ity, and t6 = its. Hence,

,8 = ’lt261 + t262 + 11463 '1' L164 '1' ’lt765 + #366
(114) +t767 + t868 + Zgnm=l SmnEmn

= “201 + 71402 + “703 + ”304 + £31m=1 SmnEmn-
Therefore, B E I and hence Ann5(I) g I . Since 12 = 0, I ; Ann5(I). Thus,
Ann5(I) = I.

Notice that A = {114 + 1,61 + I, 63 + I, 65 + 1,66 + I} is a k-vector space basis of
S/I. Since dimk(I) = 8 and dimk(8) = 13, we have dimk(S/I) = 5. Since ion 6 I

54

for all n = 1,2, 3,4, we have

62+I=—i61+1
64+I=—i63+1
67+I=—i65+I
68+I=—i65+l.

(115)

Let 0 be the k-vector space homomorphism from S / I to 5 defined as follows:

0(114 + I) = 1,,

6(61 + I) = 161 + 5‘62
(116) 0(63 + I) = §63 + $64

(9(65 + I) = §65 +1167

6(66 + I) = 1,66 +1168.

Then,

(9(62 + I) 6(— )= —i9(61 + I ) = §62 — #6,
0(6., + I): 6(— 2'63 + I): —i6(63 + I) = §64 — 5163
0(67 + I): 6(— —i65 + I) —i6(65 + I ) = §67 — £165
6(68 + I): 0(— 456 + I) = 40056 + I): §68 — 12°66

—’l(51+I

(117)

Furthermore, 6 is a k-algebra homomorphism. To see this, we proceed as follows. Let
7,7’ E 8/]. Then, 7 = (tIM + a) + I and 7’ = (t’IM + a’) + I for some t,t’ 6 k and
a, a’ E J(S). Note that

9(77’) = “(@114 + a) + I)((t'114 + 0’) + 1))
= 0(tt’Il4 +ta’ +at’ + aa’ + I)
= 0(tt’Il4 + ta’ + at’ + I)
= 6(tt’Il4 + I) + 0(ta.’ + I) + 6(at’ + I)
= tt’0(Il4 + I) + t6(a’ + I) + t’0(a + I)
= tt’Iu + t0(a’ + I) + t’0(a + I).

(118)

55

and
9(7)9(’Y’) =6((t114+a)+I)0((t’Il4+a’)+I)
=(0 (tI14+I)+0(a+I))(0 (t’I14+I)+6(a’+I))
= (tIM + 6(a + I))(t’Il4 + 0(a’ + 1))
=tt ”Il4+t6(a +I)+t’0(a+I)+9(a+I)6(a’ +1).

Thus, it remains to show 6(a + I)6(a’ + I) = 0. Let a = 2:1 un6n + an,n=1vmnEmn
and a’ = _1 u’ 6,, + Zmn_ lvmnEmn,un,ufi,,vmn,vfi,m E 1:. Then, we have
a + I = (U1 — 2U2)61 + (U3 — 2U4)63 '1" (U5 — 2U7)65 '1' (U6 — 2U3)65 + I

(119)
a’ + I= (u’l — iu2)61+(—iuf,)63 + (113 — iu§)65 + (112s — iu’8)65 + I

Therefore, by ( 116)

0(0 + I) ='2-(U1 — 2U2 (61 + 262)+ 3(U3 — 2U4)(63 + 264)

)
+3(u u5 — iu7)(65 + i67)+ 3(u6 — iu3)(65 + i68)
(120)
0(0’ + I) = 3(U’1—2U’2 (61+262)+3(U§—2UQ)(63 + 264)
(

+3(u’5 — iu ’) 65 + i67) + 3(14; — iué)(6t + i621)-
By using Table 1, we have

6(0. + [)0 (a + I): 3((U1 — 2U2)(61 + 262) '1" (U3 — 2U4)(63 + 264)
(121) +(u5 — iu7)(65 +167) + (06 — 2U3)(65 + i68))((u’1 — iu’2)(61 +162)
+(ug-iuf1)(63 + i6.;) + (u;3 — iu7)(65 + i67) + (u;5 — iug)(65 + i63))

=0

Recall u : S ——> 8/ I is the natural homomorphism defined by v(r) = r + I for r E S.
Then,

v0(114+I) =v(Il 4)=114+I

v9(61+I) — (361+ 3i62)= 361+3i62+I
-(361+1-i62)+ (-361—3i62)+1
—61+I '

(122)

56

06(63 + I) = 11(363 + 3i64) = 363 + 3i64 + I
= (363 '1' 37:54) + (353 ‘ 3254) + I
= 63 + I

226(65 + I) = ’U(365 + 3267) = 365 ‘l' 3267 + I
= (365 + 3167) + (365 — 3267) + I
= 65 + I

210(65 + I) = 21(366 + 3268) = 365 + 3263 + I
356 + 3268) + (356 — 3268) + I
= 66 + 1.

Thus, it is easy to check u0(r + I) = r + I for all r E S. This implies the exact

sequence
(123) O——*I—-18-—>S/I—-+O

splits as k-algebras. Therefore, the ideal I of 8 satisfies the two conditions in

[3:Theorem 4] and S is a (B, N )-construction. C]

Theorem 3.5 and 3.6 show that the question: “When is (R, J, k) E Q a (B, N )-
construction?” depends on the field 1:. From Theorem 3.6, one could conjecture that
every (R, J, k) E Q is a (B, N)-construction if k is an algebraically closed field. At

present, this conjecture is still opened.

Next we show 5 is not k-algebra isomorphic to C. In what follows, we will need a

multiplication table for C. Let

02 o o
(124) 1,: A,- 010 o ,i=1,...,8.

O B, 02

57

 

 

 

 

 

 

 

 

 

 

 

Here,
(In (02 on (on
02 12 02 02
A1 = 02 1 A2 "" 02 a A3 — [2 a A4 — 02 a
02 02 02 12
w.) \02 02) \02/
r02 on (02 (on
()2 ()2 ()2 ()2
A5 "' 02 , A6 = 02 3 A7 _ 02 7 A8 = 02 a
02 02 02 02
\ E11 E12 J \ E21 \ E22 )
and

Bl = (02 02 02 02 E11), B2 = (02 02 02 02 E12)
Ba = (02 02 02 02 E21), B4 = (02 02 02 02 522)
B5 = (E11 02 E21 02 02), 35 = (E12 02 E22 02 02)
B7 = (02 E11 02 E21 02), Ba = (02 E12 02 E22 02).

