Page 1
JOURNAL OF ALGEBRA
85, 41Wl23 (1983)
Nilpotent Elements in Integral
Orders in Group Algebras
of Prime Order
Representation Rings
of HopfAlgebra
A. L. JENSEN
Department of Mathematics, Mundelein College, Chicago, Illinois 60660
Communicated by I. N. Herstein
Received December 11, 1981; revised July 5, 1982
In this paper we shall find necessary and sufficient conditions for integral
representation rings of Hopfalgebra orders to have nonzero nilpotent elements,
when the order is a module over a discrete valuation ring, and an order in a group
of prime order.
1. INTRODUCTION
Let R be a discrete valuation ring of characteristic 0 with field of quotients
K. Let A@, r, d, F) be a Hopfalgebra over R. A is called a Hopfalgebra
order in a Hopfalgebra H over K if K OR A N H. An element /i of A is
called a left integral if aA = E(A) n for all a E A. Let L, be the ideal of all
left integrals in A. The ideal &(LA) gives much information
of H and A. It plays a role similar to that played by the order of the group in
the representation theory of a finite group. In fact, it is always a divisor of
dim H. R. G. Larson [6] has used properties of &(LA) to apply the theory of
Hopfalgebra orders to the representation theory of finite groups and has
obtained a new bound on the degrees of the absolutely irreducible represen
tations of a finite group.
Integral representation theory developed from matrix representation of
associative algebras and number theory,
Methods of homological algebra have played an increasingly important role
in recent years.
By definition all integral representation rings are free Zmodules. Let M
denote a finitely generated Amodule which is Rtorsion free. Let [M] denote
the isomorphim class of the Amodule M. The integral representation ring
,3’ is the free Abelian group with generators [Ml,
[M] + [N] = [M @ N] and multiplication
The action of A on M@ N is given by a(m @ n) = C a,,,m @ aC2)n.
410
0021.8693/83 $3.00
on the structure
especially from ideal theory.
addition defined by
defined by [M] [N] = [M @ N].
Copyright
All rights
0
of reproduction
1983 by Academic Press, Inc.
in an> form reserved.
Page 2
REPRESENTATIONS OFHOPFALGEBRA ORDERS
41 4
I. Reiner [791 has shown that when G is a cyclic grou
maximal ideal P and the KrullSchmidt theorem holds for
RGY contains at least one nonzero nilpotent element i
then RGT contains no nonzero nilpotent el
Let p be a fixed rational prime and A a
show in this paper that if p is odd then J’
element if e(LA) E P2, and J’
s(LA) S& P2. This is a generalization of Reiner’s result for FZ =p, since for the
group ring s(LRG) =pR, the ideal generated by th
Tate an rt [13] have shown that when A is a
KZ,
and P = (D), then A has a basis { 1, X
“X for some (Y > 0 and some oP, a unit in
as the “Tate Oort basis.” It will be used t
The ideal of integrals in A is a rank 1 free Rmodule with basis AA = Xp’ 
wpIP and &(A,) = w,lF.
R is assumed to be a discrete valuation ring of characteristic 0 with
maximal ideal P, field of quotients K and residue class field k.
Ml Amodules are assumed to be finitely generated
modules. K is assumed to be a splitting field for 6, and so the
theorem holds for Amodules [ 11.
contains a nonzero nilpotent
contains no nonzero nil~~te~t elements if
2. THE CASE WHERE p Is ODD AND e(&A)~B2
In this section we prove that J
&(La> E P2. We shall use the following test due to I. einer [7] when a non
zero element is nilpotent.
contains a nonzero nilpotent element if
LEMMA 2.1. If X and Y are a pair of ~on~sornorph~~ Amodules
satisfying PA G XL A, PA G Y c A and XfPX N Y/PY as Al~Arnod~~e~,
then [X]  [Y] is a nonzero nilpotent element of,.3’.
THEOREM 2.2. Let p be an odd prime and let A be a copyalgebra order
in KZpa ilC 7 contains a nonzero nilpotent element $sE(L~) i P2.
