MITTEILUNGEN aus dem

MATHEM. SEMINAR GIESSEN

Herausgegeben von den Professoren

des Mathematischen Instituts der Universität Giessen

Geschäftsführung: D. Gaier, G. Pickert

Sonderdruck aus Heft 163

Coxeter - Festschrift

Teil I

GIESSEN 1984

SELBSTVERLAG DES MATHEMATISCHEN INSTITUTS ISSN 0373 - 8221 CODEN: MMUGAU

Fachbereich Mathematik, Universität Duis- burg, 4100 Duisburg, Federal Republic of Germany

Department of Mathematics, Uiiversity of Alberta, Edmonton, Alberta, Canada

Our investigation of prime ideals in right chain rings

has two different roots. During a classification of

Hjelmslev planes according to their ideal type [?I the

third author came across the following at that time

hypothetical case: a Hjelmslev ring with just two

proper ideals, namely the Jacobson radical J and (01, and

J* = J in contrast to the case of the well known uni-

form planes. Also the question about the existence of

a special Hjelmslev ring with few two-sided ideals was

left Open in a module classification by Osofsky [:I.

E s s e n t i a l i n both s i t u a t i o n s was t h e q u e s t i o n whether

i n cha in r i n g s t h e r e e x i s t prime i d e a l s which a r e n o t

comple te ly prime. For t h e i n t e r n a l unde r s t and ing o f

cha in r i n g s t h i s q u e s t i o n is c r u c i a l a s t h e s t r u c t u r e

of t h e l a t t i c e o f prime i d e a l s is c l o s e l y r e l a t e d t o

t h e inva r i ance p r o p e r t i e s of t h e r i n g .

Recent ly Dubrovin c o n s t r u c t e d examples o f t h e type

above i n two Papers [ 4 , 5 ] which became t h e s t a r t i n g

p o i n t f o r [ I ] and t h i s a r t i c l e .

The f i r s t s e c t i o n c o n t a i n s some u s e f u l p re l imina ry

r e s u l t s and demonst ra tes t y p i c a l arguments f o r cha in

r i n g s . I n t h e fo l lowing s e c t i o n w e o b t a i n c r i t e r i a t o

dec ide whether c e r t a i n i d e a l s a r e comple te ly prime.

Moreover, it i s shown t h a t i n cha in r i n g s f i n i t e l y

genera ted non-zero prime i d e a l s a r e maximal, e i t h e r a s

comple te ly prime i d e a l s o r a s n o t comple te ly prime prime

i d e a l s . Sec t ion 3 i s devoted t o t h e a n a l y s i s of t h e

Segment between t w o prime i d e a l s . I t t u r n s o u t t h a t a

prime i d e a l which i s n o t comple te ly prime is always

c l o s e l y connected t o a comple te ly prime i d e a l .

Also, t h e e x i s t e n c e o f a n o t comple te ly prime prime

i d e a l i m p l i e s s p e c i a l p r o p e r t i e s o f t h e l a t t i c e o f

r i g h t i d e a l s above it. I n s e c t i o n 4 w e Sketch t h e

arithmetic of not completely prime prime ideals, and

in section 5 we describe the Segment between a not

completely prime prime ideal and the next completely

prime ideal (or (0)) below it for chain rings. The

paper ends with a section in which we construct an

example of a not finitely generated not completely

prime prime ideal, using a result of Dubrovin [? I .

1. Preliminaries

All rings R are not necessarily commutative and

have a unit 1 E R. We denote by J(R) = J the Jacobson

radical; U(R) = U stands for the group of units.

DEFINITION 1.1: A & o c a t h i n g R i b c a l l e d neah.ty

b i m p l e i 6 J und (0) ahe t h e o n l y t w o b i d c d i d e a l b

0 6 R.

DEFINITION 1.2: A h i g h 2 c h a i n h i n g R i b a e i n g w i t h

aR 5 b~ o h b~ c aR doe any t w o e lementb a,b i n R. 16

d b 0 Ra _c Rb o h Rb c Ra h o l d 6 6oh a l l a,b E R, R i b

c a l l e d a c h a i n h i n g . A k i g h t c h a i n h i n g R i b a a i g h t

H j c l m b l e v h i n g (bomet imeb a d d i n e H j t?hd&V h i n g 1 i d

euch e l c m e n t i n J i b a t w o b i d e d Zeh0 d i v i b o h .

AnaCogoubly , we 6peak 0 6 a HjeerndLev h i n g (bomet imeb

p h o j e c t i v e H j e l m s l e v h i n g ) .

The following results will 'be used repeatedly

LEMMA 1.3: L e t R be a t i g h t c h a i n k i n g , a E R , r E J.

Then t h e k e e x i b t e tementb rl E U, r2 E J w i t h

r = r r = r2rl and r2a = ar ' 6oh bome r2' E R. 2

PROOF: If ra = ar2' we are done. Otherwise ras = a for

some s in J and ra(l+s) = (l+r)a. But r,s in J inplies

that (l+r)-' and (I+,)-' exist and (l+r)-lra = a(l+s)-I.

The lemma follows with rl = l+r, r2 = (l+r)-Ir.

COROLLARY 1.4: L e t R be a k i g h t c h a i n k i n g . Foh x,y

w i t h iix 5 yR we have RX 5 YR.

LEMMA 1.5: L e t 1 be a k i g h t i d e a l i n a i r i g h t c h a i n h i n g R ,

Then I1 = U uI i b t h e m i n i m a l t w o b i d e d i d e a L c o n t a i n i n g

I und I* = n UI i b t h e maximaL t w o b i d e d i d e a L c o n t a i n e d

i n I whehe U kUnb t h k o u g h U.