Then, C = k[/\1, . . . , A3, B11, B12, B21, B22] and the multiplication table for J (C) is as

follows:

Table 2: Multiplications of A,’s

 

 

 

 

 

 

 

 

 

A1 A2 /\3 A4 A5 A5 A7 A8
A1 0 O O 0 E1 B12 0 0
A2 0 O O O O 0 B11 B12
A3 0 O O 0 B21 B22 0 0
A4 0 O O O O 0 B21 B22
A5 E11 0 E21 0 O O O 0
A6 B12 0 B22 0 O O O 0
A7 0 E11 0 E21 0 O O 0
A8 0 B12 0 B22 0 O O O

 

 

 

 

 

 

 

 

 

 

 

We don’t include the multiplications for Bij’s since EU J (C ) = (O) for all i, j = 1, 2.

58

We will now prove that the set 9 = {(R, J, k) E ./\/114(k)|dim;c R = 13,i(.I) = 3}

has at least two isomorphism classes [8] and [C].

Theorem 3.7: Let (S, J, k) E Q be the k-algebra given in Theorem 3.4. Then, S

is not k-algebra isomorphic to the Courter’s algebra C.

Proof: Let X, . . . ,X be indeterminates over It and set A = k[X1, . . . ,Xg]. Let I

be the following ideal (in A):

I = (X12,X§,X§,X},X52,X§,X72,X82,X1X2,X1X3,X1X4,X2X3,
X2X4,X3X4, X5X6,X5X7,X5X8,X6X7, szs, X7X8, X1X7,
X1X8,X2X5, X2X6, X3X7. X3X8,X4X5, X4X6, X1X5 — X2X7,
X1X5 — X2X3, X3X5 — X4X7, X3X6 — X4X3).

(125)

Let 1r : A ——+ C be the map defined by 7r(X,) = A, for all i = 1,. . .,8. From Table
2, it is easy to check 7r is a surjective, k-algebra homomorphism. Table 2 also implies
I g kervr. Thus, the map 7'r : A/I —, C defined by 1'r(f + I) = 7r(f) is a well-
defined k-algebra epimorphism. Let m = (X1, . . . ,Xg). Then, m3 (_I I and hence
{1 + I, X1 + I, . . . ,X8 + I, X1X5 + I,X1X6 + I,X3X5 + I, X3X6 + I} is a k-vector
space basis of A/I. Thus, dimk(A/I) = 13. Since dika = 13, 1? is a k-algebra

isomorphism. Thus, C E’ A / I as k-algebras.

Let L be the ideal of A defined as follows:
(126)

L: (X52,X62,Xg,X3,X1X2,X1X3,X1X4,X2X3,X2X4,X3X4,X5X6,
X5X7,X5X8,X6X7, X6X8,X7X8,X1X7,X1X8, X2X5,X2X6,X3X7,X3X8,
X4X5, X4X6, X? - X3, X12 — X§,Xf — X3, X1X5 — X2X7,X1X5 — X2X3,
X3X5 - X4X7, X3X5 — X4X3, X12 — X1X5 - X3X6).

'Let 7r1 : A ——» S be a map defined by 7r1(X,-) = 6,- for all i = 1,...,8. Then, 7r is

a surjective, k-algebra homomorphism. Using Table 1, L g ker7r. Hence, the map

59

in : A / L —> 8 defined by 7T1 (g+L) = 7r1(g) is a well-defined k-algebra epimorphism.
Since m3 g L, {1 +L,X1+L,...,X8+L,X,2+L,X1X5 +L,X1X6+L,X3X5+L}
is a k-vector space basis of A/ L. Thus, dimk(A/ L) = 13. Since dimkS = 13, 7T1 is a

k-algebra isomorphism. Hence 8 g A / L as k-algebras.

Suppose 8 is k-algebra isomorphic to C. Since 8 E A/L and C E A/I, there
is a k-algebra isomorphism go : A/ I —) A/ L. Notice that A/ I , A/ L are standard
graded rings. Since J(C)3 = (0) and J(S)3 = (0), we have A/I = Co 69 C1 69 C2 =
k EBCI Elan and A/L = So €981 6982 = It €981 €982. Here, Cu and 8,, are the n—th
homogeneous components of A/ I and A/ L, respectively. Since C and S are local
rings, J(A/I) = m/I and J (A/L) = m/L. Since w is a k—algebra isomorphism,
cp(m/I) = <p(J(A/I)) = J(A/L) = m/L. Thus, <p((m/I)“) = (m/L)" for all n > 0.
We can now define amap 2p : grm/1(A/I) —> grm/L(A/L) given by 1/)(a+(m/I)"+1) =
<p(a) + (m/L)"+1 for all a 6 (m/I)". Since <p((m/I)”) = (m/L)“ for all n > 0, d)
is well-defined, k-vector space homomorphism. Since (p is surjective, w is surjective.

Hence, 1,!) is a k-vector space isomorphism.

Next we will show 1/J is a k-algebra homomorphism. Let al 6 (m/I)" and let
(12 E (m/I)‘. Then,
(127)

112((01 +(m/1)"+‘)(a2 + (Tn/Um» = Waiaz + (m/I)"+‘+‘)
= M01102) + (m/L)"+‘“
= Mam/9&2) + (m/ 10W“
= (Mal) + (m/L)"+1)(<P(a2) + (Tn/LY“)
= 10(01 + (m/I)"+‘)¢(az + (Tn/IV“).

Thus, 2,!) is a k-algebra isomorphism.

Let P1 3 k 9C1 €902 --* ng/1(A/I) be a map defined by p1(a,,) = a,, + (m/I)"+1

for all an 6 C", n = 0,1,2. Then, it is easy to check that p1 is a k-vector space

60

isomorphism. Let an, 6,. E C... Then,
(128)
101((00 + 011+ a2)(50 + 51+ flail) = P1(aofio + C1031 + 01150 + 0032 + 01250 + 01151)
= P1(Oofio) + P1(0051 + 0150) + 101(0052 + a2fio + 0151)
=¥ C1030 + (00131 + 01150 + (m/I)2) + (01032 + 02130 + 0151
+(m/I)3))
= (ao +(01+(m/1)2)+(02 + (m/1)3))(fio +(fi1+(m/1)2)
+(52 + (m/I)3))
= 01(010 + 011+ 02)P1(30 + 51+ 52)-

Thus, p1 is a k-algebra isomorphism.