Prooj
Lemma 2.1. Let X be the TateOort generator for A. Since E(E,) E P2 we
have that X* = w,PX, where a > 2. Let M = A’ + IIA, where A’ is the
augmentation ideal of A. M has an Rbasis m, = X, m, =X2,...? rnp 1 = Xpp iy
mP = 17. Let N = RA + DA, where A is the integral of A. N has an basis
II, n2 = IIX ,..., npp, = IIXpe2, mp =A.
bserve that PAGMGA and PAENGA.
denote N/PN. By Lemma 2.1, to show [M]  [N] is a manzero nilpotent
element of ,J” we have to show that (a) ic? N # and (b) M 71: PC To prove
We shall construct an element that satisfies the conditions of
Let
denote M/PM and P7
Page 3
412
A. L. JENSEN
this consider the action of X on M. With respect to the given basis this is
given by
/
i
\
0 1o.v.o 0 .a* 0 o.+pa 0 l7 0
\
OO... 0 Ol.*.O :
OO... 0 0 Ol.*.O 1o.v.o :
0 :
0 0 :
***
***
.a* 0 0 i 4 o.+pa
0 i
0 0 0 0
0 0 0 :
l7
0 0 0 0
0 0 0 :I*
pxp pxp
This
moreover, TrFii = B,, r for 1 < i <p  2 and XCi, r = 0, since a > 2.
The action of X on N with respect to the given basis is given by
shows that &!i = U@ RrF$,, where z?ii,=O and U=Cf:;Rtii;
i i 0 0
Ol... ii’... Ol... ii’... 0 IO...
: 0
0 : **.
... 0 IO...
: 0
0 : **.
... qJPt’ qJPt’
17 0
4 4
0 0 0 0 0 0 0
0 0 0 0 17 0
0 0 0
pxp pxp
Hence N= w@R_n,b, where XEP = 0, I?‘= Cf:’ Rfii,
1 < i <p  2 and XtiP,_ r = 0. So MN fi and (a) has been proved. We must
show that A4 and N are not isomorphic as Amodules.
Now U= Cf:,’ Rmi is an R direct
isomorphic to the trivial Amodule.
submodule V of N such that V is an Rdirect summand of N, N/V
K@, VcrK@, U.
U is the augmentation ideal of A, hence Hom,(U, R) = 0. Let p be any A
module homomorphism from M to N and consider the following
with exact rows:
xgi = @,+ r for
summand of M and M/U
Suppose we have constructed
is
a
R and
diagram
OUfMRO
OVtNAR0
Here qi = 0 since Hom,(U, R) = 0. Therefore a, induces p’ : U + V and
p” : R + R such that the following diagram commutes:
OULMR0
I’ I lo”
0VtNRO
Page 4
REPRESENTATIONS OF HOPFALGEBRA ORDERS
413
By the 5lemma, 9 is an isomorphism
isomorphisms. In particular,
modules and 0~ p as xmodules.
isomorphic to v by showing that the trivial xmodule I? is not an xdirect
sum of 0, but we shall construct p such that I? is an xdirect summand of
v.
Looking at the representation of l? it is obvious that ki has no pro
direct summands by the uniqueness of the Jordan canonical form.
Now let Y be the Rmodule with basis
if and only if q~’ and q~” are
if A4 N N as Amodules then Ur
We shall now show that &7 is not
V as A
v1 =17x,
v2 = 17x2,..., vpu2 =
p2 9
V pp1 =A + q,lIa =XpI.
Y is Rpure and R is a principal
summand of N. K OR V is the augmentation ideal of KG, so K OR VZ
K@, U and N/VR
since K@N/VZKGJN/K@
e see that
ideal domain, hence V is an Rdirect
VKG/K@ UK.
xv,=v,+,
for 1 <g<p3,
XV p2=17xp1 =nvppl,
xv pl=X*=WpnaX=W

Therefore, since o > 2 Xv, _ 1 = 0 and v = C
xmodule Ih is a direct summand of VT so v
the proof of (b). Hence N and M satisfy the conditions for Lemma 2.1 and
we may conclude that ]M]  [N] IS a nonzero nikpotent element of A3.