PROOF: by Lemma 1.3 we have I1 = RI. This is clearly

the minimal twosided ideal over I. Any twosided ideal

contained in I 1s contained in uI for every U E U.

That I2 is twosided follows from Lemma 1.3.

2. Prime ideals in right chain rings

DEFINITION 2.1: Let R be a h ing and P a h i g h t i d e a l .

( i] P i 6 c a l l e d comp le t e l y phime ( c . p . 60h b h a h t ] i 6

and a n l y 16 xy E P i m p t i e n X E P a h y E P.

( L i ] P L6 c a l t e d phime i 6 and onLy i d xRy S P i m p l i e h

X E P O I L Y E P .

In [gl it was proved that x2 E P implies X E P

is necessary and sufficient for a prime ideal P to

be completely prime. For right chain rings we even

have the following:

LEMMA 2.2: Let R be a h i g h t cha in h ing and P a twonided

i d e a l . Then tue have:

P i 6 a campLeteTy phime i d e a l i 6 and o n l q i 6 x2 E P

PROOF: If x,y are not in P and xy is in P we have either

X = ys or y = xs for some s in R. In the first case

xL = xys is in P and hence X in P. In the second case

we have y2 = yxs in P if yx is also in P and y in P

follows. This last assummption is always satisfied

since (yx)2 is in P.

With the following result we have a method at hand

to get completely prime ideals in right chain rings.

THEOREM 2.3: Let R be a k i g h t chain a i n g .

(i) Nonzeho i d e m p o t e n t i d e a l b ahe c o m p l e t e l y phime.

( i i ) l d I i b an i d e a l which i n no t n i l p o t e n t t h e n

n In i b a c o m p l c t e l y phime i d e a l .

( i i i ) 1 6 t E R i d no t n i l p o t e n t , t h e n P = $ t n ~ i b a

phime h i g h t i d e a l . Moheoveh,i6 P i b a t w o - b i d e d

i d e a l , t h e n P i b c o m p l e t e l y phime.

2 PROOF: (i) L e t (0) * I = I . Suppose a 6 I b u t a 2 E I.

2 Then I 5 aR and t h u s I = I2 5 a I 5 a R 5 I. Hence

I = a L ~ . But nonzero idempotent i d e a l s a r e never

f i n i t e l y gene ra t ed a s r i g h t i d e a l s . Th i s c o n t r a d i c t i o n

shows t h a t I i s c.p.

(ii) S e t P = f i In. I f t 4 P, then t h e r e e x i s t s n E IN n

2 wi th In 5 t R . Suppose t E P then we o b t a i n

i s idempotent and t h u s by ( i ) c . p . a s I is no t n i l p o t e n t .

(iii) For t h e f i r s t a s s e r t i o n it s u f f i c e s t o show t h a t

xRx 5 P i m p l i e s X E P. Suppose n o t , then t n ~ 5 xR f o r

some n , hence t 2 " ~ 5 t n x ~ 5 xRxR 5 P. But t2" 6 P ac

t i s n o t n i l p o t e n t , c o n t r a d i c t i o n . Now assume t h a t P

is a twos ided i d e a l . Le t X E R wi th x2 E P. I f x 6 P

then tn = xa f o r some n , where a xR because tn 4 P.

Hence X = a r f o r some r E JxP. Now f o r some m we have

2 tm = rs wi th s E R and t h u s tn+m = x a r s = X s E P - c o n t r a d i c t i o n .

REMARK 2 . 4 : I t i n easy t o phove t h e 6o t tow ing :

[i) 16 I i n an i d e a t 0 6 R , t h e n t h e h e i n no phime

i d e a t P w i t h n I" 5 P 4 I. n

[ L i ] 1 6 t E R t h e n t h e h e bn no compte teey phime i d e a l

P w i t h n t n ~ 5 P 5 t ~ . n

REMARK 2 . 5 : The exampte i n 1 4 1 ( 5 e e a tno n e c t i o n 6 )

nhown t h a t n t " ~ i n i v t genehat n o t a C . p . h i g h t i d e a t n

doh t E R n o t n i L p o t e n t .

3 . The i d e a l l a t t i c e between two neighbour prime i d e a l s

Let R be a r i g h t cha in r i n g and P 2 Q neighbour prime

i d e a l s , t h a t means t h e r e a r e no f u r t h e r prime i d e a l s

d i f f e r e n t from P , Q and l y i n g between P , Q. I f t h e r e

e x i s t s a twosided i d e a l I wi th P =I I 3 Q we g e t a

cha in o f twosided i d e a l s

n P = I = 1 ' a... = I =... =i l 1" = Q , h e n c e Q i s

comple te ly prime!

Cven more i s t r u e :

PROPOSITION 3.1 L Let R be a h i g h t chain h i n g , P = P 2

a comple t e lq phime i d e a l . Then

( i l Foh anq i d e a l I 5 P which i n n o t phime, P/I i n n o t

s i m p l e .

(ii] 1 6 P/Q i b no t d i m p l e , t h e n we have: ,$oh any

X E P\Q t h e h e e x i n t n an i d e a l I w i t h x R 5 I 5 P.

PROOF: ( i) Suppose I c P is a twosided i d e a l w i th P/ I * simple. Now l e t X , Y be i d e a l s o f R w i th XY 5 I . I f

X I and Y I , then we must have P 5 X and P 5 Y a s

P / I is simple. Hence P = p2 E XY _c I. Con t rad ic t ion .

