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Finite Homogeneous Rings of Odd Characteristic
201
FINITE HOMOGENEOUS RINGS OF ODD CHARACTERISTIC Dan Saracino and Carol Wood Colgate University, Hamilton, NY 13346 Wesleyan University, Middletown, CT 06457
0.
Introduction. We complete here the classification of finite homogeneous
rings of odd characteristic begun in [ 5 1 and based on work of Berline and Cherlin in [ l ] .
A countable ring R
is homogeneous if every isomorphism
between finitely generated subrings of
R
extends to an automorphism of R.
Although we draw heavily on results in [ l ] and C51 for o u r present analysis, we give (without proof) enough background results in Section 1 to allow the present work to be read independently. In C11 the classification of all quantifier-eliminable (QE) rings is reduced to the classification of the Jacobson radicals of such rings. Assuming finiteness (hence QE is equivalent to homogeneous), we listed in [ 5 l all possible radicals for the characteristic p2
case, p
an odd
prime.
Previous work [ 2 ] provided a list for characteristic p.
prime.
We now extend this to finite homogeneous rings of characteristic
pn, n > 2 and
p
odd.
p
any
Since any QE ring of finite characteristic is a
product of QE rings of prime power characteristic, this gives a list of all finite QE rings of odd characteristic. characteristic 2m, m
odd.)
and we do not include it here. done for
(Actually, using [ 2 ] we can get
As usual, the prime
2 behaves differently,
Much of the analysis in C51 has however been
p = 2, and we expect to complete the picture for finite
homogeneous rings in a subsequent note. In light of fascinating developments involving homogeneous structures (e.g., [ 3 1 , C41, and unpublished work of Cherlin on homogeneous digraphs) the present enterprise might become part of a special case analysis of some quite general phenomenon involving homogeneous structures. It is our hope
D.SARACINO and C. WOOD
208
t h a t a t l e a s t w e may have found some c l u e s t o u n d e r s t a n d i n g e i t h e r QE o r homogeneous r i n g s .
In S e c t i o n 1 we s t a t e o u r main t heorem and background r e s u l t s .
The
a n a l y s i s s p l i t s r o u g h l y i n t o two cases ( r o u g h l y , l a r g e and small r i n g s , b u t with a special r o l e i n the latter f o r
p
In S e c t i o n s 2 and 3 we
3).
=
o b t a i n b a s i c i n f o r m a t i o n a b o u t t h e two cases, r e s p e c t i v e l y , and i n t h e f i n a l two s e c t i o n s we o b t a i n t h e c l a s s i f i c a t i o n . We thank Greg C h e r l i n f o r g i v i n g o u r m a n u s c r i p t a c a r e f u l r e a d i n g and
o f f e r i n g v a l u a b l e comments. Background and main theorem.
1.
Throughout t h i s p a p e r , teristic
pn
p
where
n 2 1,
is an odd prime and
Jaco b so n r a d i c a l .
W e let
t h e f i e l d with
elements,
q
w i l l d e n o t e a QE r i n g ( w i t h 1 ) of c h a r a c -
R
and
be t h e a n n i h i l a t o r of
R1
a power of
q
w i l l denote its
J
pn-l
i n R , and
Fq
p.
c21, [ 5 l which w i l l be used t o
We b e g i n by l i s t i n g f a c t s from [ l ] , r e d u c e t h e problem a t hand. F a c t 1 (odd p case o f c 2 1 ) . p, p
odd, t h e n
53
with a l l products 0)
cp
=
:
i n Fp, and D p ( t ) 1
-
px
or
or
(i.e.,
x3
=
y3
=
xy
yx
=
y2
+
tx2
< x , y : px
=
py
=
x3 = y3
=
xy
=
0 , yx
=
Zp 8
Q
*.-
Zp,
D p ( t ) , where
is
py
Cp
is t r i v i a l
J
r i n g of c h a r a c t e r i s t i c
QE
J
=
=
and e i t h e r
0
=
is a f i n i t e
R
If
=
o>, t a nons qua re
=
=
x2, yz
=
tx2>,
4 t a non-square. H e n c e f o r t h we assume
n > 1.
F a c t 2 (Theorems I , 11, [ l ] ) . (i) QE
(ii) p , and
J
odd, n
(p
is n i l p o t e n t o f o r d e r a t most
f o r t h e l a n g uage based on €31
-
J
or
J b Fq
C(X,x; Fq,O)
vanishing a t
x
E
X, X
or
(0.
p.
>
1)
2n + 1 +,
J
QE(p), i.e.,
has
-1.
J @ C ( X , x ; Pq,O)
is t h e s u b r i n g of
and
C(X; Fq)
where
q
is a power of
c o n s i s t i n g of f u n c t i o n s
a Boolean s p a c e w i t h o u t i s o l a t e d p o i n t s .
Finite Homogeneous Rings of Odd Characteristic ( i i i ) R/R1
I
209
F ~ .
Thus F a c t 2 r e d u c e s t h e c l a s s i f i c a t i o n o f
QE
rings of characteristic
pn t o t h a t of t h e i r r a d i c a l s ( a n d similar r e s u l t s h o l d f o r
n o t s e m i s i m p l e , one f i n d s many s u c h
( c f . [2],
J
where
2).
=
For
[ 6 1 ) , h e n c e many
t h e s t r u c t u r e of s u c h r i n g s is a t p r e s e n t u n c l e a r . however a v a i l a b l e from t h e a n a l y s i s i n [ l ] ,
p
R
R , and
Some i n f o r m a t i o n is is s t u d i e d " i n
J
layers".
Notation ( [ l ] ) . i n J.
(Notice
( i i ) Let
k
(i)
For
is
Ji
QE ( P " - ~ )
= (X
J 1 : Xz
E
noteworthy o c c u r s .
=
, and J1 ). x . . z2 . .