L€t P2 1 ($6331 6352 —-* grm/L(A/L) be a map defined by p2(7,,) = 7,, + (m/L)"+1
for all 7,, 6 8n, n = 0,1,2. Then, by the same arguments above, p2 is a k-algebra

isomorphism.

Let a = p;1 oz/xopl : kEBCl €962 —» k®816952. Then, a is a k-algebra
isomorphism with (J(Cn) = 8,, for all n = O, 1, 2. Thus, a is a k-algebra isomorphism
which is homogeneous of degree 0. Therefore, we may assume (,0 : A/ I —> A/ L is a
k-algebra isomorphism which is homogeneous of degree 0. Since <p(C1) = 81,
<p(X,- + I) = L1 ainj + L for some (1,-,- 6 k, i,j = 1,. . . ,8. Here, det(a,-,-) 75 0. Let
77 : A —» A be a map defined by n(X,-) = 23;, (1,-ij for all i = 1,. . .,8. Then, by

[10, Corollary 2, p137], n is an automorphism.

Let Go : A —+ A/I and 91 : A —> A/ L be natural homomorphisms. Then the

following diagram commutes.

A .91. A/I
(129) 771 190
A _61_, A/L

61

To see this, we proceed as follows. The four maps in (129) are homogeneous of degree
0. Hence, it suffices to show that <p60( f ) = 9117( f ) for any homogeneous form f in A.
Let f(:t:1, . . . ,axn) be of d-form in A. Then,

v(00(f($1, . . . .2371)»

(:1+I) "290(3371'1’1»

W(f($1+1,.. w$n+1))
(4p
(28

f
f: _1 aux, + L. ..,:ng agjxj + L)
f(2§_1a1jzj, . . . , 22:1 agjxj) + L

f

77

01

(130)

( ( ) ---,17($e))+L
(f($1,-.$3))+L
n(f($1,-- 9311))

Thus, the diagram in (129) commutes. Hence, (p can be lifted to an automorphism n

which is homogeneous of degree 0.

Notice A/I is an A-module via 60, 1‘(a + I) = 00(r)(a. + I) = ra + I for r, a E A.
Also, A/L is an A-module via 6117, Th = 61(n(r))b for r E A, b E A/L. If r E A
and a E A/I, then

(131) 90(7‘0) = <p(0o(r)a) = <p(00(r))s0(a) = 01(n(7"))<p(a) = 7Ma)-

Thus, cp is an A-module isomorphism homogeneous of degree 0.

The minimal free resolution of the A-modules A / I and A/ L are as follows:

(132)
0—2A4—)A27—)A92—>A204-+A296->A266—>A136—>A32—)A—)A/I—+O

O—>A4-¥A26->A87—+A197—2A293—)A266->A136-+A32—>A—’A/L—+0.
These resolutions were computed using Macaulay. Notice the resolutions have
different betti numbers. By [5: Proposition 1.5.16], this is impossible. Thus, A/ I
is not k-algebra isomorphic to A / L. We conclude that S is not k-algebra isomorphic

to C. C]

62

By Theorem 3.7, we conclude that the set 0 = {(R, J, k) E M14(k)|dimkR =

13, i(J) = 3} has at least two isomorphism classes [C] and [8].

Since 8 is not k-algebra isomorphic to C, (8, V1) is not (0‘, T)-isomorphic to (C, V2)
for any finitely generated, faithful, S—module V1 and for any finitely generated, faithful
C—module V2.

In our last theorem in this thesis, we will prove that (S, k”) is not a
Cl-construction. Let B be the Schur algebra of size 4 given in (2). Let B be a
k-algebra which is k-algebra isomorphic to B. Suppose f : B —* B is a k-algebra
isomorphism. Let N be a finitely generated, faithful, B-module. Then, N is a finitely
generated, faithful, B—module via f. Hence, we can form the pairs (B x N‘, 8‘ 69 N)
and (B x N‘,B‘€BN) in MX.

Theorem 3.8: With the notation given above, suppose dimk(N) = 4. Then, f
induces a (a,T)-isom01phism(a,7') : (8 IX N‘,B‘ 63 N) -2 (B x N‘,B‘ GB N).

Proof: Recall N is a B—module via nb = nf(b). Let a : B o< N‘ —-> B l>< N‘ be the

map defined by

(133) a(b,n1,...,ng) = (f(b),n1,...,ng).

Then, it is easy to check a is a k-vector space isomorphism. Let (b, n1, . . . ,ng),

(c,m1, . . . ,mg) 6 8 IX N‘. Then,

0((b, n1, . . . ,ng) (c,m1, . . . ,mg)) = a(bc,m1b+ nlc, . . . ,mgb + 72(6)
= (f(bc),m1b+ nlc, . . . ,mgb+ 12(0)
(134) = (f(b)f(C), m1f(b) + n1f(C),- - . ,mzf(b) + nef(C))
= (f(b),n1,-..,ne)(f(C),m1, . - . ,me)

= a(b,n1, . . . ,n()a(c,m1, . . . ,mg).

Thus, a is a k-algebra isomorphism.

63
Letr:B‘EBNHBZEBNbeamapdefinedby
(135) T(b1,...,b£,n) = (f(b1),....,f(bg),n)

Then, it can be easily checked that r is a k-vector space isomorphism. Let

(b,n1,...,ne) 68x Ne and let (b1,...,bg,n) EB‘EBN. Then,

T((b1, . . . ,b(, n) (b, n1, . . . ,ng)) = r(b1b, . . . ,bgb, nb + 25:1 nibi)
= (f(b1)f(b)a - - - a f(b£)f(b), nf(b) + Zi=1nif(bi))

(136)
= (“51), ° ° - af(bl)1n)(f(b)inla ° - ° 1”!)
= T(b1,...,bg,n)a(b,n1,...,ng).
Thus, we conclude (B t>< N‘,B‘ 69 N) 20”) (B o< N‘, B‘ 69 N). D

We can now prove that (8,1914) is not Cl-construction by using the result in

Theorem 3.8.

Theorem 3.9: Let (S, J, k) 6 Q be the k-algebra constructed in Theorem 3.4.
Then, (S, k“) is not a Cl-construction.