Thus X 5 I o r Y I. But t h i s shows t h a t I i s a prime

i d e a l .

(ii) Suppose n o t , t h a t i s , f o r any i d e a l X w i th

Q 5 X 5 P w e have X g xR. S e t I = U X c x R c P $

X 4 R , XcxR

As R is a r i g h t cha in r i n g , I is an i d e a l 3f R and

Q 5 I a s P/Q i s n o t s imple . Now by ( i ) , P/I i s n o t

s imple , hence t h e r e e x i s t s an i d e a l I ' wi th I 5 I ' 5 P.

Moreover, by ou r assumption I ' 5 xR. But t h e n , by

d e f i n i t i o n o f I , I ' 5 I. Con t rad ic t ion . Thus t h e r e

must be an i d e a l I wi th xR 5 I 5 P.

COROLLARY 3.2: Let R be a h i g h t cha in h i n g , P , Q

neighbouh phime i d e a l n und P/Q n o t s i m p l e . Then doh

anq x E PxQ we have ll X"R = Q. n

2 PROOF: I f P + P , then f o r a n y X E P x Q w e h a v e

il xn:i n P" = Q w h e r e t h e l a s t equat ion f o l l o w s f r o m n - n t h e f a c t t h a t i! P" i s the m i n i m a l p r i m e ideal b e l o w P .

n M o r e o v e r , Q = f l P" i s a c o m p l e t e l y p r i m e i d e a l b y

n n

T h e o r e m 2 . 3 . T h e r e f o r e Q 5 X"R, h e n c e Q = i l X R. n n

2 T h u s w e c a n n o w a s s u m e P = P . L e t X E PxQ. B y

P r o p o s i t i o n 3.1 t h e r e ex i s t s an ideal I w i t h

x R I C P . H e n c e Il X"R 5 fl 1" and n 1" = Q , ac * n n

n 1" is t h e m i n i m a l p r i m e idea l b e l o w I . A g a i n , Q is n

a c o m p l e t e l y p r i m e idea l and t h u s w e also have

Q c n X"R. - n

Now w e t u r n o u r a t t e n t i o n t o t h e case w h e r e P / Q i s

s i m p l e . N o t e t h a t t h e r e are c h a i n r i ngs R c o n s t r u c t e d

b y M a t h i a k [ g ] and D u b r o v i n [ & I w i t h P = J and Q = ( 0 )

c o m p l e t e l y p r i m e , hence n e a r l y s i m p l e .

PROPOSITION 3 . 3 : L e t R b e a h i g h t c h a i n h i n g , A 5 B

2 i d e a L n o d R w i t h B/A h i m p l e . Lct X E BxA w i t h X $ A

und x R 9 B. Then t h e n e e x i o t o a u n i t U w i t h

n A 5 XR 5 " u x ) ~ R , i n p a h t i c u i ? a h aLoa A 5 Il ( x u ) R.

n n

PROOF: L e t X E BxA, U E U. I £ u x E x R , t hen ux2 E x R .

I £ X = u x w for s o m e W E R, w e consider t h e f o l l o w i n g

t w o cases:

(i) X =wq. Then uxL = uxwq = xq E xR. 2 2

(ii) W = xq. Theri X = uxw = ux q = ux(ux q)q =

( w ~ ~ ~ ~ ~ = (ux)"xqn for all n E W . Thus

Hence in this case A xR 5 n (ux)"~, and as A is a n

twosided ideal also A c n (xu)"~. But if this cace does 9 n

not occur we get ux2 E xR for any u E U and thus

2 Rx R 5 xR. By assumption x2 4 A and xR 9 B, hence

2 A s Rx R 5 xR c B contradicting the fact that B/A is

9

simple.

COROLLARY 3.4: Let R be a h i g h t c h a i n h i n g , P,Q neighbouh

pkime ideaLs w i t h P/Q s impLe . Then 6oh any X E PxQ t h e h e

e x i d t u n i t d u,v E U w i t h Q 5 xR 5 n (uxv)"~ and n

Q 5 n (XVU)~R. 16 Q 16 compLeteLy phime we can choose n

V = 1.

PROOF: First of all, note that p2 = P, since otherwise

P/Q is not simple. This implies that xR 5 P for any 2

X E P. If Q is c.p., then clearly X 4 Q for any X E PxQ. If Q is prime but not c.p., then for any

X E b Q there exists a unit V with (xv)~ 4 Q, since otherwise xRx 5 Q, contradicting the fact that Q is

prime. Now apply Proposition 3.3 to X or xv, respectively.

See Example 6.5 where this "ef fect' is "visualized".

The c a s e P/Q s imple is t h e normal s i t u a t i o n i f Q is

prime, b u t n o t comple te ly prime.

THEOREM 3.5: L e t R be a h i g h t c h a i n h i n g , Q a phime

i d e a l w h i c h i b n o t c o m p l e t e l y phime und P t h e i n t e h -

b e c t i o n 0 6 aLL c o m p l e t e t y ph imc i d e a t b c o n t a i n i n g Q.

2 Then P = P und t h e h e ahe no t w o b i d e d i d e a b between

P und Q d i d d e h e n t dhom t h o b e t w o i d e a l b . Moheoveh,

Q $ (0) impL ied Q' $ Q und Q i b n i l p o t e n t o h n Q" i d

compLeteLy ph ime.