V'
= 0).
l a y e r of
kth
+
F a c t 3 (Main Theorem, [ S l ) . Then
Let
J.
I.
J1
=
V
=
11.
~1
=
v
8 , v
111.
J1
=
V 8 B < b > ,
and
J1
a p p e a r s a g a i n and
V'
~1
=
-
be a f i n i t e
Fp
.
QE(pn-l) r i n g o f charac(Here
a nd d e t e r m i n e t h e m u l t i p l i c a t i o n ;
B
of t y p e
V
-ba ,
0
I , v-a
of type
V'.
J1 :
II
E
Fp, a 2
a-v
=
=
I,
V-a
=
pn-l,
0, a2
=
pn-1
V-b
=
a.V
=
a nd b2
=
i
=
or
tpn-1.
b-V
=
0,
p
E
tpn-l, pn-l,
IV.
are
Jk
F p. )
=
The s e t
J1
is i s o m o r p h i c t o o n e of t h e f o l l o w i n g :
J1
is a f i x e d n o n s q u a r e i n
Lpn-l
J k - < J k - 1 , pn-k>.
t h e r e is o n l y one t y p e r e a l i z e d i n
R
J1 as a vector space over
=
=
is t h e l a s t p l a c e where a n y t h i n g
J
We f o c u s f i r s t on t h e b o t t o m l a y e r
t e r i s t i c p , p odd.
Our a n a l y s i s s u c c e e d s by showing t h a t
a g a i n , s i n c e f o r homogeneous
ab
pi.)
Jk.
v -
=
t h e only r e l e v a n t l a y e r s f o r determining
t
pi
be t h e least i n t e g e r such t h a t
Intuitively, the
we write
b e t h e a n n i h i l a t o r of
Ji
of c h a r a c t e r i s t i c
J ( i i i ) Let V
let
1 5 i 6 n-1,
B
=
ajai
=
0, i
*
j,
p
E
1 mod 4
3 mod 4 .
D. SARACINO and C. WOOD
210
V.
VI. b2
=
J1
-V
=
-
aibi
=
(p
3 only) J1
+3.
=
pn-l
If k > 1
-
1
aibj
-biai, 1 5 i 5 n, and =
Moveover, any
Thus if k
B
<3"-1>
ai
B B where a2
as in I - VI
J1
=
we are done:
J
=
J1
+
=
bi
=
where =
bjai
0, ab = 3
0, i
= =
or trivial).
If k
>
j.
-ba,
is QE.
is known by Fact 3.
we can get additional information from [l]:
Fact 4 ([l]
i
1 then pk-l Jk
=
V and so VJ1
=
J1V
-
0.
Moreover V 2
If
k
=
J1V
=
0, so we have
> 1, then J1 is of type I, 11,
or 111.
(We shall see that all three possibilities can occur.)
is
Additional technical information about multiplication on Jk summarized in: Fact 6.
(113 or trivial).
If k
>
2 1 then Jk
We remark that more is claimed in [l] 2
that Jk
5 Jk-1
whenever
v
t
5 (Jk-1, pn-k>. If in addition
than what we state above, namely
p. 150 of 111, the additional hypothesis that p > 3
is required when V
is two- dimensional over Pp. As
2
Jk
we see in the theorem below, it can in fact happen for p
$ Jk-l * The goal of this paper, then, is the following:
=
3 that
Finite Homogeneous Rings of Odd Characteristic
Classification Theorem.
Let
J
be the Jacobson radical of a finite QE ring
of characteristic pn. n > 1 , p odd. Then J
21 1
Suppose k > 1
(i.e., J
i
J1 +
is isomorphic to one of the following (all of which are QE):
>
(Fp-dim V
A.
XiJk
v
=
JkXi
=
> 3)
2 or p =
J
=
0, J1 Of type
..., xs,p>, where
I, 11, or 111, k S n/2,
e
pk-1 xi, i
=
...,
I,
=
S.
or
B.
(Fp-dim V (i)
=
2, p
( J Z
3)
=
J
0,
< J ~ ,x, 3>, x2
=
j3n-1, j
J1 type I, 11, or 111, k S n/2, V
=
<3k-1x> 8 <3"-l>.
xJ1
=
=
1, or - 1 ,
J ~ =x 0,
=
or (ii) (5:
xJ1
$
<3"-l>)
=
J1x
=
x2
=
0, k
=
2, J1 of type I, 11, or of type I11 with the
additional restriction that ab
ba
=
=
E
<3n-2>
-
J
0, V
=
<3"-1>,
<3x> f3 <3"-1>.
This, then, together with Facts 1, 2, 3 provides the classification of finite QE rings of characteristic pn, hence of all odd characteristics. The proof breaks into cases in two different ways: size of V
and according to whether or not
p
=
according to the
3. There is considerable
overlap, but at little or no cost, in some of the cases--in particular, for small V
and
p > 3.
We choose to include arguments in some generality,
with an eye to possible future analysis. 2. Case A (F,-dimension
Let u
>
v
2 or
over
V
>
2
be the dimension of
p > 3).
or
V
p > 3, and so by Fact 4,
over v
Fp; by Fact 4, v 2 2.
We assume
is also the dimension of Jk/Jk-l
Fp.
We assume k > 1, J
finite as before.
show that in Case A , multiplication on Jk trivial.
Our goal in this section is to goes into
We work our way through the layers of
J
via a series of lemmas.
D.SARACINO and C. WOOD
212 Definitions.
( i ) Let
are i n d e p e n d e n t over
x, y
pk-lx,
if
("real
Jk'
E
pk-ly,
a nd
x and y
We s a y
Jk-elements").
are l i n e a r l y i n d e p e n d e n t
pn-l
Pp. ( i i ) Given x , y ( i i i ) G i ven
J,
E
-
x y = xy
x
XI,...,^,., y 1 ,
...,yr
yx.
