10
M. L. RACINE
(14)
(15)
(16)
M
2
G C M
2
.
M 2 U M
2
C M 2 •
o[x]U
M 2
C M
2
.
Let M = o[x] + M ; 1 € M and M is an olattice of ^. We wish to show
that M is an order. Let y, z e o[xl, a, b e 1VL. Since U ,. = U + U . + t l ,
L •" ' 2 z+b z z, b b
by (14), (15) and (16) to show (a + y)U € M it suffices to consider aU
Z
"TiD
Z ,
KJ
and yUz
.
Butby (3) and
(14),
yUz
= {zyb}= {byz} =
(bo
y)
o
z
 {ybz}=bU
I T

bU
e M .
By
(14)
and
(16)
aU
, =
l , y l , z y, z 2 z, b
{zab} = (z o a) o b  {azb} = ( a l ^
z
) ° b " z U
a b
= l U
a U
b " z U a b € M 2
l,z *
(note that 1 c o[x] so by (14) aU. € M . and by (16) 1U
TT
, € M_).
1, z L a u , , D £
1, z'
q . e . d .
THEOREM 1. If p contains a maximal order M then $ is s e m i 
simple.
PROOF. Let 9 be rad #, fl' = 9, 9, = «?U
r
_
i y
j € S , j 0,
the Penico series of 9. Since # is finite dimensional (5 ? is nil (McCrimmon
[28] p . 678). But 9 finite dimensional nil implies 9 is solvable and hence
9 is Penico solvable ([33]). That is there exist s r € Z , r 0 such that
(r) (r1)
® = {0}, that is z e 9 implies z is an absolute zero divisor. Let
(r1)
L = M n 9 . Since L is a submodule of M it is finitely generated and
KL = KM n ^ ( r "" 1 ' = £ n 9^'1 ' = 9{T~l \ So L is a lattic e of fl*1""1 \ Pick
a € K, orffc 0 such that a / 0, * e 0. Then L CffL and M' = M+*L 3 M .