PROOF: Le t I be a twosided i d e a l w i th Q 5 1 5 P. As

Q is prime, Q 5 f o r a l l n , hence Q ll 1". But n

ti 1" is a c .p . i d e a l by Theorem 2 .3 , hence n

2 P = ti 1" c I* I 5 P. T h i s i m p l i e s I = P and P = P .

n - The l a s t a s s e r t i o n fo l lows from Theorem 2.3.

W e n o t i c e t h a t non-completely prime prime i d e a l s a r e

always p a i r i n g wi th a comple te ly prime i d e a l .

The f i r s t example of a n o t comple te ly prime prime

i d e a l i n a cha in r i n g was g iven by Dubrovin [?I (see

s e c t i o n 6 I .

LEMMA 3.6: L e t R be a t i g h t c h a i n h i n g , P 3 Q n e i g h b o u h

phime i d e a e b und Q n o t c o m p l e t e l y ph ime. Fuh theh L e t

X E P\Q be n o t Q - n i e p o t e n t . Then Q 5 ti X"R. n

PROOF: As X is not Q-nilpotent, Q 5 il xnR. I£ Q = fl xnR, n n

then Theorem 2.3 (iii) would imply Q completely prime,

contradicting our assumption on Q.

It is natural to ask the following question:

PROBLEM: Let P/Q be b imple und Q comple t e l y phime.

Doeb t h e h e alwayb e x i n t an e lement X € P\Q w i t h

n X ~ R = Q? n

By Corollary 3.4 elements which are not Q-nilpotent

are "everywhere" between P and Q; the Same is true for

Q-nilpotent elements.

PROPOSITION 3.7: Let R be a i r ight cha in h i n g , P =I Q

ncighboua phime ideaeb und Q n o t comp le t e l y phime.

(i] Foh euch X E b Q e x i b t n a t Leas t one u n i t V w i t h

(xv12 E Q.

( L i ) Let X E P\Q be Q - n i l p a t e n t . 16 Q i b n i l p o f e n t ,

t h e n n X"R = {O). 1 4 Q i n n o t n i e p o t e n t , t h e n n

li xnR = ll Q" i b a compeete ly phime i d e a l . n n

PROOF: (i) Let X be in P\Q and Q = fl uxR, where U runs

through U, follows from Lemma 1.5. I£ xUx is contained

in P\Q we can conclude that Q is completely prime:

Let y be not in Q. Then there exists s E J, U E U with

ys = ux.

Let sl E U with s sy = sy' (Lemma 1.3). Using s, E U, 1

ux E P\Q and our assumption we have uxs ux = yss ys = 1 1 2 ys,sys = Y s's in P\Q and y2 is not in Q. Lenuna 2.2

shows that Q is completely prime. The contradiction

shows that our assumption xUx in m Q for X E is

wrong and proves statement (i).

(ii) If X is Q-nilpotent then n xnR 5 n Q". For n n

nilpotent Q the assertion is clear. I£ Q is not

nilpotent then n - Q" is a completely prime ideal. Now n

n X"R s n Q" would imply xmR 5 n Q" for some m, hence n n n - as n Q" is completely prime - X E n Q" 5 Q.

n n Contradiction. Thus n xnR = n Q".

n n

4. Arithmetic of not completely prime prime ideals

In the following, Q will always denote a prime ideal

which is not c.p., and P the minimal c.p. ideal

containing Q.

PROPOSITION 4.1: L e t R bc a a i g h t c h a i n h i n g . T h e n we

h a v e :

(ij L e t s 4 p . T h e n ts E Q i m p L i e s t E Q.

(ii) PQcQP.

PROOF: (i) As s 6 P, P 5 SR. Thus tP 5 tsR 5 Q.

NOW Q is prime and P & Q, hence t E Q. (ii) Set I = (X E. P ~ X Q 5 QP]. As Q is a twosided ideal.

so is I.

Obviously, Q 5 1 5 P. As W is simple (by Theorem 3.5)

we must have I = Q or I = P. Using Proposition 3.7 we

get X E P\Q with X' E Q. Now let z E Q. As R is a right

chain ring, z = xa for some a E R. By (i) a E P, hence

2 xz = X a E QP. Thus I = P.

COROLLAIlY 4.2: L e t R be a c h a i n h i n g . Then we h a v e :

( i ) L e t s $ P und st E Q, t h e n t E Q.

liil PQ = QP.

PROBLEM: 16 R i b a c h a i n h i n g , do we a lwayb h a v c

PQ f Q?

It seems that only under further conditions a

stronger result (see Proposition 4.5) can be obtained

for right chain rings.

First a general lemma:

LEMMA 4.3: L e t Q be a nonzeho i d e a l 0 4 a t 0ca . t h i n g R

w i t h Q = aR = Rb. Then we aL6o have Q = Ra = bR.

PROOF: There exist r,s E R with a = rb, b = as.

Hence a = ras = ar's for some r' E R, as ra E Q = aR.

If r's E J(R), then we must have a = 0, contradiction.

Hence r' s E U and so s E U. Similarly, r E U. So

aR = Rb = Rrb = Ra and bR = asR = aR = Rb.

PROPOSITION 4 . 4 : L e t R be a h i g h t c h a i n h i n g w h i c h i b

a domain und Q d i n i t e l y geneha ted ab l e d t und h i g h t

i d e a l . Then s t E Q , s q P i m p l i c n t E Q.

PROOF: By Lemma 4 . 4 there e x i s t s q E R w i t h Q = q R = Rq.

