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,...,