3, THE CASE WHERE p Is ODD AND c(LA) 2 %a
We show that if p is an odd prime, and A is a Hopfalgebra order in KZ, 1
then ,,J contains no nonzero nilpotent elements when E(LA) 3
case where &(LA) = P we shall first find all ~ndecom~osabl~ Amodules. Hf
= IIR, and X is. the TateOort
[X]/(Xp  oJlX). Let S denote the ring R [X]/(XpP1  p> 5
generator for A, then Xp =
LEMMA 3.1.
If m is a nonzero element of M then m, mX>..., mXPe2 are
~~~e~e~~e~t.
Let S = R[X]/(XP’ p> and let M be a Rfree
Proof. Suppose 0 = Cfz;
x?: a$? as an element of K OR S. M OR S is a field since XpP’ p
a,mX’ = m CF:: a,X’ for some ai E R. View
is
Page 5
414
A. L.JENSEN
irreducible
implies ai = 0 for all i = 0 ,..., p  2, so m, mX ,..., mXp2 are Rlinearly
independent.
by Eisenstein’s criterion. Hence m . 2::: a,X’ = 0 and m # 0
Notation.
maximal Ssubmodule of M subject to the conditions that 5 E N,,, and N,
has a Rbasis of the form m, mX,..., mXp2 for some m E N. Note that 6 is
not necessarily equal to m, for example, if rii = rm, r E R, m E N, then in
order to get the maximal submodule
m, mX,..., mXpe2 instead of 6, liiX ,..., riiXp2.
Let M be a Rfree Smodule. For G E M, let N, denote a
we must choose the basis
LEMMA 3.2. Let N,,, be as above. N, is an Rdirect summand of M.
Prooj
pure. It suffices to show that for any m’ EM, Ilm’ EN,
m’ E N,. So choose m’ E M such that Ilm’ E N,.
are elements ai of K for 0 < i <p  2 such that
Since R is a principal ideal domain we must show that N, is R
implies that
Since Llm’ E N, there
P2
m’= 2 aimXi.
i=O
(*>
Since R is a discrete valuation ring we may assume that each ai is of the
form ai = ui/17 or ai = ri, where ui is a unit in R and ri E R. If czi = ri for all
i then m’ E N, and the lemma is proved.
We shall prove this by contradiction.
and let i, be the smallest i with ai 6C R, i.e., a0 ,..., aiP1 are in R and
ai = u,/fl. Using (*) we get
Suppose ai is not in R for some i,
Xp2iml =ai+lmp +q+,mpX+
+ aomXp2pi + . . . + aifl*2
.a. + ap2mpXp3i
and we let
VEM
Xv = XaimXp2 = crimp = wpuim. Hence m is an element of the S
submodule V of A4 generated by v. But v = (u,/I7) mXPp2 & N,, therefore
N,,, & V. This contradicts the maximality
This proves Lemma 3.2.
since
lIajER
and a0 ... aiPl are
elements of
R.
of N,,,, so we conclude that ai E R.
PROPOSITION 3.3. Let M be an Rfree Smodule. Then M is a free S
module.
Proofi
The proof is by induction on the Rrank of M. Assume that the R
rank of M is less than or equal to p  1. Take 0 f 2 E M. Then &, 6X,...,
Page 6
REPRESENTATIONS OFHOPFALGEBRAORDERS
415
JiiXpe2 are Rlinearly
Lemma 3.2 we see that for some m E M and some R
M = N, @ U as Rdirect sum, but the Rrank of N, is p  1. Therefore the
Rrank of U is zero. Hence M = N,) but N, N S, so M is a free Smodule.
Now let the Rrank of M be t, and assume that every Rfree Smodule
with Rrank less than t is Sfree. Choose N, as before, i.e., N, is a pure
submoduie of M with Rbasis m, mX,..., mXpe2 for some m E M, and N, i
Ssubmodule of M. Hence we have a short exact sequence of Smodules
independent, so the Rrank of
O+N,+M+M/N,+O.
B is a free Rmodule
as Rmodules and the Rrank of B is less than the
so by induction B N 0’ S, i.e., B is a free S
+ 0 splits
as Smodules,
M 1 @ l+ ’ S. This proves that M is a free Smodule.
since N, is R
le, hence ON,+
i.e.,
PROPOSITION
indecomposable Rfree Amodules are the trivial
tation ideal A + and A itself.