J , we s a y ( X I ,
E
" h a s t h e same t y p e a s " ) p r o v i d e d t h e map x i
(read
+
-
...,x r )
yi, i
1,.
=
(y1,
...,y r )
...,r ,
p
+
p
d e t e r m i n e s a n i s om or p his m of t h e c o r r e s p o n d i n g s u b r i n g s o f J ( g e n e r a t e d by
IP, x i ,
..., X r ) ,
I f t h e r e is
well-defined. E
>
(v
Proof:
or
2
> 3).
p
Suppose n o t .
But now f o r any Z
Jk*
Z
= y + C, C
V'
Suppose
Let v
pk-lx E
x2
6
p
=
pk-2z2
3 we use
pk-2(x + y ) 2 and
pk-2
=
Jk' w e have y
E E
Jk-1,
=
and
E
k 2 3.
v
>
2
then
$.
Th u s
=
-9.pn-1,
E
Jkl
then
x2
E
V', p k - 2 ~ 2 = Qpn-1, Q k 0 mod p . y
E
J ' , pk-ly
J k ' w i t h pk-l
=
p k - 2 ~ 2 + pk-z (y c
=
pk-2y2
(y-2)
=
v , pk-2y2
= 0 , pk-2y2
~k-2. Then Qpn-l.
=
Lpn-'.
=
SO
t o choose =
x
If
E
for all
Lpn-l
pk- 2( x -y)2
(-x I y )
both cases.
~ k - 2+ < p n - ( k - l ) >
0
z
E
Lpn-1
=
x
If
J;.
and
pk-2x2
contradicting
+ cy + c 2 )
by F a c t 6 .
c o n t r a d i c t i o n by compa ring z a nd 22 : p k - 2 ( 2 2 ) 2 If
is
Jk-1/Jk-2
w e g e t t h a t e v e r y coset o f J ~ - I / J ~ - Z
V' t h e r e is
p k - 2 ~ 2 = pk-2(y + c ) 2
This gives
with
to
Jk/Jk-l
< ~ " - ( ~ - l ) > is i n t h e image o f
by h o m o g e n e i t y , f o r e v e r y
Thus
E
from
r$~
I 2p, a c o n t r a d i c t i o n .
S pv/2. p v
Lemma 2 .
x
V' a n d by h om oge ne ity o f
e x c e p t p o s s i b l y ones o f pv-p
resp. ).
By F a c t 6 , t h e S q u a r i n g map
Proof:
pk-2 x2
..., y r ) ,
Ip. Y ,
y =
> 3 we g e t a n i m m e d i a t e
p =
l l p k - 2 ~ 2 = 4P.p"-1
independent.
pk-2y2.
P. &
o
Lpn-1.
#
Now
*
T h u s pk-2(,
mod p . T h u s
x2
y) E
=
-1pn-l
~ k - 2 in 0
Finite Homogeneous Rings of Odd Characteristic Lemma 3 .
>
(V
Pro0
2
v = 2
If
S u p p o s e now
>
v
t h e n
> 3).
p
Or
and
If
> 3,
p
n
pk- l x>
n
p k - 2 ~ ~
pk-ly>
pk-lXr
pk-2 xy'
=
elements the
pv
pk-lx,
=
pk- 2x1y'
=
p2
v.
=
a contradiction.
v.
xy
independent, (pk-lx, (pk-lx,
pk-ly,
-
y)
pk-2Xy'
=
f o r independent
so are
x, y
x . pn-k
E
Jk-2.
and
+
pn-k
y
+
E
E
y
x
V
pk-'y) (pk-lx,
p k - 2 ~ ~ cpn-1,
v
B u t now
pk-2xc
Let
=
pk-2xy
(pk-lx, 2pk-ly,
x pn-k,
a nd
y.
X2
E
y'
Since
Jk-2.
x E
E
If
~ ~ - 2x ( ,y
Thus
X'Jk
+
5
y.
x
and
y
y'
pk-ly'
x, y pn-k) Jk-2.
mod J k - 1 ,
and
are with
2pk-ly
=
2y mod J k - 1 ,
Jk'.
Jk-1
x
9, c 0 mod p , a n d
Thus
xy
5
p".
6 y'
x + y'
T h u s t h e r e is
y'
>
p)
e x i s t s mod
for i n d e p e n d e n t
Now f i x a ny
-
2(pv
while
~pn-1.
SO
T h e s e elements must c o v e r
0.
Since
2pk-1y).
a nd we g e t t h a t
By Lemma 2 ,
=
pk-2xy'
=
BY
t h e r e are a t most p v
y'); in particular
p k - 2 ~ * 2 y= 2f,pn-l.
=
x ' mod J k - 1 ,
5
x , e x a c t l y one
E
x
a nd c o u n t y" s :
and
pk-lx>.
pk-lx> t h e r e e x i s t x' a n d y '
-
pk-2xy
J k-2.
say
J k-1,
E
=
We f i x
c
independent.
The f i r s t e q u a t i o n i m p l i e s
p k - 2 ~ ~ =1 ~ p n - 1 . !Lpn-l
pk-2 xy
B u t P ~ - ~ x ( +x y ' ) Thus
W e now show
If
v.
e l e m e n t s of
pk-2xy'
a nd
=
i t must b e t h a t f o r a g i v e n with
eas .y.
pk-2xty'
p k - 2 ~ ~ 1 since , for
-
5 Jk-2
-
E
5 Jk-2. 2 Jk
V
v
Jk
t h e n Lemma 2 i m p l i e s
homogeneity, t h e n , for e v e r y so t h a t
x
2 , and c h o o s e
2
then
k 2 3
213
and
so Xy
SO
E
Jk-2
are i n d e p e n d e n t E
~ k - 2 implies
A150
X +
pn-k
are i n d e p e n d e n t , s o
(X + p n - k ) ( y + pn-k)
E
T h i s p r o v e s t h a t a l l p r o d u c t s from
J k-2,
giving
( p n - k ) 2 E Jk-2.