Proof: Suppose (S, k“) is a Cl-construction. Then, (S, k”) is (a, r)-isomorphic to
(BKN‘, B‘®N) for some (B, N) E X and E E N. Let d = dimk(B) and n = dimk(N).
Since 8 is k-algebra isomorphic to B x N t and k” is k-vector space isomorphic to

B‘ 83 N, dimk(8) = dimk(B D< Ne) and dimk(k”) = dimk(B‘ G} N). Thus, we have

13 = d + Zn
(137)
14 = Ed + n.
The only solution (d, n,€) E N3 for Equation (136) is d = 5,n = 4,2 = 2. Thus,
(5, k“) Ea”) (B x N2,B2 69 N).
Notice that J(B o< N2) = J(B) x N2. From this, it easily follows that

i(J(B x N2)) = i(J(B)) + 1. Since 8 is k-algebra isomorphic to B x N2 and

 

 

64

i(J(S)) = 3,i(J(B)) = 2. Since dimk(B) = 5 and i(J(B)) = 2, B is k-algebra
isomorphic to B. Theorem 3.8 implies (B l>< N2, B2 69 N) afimm) (B o< N2,B2 EB N).
Thus, (S,k14) Em”) (B D< N2,B2 69 N), where 02 = 01 o a and 7'2 = 7'1 0 7'.
By [4, Proposition 1], (B x (lc“)2,B2 EB 16’) 910,333) (C, k”) and by Theorem 2.12,
(B x N’",B2 EB N) Em“) (B M (lc“)2,B2 EB k4). Thus, (8, k”) E’ww) (C, k”), where
0’ = 03 o 04 o 02 and r’ = 7'3 0 r4 0 r2. Therefore, 8 is k-algebra isomorphic to C. This
is impossible by Theorem 3.7. We can conclude that (S,k14) is not Cl-construction.

D

Theorem 3.9 implies that if (R, J, k) E Q, then we can not conclude (R, It”) is a

Cl-construction.

 

 

Appendix

 

65

Appendix

We will prove Subcase 1 through 9 in Case 4 of the proof of Theorem 2.9.

Subcase 1: Suppose dimk(Ann3(a,-)) = 1 for i = 1, 2, 3. Let dimk(Ann3(a,-)) =

S] O 0
ks,,s,~ E J(B),i = 1,2,3. Then, 0 , $2 , 0 e kerz/J. Let
0 0 83

$2 $3 $4 $5
(138) 55 = 92 ’66 = 213 ,57 = 314 ,58 = y5
22 Z3 Z4 Z5

be a basis of kercp. Here, z,,y,~,z,- E J(B). Since dimk(J(B)) = 4, {$1,11,$2,$3,.’II4,

x5,$6,x7,x8} is linearly dependent. Thus, there exist d,c,- E k,i = 1,...,8, not
all zero such that d31+ z§=,c,z,- = o. If c.- = 0 for all i = 1,...,8, then at 7A 0
and dsl = 0. This implies 31 = 0. This is impossible. Hence, some c,- is not
zero. We can assume cs 96 0. Thus, $3 6 L(sl,:1:1,...,:1:7). We can repeat this
argument four times and assume $4, $5,186,137,“ 6 L(sl, $1,232, 2:3). Therefore, :84 =
dsl + c1231 + 02.732 + 03233 for some d,c1,c2,c3 E 1:. Since {61,64,65,66,67} is linearly

0
independent, dbl +c164 +c265 +c366 — 67 = ( u; ) 79 0. Since {61, . . . ,67} is linearly
’01

 

66

0
independent, ( ul ) ¢ kbg + [£63. Thus,

91
81 O 0 $1
61: 0 :62: 32 $63: 0 ’64: 3’1
0 0 83 231
- $2 $3 0 $5
(139) 55 = 92 :66 = 93 ’57 = “1 ,58 = 95
22 23 v1 25
$5 $7 $8
59 = 96 ,510 = 97 ,511 = 98
26 27 23

is a basis of kergo. We can repeat this argument four times and assume
81 0 0 $1
51: 0 ’52: S2 ’53: 0 ’54: 91
O 0 83 251
$2 $3 0 0
(140) 55 = 92 , 56 = 93 ,57 = 91 :68 = “2
2'2 Z3 ’01 ' ’02
0 0 0
59 = Us , (510 = U4 ,511 = Us
’U3 U4 U5

is a basis of kercp.

Since dimk(J(B)) = 4, {32,u1,...,u5} is linearly dependent. Thus, there exist
d,c,- E k,i = 1,. . . ,5, not all zero such that (182 + 2L1 cm,- = 0. We can assume

0
Cs 7&4 0. Since 0162 + c167 + 0268 + c369 + 04610 - 611 = ( 0 J for some d, c,- E k,i =
v

1, 2, 3, 4, v E J(B) and (62,67, 63, 69,610, 611} is linearly independent, v 75 0. If v = ts3
for some t E k, then d62 + c167 + c268 + c369 + c4610 — 611 — t6;; = O. This is impossible.

0 0
Thus, v ¢ ks3. Therefore, ( 0 ) E ker<p\k63. This implies ( 0 ) ,63 E Ann3(a3).

’U ’U

 

 

67

This is a contradiction since dimk(Ann3(a3)) = 1.

Subcase 2: Suppose dimk(Ann3(a1)) = 2 and dimk(Ann3(a,-)) = 1 for i = 2, 3.

Let Ann3(al) = k51+ ksg,AnnB(a2) = ks3, and Ann3(a3) = ks4,s,- E J(B),i =

81 82 0 0
1,2,3, 4. Then, 0 , O , s3 , 0 E kergo. Let
0 0 0 84

$1 $2 $3 $4
(141) 65 = yl ’66 = y2 167 = 3’3 ’68 = 314
2'1 22 Z3 24

be a basis of kercp. Here, 23,-,y,, z,- e J(B). Since dimk(J(B)) = 4, {31, 52,2:1,a:2,a:3,

2:4, 2:5,z6,$7} is linearly dependent. Thus, there exist d1,d2,c,- E k,i = 1, . . . ,7, not
all zero such that dlsl + d232 + 23:10:93: = 0. If c,- = O for all i = 1,...,7, then
dlsl + d282 = 0. This implies d1 = d2 = 0. This is impossible. Hence, some c,- is
not zero. We can assume c7 75 0. Hence, :57 6 L(31,32,$1, . . . ,ze). We can repeat
this argument four times and assume x3,$4,2:5,:56,x7 E L(sl, 32,11,232). Therefore,
233 = dlsl + (1232 + c1221 + 022:2 for some d,,c,- 6 k,i = 1, 2. Since {61,62,65,66,67} is