3.4. If e(LA) =L!R then up to isomorphi~m the only
Amodule
ProoJ
R [X]/(Xp
Let M be an Rfree Amodule. Set M’ = {m E M 1 Xm = 0).
Rmodule.
We have a short exact sequence of Amodules 0 + M’ + M +
M/M’ is a free Smodule by Proposition 3.3.
Hence M is an extension of a free Rmodule by a free Smodule. Let us
calculate Ext:(S, R). We have a short exact sequence of Amodules
Since
’  p) = S, as Amodules.
s(L,) = IlR,
A = R [X]/(Xp  pX)
and
A v
’ is a free
’ + 0.
O1 (XpP1 p)A+A+A/(XP’ p>A+O.
So since A is a projective Amodule and A/(XP]  p) A N S, we have
OtHom,(S,R)+Hom,(A,R)~Hom,((XP~’p)A?R)
) Ext;(S,
R) + 0
is exact. Hence
Ext:(S, R) N Hom,((XP’
N R/pR ‘v k.
p) A, R)/J’m(Hom,(A, R))
Let s be the Srank of M/M’
Ext(M/M’,
M’) N Ext(@$ S, @‘R)) N (Ext(S, R))rxs = (k)‘xs, where (k)“’
and Y the Rrank
of M’.
NOW
[I4]
Page 7
416
A. L.JENSEN
denotes r by s matrices with entries in k. Thus M corresponds to an r X s
matrix with entries in k. Also Horn,@, S) = 0, since S is isomorphic to the
augmentation ideal of A. Hence the isomorphism classes of M’s correspond
bijectively to isomorphism classes of r x s matrices with entries in k under
the actions of GL(r, R) and GL(s, AutA(S)) E GL(s, R) [lo]. Now k is a
field so every matrix in (k)‘xS can be diagonalized by elementary row and
column operations. Since R is a principal ideal domain each such operation
comes from one in GL(r, R) or GL(s, R), so each isomorphism class
contains a diagonal matrix. Thus the indecomposable Amodules are R, S
and nonsplit extensions, M : R + M+ S, corresponding to a nonzero
k, E k. The isomorphism class of M is determined by the isomorphism class
of k, under the action of the units in R. But k is the residue class field of R,
hence there is only one isomorphism class. This isomorphism class
corresponds to the nonsplit extension A : R + A + 5.
In order to find the multiplication table for AY we need the following
lemma and proposition.
LEMMA 3.5. Let M be an Amodule. Assume M has an Rbasis of
cardinality s. Then A OR M N 0” A.
Note. Lemma 3.5 is valid for any Hopfalgebra order A.
ProoJ: Let N, = A OR M with Amodule action given by
b(a @ m) = ba @ m.
N,p@A.
Define f:N,tN2
f’ : N2 f N, is given by b 0 m + C b(,, 0 S(b& m. Hence N, N N, and
J
A
by barnt2 b(,, @ bC2)m. The inverse of f,
Let N, = A OR M. The Amodule action on N, is given by b(a 0 m) =
C bu,a 0 bc,,m.
PROPOSITION 3.6. S@SP@~*A@R.
ProoJ: A k[ Y]/(p),
Y @ 1 + 10 Y. We will compute S @ S by computing S @ S. We can do
this since S @ S is a direct sum of indecomposable Amodules, and S, A and
k are not isomorphic. Hence
where k is the residue class field and AY =
 
S@S= (+ (&R)@(&A)
Page 8
REPRESENTATIONS OF HOPFALGEBRA ORDERS 447
I’dow s is the augmentation
I”@Y’, I<s<p1
ideal of A. Thereore .?@ s has a kbasis
and l~t~pl.Letvj=Y’OYf~r %<i<i;2.~
Claim:
{Yjv,,w1O<j<p l,l<i<p2},
where w = cp::
Since Y ~ w = 0 this will
(1) ‘+ l Y’ @ Yp’ is also a k
prove that ,?@
rove the claim
1< i <p  2 and w = Cf::
Yjvi can be expressed in terms of Y” 0 Y”, where s + t = i ij + 1, so it
suffices to prove that for fixed c,
we must prove that
(1) ‘+’ Y’ @ Yp’ are linearly i~de~e~de~t.