Jk
are i n
J k- 2 ,
as desired.
0
D. SARACINO and C. WOOD
214 Lemma 4.
Jk-1
Proof: For
i
=
J1
We proceed i n d u c t i v e l y t o show t h a t t h i s is t r i v i a l :
= 2
M u l t i p l i c a t i o n by hence
Ji
>
>
(v
Proof: k
p
PJi+l
=
Lemma 5.
For
PJk.
+
k
=
JiIJi-1
Ji
PJi+i.
When
=
+
J C ~~1
Lemma 6 .
xy
J1
E
>
(v
2
or
p
By F a c t 6 ,
Proof: V
for a l l
x, y
we g e t
JEcV.
JlJk
V,
Jk.
E
u
E
J~
ux $ < p n - l , pk-lx>.
t h e r e is y from
Ji-1
=
Ji,
o
. For
k = 3 , i t is Lemma 3.
Since
Jk-1 v
with = =
Uy =
J1 + PJk
V,
-
3
Jk
SO
5 JlJk 5 V
VJk
xy
E
V, x
E
-
0
=
-v, a c o n t r a d i c t i o n .
xyxy
=
Thus 0.
Since
V.
Jk'
pk-ly.
t h i s implies
by Lemma 5.
by F a c t 6 ,
with
ux
In t h i s case for a l l
pk-lX
e t c . , u n t i l we g e t t o
We omit t h e g o r y d e t a i l s .
J1.
> 3).
by Lemma 5 , t h i s shows
Suppose t h e r e is Case 1 :
+
we a r e f i n i s h e d .
i = k
< ~ k - 3 , pn-(k-2)>,
t o squares i n
J: 5 Jk-2
JZ 5 ~ k - 3 and t h e n r e p e a t down t o
E
PJi+1
and so
pJi.
+
3 , we r e p e a t t h e p r o o f s of Lemmas 1-3, u s i n g Lemma 4 f o r t h e i n d u c t i o n ,
t o go from
xyx
Ji-1 = J1
t h i s is c o n t a i n e d i n F a c t 6.
2
J1 + p J i . 2 5 i S k.
=
Suppose
onto
PJi
+
J1 + pJ2.
=
Ji+l/Ji
P > 3).
or
2
For
maps Ji
+
J1
Ji-1
UX z
But t h e n
d
E
X E
V
-
< p n - l , pk-lx>
y mod J k - 1
Uy mod < p n - l > .
Thus Case 1 c a n n o t o c c u r .
Choosing
and V * -UX
Finite Homogeneous Rings of Odd Characteristic F o r a l l u a n d x , ux E < p n - l , p k - l x > .
Case 2:
mod < pn- l > .
Then f o r a l l
mod < pn- l > .
Since
-!Lpk-lx
I
(-u)x
Let
with
pk-ly
y
ux
u
by
uz
(since u
-u
!Lpk-lz mod
5
0 mod p , a nd so
5
ux
!2pk-lx
=
v , uy
=
t h i s s a y s t h a t whenever
i m p l i e s (-u)z T h u s 9.
Lpk-lx.
5
t h e r e is
V'
B u t now r e p l a c e
(-u)z d
again that
E
5
uJk-1
Ilpk-lz mod < pn- l >.
uz
v
F i x u.
215
-
L Pk-l Y
-u)
This says
< p n - l > , showing
E
5
JlJk
Similarly,
JkJl
5
hence
5
JkJk-1
0
Now we are r e a d y t o p r o v e Proposition A.
Proof:
>
(v
Case 1 :
pk-lx,
pk- l y>
xf,yl
with
2.
t h e n for a l l (pk-lx,
pk-ly,
=
v',
X'
x'y'
I
xy
mod < pn- l >.
xy
>
v
x'y'
v
or
2
E X
mod J k-1,
=
x (-y )
5
0 mod p.
a l l independent
x
y'
I
Thus
y
mod J k-1.
v'
=
v
xy
E
< p n - l , pk-lx,
B
y.
Now
I
=
~ ( p " - +~ y ) similarly T h i s s how s
E
< pn- l >
implies
xpn-k
E
s i n c e a l l m u l t i p l e s of
From x
pn-kx
lie outside
a n d so k 5 n / 2 , f i n i s h i n g Case 1 .
-apk-lx
+ E
xy
E
E
E
for
and
x + y
Likewise
This gives
hence
This
6pk-ly.
x)
there exist
t h i s implies
E
5i.x c < p n-l>. T h u s pn-k(pn-k
J i 5 < pn-l>.
-xy
x2
xy .k < p n - l ,
v', x', y'),
(x,-y),
0 mod p , a nd so
are a l s o i n d e p e n d e n t , a n d t h i s i m p l i e s
=
pk-ly>, say
-
x(x + y)
v
by LenUIIa 8 we h a v e
(x,y) =
If
n/2.
we see t h i s is i m p o s s i b l e :
pk-lx
+
Since
NOW
s
k
pk-ly>
(pk-lx, pk-ly,
But x(-y)
Similarly and
-
v, x, y)
a p k - l x + Bpk-l(-y).
implies a
and
v' E V -
apk- l x + Bpk-ly mod
5
T a ke x a n d y i n d e p e n d e n t .
But f o r
v + pk- l x mod < p n-l>.
!j
2 Jk
> 3).
p
X-Jk
5
E
we d e d u c e t h a t pn-kx = 0 , except
0.
