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The ore condition for polynomial and power seriesringsFerran Cedó a & Dolors Herbera aa Departament de Matemátiques , Universitat Autónoma de Barcelona , Bellaterra,08193, SpainPublished online: 27 Jun 2007.

To cite this article: Ferran Cedó & Dolors Herbera (1995) The ore condition for polynomial and power series rings,Communications in Algebra, 23:14, 5131-5159, DOI: 10.1080/00927879508825524

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COMMUNICATIONS IN ALGEBRA, 23(14), 5131-5159 (1995)

THE ORE CONDITION FOR POLYNOMIAL AND POWER SERIES RINGS

Ferran Ced6 and Dolors Herbera

Departament de VatemAtiques

Universitat Autbnoma de Barcelona

03193 - Bellaterra (Barcelona), Spain

Introduction

Let R be a rmg (with unity). A non-zero element of R is said to be regular

if it is not a zero-divisor. .A ring R is a right Ore ring if for e v r y u. b E R,

with b regular, there exist c. d E R, with d regular, such that ad = bc. It is

well-known that R is a right Ore ring if and only if it has right classical ring of

quotients Q,T,(R). We say that a ring R is classical if ever1 regular element is

invertible. Clearlk R is classical if and only if it coincides with its right (and

left) classical ring of quotients.

In this paper we study the Ore condition in polynomial and power series

rings. It is \yell-known that if R 1s a right Ore domain then so is the polynomial

ring R [ z ] . Kerr In [ lo] constructs an example of a right Ore domain such that

the power series ring R[x] is not a right Ore domain. In the first section we

Copyright O 1995 by Marcel Dekker, Inc

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5132 CEDO AND HERBERA

construct a classical ring R such that R[.z] is a right and left Ore ring and

R[.r] is neither right nor left Ore.

In the second section rve give some necessary conditions on R for R[z] to be

a rigllt Ore ring. In pr t icu lar t,he matrix ring JI ,(R) must be direct'ly finit'e

for all n 2 1. 1% do not knox whether R is right Ore for R[x] to be right Ore.

In contrast we give an example of a ring R mhich is neither right nor left Ore

and R[z] is right and left Ore. On the other hand, (von Xeumann) regular

rings are classical. and we study whether the polynomial ring over some classes

of regular rings are right Ore.

In section 3, bye study the monic localization of polynomial rings. In [11]

Resco, Small and Stafford show that if R is a right Noetherian ring then the

set of monic polyomials of R[r ] is a right Ore set in R[I]. They comment

that although the converse of this result is false, the precise class of rings for

mhich the conclusion holds is far from clear. Later in [6] Goodearl defines right

repetitive ring and he sllows that the matrix ring .l.I,(R) is right repetit,ive for

all n > 1 if and only if all surjective endomorphisms of finitely generated right

R-modules are automorphisms. We show that this condition is also equivalent

to the conclusion of the result of Resco. Small and Stafford, i. e. that the set

of nlonic polynomials of R[z] is a right Ore set in R[x].

In section 4, we study the Jacobson radical of QT,,(R[x]) (when R[x] is a

right Ore ring). Lye see that in many cascs (for example: R commutative or

2-' E Qi,(Rj) there exists an ideal I of R such that J (QZl (R[x] ) ) f l R[s] = / [ a ]

and in this case lye prove t'tiat if Q:i(R[.r]) semilocal, right (left) perfect or

semiprinlary, then Q:,(R) exists and it is semilocal. right (left) perfect or

semiprimary respectively.

In section 5. we study t,lie Ore condition on the polynomial ring over an

algebra R over a field Ii with dimriR < IIil. This study was made with the

collaboration of Professor L. Kowen. First we characterize the regular elements

of R[s] . An easy consequence of this is that if R[r] is a right Ore ring then

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POLYNOMIAL AND POWER SEIUES RINGS 5133

R is right Ore. MTe do not know whether the converse is true, but we prove

that if R is classical then R[z] is an Ore ring and Q:, (R[x] ) = R[r]S-', where

s = IC [ X I \ ( 0 ) .

1 Trivial extensions

Let R be a ring and let hi' be an R-bimodule. We denote the trivial extension

of llil by R by R cx M. Recall that R cx. M is R e 121 as abelian group and the

multiplication is defined by:

The first part of the following result is implicit in [ll]

Proposition 1.1 Let R = S cc ( D I S ) , where S is a subring of a division ring

D . Then

(z) (Menal) R is classical.

(zz) R [ x ] zs rzght and left Ore rzng and zts classzcal rzng of quotzents 1s t he

Laurent power serzes rzng R([x,x- '].