0
linearly independent, dlbl + d262 + c165 + c265 — 67 = ( ul ) 74 0. Since {61, . . . ,67}

’01

 

 

68

O
is linearly independent, ( ul ) ¢ 1:63 + I664. Thus,

vi
31 32 0 0
(51: 0 ,62= 0 ,63= 83 ,64= 0
0 0 0 34
2:1 2:2 0 2:4
(142) 55 = 92 , 56 = 93 ’57 ui , 58 = 94
22 23 ’01 24
335 $6 $7
59 = 95 ,510 = 96 .511 = 97
25 26 Z7
is a basis of kercp. We can repeat this argument four times and assume
81 82 0 O
(51: 0 ,62= 0 ,63= S3 ,64= 0
0 0 0 34
$1 $2 0 O
(143) 65 = 91 166 = 92 a 57 = “1 ,58 = 92
21 22 v1 ’02
O 0 0
59 = Us ,510 = 114 1511 = “5
213 v4 ’05

is a basis of kergo.

Since dimk(J(B)) = 4, {s3,u1,...,u5} is linearly dependent. Thus, there exist
d,c,- E k,i = 1, . . .,5, not all zero such that d83 + 2:21 em, = 0. We may assume

0
CS 7é 0. Since (163 '1' C167 + 6268 + C369 + C4610 — 611 = ( 0 ) for some d, Ci 6 [6,2 =
v

1, 2, 3, 4, v E J(B) and {63, 67, 68, 69,610, 611} is linearly independent, v 74 0. If v = ts4
for some t E k, then d63 + 0167 + 0268 + c369 + c4610 - 611 — t64 = 0. This is impossible.

0 0
Thus, v ¢ ks4. Therefore, ( O ) E ker<p\k64. This implies ( O ) ,64 E Ann3(a3).

'U ’U

 

 

69

This is a contradiction since dimk(Ann3(a3)) = 1.

Subcase 3: Suppose dimk(Ann3(a,-)) = 2 for i = 1, 2 and dimk(Ann3(ag)) = 1.

Let Ann3(al) = ksl+k32,AnnB(a2) = k33+ks4, and Ann3(a3) = 1:35, 3,- E J(B),i =

31 82 0 O 0
1,2,3,4,5. Then, 0 , 0 , 33 , s4 , 0 E kergo. Let
0 0 0 0 85

$1 $2 $3
(144) 65 = $66 = 91 967 = 92 568 = 93
S5 21 22 23

be a basis of kercp. Here, z,,y,-, z,- E J(B). Since dimk(J(B)) = 4, {51, 82,$1,$2,$3,

x4,x5,xs,} is linearly dependent. Thus, there exist d1,d2,c,- E k,i = 1,...,6, not
all zero such that dlsl + d282 + z?=,c.-x.- = 0. If c,- = 0 for all i = 1,...,6, then
dlsl + d282 = 0. This implies d1 = d2 = 0. This is impossible. Thus,c,- 91$ 0 for
some i. We can assume c6 95 0. Hence, £6 E L(sl,32,xl,...,a:5). We can repeat
this argument three times and assume x3,x4,$5,$6 E L(sl,32,:1:1,:1:2). Therefore,
33 = dlsl + d232 + 611:1 + c2x2 for some (1,, c,- E k,i = 1, 2. Since {61,62,66,67,63} is

0
linearly independent, 61161 + (1262 + C165 '1' C267 — 63 = ( U1 ) '7'é 0. Since {61, . . . , (53}
’01

 

 

70

0
is linearly independent, ( ul ) E 1963 + 1964 + 1665. Thus,

91

0 $1 $2 0
(145) 55 = 0 ,56 = 91 ,57 = 92 168 = 91
35 21 22 '01

$4 $5 $6

59 = 94 1510 = 95 ,511 = 96

24 25 26

is a basis of kergo. We can repeat this argument three times and assume
S1 82 0 0
51: O ’52: 0 ’53: 33 154: 34
0 0 0 0
0 2:1 3:2 0
(146) 65 = O 966 = 91 a 67 = 92 :68 = 221
35 21 22 91

is a basis of kergo.

 

Since dimk(J(B)) = 4, {53, 34,121, . . . ,u4} is linearly dependent. Thus, there exist

d1,d2,c,- E k,i = 1,...,4, not all zero such that (2153 + d234 + ELIQU, = 0. We

0
may assume C4 3'5 0. Since d163, + d264 ‘1" C163 ‘1" C269 + C3610 — 611 = ( 0
'U

) for some

d1, d2, c,- E k,i = 1, 2, 3, 4, v E J(B) and {63, 64, 68, 69, 610, 611} is linearly independent,

v 79 0. If v = tss for some t E k, then d163, + d264 + c163 + 0269 + c3610 — 611 - t65 = O.

O

This is impossible. Thus, v E 1:35. Therefore, ( 0 ) E kercp\k65. This implies

’U

71

O
( O ) ,65 E Ann3(a3). This is a contradiction since dimk(AnnB(a3)) = 1.
v .

Subcase 4: Suppose dimk(Ann3(a,)) = 2 for i = 1, 2, 3. Let Ann3(a1) = ksl +

ksg,Ann3(ag) = ksg + 1:34, and Ann3(ag) = [:35 + kss,s,- E J(B),i = 1,2,3, 4, 5, 6.