Thus
n
-
k 2 k,
D.SARACINO and C. WOOD
216 Case 2: v
p
If
v = 2.
t h e r e is
V'
E
> 3,
y
with
pk-ly
we c o n c l u d e t h a t f o r a l l Y
z
E
z2
=
Y2apk-1,
w = 2 , J k = < x , pn-k,
Since
=
Jk',
E
Fp*, t h e n f o r a l l
v , y 2 z a pk-ly mod
=
apk-1,
By Lemma 8
But f o r z
mod
=
Y(crpk-lz) p a p k - l z mod
f o r a ny
Jk-1>
J i 2
follows readily t h a t
x
a n d u s i n g Lemma 8 i t
Jkl
E
k 2 n / 2 is a s i n
T he a r g u m e n t t h a t
Case 1 .
3.
YX,
=
we g e t a c o n t r a d i c t i o n :
Fp - ( 0 . 1 1
E
x2 s a p k - l x mod < p n - l > , a
0
Small V. I n what f o l l o w s we c o n s i d e r J k s u c h t h a t
v
t h e d i m e n s i o n of V
2, v
=
Pp, and show t h a t a l l p r o d u c t s from J k l i e i n t h e p r i m e s u b r i n g , h e n c e
over
< pnW k> . In p a r t i c u l a r , we p r o v e t h i s i n c a s e v = 2 , p
in
we r e a l l y a d d , i n l i g h t o f P r o p o s i t i o n A . a p p a r e n t l y no h a r d e r t h a n for sh o wi ng t h a t f o r
p
> 3
and
p
3, w h i c h is a l l
=
The p r o o f f o r a r b i t r a r y
p
is
3, a n d d o e s g i v e a n a l t e r n a t e r o u t e t o
=
2 Jk
v = 2,
5 < p n - l > , as a n e a s y c o r o l l a r y t o
J i c
N-o t a t i o n . V
We f i x some n o t a t i o n f o r t h i s s e c t i o n .
< pn- l >
=
@I
x t Jk'
with
pk-lx
J 1 is o f t y p e I , 11, o r 111, a nd we write
as i n F a c t 3 , h e n c e
Jk
=
or
J1
=
Since
or
V
=
2 , we write
U s i n g F a c t 5 we h a v e
v.
=
w
or
Th ro u ghout t h e p r o o f s i n t h i s s e c t i o n we a r g u e f o r t h e t y p e I11 case o n l y , a n d n o t e t h a t t h e o b v i o u s m o d i f i c a t i o n s (whe re
-ba
=
Lemma 1
=
(v
Proof:
axa
E
Lpn-l, =
2).
Since
ax, e t c .
w he re b2
Thus
=
x a , a x , x b , bx pa
=
we know
0
xaxa
=
o r pn-l
tpn-l E
and
b
are
W e s e t a2
=
pn-l,
according to p
E
1 o r 3 mod 4 .
a b s e n t ) g i v e t h e p r o o f f o r t h e o t h e r two t y p e s . ab
a
or both
b
V.
pxa
=
0, giving t h a t
0, hence xa
E
V.
xa
E
J1
a n d so
Similarly for 0
Finite Homogeneous Rings of Odd Characteristic Lemma 2
(v
=
Proof: exists pk-ly
Suppose
y E Jk =
with
xa
Thus
V'.
xa
6ba
+
(Y
=
( Y +6!L)pn-1 = 0.
P r o p o s i t i o n B.
Now v
-
=
=
Jk-lJk
contradicting
2pn-kx
E
+
a
d
k 5 n/2
x2b
p bx2
2
5
Jk
y
x2
and
=
x2a
-
(v,x) a(-v)
+
6
=
Then
we l o o k a t
(-v,y),
(x
+
E
0
0. so
=
E
By a
x2 = a p r x + ~ ? p " - ~ ,where
pk-r-1x2
From a(-v)
y2 y
=
pk-r
+
getting
gpn-r-l.
x2
4
pn-k>.
0 , a n d so x2
Suppose
This implies
Thus
2, a
E
Y t 0 mod p.
+ p J k ) J k = pk-r-lJIJk
mod < P " - ~ > . 0 mod p.
5
k I n/2.
1 S r 5 k - 1.
with
E
a , 5, Y , 6
2 Jk
2Y, hence
E
BY Lemma 2.4,
a x , x b , bx
where
we h a v e
0
=
and
by Lemma 2 , and so
0
=
divides
=
pk-r-l(J1
=
ax2
z
=
=
av
2 Jk
+
BP"-~-~.
-
apry
+
We
BP"-~,
-x mod Jk-1
we have
a v mod < P " - ~ > ,
pn-k)2
x2
E
and c o n c l u d e
n - k h k , so
But t h i s c a n o n l y happen i f
By Lemma 2 . t h e n , i t follows from
J i 5
Ya + 6 b
6!Z)pn-l.
Also
xa
=
mod < p n - l > , c o n t r a d i c t i n g
-
pkwr-ly2
P k-r-1x2
To see
s x(-a)
y(-a)
x mod Jk-1.
(v = 2 ) .
-v, a n d
pk-r-1y2
=
y
E
This gives
to produce
-v
y(-a)
=
2, a f 0 mod p. and
pk-r-l
pk-ly
xa
F i r s t we show t h a t
Proof: E
In particular,
< p n - l > ; we a r g u e s i m i l a r l y f o r
similar a r g u m e n t w i t h
a , 13
,., ( - a , v . y ) .
we c o n c l u d e t h a t t h e r e
(-a,v),
r 2 1 , s i n c e by F a c t 6 we know t h a t
Here
Ya2
-
(a,v)
x2 = a p r x + f3pn-k
Now by Lemma 2 , x a =
Since
V'.
(a,v,x)
E
Write
Proof:
0
E
p J k , a n d so
+
mod p.
xa
E
v ; from t h e l a t t e r we c o n c l u d e t h a t
Jk-1 = J 1 E
x a , a x , x b , bx
2).