Yjv, for 0 <j <p  1,
(a)
(b)
Yjz,, for j + i + 1 = c and 2 < c <p  ! are linearly i~de~en~e~t~
Yjvi for j + i + 1 =p and w= Cf:i
linearly independent,
(c) Yjjvi for j+ i+ 1 =c and p + I <c<
independent.
(I”’ Y’@ YPP’ are
2(p 1) are linearly
For (a) the representation of the YCPi‘vj for 1 < i < c  I, 2 < c <<p  4
in terms of the Yk 0 YCPk for 1 < k < e  1 is given by
Cl
Yc‘59, = z, ski Yk @ Ycek,
where
ZZ Q,
k < i.
Hence the Yc‘“vi
For (b) the representation of the YPiP1~i for I < i <JJ  2 in terms of
Yk@Ypkfor l<k<pl isgivenby
are linearly independent in case (a).
Pl
)J aik Yk @ Ypk,
k=l
ypilvi =
Page 9
418
A. L. JENSEN
where
= 0, k < i.
Obviously Yp‘uI,
Yp2v 1,.e9
yf ai pil
that (1) Cf:f a, =p  1. We shall show by induction that ai = 1 for all i.
Hence (1) yields p  2 =p  1, which is a contradiction, and we can
conclude that w Yp‘v I,..., yv,2
Assume that a, L a2 = . . . =aiel= 1 and consider
YpP3v, ,..., Yv,, are linearly independent. Hence if w,
are dependent, then there exists ar,..., ape2 in k such that

v,  w. Looking at the coefficients of Yp’ @ Y, this implies
yv,2
are linearly independent. Clearly a, = 1.
a,+ (;I;)+ (pr:)+...+(“‘:“)+(“ri).
We will show that
0
if i is odd,
=2

P
if i is even.
Now
(j;; 1 )=
(ps)!
I)! (pi (is+ I)!
= (P4 a (P s>
(is+ l)!
7 (l)if+l
i!
I)! (sl)! (is+
(s 4 1 ). = (q1
Hence
I
z(
ps
i
s=2 is+1
i
T $J2 (l)is+l
( 1
sl
il
= z1 (1)if i
0 t
Page 10
REPRESENTATIONS OF HOPFALGEBRA ORDERS
419
= (1)’ i (I)’ (J  (l)i  (l)i (I)i
t=o
z(l)‘ 1
=2 if i is even,
=o
if i is odd.
ence if i is even ai  2 asp  1, that is, ai zp 1, and if i is odd ai p 1.
P’or (c) the representation of the Ycp’“u, for c p < i <p  2 in terms of
Yk @ Yek for c  (p  1) < k <p  1 is given by
P1
ycp’lui=
c
ski Yk 0 Ycmk,
k=c(p1)
where
= 0, k < i.
A direct computation shows that
det(ak,) = (P  l)(P  2) ... (P  (2P  c  1))
I
(2pcl)! 9
which is clearly different from zero for c =p + 1, p + 2,..., 2p  2. Since the
Yk @ Yck are linearly independent, this implies that the YCii~i
linearly independent. This concludes the proof of Proposition 3.6.
are
THEOREM 3.7.
in KZD. ,4i 7 contains no noytzero nilpotent elements 8 F(k,) 2 P.
Let p be an odd prime and let A be a ~op~~~geb~a order
Proof.
has no nilpotent elements [2].
If &(LA) = P, then using Proposition
multiplication table for A.7 with respect to the Zbasis [I?], [S]> [A]:
If e(LA) =R, then A is separable [5] and so aT = KGY, which
3.6 and Lemma 3.5 we have the
PI
PI
[Sl
ISI
!Sl
[A 1
PI
is1
IA I [A !
(P2)[Al+ [RI
(P  l)[Al
Page 11
420
A. L. JENSEN
Q oZ Ar has a basis of orthogonal idempotents E, = l/p [A], E, = +([R] +
[S]  [A]) and E, = $([R]  [S] + (p  2)/p [A]). Therefore Q OZ J’
commutative semisimple ring. Hence J’
nilpotent elements and the proof of Theorem 3.7 is complete.
is a
c Q aZ AJ’ has no nonzero
4. THE CASE WHERE p = 2
Let G be the group of order 2 and as before let A be a Hopfalgebra order
in KG containing RG, where E(L*) =pR, p = WAXY. Also let S denote
R[Xl/(Xp). First we classify the indecomposable Amodules. Then we
compute the multiplication table for the integral representation ring and,
finally, we prove that the integral representation
nilpotent elements if p = 2.
ring has no nonzero
PROPOSITION
for each i, i = 0, l,..., a  1, a module Mi. Note A =M,.