(zii) If R[x] is right Ore then S is right Ore.

PROOF. (i) Let ( s , d ) be a regular element of R. Since (0 . d )2 = 0 ,

s # 0. Thus s-' E D . Now (s.d)(0,?) = 0. Hence s-' E S and then

( s - l . -s-Ids-') E R is the invers of ( s , d). Therefore R is classical.

(ii) In order to prove the statement, it suffices to show that for any regular

element a E R [ z ] , there exist cul.cu" E R[x] such that cula = aa" = xn for

some n.

Let a = C,",,(a,.&)xn E R[zl be a regular element. If a , = 0 for

all n then cu2 = 0. Thus there exists a , + 0 . Let z be the smallest non-

negative integer such that a, # 0 . Consider CF=;=, b,xn = (CT=, a,+,xn)-' E

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- D[.r] . iYow a ~ , ' = ~ ( ~ , b , ) . z " = (0. i ) x Z = 0 and bo = a;'. thus a;' E S. Let

.j = (~$o '= , ja ,+ , , . cl,+n)x71)-1 E R [ x ] and consider 7 = ( ~ i ; ; ( a , . (7,)sr73 and

7' = 3 ~ k ; \ ( n , . d , ) s ~ . Non x2' = a,3(rZ - y ) = ( x ' - y 1 ) 3 0 .

(iii) Let cc. b E 5' \ (0). Slnce R[x] is a right Ore ring, there exist p ( x ) = -

(p" ,ph ) + . . . + ( p m . ~ ) 3 n z and q ( z ) = (go qb) + . . . + (q , ,Z) . zn in R [ z ] . with

q j r ) regular. such that

Since q ( x ) is regular qo + . . . + y,,xn E S [ x ] \ (0). b'e may assume that q, # 0.

Ti ow

(1 + a.c)jpo + . . . + p , - l r n - l ) = b(qo + . . . + qiLxn).

Hence apn-l = bq, # 0, thus S is a right Ore domain.

Corollary 1.2 There e x ~ s t s a classzcal rzng R s l ~ c h that R [ x ] 7s a rzght n n d

l ~ f t Ore r m g and R [ x ] 1s n u t h e r rzght n o r left Ore.

PROOF, It follows from Proposition 1.1 and the existence of domains S

that arc neither right nor left Ore. but can be embedded in division rings, for

example the free algebra I< (n . b) over a field K . 8

2 The Ore condition on R[x]

In this section we obtain some necessary conditions on R for R[x] to he a right

Ore ririg. Recall that a ring R is directly finite if ab = 1 implies ba = 1. for all

a .b E R.

Lemma 2.1 Let R be a rzng such that for all a , b E R there exzst p ( x ) . q ( x ) E

R [ x ] , wzfh q ( x ) regular, such that (1 - a r ) p ( x ) = b q ( x ) . Let a , b E R.

( 1 ) If for all posztzve tnteger n , a'% R an- lbR = 0 , then there erlsts m 2 1

such that amb = 0 .

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POLYNOMIAL AND POWER SERIES RINGS

(ii) If r ~ ( a ) = 0, then aR <, R .

(iii) If ~ ~ ( a b ) = 0, then rR(a) = 0.

(zr) R is dzrectly fintte.

PROOF. (i) There exist p(x), q(x) E R[x]. with q ( s ) = qo + q1.r + ... + qTxT

regular, such that (1 - ar )p(x) = bq(r). Thus p(x) = (C:, a'xz)bq(x) E R[s] .

Hence there exists s > 1 such that

Let m = T +s. Since anRnan- 'bR = 0 for all n > 1. we have that ambq(x) = 0

and so amb = 0.

(ii) Let c E R such that a R fl cR = 0. Since r R ( a ) = 0, a n R n an-'cR = 0

for all n > 1. By (i), there exists m > 1 such that a m c = 0. Smce rB(a) = 0,

c = 0. Thus aR <, R .

( ~ i i ) Smce rR(b) = 0, by (ii) bR <, R. Clearly bR f l rR(a) = 0. Thus

r R ( a ) = 0.

i ~ v ) Suppose that ab = 1. By (iii) r ~ ( a ) = 0. Since r ~ ( b ) = 0. clearly

r ~ ( b a ) = 0. Since ba(1 - ba) = 0, ba = 1.

Note that in Lemma 2.1. ( 1 ) + ( z z ) + ( 1 1 2 ) + (zu). It is eas? to see that

if R a is right Ore rmg then the condition (127) implies that Qz,(R) is directly

finite. Clearly if QLI(R) 1s directly finite then R is directly finite. The next

example shows that the conberse 1s false.