217
E
and
k I n/2. k 5 n/2
that U
Use
D. SARACINO and C. WOOD
218
4. Large V Classified.
>
w
We assume in this section that
2, and we show the Classification
Again we use the notation of Fact 3 f o r
Theorem's claims for this case. possible J1 ' s . Lemma 1
VJk
(v
>
=
0
Jkv
=
Choose x
-
x
But
y
x'y'
=
xy, so =
=
xy
=
0, giving
=
0.
JkJk-1
=
L
PJk
=
0 , by
-pk-ly, spn-l), there exist
pk-lx, pk-1~' = -pk-ly, xy
=
=
independent, and let xy = spn-l.
,.- (pk-lx,
Jk-1, hence x(y' + y)
E
x(x + y)
so
and
(pk-lx, pk-ly, spn-1)
XI, y' with pk-lxl x'
2
is of type I, then Jk
J1
by Fact 6. Therefore Jk-lJk
Proposition A . Since
If
2).
(x' - x)y
0. Now X2 = 0.
x
and
+
2
Thus Jk
This settles the picture for J1
0, giving
=
x
x'y'.
=
y
Now x'y'
y' + y =
xy'
=
and x(-y).
are also independent,
= 0.
0
of type I completely. We turn to types
I1 and 111: Lemma 2. x2
and
>
(u
2).
If J1 is of type 11, then there is x
E
Jk' with x x J1
z x J1
y x z any
=
J1
t
=
0. By adding a suitable multiple of y to z we can assume that
=
If either y2 or z2 is 0. fine.
0 also.
spn-1
for
s
+ o mod
p as w2 where w
=
ay
Otherwise we can represent +
E
Jkf for some a, 6.
BY
definability of J1, this says that all elements w of Jk' with nonzero square satisfy w (w
+
i
J1
=
0. Taking w with w2
a) x a = 0. But (w
Lemma 3 .
0
0.
=
Proof: Since w > 2 we can find y. z independent so that y =
=
(w
> 2).
if and only if
x2
+
a) x a
=
a
=
a2 we get (w
i
a
=
2a2
f
If J1 is type I1 or I11 and x =
0.
+
a)2
=
2a2
f
0 and so
0, a contradiction. E
Jk' then x x J1
=
0
0
Finite Homogeneous Rings of Odd Characteristic Case I :
Proof:
of type II. Here we know
~1
x
x
J1
y
I
J1 = 0 with y2
0 by Lemma 2 and homogeneity.
=
0. Since
f
(again by homogeneity) that y2 If
(y
Then
> 3
p
If
p
(t
=
l)a2
+
3 then y2
=
=
=
a2
and
says that +
(x
t
and
2a2
=
(y
x2
spn-1 * 0.
=
a)2
=
0, giving
Suppose there is x
(x')~ = 62spn-1
Since Jkv a2a2
+
=
VJk
fi2b2.
+
=
If
o
( x ' ) ~= x2 x2
o
=
z
+ i
p
=
3 we use
(x
-
o
have
0
imply
=
(x
+
t + 1 =
0, a contradiction.
2a2 0
Jk', x x a
are non-squares.
a x (y
=
6
x x b
=
0,
+
fib that
=
aa
+
0.
62 @ 0, 1 mod p
with
Necessarily a and 6 are not both 0. But
x , so
v
> 2 to produce x, y independent so that
=
0
=
=
and x' x b are not both
x' x a
a)
+
implies y2
-
0. Recall that here
=
Z x J1
But
-
a)2
~1
E
=
p > 3 we can find
x' x a
0 we know
f
x' x b
0, a contradiction.
=
Thus
p > 3.
x x J1 = y x J1 (x
0, 1, and so
implies x'
if
If
i
Jk* implies
E
we know for any x' = 6x
and solve for a t 6 so that (x')~ = spn-l. zero, since
x
a) x J1
+
a) x a
+
by homogeneity, again a contradiction. Thus y x J1 Case 2: J1 of type 111.
and
a non-square mod p.
ta2, t
-a2, so (y
=
aI2
+
o
=
To prove the other direction, suppose
so that both
t
we can choose
a)2
+
(x
x2
219
0,
yI2
=
+
a
+
bl2
z2
=
-3"-1.
Z2
x2
=
=
3"-l.
a2
b2
=
=
3"-'.
If x2
-3"-'
=
3"-l, so the Only elements of Jk' Similarly, if Notice that
says x x y = -x2
and
x2 = 3"-1
X f y
(x
-
E
Jk',
then z (X f
we get
which satisfy E
Jk* and
y) x J1
=
0.
y)2 = x2 says x x y = x2.
0 is the only possibility when x x J1
=
0.
This will complete our argument for Case 2 provided we can find some x
E
Jk'
This contradicts x2
with
x x J1
Jk/pJk:
0, and so x2
+
1)-dimensional subspace, and so the *-annihilator of J1
has dimension at least X
E
=
0. To do this we intersect the *-annihilators of a and b in
each is a ( v
in Jk/pJk suitable
=
i
Jk'.
v.
Since
v
>
2 , this subspace gives us 0
220
D. SARACINO and C. WOOD
then xJ1
>
(v
Lemma 4.
J1x
=
Proof:
Then f o r any v'
-
(or (v, a, pn-1) xa, (x'b
=
(v, a, pn-')).
Thus there is
xb), ( x ' ) ~= 0.
=
and so (x - x1I2
By choosing
xJ1
x2
y2
=
Proof: hence (x
>
(v
Lemma 5. with
0. Similarly
=
0, then
Take y)'
+
=
J1x
=
Notice x
-
v' e!
Jk'. =
pk-lxl
But (x -
0
=
v',
it
XI)
J1
0,
=
-
0.
=
x 0
x 0
and
y
*
y.
x
=
xy
yx
=
=
0.
a s above and notice that Let pk-lx
=
v, pk-ly
=
(x
v'.