4.1.
The Rfree indecomposable Amodules are R, S and
Proof.
a free Rmodule. We have a short exact sequence of Amodules 0 f M’ f
M+MfM’+O.
M/M’ is a free Smodule, since it is a finitely generated torsionfree R
module and S is isomorphic to R. Hence M is an extension of a free R
module by a free Smodule. To calculate Ext:(S, R) consider the short exact
sequence of Amodules
Let M be an Rfree Amodule. Set M’ = {m E M 1 Xm = 0). M’ is
0+(Xp)A+A+A/(Xp)A+O.
Now A/(X  p)A IV S and A is a projective Amodule. Hence
O+Hom,(S,R)+Hom,(A,R)tHom,((XpA,R)
+ Ext:(S, R), 0
is exact, and
Ext:(S, R) N Hom,((XpA, R)/3’m Hom,(A, R) N R/pR).
Let s be the Srank of M/M’
Ext;(M/M’, M’)
(RIPR>'~~ denotes r by s matrices with entries in R/pR. Thus M corresponds
to an r x s matrix with entries is R/pR. The same argument as the one used
in Proposition 3.4 shows that there is a onetoone correspondence between
isomorphism classes of
nonsplit
isomorphism classes of nonzero elements r E R/pR under the action of the
group of units in R. Recall p = ~+lI7~. Thus for each i = 0, l,..., a  1 we get
an isomorphism class represented by a nonsplit extension Mj: R + Mi + S,
and r the Rrank
S, @‘R) N (Ext:(S, R))rxs
of M’. Now [14]
cz Ext;(@ = (R/PR)‘~~.
extensions
M : R + M + S
and
Page 12
REPRESENTATIONS OF HOPFALGEBRA ORDERS
421
and up ts ~somorphi§m the Rfree indecomposable ~rnod~~e§ are
the Mi)s for i= 0, I,..., a  1. Mi has an Rbasis uil an
Xvi1 = 0 and Xui, = Ilnvi2 + ulTivil,
basis 1 and X* To see that ikl, N A change this to
~,=I+X~thenXu,=OandXv,=II”u,+v,.
where 21 is a unit in
WI”, WI1 iI1
2[M,] ...
2w, I 2Wol
Pfol Pfol iM,l
2[M,] .‘.
ProoJ
lOX+XO1+bXOXanda.b=2.
A3 is commutative and [R] is the identity, so it suffices to show that
(a> ISl[Sl= PI,
(b) [S] [Mi] = [“i], and
(‘1 i”jl I”il
= 21Mmin(j,i)l.
(a> Let s be an Rgenerator for S, then S @ S is generated by s 3 s
and
The TaleOort basis for A is I, X, where X2 = aX and AX =
dX(s @ s) = a(s 0 s)  a(s @ s) + db(s 0 s)
= 0,
since ab = 2.
(b) S @ Mi has Rgenerators s @ vi1 and $0 viz4
X(s @ VJ = a(s 0 Vi&
X(s @ vi*) = a(5 @ VjJ + s @ (avi2 + 2L11’vJ
+ b(as @ (aviz + u47’ql))
= (a a + a2b)(s @ viz)
+ (~17’  ab&)(s @ vi!)
= zdI’(s @ VJ,
since ab = 2.
Page 13
422
A. L. JENSEN
Choose j and u, such that u,ll”j = a and an Rbasis
x, = s @ vi1  uu,IP(s @ VJ,
X2=SoVi2,
for SOMi.
NOW XX, = 0 and XX, = u&
Mi @I Mj has Rgenerators Vii ~3 Vjl) Vi, 0 Vj2, Vi2 0 Vjl) Vi2 0 Vj2*
 uIFX, . Hence [S] [Mi] = [Mi].