A s P i s c o m p l e t e l y p r i m e , s t E Q E P i m p l i e s t E P . By

3.7 w e can a s s u m e t h a t t2 E Q. T h e r e ex i s t s r E R s u c h

t h a t s t = rq, b e c a u s e s t E Q = Rq. Now s u p p o s e t 4 Q.

T h e n q = t a f o r s o m e a E P and qa = bq for s o m e b E R.

Now sq = s t a = rqa = rbq a n d as R i s a d o m a i n t h i s

i m p l i e s s = rb , so b 4 P . H e n c e , s t = r b t = rq, so

q = b t . T h u s q2 = b t t a = bcqa f o r s o m e C E R, as t2 E Q,

a n d w e get q2 = bcbq, hence q = bcb. S i n c e b 4 P , b y

4 .1 ( i ) w e obtain bc E Q. So bc = qd for s o m e d E R.

T h e r e f o r e q = qdb and t h u s 1 = db. B u t n o w t = dbt =

dq E Q. C o n t r a d i c t i o n .

I n c h a i n r i n g s p r i m e ideals are " s e l d o m " f i n i t e l y

generated :

THEOREM 4.5: L e t R be a c h a i n h i n g , P a phime i d e a l 0 6

R. 16 P i b d i n i t e l y g e n e t a t e d ab h i g h t i d e a l , we have

one od t h e d o l l o w i n g b i t u a t i o n b :

i i l P = ( 0 )

( L i ) P = J und P = R a = a R doh bome a E R

( i i i ] P L6 t h e maximal phime i d e a l be low J, P i b n o t

c o m p l e t e l y ph ime und P = R a = a R doh bome a € R.

PROOF: Assume 0 * P * J and P = aR. Let X E J\P. Then

there exists r E J with a = rx. I£ P is c.p. then r = as

for some s, hence a = asx and sx E J implies a = 0,

contradiction. Thus P is not completely prime. Now let

Q be the minimal prime ideal above P; this is c.p. by

Theorem 3.5. Suppose Q * J. Let X E J\Q. Then a = rx

for some r E R. By Proposition 4.l(i), r E P, say r = as.

But now a = asx and sx E J implies a = 0 - contradiction.

Thus Q = J. It remains to show P = Ra if P + 0. Clearly, Ra E aR. If ar $ Ra for some r E R, then a = sar for

some s E J. For P = J s = at for some t E R and hence

a = atar. As tar E J this implies a = 0. Contradiction.

I£ P * J, the above and Proposition 4.l(ii) gives sa E aJ, say sa = at with t E J. Now a = atr with tr E J

again implies a = 0. Contradiction.

REMARKS 4.6: ( U ) T h i n theohem i n n o t t t u e doh h i g h 2

c h a i n t i n g o [ I ] .

( 6 ) I n S e c t i o n 6 we b h a e t d iocubn a n exampee doh

b i t u a t i o n ( i i i ) g i v e n by Dubhov in 151.

( C ) The phood above nhown: i6 J = aR i n a e e d t c h a i n

h i n g , t h e n aebo J = Ra.

5. Investigation of Q/n Q" n

It is plausible that the fact that there are no

ideals between a not completely prime prime ideal Q

and the minimal completely prime ideal P containing

Q has consequences for other parts of the ideal lattice,

in particular for the following Segment. Ac the

Situation for chain rings is much clearer we restrict

ourselves to this class of rings. First an observation:

LEMMA 5 . 1 : Let R be a h i g h t chain h ing , Q pkime but not

c omp le t e l y phime, P t h e minimal comple t e l y phime i d e a l

OVeh Q.

( i ) Foh any a E Q thehe ahf? x,y E P\Q w i t h

aR 5 xyR 5 Q. 2 ( L i ) 1 d X, y ahe i n P\Q w i t h xy E Q, t h e n xy E Q\Q L6

Q~ * (0).

PROOF: (i) By Proposition 3.7 there exists X E P\Q

with xL E Q. NOW if a E Q then a = xy for some y E P

by 4.1. If y 6 Q we are done. So we can assume y E Q. But then y = xs for some s E J and we obtain

2 2 aR = xyR = X SR 5 X R with X E PLQ.

(ii) Suppose xy = ab with a,b E Q. Then a = Xi for

some r E R, and xy = xrb. Now y - rb 4 Q as y 4 Q but b E Q. Hence Q* _c xQ - c x(y-rb)Q = (0). Contradiction.

LEMMA 5 .2 : Let R be a n i g h t cha in n i n g , I a non-zeno

i d e a l . Then t h e dollowing a t e e q u i u a l e n t :

l i l I l b n o t d i n i t e l y gene ta t ed ah h i g h t i d e a l .

(ii) I = IJ

PROOF: I f I i s f i n i t e l y g e n e r a t e d as r i g h t i d e a l , s a y

I = qR, t hen obv ious ly q $ q J = IJ. So ( i i ) i m p l i e s ( i ) . I f

I is n o t f i n i t e l y gene ra t ed a s r i g h t i d e a l , t a k e a E I.

Then t h e r e e x i s t s b E I w i t h aR 5 bR, hence a = b s w i t h

s E J. Thus a E IJ.

PROPOSITION 5.3: Let R Je a chain t i n g , Q a n o t

comp le t c l y ptime ptime i d e a l w i t h J t h e minimal

comp le t e l y pnime i d e a l con ta in ing Q . 16 i i b a twonided

i d e a l w i t h Qn c 1 c Q"-I o t (0) c I c Qn-' w i t h maximal

n , t h e n one 0 6 t h e 6ollowing h o l d s :

n- 1 (il I = Q J

(ii] I = SR w i t h Js 5 sJ = Q" 05 JS = sJ = ( 0 ) .