(v',v,x',y').
xy, ( x ' ) ~= (y'I2
+
y) x J1
Then (v,v')
5
x mod Jk-1. and so by Lemma 4 and Lemma 2.4,
x'
=
y mod Jk-1, y'
xy
=
x'y'
yx
=
0 also.
-xy.
=
Thus xy
0, and from
=
x
i
=
y
=
0,
-
(v',v),
In particular,
v, x'y'
=
=
0. But now x - x'
=
yx
J1
we can be
v', pk-ly'
=
*
(v', a, b, pn-l)
so that
x'
Thus x - x'
and so there are x', y' with (v,v',x,y) pk-'xr
= 0,
If J1 is of type I 1 or 111, and x and y are independent
2). =
pk-lx.
Moreover (x - x')J1
0 by Lemma 3 .
=
=
V' we have (v, a, b, pk-')
E
sure that x and x' are independent.
gives
Jkl such that x2
E
0.
=
Take x as in the hypothesis, and let v
by Lemma 3 .
x'a
is of type 11 or 111 and x
If ~1
2).
0. But then
=
0
it follows that 0
We now prove the Classification Theorem, Part A , for the Case I I =
Fp-dim V > 2.
For
J1
=
V , Lemma 1 is all we need.
111, we see that Jk is generated by modifying any element and
b)
to get (y
+
y aa
+
For J1
~ 1 pn-k, , and elements with square 0, by
of
Jk'
by a suitable multiple of
6b)
*
=
J1
of type I1 or
a
(or of
a
0, then applying the previous lemmas.
The verification that the resulting J's
are homogeneous is routine.
Finite Homogeneous Rings of Odd Characteristic
5. Small V
221
Classified.
In this final section we consider V
of dimension
over
2
Fp, using
results in Sections 2 and 3 to obtain the rest of our Classification Theorem. 2
Recall that from Proposition B we know that Jk f
If p > 3
0.
=
Jk
(v
Lemma 1
Jkv
=
0.
2
J1 = V (type I) then Jk =
V
=
=
2x
mod Jk-1
+
From
Jk' such that (pk-lx, x2, x)
E
implies y x2
If
Notice that Jk-1
suffices to show x2 y
0.
=
> 3).
2, p
=
Proof.
=
we known more by Proposition A , namely
pJk
0.
=
annihilates Jk.
Let
x
Jk'.
E
It
(x2,pk-lx) * (x2,2pk-lx) we know there is ,- (2pk-lx,
and so
x2
=
y2
x2, y). 4x2.
=
But then pk-'y p > 3
For
=
2pk-lx
this implies
0.
0
Thus we have again that for type I J1's. multiplication is trivial, and
so we turn to types I1 and 111. (v
Lemma 2 such that
x x J1
Proof. x + a If
-
a(x + a)
(ii)
exists c
=
0.
of type 11: Let J1
=
=
I
x
=
-
-
a), then a2
=
Jk'
E
0. Since +
a)
=
+a(x - a).
-a2, which is impossible. From
a) we conclude that ax
-ax, hence ax = 0. Now a x x
=
=
0
0 as well.
=
J1 E
Then J1x
(a,-a) is definable modulo V, we get a(x
a(x
=
-a(x
=
gives xa
=
J1
(i)
is type I1 or 111, and suppose x
Suppose J1 xJ1
= 0.
x - a and
a(x + a)
J1
2, p > 3 ) .
=
J1
of type 111: Suppose J1
-
V
such that
c2
or
=
tc2
0.
=
tc2.
and so if d2
mod
=
cx
*
x
c2
(and so p
E
0. Since J1x
Choose d
Since =
=
v =
then
1 mod 4).
E
J1
2, fx dx
=
5
with
c x d
there =
0,
is definable
0 also, giving
J1x
=
Now every element of Fp*
0.
D. SARACINO and C. WOOD
222
u2
c a n be w r i t t e n as u,v
i
uc
with
0
giving
J1x
vd
+
v2t
-
pv
with
But t h e n
C.
0 , and i n p a r t i c u l a r t h e r e e x i s t
f
(uc
We are now r e a d y t o p r o v e t h e small
(v
>
2, p
=
From Lemma 2 we know
*
1, -I,
0,
-
(pk-lx,x2,x)
Write y
it
y
a
=
Ya
8
xa
=
dx
SO
xJ1 = 0
0,
=
also.
0
p a r t of Case A o f our Theorem
V
+
x2
so
0,
=
x2
Thus
for
Y, 6
y = Bx, x2 J
=
E
y2
=
< a , x , p>
y
Jkl,
E
8pk-lx, y2
=
Fp.
Then 6
82x2.
=
Fp, and c h o o s e
E
t h e r e is
8pk-lx,
is o f t y p e 111, t h e n c h o o s e
J1
a
=
T h i s g i v e s pk-ly pJk>
0.
=
-
pk-lx
6x mod
Y
J k ' s o t h a t a x x = 0 , J1 =
E
Let
0.
=
(gpk-lx,x2,y).
conclude t h a t If
ax
F ~ .Since
E
we g e t
0
=
and
0,
=
3):
I f J1 is o f t y p e 11, t h e n c h o o s e x
8
vd)x
+
I t f o l l o w s from J1 x x = 0 t h a t
0.
=
+
x2, y x a
=
=
0.
B and from
=
b2 * 1 , w e
Since
as i n t h e theorem.
*
x E Jk', x
J1
and a r g u e as
0
=
above, a g a i n u s i n g Lemma 2, t o g e t t h a t J
=
J k X = 0, k 5 1112.
=
Again, t h e c h e c k s t h a t t h e s e are homogeneous are r o u t i n e and we o m i t them. Notice t h a t a complication arises
We now t u r n t o P a r t B o f o u r theorem. h e r e s i n c e we do n o t know t h a t
VJk
=
Lemma 3.
x
Jk'.
ax
(v
=
(11) I f
J1
=
xa
=
bx
xb
=
(i) If
Vx
=
=
xv
=
(v, a, x)
a2.