31 $2 0 0 0 0
Then, 0 , 0 , 33 , s4 , 0 , 0 E kercp. Let T
o 0 o 0 55 36 '

 

0 $1 $2
(147) 55 = 166 = 0 ’57 = 91 ’68 = 92
S5 85 21 252

$3 $4 $5

59 = 93 ,510 = 94 ,511 = 95

Z3 Z4 25

be a basis of kercp. Here, :c,,y,~,z,- E J(B). Since dimk(J(B)) = 4, {31, S2,$1,$2,$3,

$4,235} is linearly dependent. Thus, there exist d1,d2,c,- E k,i = 1,...,5, not all
zero such that (1131 + dgsg + 23.110193: = 0. If e,- = 0 for all i = 1,...,5, then
dlsl + (1232 = O. This implies d1 = (12 = O. This is impossible. Thus,c,- 79 0
sor some i. We can assume c5 79 0. Hence, 3:5 E L(sl,52,$1,...,a:4). We can
repeat this argument two times and assume $3,214,225 E L(sl, 52, 2:1, 232). Therefore,
233 = dlsl + £12.92 + 012:1 + C2222 for some d,,c,- E k,i = 1,2. Since {61,62,67,68,69} is

' 0
linearly independent, dlbl + 62262 + c167 + c268 — 69 = ( ul ) 74 0. Since {61, . . . ,69}

'01

 

72

0
is linearly independent, ( ul ) E 1563 + 1664 + 1665 + [£66. Thus,

vi
31 S2 0 0
51: 0 ’52: 0 ’53: 33 ’54: S4
0 O 0 0
l 0 0 $1 $2
(148) 65 = 0 166 = O 167 2 3’1 168 = y?
35 36 21 22
0 $4 $5
59 = “1 .510 = 94 ,511 = 95
v1 Z4 25

is a basis of kercp. We can repeat this argument two times and assume
S1 S2 0 O
51: 0 152: 0 ’53: 33 ’54: S4
0 0 0 0
0 0 $1 $2
(149) 65 = 0 :66 = 0 ,57 = 91 968 = 92
85 36 Z1 z?
0 0 0
59 = “1 ,610 = “2 ,511 = “3
’01 v2 ‘03

is a basis of hemp.

Since dimk(J(B)) = 4, {33, 34,111, . . . ,u3} is linearly dependent. Thus, there exist

d1,d2,c,- E k,i = 1,2, 3, not all zero such that d1s3 + d234 + 2:10:91 = 0. We

0
may assume c3 # 0. Since d163 + d264 + 0169 + c2610 - 611 = ( 0 ) for some
v

d,,c.- E 12,2 = 1,2,2) 6 J(B) and {63,64,69,610, 611} is linearly independent, 'U 74 0. If
'U = 2135+t286 for some t1, 22 E k, then d163+d264+6169+02610—611—t165 —t265 = 0.

O
This is impossible. Thus, 12 E 13.95 + kss. Therefore, ( O ) E kergo\k65 + 1966. This
12

 

 

73

0
implies ( 0 ) ,65, 66 E Ann3(oz3). This is acontradiction since dimk(AnnB(ag)) = 2.
v

Subcase 5: Suppose dimk(Ann3(al)) = 3 and dimk(Ann3(a,-)) = 1 for i = 2, 3

Let Ann3(al) = ksl+ks2+ks3,AnnB(a2) = 11:34, and Ann3(a3) = kss, s,- E J(B),i =

l 31 32 83 0 0
1,2,3,4,5,. Then, 0 , 0 , O , s4 , 0 ,E kercp. Let
0 0 0 0 35

$1 $2 $3
(150) 55 = ,55 = 91 ’57 = 92 :68 = 93
85 21 Z2 Z3

be a basis of kercp. Here, 23,-,y,-, z,- E J(B). Since dimk(J(B)) = 4, {S1, 32,x1,$2,$3,

24,35,236} is linearly dependent. Thus, there exist d1,d2,c,- E k,i = 1,...,6, not
all zero such that dlsl + (1282 + Z?=1c,-$,~ = 0. If c,- = O for all i = 1,. . . ,6, then
d181+d282+d383 = 0. This implies d1 = d2 = 42;; = 0. This is impossible. Thus,c,~ aé 0
for some i. We can assume c6 91$ 0. Hence, 1:6 E L(S1,S2,$1, . . . ,z5). We can repeat
this argument four times and assume $2,333,154,:r5,a:6 E L(sl, 32, 33,11). Therefore,
2:2 = dlsl + d232 + d333, + can for some d,,c E k,i = 1, 2, 3. Since {61,62,63,66, 67} is

0
linearly independent, d161 + c1262 + d363 + C65 — 67 = ( ul ) 91$ 0. Since {61, . . . ,67}

91

 

 

74

0
is linearly independent, ( ul ) ¢ [664 + 1665. Thus,
91

31 32 S3 0
61= 0 ,62= O ,63= 0 ,64= S4
0 0 0 0
0 2:1 0 2:3
(151) 65 = 0 ,66 = 311 ,67 = 221 ,63 = y3
s5 21 U1 23
$4 $5 $6
59 = 94 1510 = 95 ,511 = 96
Z4 Z5 26

_ is a basis of kergo. We can repeat this argument four times and assume
31 32 33 0
61: 0 ,62= 0 ,63= 0 ,64= S4
0 O 0 0
0 $1 0 0
(152) 55 = 0 :66 = 91 157 = U1 ~ ’68 = U2
35 zl v1 ‘02
O 0 O
59 = Us 1510 = U4 1511 = U5
v3 v4 Us

is a basis of kergo.

Since dimk(J(B)) = 4, {s4,u1,...,u5} is linearly dependent. Thus, there exist
d1,c,- E k,i = 1,. . .,5, not all zero such that d134 + 2:15:101Ui = 0. We may assume

0
c5 79 0. Since c1164 + C167 + c268 + c369 + c4610 — 611 = ( 0 ) for some d1,c,- e k,i =
v

1, 2, 3, 4, v E J(B) and {64, 67, 68, 69,610, 611} is linearly independent, v 7e 0. If v = tss
for some t E k, then d164 +clb7+0268 +6369+C4610 — 611- t65 = 0. This is impossible.

0 0
Thus, v ¢ 1635. Therefore, ( O ) E kercp\k65. This implies ( 0 ) ,65 E Ann3(a3).
v v

 

 

75

This is a contradiction since dimk(Ann3(a3)) = 1.