PROOF: We can assume t h a t Q" = ( 0 ) . Suppose I Q"-'J.

n- 1 Then t h e r e e x i s t s a E Q JNI w i t h a J I , s i n c e i f

aJ 5 I f o r a i l a E Q"-'J-I we would g e t I ( Q n - ' J ) J =

Qn- 1 J - c o n t r a d i c t i o n . Now t h e r i g h t a n n i h i l a t o r

o f I i s a twos ided i d e a l w i t h Q 5 1' 5 J. By

Theorem 3.5 w e g e t ir = Q o r = J.

CASE 1: = Q. Let b E Q"-'J\I. AS I ~b c Q"-' we

have Q 5 ( Q n - l ) I c - ( ~ b ) ~ c - Ir = Q, hence Q = ( ~ b ) ' = Ir.

2 Now l e t X E J q . By 3.7 t h e r e e x i s t s U E U wi th (xu) E Q.

I f a x 4 I , t h e n a l s o axu 4 I and by t h e above

Q = ( ~ a x u ) ~ . Thus axuxu = 0 s i n c e ( x u ) ~ E Q = ( ~ a ) ~ ,

b u t xu f Q. Con t rad ic t ion . Hence a x E I f o r any X E J\Q

and s o aJ 5 I 5 aR which i m p l i e s I = a J , c o n t r a d i c t i n g

t h e cho ice o f a .

CASE 2 : Ir = J. Suppose I is n o t f i n i t e l y gene ra t ed a s

r i g h t i d e a l . Le t 0 $ r E I , t h e n t h e r e e x i s t s s E I wi th

r R 5 SR, hence r = st wi th t E J and t h u s r = 0 ,

c o n t r a d i c t i o n . The re fo re I i s f i n i t e l y gene ra t ed a s

r i g h t i d e a l , s a y I = SR, and, of Course, SJ = ( 0 ) . I n

p a r t i c u l a r , t h i s i m p l i e s t h a t f o r U E U w e have us = SV

wi th a u n i t V. Le t X E J\Q. Then t h e r e e x i s t s a u n i t

W E U w i th (xw12 E Q. I f x s = s y wi th y E U , t h e n

xwxws E s U , c o n t r a d i c t i n g t h e f a c t t h a t Qs = (0). Thus

J S 5 SJ .

R E ~ ~ A R K : We h a v e u b e d t h e d a c t t h a t R i b a C e d t c h a i n

h i n g o n l y t o g e t I c R a c Qn-'. 1 t L6 p o b b i b t e t o

h e w h i t e t h e p a o o d 6 0 a b t o U b e t h i a d a c t a t a n o t h e h

p o i n t b u t w e c o u L d n o t g e t a i d 0 6 t h i b a b b u m p t i o n .

COROLLARY 5.4: 1 6 Q i n d i n i t e e y gcneitated an t i g h t

i d e a l , then aiiy i d e a l n Qn 5 I 5 Q i h 0 6 t h e dotm n

i Qi oa Q J.

PROOF: I f Q is f i n i t e l y g e n e r a t e d as r i g h t i d e a l , t h e n

i s o is e v e r y Q and hence Q~ $ Q ~ J . But by P r o p o s i t i o n

5 .3 t h e e x i s t e n c e o f an i d e a l which is n o t o f t h e form

k Qn o r Q ~ J i m p l i e s Qk = SJ = sJ2 = Q J f o r some k ,

c o n t r a d i c t i o n .

I f we assume t h a t Q is f i n i t e l y g e n e r a t e d as l e f t i d e a l

w e can g e t a more g e n e r a l r e s u l t even f o r a r i g h t c h a i n r i n g :

PROPOSITION 5.5: Let R be a t i g h t chain hing, Q a ptime

i d e a l which i n not comple te ly ptime und P t h e minimal

comple te ty phime i d e a l containing Q. 1 6 Q i h 6 i n i t e l y

genehated as l e d t i d e a l then thehe i d no twonided i d e a l

i wi th an F I 5 Qn-'p 6ok any n E N .

PROOF: Suppose I i s an i d e a l w i t h Qn F I s Qn-'P. Let $

L = ( X E P ~ Q " - ' X 5 11. Then L is an i d e a l o f R and

n- 1 Q c L _ c P . By Theorem 3.5 L = Q o r L = P. As 1 5 Q P

w e o b t a i n L = Q. Ac Q is f i n i t e l y g e n e r a t e d a s l e f t

i d e a l , Q = Rq f o r some q E R. Now q n - ' ~ : I s i n c e

o t h e r w i s e Qn-'P = R ~ ~ - ' P 5 I. Le t z E l \Qn . Then

n- 1 n- 1 z = qn-lr w i t h r E P w . Hence Q r = Rq r = Rz 5 I ,

and by d e f i n i t i o n o f L, r E L = Q. C o n t r a d i c t i o n .

6 . Examples

In 111 t h e concept of r i n g s with weak decomposition

proper ty was in t roduced which was i n a modified form

a l ready i n [z]. Before we w i l l Sketch it f o r t h e Spec ia l

case of chain r i n g s w e s t a t e some d e f i n i t i o n s . Let T be

+ a group. A m u l t i p l i c a t i v e semigroup r i s c a l l e d a

genera l i zed p o s i t i v e cone i n i', i f a E Tkr+ impl ies

- 1 + a E T + , and r ll (r'1-l = { I } . Then a l e f t o r d e r i s

def ined by S e t t i n g a SE B ( a Sr ß, r e s p e c t i v e l y ) i f

a - l ß E I'+ (ßa-' E r+, r e s p e c t i v e l y ) . With t h i s d e f i n i t i o n

we ge t :

a S, ß impl ies ya SE yß f o r a l l y E T

a s ß impl ies ay Ir ßy f o r a l l Y E I'. r

Then r i s a l s o s a i d t o be a l e f t - ( r e s p e c t i v e l y , r i g h t - )

ordered group ( see Conrad [? I ) .