-
-
(v, v
I
we have m
kv
=
+
vx
!La
1 mod
+
Vx
*
a
then
ax
x x b
=
0
=
x x a
0, then
=
ab
ba
=
xv
-
for all (v,v-a),
v
E
V.
y
xa
=
0.
0, then 0.
I f not, then
Choose
there exists
=
=
v E
E
V'
with
J k with
a, y). hence
mx
3.
=
(v.a)
vy y
it
we a r g u e e x a c t l y a s i n Lemma 2.
0
Since
x
is o f t y p e 111, and Moreover, i f
0.
=
<3n-1, 3k-1x>, so
=
Now
E
J1
=
vx
Let
(i) I f
Proof: V
2, p = 3 ) .
=
0.
=
vx, ( v
-
a)y = ax, (v
mod <3n-k> f o r some Thus from ( v
-
a)y
=
-
a) x y
k , II,
m
= E
0.
Z.
S i n c e vy
ax it follows t h a t
=
vx
Finite Homogeneous Rings of Odd Characteristic
vx
-
-
!La2
we g e t
*
v
-
x
!La x a
-
a2
a x = a x , hence
!La2 = -ax.
0 , a 2 -!La2
=
Thus
ax
( i i ) As i n ( i ) we need only c o n s i d e r when
vx
xv
=
Then
a2.
=
( v , v - a , -b, (v - a ) x y
y)
-
-
a , -b)
(v, a, b, x). y
0, -b
=
-
(v, v
0 , -by
=
bx.
Writing y
=
1 mod 3 we g e t ( v - a ) y
while (v
=
0 y i e l d s a2
so bx
s
0 mod 3.
xb
=
y
x
.a2
sb
and so a2
-
.a2
Now b x y
0.
=
Combining a l l t h i s g i v e s
- by
bx
=
ax
+
a ) y = ax, +
mx
- sab
xa
0.
=
-ax,
=
0 implies sb x b
=
0 , hence
=
with
0,
=
Similarly
we get
0, r
=
it follows t h a t
ab
-ba
a l s o , when
0
=
Vx
s b 2 = -bx.
-bx, hence
= #
We now p r o v e Case B of o u r theorem.
a s i n Case
-
bx, -bra
=
t h i s becomes
1
=
- bx
- sb2
-bra
s
+
kv
V'
with
ra
=
0.
= E
0
0.
=
Now from
Since
-
y
-
=
ax
=
v
vx, ( v
I n p a r t i c u l a r , vy =
xa
Choose
0.
#
( v , a , b ) , so there is
mod <3n-k> w i t h m
- a)
Vx
0 , and so
=
-
(v - a) x y
F i n a l l y , from
0.
=
223
ba
a x b
-
0
0.
If
J;
5 <3"-l>
then t h e d e s c r i p t i o n x2
( i ) is e x a c t l y a s f o r Case A , e x c e p t t h a t now
€3
From
0.
=
*3"-'
=
is
a l s o possible.
J ~ <3"-1> C
If xa
=
If r
xb
=
>
ax
=
bx
=
consider
2
Observe t h a t
so y
=
mx
and so r
(1
2 , x2
=
by Lemma 3 .
Let
0
m
-
-
3r-2
=
-
( 3 m , a, b )
=
-
v.
-
Thus x2
=
y2
E
=
y2
=
=
m2x2
+
m2) = 0 mod <3n-1-(n-r)>,
~ 3 " - ~a,
4x2
x x a
with
-
<
~ 3 " - ~ 1,
1 mod 3r-2
but
=
x x b = 0, hence
r S k , a $ 0 mod 3 . m2 f 1 mod 3 r - l .
and so there is y w i t h
z
ay
=
J1, and from ya
2mxv.
=
by = 0.
=
yb
=
0, e t c . we g e t y
But now ( 1
contradicting
-
m2)x2
&
m2
=
-
mx
E
V,
2mxv E <3"-'>
1 mod 3r-1.
Thus
0 mod 3 .
Now a p p l y t h e above argument w i t h x2
B
x2
x 2 , 3y = 3mx, ya = yb
3mx we g e t y - mx +
1 : m2
Jk*
E
(3mx, a. b , y ) , hence y2
=
x
(3x, a, b )
(3x, a , b, x )
From 3y
t h e n we choose
+ ~ X V , -3x2
=
4xv.
If
t o get
m = 2
4xv
-
0
then
y
=
2x
-3.3"-2
+
-
v
such t h a t
0, contradicting
D.SARACINO and C.WOOD
224 characteristic 3".
If 4xv
hence n - 2
=
+
k
- 1
#
n - 1.
0 then so k =
=
?3"-l, and so 3k-1*a3n-2 = t3"-l,
2. Thus we have k
=
2, x2
E
<3"-*> - <3"-l>
The rest of Case B (ii) follows from Lemma 3. Again we omit verification of homogeneity, as tedious and routine. This proves the Classification Theorem, and thus completes o u r description of finite homogeneous rings of odd characteristic.
References:
C. Berline and G. Cherlin, QE rings in characteristic pn, J . Symbolic Logic 48 (1983), 140-162. C. Berline and C. Cherlin, QE rings in characteristic p, in Logic Year 1979-80 (Storrs), Lecture Notes in Math. 859 (Springer, Berlin, 1981). G.
Cherlin and A. Lachlan, Stable finite homogeneous structures, preprint.
Lachlan, On countable stable structures which are homogeneous for a finite relational language, preprint.
A.
D. Saracino and C. Wood, Finite QE rings in characteristic p2, to appear in Annals of Pure and Applied Logic. D. Saracino and C. Wood, QE commutative nilrings, J. Symbolic Logic 49 (1984), 644-651.