Subcase 6: Suppose dimk(AnnB(a1)) = 3,dimk(AnnB(a2)) = 2, and

dimk(AnnB(a3)) = 1. Let Ann3(a1) = ksl + 1632 + 1633, Ann3(a2) = 1634 + kss, and

31 32 S3 0 0 0
Ann3(a3) = 1666,55,- E J(B). Then, 0 , O , 0 , s4 , 35 , 0 E
0 0 0 O O 36

kercp. Let

0 0 $1 $2
(153) 55 = 35 :66 = 0 167 = 91 168 = 92
0 85 Z1 22

$3 $4 $5

69 = 93 1 610 = 5’4 1 611 = y5

Z3 Z4 25

be a basis of kergo. Here, x,,y,-,z,- E J(B). Since dimk(J(B)) = 4, {31,32, 33,351,32,

23,14,235} is linearly dependent. Thus, there exist d1,d2,d3,c, E k,i = 1,. . . ,5, not
all zero such that dlsl +d2S2 +d333 +Zf=1ciazi = 0. If e,- = 0 for all i = 1, . . . , 5, then
dlsl +d232+d333 = O. This implies d1 = d2 = d3 = O. This is impossible. Thus,c,- 79 O
sor some i. We can assume as 51$ 0. Hence, 135 E L(sl, s2, 33, 2:1, . . . , 3:4). We can repeat
this argument three times and assume $2,$3,:c4,$5 E L(sl,32,33,$1). Therefore,
2:2 = dlsl + d232 + d333 + can for some d,,é E k,i = 1, 2, 3. Since {61,62,63, 67,68} is

O
linearly independent, (1161 + d262 + d363 + C67 — 68 = ( 111.) # 0. Since {61, . . . , 68}
U1

76

O
is linearly independent, ( ul ) ¢ [:64 + 1665 + 1666. Thus,

111
31 32 33 0
61: 0 ,62= O ,63= 0 ,64= S4
0 0 O 0
0 0 $1 0
(154) 65 = 35 , 63 = 0 ,67 = y1 ,63 = 21.1
0 86 21 ”1
$3 $4 $5
59 = 93 1510 = 94 , 511 = 95
Z3 Z4 Z5
is a basis of kercp. We can repeat this argument three times and assume
31 32 33 C)
61: 0 ,62= 0 ,63= 0 ,64= S4
0 O O 0
0 0 $1 0
(155) 65 = 35 166 = 0 167 = yl 1 68 = U1
0 86 Z1 ‘01
O O 0
59 = U2 1510 = Us 1511 = U4
v2 '03 U4

is a basis of kercp.

Since dimk(J(B)) = 4, {s4, 35, 141,. . . ,u.;} is linearly dependent. Thus, there exist

d1,d2,c,- E k,i = 1,...,4, not all zero such that c2184 + d235, + 23:16:12,- = 0. We

0
may assume c4 79 0. Since d164 + (1265 + c163 + 0269 + 03610 — 611 = ( 0
v

) for some

d1,d2,C{ 6 [13,2 = 1,2,3,?) 6 J(B) and {64,65,63,69,610, 611} is linearly independent,

v 7é 0. If v = tss for some t E k, then d164 + d265 + c163 + c269 + c3610 - 611 — t65 = O.

0

’1)

This is impossible. Thus, v ¢ kss. Therefore, ( 0 ) E ker<p\k66. This implies

 

77

0
( 0 ) , 66 E Ann3(a3). This is a contradiction since dimk(Ann3(ag)) = 1.
v

Subcase 7: Suppose dimk(AnnB(a1)) = 3, dimk(Ann3(a,-)) = 2 for i = 2, 3.

Let Ann3(al) = ksl + kS2 + k33,Ann3(a2) = 1634 + ks5, and Ann3(ag) = 1686 +

31 S2 33 0 0 O 0
ks7,s,- E J(B). Then, 0 , 0 , O , s4 , s5 , 0 , 0 E
C) C) C) C) C) 86 87

kercp. Let

0 0 $1
(156) 65:=: 35 166:=: C) 167:= 168:=: yl
C) 86 S7 21

be a basis of kercp. Here, x,,y,-, z,- E J(B). Since dimk(J(B)) = 4, {31, S2, S3,$1,$2,

$3,234} is linearly dependent. Thus, there exist d1,d2,d3,c,- E k,i = 1,. . . ,4, not all.
zero such that dlsl + d282 + d333, + 23:1 qr,- = 0. If e,- = O for all i = 1,. . . ,4, then
dlsl+d232+d333 = O. This implies d1 = d2 = d3 = O. This is impossible. Thus,c,- 75 0
for some i. We can assume c4 aé 0. Hence, 1:4 E L(31,S2,S3,$1,$2,$3). We can
repeat this argument two times and assume 32,33,224 E L(sl, 52, 83,231). Therefore,
2:2 = dlsl + 61232 + (1333 + cal for some d,,c E k,i = 1,2,3. Since {61,62,63, 63, 69} is

0
linearly independent, d161 + d262 + d363 + C63 — 69 = ( ul ) 51$ 0. Since {61, . . . ,69}

91

 

 

78

0
is linearly independent, ( ul ) é k6,; + [:65 + [:66 + 1667. Thus,

U1
31 52 53 0
61: 0 ,62= 0 ,63= 0 ,64= S4
0 0 0 0
C) C) (9 1E1
(157) 65 = 35 ,66 = 0 ,67 = 0 ,63 = y1
0 85 S7 21
0 $3 $4
59 = U1 1510 = 93 1511 = 94
v1 23 24

is a basis of kergo. We can repeat this argument two times and assume
31 32 S3 0
61: 0 ,62= 0 ,63= 0 ,64= S4
0 0 O 0
0 0 0 $1
(158) - 65 = 35 ,66 = 0 ,67 = 0 ,63 = 12/1
0 86 S7 21
0 0 0
59 = U1 1 510 = U2 1 511 = U3
v1 02 U3

is a basis of hemp.

Since dimk(J(B)) = 4, {34, 35,211, 222, 213} is linearly dependent. Thus, there exist

d1,d2,c,- E k,i = 1,2, 3, not all zero such that d1s4 + dgss + Ella-u,- = 0. We

0
may assume c3 75 0. Since d164 + d265 + C169 + 02610 - 611 = ( 0 ) for some
v
d,», c,- E k,i = 1, 2,v E J(B) and {64,65,69,610,611} is linearly independent, v 75 0. If
'U = t1Ss+t2S7 for some 21, t2 6 k, then (L164 +d265+C169+C261o—611 —t166 —t267 = 0.