Recall that a ring R IS (con Neumann) regular provided that for ecery

a E R there emsts b E R such that a = nba. It is well-known that any regular

ring is classical.

Example 2.2 There exxts a dzrectly finzte rzght Ore rmg R such that QZ,(R)

ts not dlrectly 5nzte .

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5136 CEDO AND HERBERA

PROOF. Let R be the ring of all countable infinite matrices over Z with

only finitel! many non-zero elements in each column, and with only ?\.en off-

diagonal entries. \Ye shall see that R is right Ore and that QEI(I1) is the ring

Q of all countahle infinite matrices over Q with only finitely many non-zero

elements in each co lu~nn .

Let ( q , , ) E Q. It i~ easy to see that for each n 2 1 there eviits b,, E Z \ { 0 }

such that the diagonal matrix D = diag(bl, b2,. ., h,. ...) satisfies that

Thus in order t o s h o ~ that R IS light Ore and Q = QLl(R), it i~ suffic~ent t o

prole tha t Q IS classical. In fact, Q E End(17). where V is a bector space o v a

Q of countable infinite dimension, and by 15, Corollary 1 231, Q is regular. It

is nell-known that Q is not directly finite.

Now vie shall pro\-e that R IS directly finite Let (a,,), (b,,) E R such that

(a,,)(b,,) = 1. In part~cular , C;, a,,b,, = 1. Since the off-diagonal entries

are even. a,,b,, = 1 (mod 4) . Suppose that the columns of (a,,) are Q-linearly

dependent. Then there e x ~ s t s n > 1 such that the matrix

is not invertible in All,jQ). but since a,, is odd for all 1 and the off-diagonal

entries are even, we have that

a contradiction, therefore the columns of (a,,) are Q-linearly independent.

Thus r ~ ( ( a , , ) ) = 0 . Since (a,,)(l - (b , , ) (a , , ) ) = 0. (b,,)(a,,) = 1 and R is

d11ectl~- finite.

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POLYNOMIAL AND POWER SERIES RINGS 5137

It is clear that if R is a ring such that R [ x ] is right Ore then, by Lemma

2. l ( iv) , R is directly finite. We shall see that in this case M n ( R ) is directly

finite for all n 2 1. First we prove some technical lemmas.

Lemma 2.3 Let R be a rzng such that R [ x ] 2s rrght Ore. Let Q = & ; ( R [ T ] ) .

Then for all '4 E K ( R ) , I , - A x , where In zs the identzty n x n mntrzs, 2s n

unzt zn M n ( Q ) .

PROOF. The result is clear for n = 1. Suppose that n > 1 and that the

result is true for n - 1. Let A = (a,,) E M n ( R ) . Let B = (a,,);;& E AU,-l(R).

c = (alnx, .... an-lnx)' and f = (nn l x , .... a,,-lx). Then

Since In-1 - B x is a unit in ? ~ f ~ - ~ ( & ) , In - A x =

o ) ( ~ n - 1 ; B,Z :) f ( I nV1 - Bx)-I 1 - a,,x - f ( In-l - B x ) - l c i I n 1

Since 1, - A s is regular in M n ( R [ x ] ) , so is in Thus 1 - a, ,x - f ( Ine l -

Bx) - ' c is regular in Q . Since & is classical, 1 - nnnx - f (In-l - Bx) - ' c is a

unit in Q. Hence I , - Ax is a unit in M n ( Q ) .

As above. in the proof of the following result we identify the rings A l n ( R [ x ] )

and Af, ( R) [ X I .

Lemma 2.4 Let R be a rrng such that R [ x ] ts rzght Ore. Let A, B E Aln (R) .

Then there ~s z s t P ( r ) , Q ( x ) E M n ( R ) [ x ] , wzth Q ( x ) regular, such that (1, -

A x ) P ( x ) = B Q ( x ) .

PROOF. By Lemma 2.3, ( I , - Ax)-' E 114n(Q), where Q = QLl(R[x]) .

Thus there exist P ( x ) E M n ( R ) [ x ] and a regular element q ( x ) E R [ x ] such

that ( I n - A x ) - ' B = P ( x ) q ( x ) - l . Hence ( I , - A x ) P ( x ) = B q ( x ) . rn

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Theorem 2.5 Lct R be a rzng such that R[T] 7s right Ore. 1 '11~n L\fn(R) LS

dirtct ly firl~tr tofor d l Y I > 1.

PROOF. It fo l low fro111 Lernnlas 2.4 and 2.l(iv).

The next result gives us some classcs of regular rings H such that R[x] is

right Ore.