The fol lowing d e f i n i t i o n s a r e r e c a l l e d from [Al.

DEFINITION 6.1: L e t r b c a ghoup w i t h g e n c h a L i z e d p 0 b i t i V e + c o n e r+. A h i n g R i b d a i d t o b e a s ~ o c i a t e d w i t h ( r , r ) , i d

+ t h e h e i b a monomohphibm 0 6 r i n t o t h e m u L t i p L i c a t i v e

monoid 0 6 t h e h i n g R, b u c h t h a t doh a n y r E W(O} t h e h e + e x i b t cx,, a 2 E r w i t h r R = u ( a l ) R und R r = Rv(a2) und

b u c h t h a t u ( r + ~ { l } ) ~ J(R).

This is Dubrovins's definition (note that in [z] the last condition is inadvertently omitted) whereas the

weak decomposition property defined in [I] is more general.

Before we shall describe the ideal structure of R

by means of semigroups we require some notations.

DEFINITION 6.2: L e t be a ghoup w i t h g e n e h a l i z e d

+ + p o b i t i v e conc r . A bubbe t n 5 r i b a h i g h t l t w o n i d e d

i d e a l od r i d 60h a l l a E n, ß E r+ : uB E n (aß,Ba E n

An i d e a l R 0 6 r+ i b c a t l c d phime i 6 boh any i d e a t n

$ , J i of r+, Q Ji _c n i m p t i e o ip 5 aor$ 5 n .

An i d e a l $2 0 6 r+ i b cornpLetety phime 1c.p. i d 60h any

a,B E r+ aß E R i m p l i e n u E 0 O J L ß E R.

PROPOSITION 6.3. [?I L e t r , r + , R,.v be ao i n O c d i n i t i o n

+ 6 . 1 . Fok any h i g h t i d e a l Q i n r d e d i n c

Gcni={r E Rlr=p(g)s 6on.aome g E n, s E R} = v(R)R.

Then L i b a b i j e c t i o n dh0m t h e b e t 0 6 a i g h t i d e a l b 0 6

r+ i n t o t h e b e t od non-zeho h i g h t i d e a l * 0 6 R, w h i c h

phebehveb i n c l u ~ i o n . Moheoveh, i 6 ( n ) i n an i d e a l o 6 R, +

th-en 3 i b a n - i d e a l - 06 r . - - - - - - -

PROPOSITION 6.4: L e t t h e n o t a t i o n be ab i n P h o p o b i t i o n

6 . 3 .

+ (i) 1 6 $2 i h a phime i d e a l 0 6 r , t h e n P(n) i s a

phime i d e a L 0 6 R.

(ii) 1 6 $2 i n a c . p . i d e a l 0 6 r+ und G(Q) i 6 a two-

- s i d e d i d e a l 0 6 R, t h e n p ( f i ) i n a c . p . i d e a l 0 6

R. Con~ehb'2Ly, i 6 ( $ 2 ) i b a c . p . i d e a l 0 6 R , t h e n +

R i n LX c .p . i d e a L 0 6 r . liii I $2 i b a 6 i n i t e L y genehated i d e a l o 6 r+ i 6 und o n l y

i d ;($2) i b a d i n i t c l y genehated i d e a l 0 6 R.

PZOOF: Straightforward. --

EXAAPLE 6.5: In [i] Dubhouin conbthuctb a hing R

abnocia ted w i t h a h ight -ohdehed ghoup buch t h a t R i s

neah ly -n imp le w i t h no zeho-d lu ibohb. N O W wc Want t o

denchibe t h e " h e i g h t " 0 6 t h e i n t e h b e c t i o n h i g h t i d e a l b

r l X"R, 6 0 h t h e sabe 0 6 b i m p l i c i t y we hen th i cR ouhbeLveb n

t o eLementb X = v i a ) . Howeueh, u can be i n t e h p h e t e d U 6

an a66ine t i n e a h 6unc t ion on Q w i t h t + at + b

lnee [LI I . To n i m p l i 6 y ouh n o t a t i o n we i d e n i i d y X w i t h a . +

Then a E i 6 a ( ~ ) 2 E 6oh a chosen i h h a t i o n a t numbeh E .

A nhoh: computa t ion bhowb

pn = (t + ant + (an-' + an-' + ... + t ) b )

CASE 1: 0 a < 1 und ae t b > E . Then lim (t + a t + b ) " 1 n-. -

= (t -. - b ) 1 -a

Thub 6oh evehy phOpeh h i g h t i d e a l yR we can 6 ind

b u i t a b l e a , b A U C ~ t h a t yR 5 n X"R, whehe n

X = p ( t -D a t + b ) .

CASE 2 : 1 < a und a ( ~ ) = aE + b > E . S e t 6 = a ( e )

8y i n d u c t i o n , a n ( € ) - E = (an-' + an-2 + ... + 1

hence l i m a n ( € ) = W. Thib meanb t h a t 6oh X = ~ ( a n+ -

tue haue n X"R = (0). n

It is obvious t h a t e lements o f t h e above t y p e s a r e

on every " l e v e l " i n t h e l a t t i c e of i d e a l s .