0
.This is impossible. Thus, v é kss + ks7. Therefore, ( 0 ) E ker<p\k66 + 1667. This
v

 

 

79

0
implies ( 0 ) , 65, 67 E Anng(a3). This is a contradiction since dimk(Ann5(a3)) = 2.
v

Subcase 8: Suppose dimk(AnnB(a,-)) = 3 for i = 1, 2 and dimk(AnnB(a3)) = 1

. Let Ann3(al) = ks} + 1632 + ks3,Ann3(ag) = 168.1 + kss + kss, and Ann3(ag) =

31 82 53 0 0 0 O
ks7,s,- E J (B) Then, 0 , 0 , 0 , s4 , s5 , 36 , 0 E
0 0 0 0 0 0 s7

kergo. Let

— — o
0
0 0 0 $1
(159) 55 = 85 166 = 86 157 = 0 158 = 91
0 0 S7 21

be a basis of kercp. Here, $1,311, z,- E J(B),i = 1, 2, 3, 4. Since dimk(J(B)) =
4, {51, S2, 33, 2:1, $2,233,231} is linearly dependent. Thus, there exist d1, d2, (13,6, E k,i =
1,... ,4, not all zero such that dlsl + dgsg + d333, + 23:1 qr,- = 0. If c,- = 0 for all i =
1, . . . , 4, then d181+d232+d3s3 = O. This implies d1 = d2 = d3 = 0. This is impossible.
Thus,c,- 75 O for some i. We can assume c4 75 0. Hence, :54 E L(sl, 32, 33, 2:1, 2:2, 1:3). We
can repeat this argument two times and assume 2:2, 2:3, 2:4 E L(sl, 32, .93, 2:1). There-
fore, $2 = dlsl +d252+d333+c331 for some d,, c E k,i = 1, 2, 3. Since {61, 62, 63, 68, 69}

0
is linearly independent, c1161 +d262 +d363 +c68 — 69 = ( ul ) 91$ 0. Since {61, . . . , 69}
’01

 

80

O
is linearly independent, ( u; ) ¢ 166.; + 1665 + 11:66 + 1667. Thus,

U1
31 32 S3 0
61: 0 ,62= 0 ,63= 0 ,64= S4
' O 0 O 0
. 0 0 0 $1
(160) 65 = 85 156 = 36 157 = 0 158 = 91
0 0 S7 21
0 $3 $4
59 = U1 1510 = 93 1 511 = 94
’01 23 24
is a basis of kercp. We can repeat this argument two times and assume
S1 S2 S3 0
61: 0 ,62= 0 ,63= 0 ,64'—'3 S4
0 0 0 0
0 0 0 $1
(161) 55 = 85 156 = 86 167 = 0 158 = 91
0 0 37 ' z1
0 0 O
59 = U1 1510 = U2 1511 = U3
'U1 ’U2 ’U3

is a basis of kergp.

 

Since dimk(J(B)) = 4, {s4,ss,ss,u1,u2,u3} is linearly dependent. Thus, there

exist d,-, c,- E k,i = 1, 2, 3, not all zero such that (1184 +d2S5 +d386 +2:le em,- = 0. We

0
may assume C3 74 0. Since (1164 + (1265 + (1365 + C169 + C2610 — 611 = ( 0
’U

) for some

di,c1,02 E k,i = 1,2,3,v E J(B) and {64,65,65,69,610,611} is linearly independent,

’0 7£ 0.1f’U = 257 for some t E k, then (1164 +d265 + 61366 + 0169 + C2610 - 611 — L67 = 0.

O

This is impossible. Thus, v ¢ 153—]. Therefore, ( O ) E kercp\k67. This implies

’U

81

0
( O ) , 67 E Ann5(a3). This is a contradiction since dimk(Ann3(a3)) = 1.
v

Subcase 9: Suppose dimk(Ann3(a,)) = 3 for 2' = 1, 2 and dimk(Ann3(a3)) = 2.

Let Ann3(a1) = 1:31 + ksz + k33, Ann3(a2) = 19.34 + ks5 + less, and Ann3(ag) =

31 32 33 0 0 0
k37+k33,s,-EJ(B).Then,(0),(0),(0),(34),(s5), 35),
0 O O 0 O 0
O 0
(0),(0)Ekercp.Let
S7 S8

0 O 0
(162) 65=(S5),65=(85),67=( ),63=(O)
0 0 s7 38

be a basis of kenp. Here, 93,-,yi, z,- E J(B),i = 1, 2, 3. Since dimk(J(B)) = 4, {31, .92, 33,
$1,232,233} is linearly dependent. Thus, there exist d,,c, E k,z' = 1,2,3, not all
zero such that dlsl + d231, + d333 + 2L1 qr,- = 0. If c,- = O for all i = 1,...,4,
then dlsl + dgsg + d383 = O. This implies d1 = d2 = d3 = O. This is impossible.
Thus,c,- 75 O for some 2'. We can assume c3 79 0. Hence, 233 E L(sl,32,s3,:z:1,$2).
We can repeat this argument and assume $2,933 E L(sl, .32, s3,$1). Therefore, :52 =
d131+d232+d383+cx1 for some d,,c E k,z' = 1, 2, 3. Since {61,152, 63, 69, 610} is linearly

0
independent, d161+d262+d363+cc59 —610 = ( u; ) ¢ 0. Since {61, . . . , 69} is linearly
Ul

 

82

0
independent, ( 'ul ) E 11:64 + 10%, + k66 + [£67 + [668. Thus,

91

0 0 0
(163) 65=(S5),66=(86),67=( ),63=(0)
0 0 i 87 83
2:1 0 11:3
59=(91),510=(U1),511=(93)}
Z1 ’01 23

is a basis of kercp. We can repeat this argument and assume

0 0
(164) 65 = 35 .66 = 36 .67 =
0 0 S7

is a basis of lamp.

Since dimk(J(B)) = 4, {S4, 35, $5,111,112} is linearly dependent. Thus, there exist

(1,, c1, c2 E k,i = 1, 2, 3, not all zero such that d134 + d235, + d335 + clul + c2212 = 0.
v

0
We may assume 02 74 0. Since d164 + d265 + (1366 + c610 — 611 = ( 0 ) for some

(1,-,0 E k,z' = 1, 2, 3,1) E J(B) and {64,155, 66, 610, 611} is linearly independent, 1) # 0. If

’U = t187 + t283 for some t1, t2 6 k, then 61164 +d265 + d366 + 6610 - 611 — t167 — t263 = 0.

 

83

O
This is impossible. Thus, 12 ¢ [987 + kss. Therefore, ( 0 ) E kergo\k67 + 1:68. This
22

0
implies ( 0 ) ,67, 63 E Ann3(a3). This is acontradiction since dimk(Ann3(a3)) = 2.
v

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