(c)
dX(v,, @ VjJ = 0,
dX(V,, @ Vjz) = au,, @I Vj2 + Ujl7jVil@ Vjl T
dX(Vi2 @ Vjl) = avi2 @ Vjl + Uil7’Vi, 0 Vjl5
dX(Vi, @ Vj2) = (aviz + U~~‘V~~) @ Vj*
+ vi2 @ (avj2 + ujIPvj,)
+ b(au,,
+ UiIliV~,) @ (avjz + UjnjVjl)
= bn’+juiuj(vil @ Vjl)
+ (UjH  uiPab)(vi, @ Vjz)
+ (ujHj  ujZZjab)(vi2 @ vjl)
+ (a  a + a’b)(Uf, 0 vj2)
= a17j+‘uiujvil @ vi2  uiIZivi, @ vj2
 ujHjvi2 @ Vjl.
Assume
(Uj/Uj)
that [Mi] [Mj] = 2[Mj], note
j< i and
choose
vil 0 “jl) vil 0 vj2 3
UibHiVi, @ Vjl 
IIjVil @ Vj2  Vi2 0 vjl 3 vi2 @ vj2 as an Rbasis for Mi @ Ml* TO see
LlX(&!ibl7’Vi~ @ Vjl  (UJUj) IIjVi, 0 Vjz  Vi2 0 “jl)
= (uj/uj) IPj(au,, 0 Vj* + ujnjVi, 0 Vjl)
= (avi2 @ Vjl + Uil7’Vi, @ Vjl)
= a((ui/uj)
Pjvj,
@ Vj2  Vi2 @ Vjl + UibII’Vf, @ Vjl).
We can now prove the following:
THEOREM 4.3.
be a Hopfalgebra order in KG containing RG. The integral representation
ring J does not contain any nonzero nilpotent elements.
Let p = 2, let G be the cyclic group of order 2 and let A
Page 14
REPRESENTATIONS OF HOPFALGEBRA ORDERS
423
Oz ,Y has a basis of orthogonal ~~ern~Qte~ts.
e, = t pf,],
e, = #Y~l
e ail = ml
e a+2 = WI  ISI>.
 [
+ ISI  wfallh
Hence ,.J has no nilpotent elements.
ACKNOWLEDGMENTS
Part of this work was done in my thesis, prepared under the supervision of Professor R.
Larson, and submitted in partial fulfillment of the requirements for the degree of IDoctor of
Philosophy in Mathematics at the University of Illinois.
REFERENCES
1. C. W. CURTIS AND I. REINER, “Representation
Algebras,” Interscience, New York, 1962.
2. I. A. GREEN, The modular representation
(1962), 607619.
3. A. L. JENSEN, “Grothendieck
Orders,” Ph.D. thesis, University
4. R. 6. LARSQN, Characters of Hopfalgebra,
5. R. G. LARSON, Orders in Hopfalgebras,
6. R. 6. LARSON, Orders from valuations, J. Algebra 38 (1976), 4141352.
7. I. REINER, The integral representation
(1965), 1 l22.
8. I. REINER, Nilpotent elements in rings of integral representations, Proc. Amer. Math. Sac.
I7 (I966), 270274.
9. I. REINER,
Integral representation algebras,
111121.
10. I. R~NER, Invariants of integral representation,
11. I. RETNER, A survey of integral representation
159227.
12. M. SWEEDLER, “HopfAlgebras,” Benjamin, New York, 1969.
13. I. TATE AND F. OORT, Group schemas of prime order, Ann. Ecole Norm. Sup. 2 (1970)
l21.
14. G. C. WRAITH, Homological algebra notes: mimeographed lecture notes, Aarhus, 1969.
Theory of Finite Groups and Associative
algebra of a finite group. Illinois J. Math. 6
Rings and Integral Representation
of Illinois at Chicago Circie, 1981.
3. Algebra 17 (1971), 352368.
J. Algebra 22 (1972), 2Oi210.
Rings of HopfAlgebra
ring of a fnite group, Michigan Math. J. 12
Trans. Amer. Math. Sac. 124 (1966),
Pa@
theory, Bull. Amer. Math. Sot. 76 (1970).
J. Math., in press.