I n t h e fo l lowing we w i l l S t a r t from a c o n s t r u c t i o n

given by Dubrovin [I], and us ing t h i s w e w i l l o h t a i n a

cha in r i n g which h a s a prime i d e a l which i s n o t

comple te ly prime and n o t f i n i t e l y gene ra t ed . Let

2 L = <y;x i , i E N lxiy xi = y,xi+' 2 yxI+, = xi f o r a l l i E N > .

L i s a r igh t -o rde red group wi th p o s i t i v e cone Q, which

i s t h e monoid gene ra t ed by y and a l l xi , i E N (See 151).

S e t G = LxZ. Then P = [ ( x , z ) l x E Q o r x = 1 and z E N ~ }

is a p o s i t i v e cone f o r G , which induces t h e

l e x i c o g r a p h i c a l o r d e r on G. For an i d z a l S of Q, d e f i n e

s") = ( ( g , z ) E p l g E S , z E z } , t h i s i s an i d e a l o f P.

LEMMA 6 . 6 : Le t S be an i d e a l 0 6 Q.

i i) S i 6 a phime i d e a l 0 6 Q i 6 und o n l y i 6 s") i b a

phime i d e a l 0 6 P.

i i i l S i b a c .p . ideccl 0 6 Q i b find c n l y i d s") i b a

c . p . i d e a l 0 6 P.

PROOF : Straishtf orward

PROPOSITION 6.7: Let K be a d i e l d . Then KG i 6 embeddable

i n t o a d i v i d i o n hing D , und t h e h e e x i b t d a chain h ing R

i n D which conta inb KP and i b abbocia ted w i t h G .

PROOF: KG is embeddable into a division ring, since KL is

embeddable into a division ring and G/L = 2. The second

assertion follows from [ 5 , Theorem 11.

PROPOSITIOIJ 6.8: Theke e x i b t b a phime i d e a l i n R which i b

n o t c .p . and which i n n o t d i n i t e l y genehated an h i g h t

i d e a l .

3 PROOF: Set S = y Q. This is a priae ideal which is not

c.p. [?I. Hence Stl) is a prime ideal which is not c.p..

AS y3 E Z(L), (y3,0) E =(G), hence (y3,0) E Z(R).

Thus ! ( S ( ' ) ) = li(s('))~ is a prime ideal of R which is

not c.p., where is the monomorphism of P into R

according to Proposition 6.7. Moreover , S ( ' ) is not

finitely generated as right ideal. To see this, suppose

s " ) = (g,n)P for some (g,n) E P, g * 1. Now

3 (y ,n-1) E C('), hence there exists (h,m) E P with

(y3,n-~) = (gh, n+m).

1 This implies m < 0 and y3 = gh with h * 1 . Thus

3 4 y Q, so ( g , n ) E (g ,n)P C ( ' ) . Therefore, E ( s ( ~ ) )

i s not f i n i t e l y generated a s r i g h t i d e a l by Proposition

6 . 4 .

I REMARK 6 . 9 : The m i n i m a l c o m p l e t e l y pk ime i d e a l a b o v e T - - p ( ~ ( ' ) ) i b n o t J . To bee t h i b , n o t e t h a t we haue a n

i n e l u b i o n 8' ' ) (1 .2)P ~kf1) 0 6 W o d i d e d i d e a l b 0 6 P

w h i e h i n d u e e a a n i f i c l u d i o n E 1 ~ " ) ) ~ ( ( 1 . 2 ) ~ ) 5 u(Ph(11) = J 0 6 t w o b i d e d i d e a l b 0 6 R b i n e c ( 1 . 2 ) E z ( R ) ~

h c n c r by Theonem I . 5 J i b n o t t h e m i n i m a l c . p . i d e a l 1 o u e x i i ( s ( ' ) ) .

I 1 No te , t h a t t h i b d a c t a l b o i n p l i e b t h a t ~ ( 5 " ) ) i b

n o t d i n i t e l y geneha ted ab h i g h t i d e a l . L - - - - -

References --

Albrecht, U.; Törner, G.: Group r i n g s and general ized

valuat ions . To appear i n Comm. Algebra (1984)

Brungs, H.H.; Törner, G.: Chain r i n g s and prime

i d e a l s . Arch. Math. (Basel) 21 (19761, 253 - 260

Conrad, P.: Right-ordered groups. Michigan Math. J.

6 ( l 9 5 9 ) , 267- - 275 -

Dubrovin, N . I . : Chain domains. Moscow Univ. Math.

Bull . C e r . I. (1980), 56 - 60

Dubrovin, N . I . : An example of a near ly simple chain

r i n g with n i l p o t e n t elercents. (Russian) Hat . Sbornik 120 (1983), 441 - 447

Mathiak, K.: Bewertungen n i c h t kommutativer Körper.

J. Algebra (1977), 217 - 235

Osofsky, B.L.: Noncommutative r i n g s whose c y c l i c

modules have c y c l i c i n j e c t i v e h u l l s . P a c i f i c J.

Math. 25 (19681, 331 - 340

Thie r r in , G.: Sur les ideaux completements, premiers

d'un anneau quelconque. Acad. Roy. Belg. Bull . A.

Sci . (5) 0 (19571, 124 - 132

Törner, G.: Eine Klass i f i z ie rung von Hjelmslev-Rin-

gen und Hjelmslev-Ebenen. M i t t . Math. Sem. Gießen

107 (1974)