Recall that a ~nultiplicatively closet subset S of a ring R is a right Ore set

in R if. for each 1, E R nnd s E S , r S n sH # 8. Let S 11e a rnultiplicat~ively

closet suhsct cd regular clemmts of a ring R. It is well-known that there exist

a ring Q such that

( i i ) any s E S is invertible in Q,

[lii) fol eaclr q E Q there esist 1, E R and o E S such that q = r s - I .

if and only if S is a right Ore set i11 R. The ring Q is the right localization of

K at S. \I'r denote Q by RS-I.

Theorem 2.6 L F ~ R bt n q u l a r ring. I[ all the p r ~ m z t l r c factor rings of R

nrc nr.tin/au t h rn R j r ] rs n g h f and left Ore. Fui themore , Q c i ( R [ r ] ) ls the rzght

locril~zntion of R[o] nt the subsrt ,M(R[.r]) of i n o n ~ c polynomznls nnd ~t 1s ii

rcgular rozg.

PROOF. Suppose that .M(E[n.]) is not a right Ore set in R[ . r j . Then

there exist n E R[r] and r E . M ( R [ s ] ) such that aM(R[.x]) n rR[x] = 0. Set

C = { I a H / ~ ; U ( ( R l l ) [ . r ] ) n ~ ( R / l ) [ s j = f l ) . C i s partially ordered by C. IVe

shall see that C is inductive. Let { I , } he a chain in C and let 1 = U;I i . Suppose

t,hat 1 @ C . Then there exist r' E ,M(R[x]) and a' E R[x] such that ti7 = F?

in ( R / I ) [ z ] . Thus there exists b 6 I [ s ] such that art = rat + b, but t,here exists

i such that b E I , [ s ] , so ur' = m7 in ( R / I , ) [ x ] , a contradiction, therefore 1 E C

and C is inductive. BJ- Zorn's Lemma. C has a maximal element J .

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POLYNOMIAL AND POWER SERIES RINGS 5139

Suppose that J is primitive. Then R / J is an artinian simple ring. Thus

( R / J ) [ x ] is right Ore. Hence there exist c, d E R [ T ] such that ? E ( R / J ) [ s ] is

regular and nc = ~d in ( R / J ) [ x ] . By [16, Lemma 21, there exists q E R [ r ] such

that i;ii E , W ( ( R / J ) [ x ] ) and acq = rdq in ( R / J ) [ x ] , a contradiction. thus J is

not primitive.

Since J is not primitive. by [ 5 , Theorem 6.61, R I J has a non-trivial central

idempotent e. Since J is a maximal element of C, there exists s l , s2 E . M ( R [ r ] )

with the same degree and u l , a2 E R [ x ] such that u s e = ? q i ; e and ag(l- e ) =

r ~ ( l - E ) in ( R / J ) [ x ] . Thus S;e $%(I -e ) E . M ( ( R / J ) [ x ] ) and n(S;t + q ( l -

e)) = r(%t $7i;(l - t ) ) in ( R / J ) [ x ] , a contradiction, therefore , W ( R [ x ] ) is a

right Ore set in R [ x ] . B> right-left symmetr3; . M ( R [ x ] ) is a left Ore set ill

RbI. Let T be the right localization of R [ s ] at j M ( R [ x ] ) . \tTe shall see that I'

is regular. Suppose that T is not regular. Then there exists n E T such that

a @ a T a . If I is an ideal of R then it is clear that R / I is a regular ring

whose primitive factors are artinian. Thus J M ( ( R / I ) [ ~ ] ) is a right Ore set in

(R/I)[x]. \lie denote by TI the right localization of ( R / I ) [ x ] at , M ( ( R / I ) [ x ] ) .

Set C' = {I a R 1 a @ aTra) . C' is partially ordered by C. As a b o ~ e is not

difficult to show that C' is inductive. By Zorn's Lemma C' has a maximal

element J'.

Suppose that J' 1s prim~tive. Then R / J f is an artinian simple ring. Thus

Q c i ( ( R / J ' ) [ x ] ) exists and it is an artinian simple ring. Hence Q , , ( ( R / J ' ) [ . r ] )

is regular. By [16, Lemma 21, it is easy to see that Q C l ( ( R / J f ) [ x ] ) = Tjt. a

contrad~ction, thus J' is not primit~ve.

Since J' is not primitive. by [ 5 , Theorem 6.61, R / J ' has a non-trivial central

idempotent f . Since J' is a maximal element of C', there exists s E M ( R [ r ] )

and bl. b2 E R [ s ] such that a f = a K r - ' 6 f and Z ( l - f ) = aGr-la(1 - f ) in

Tjl. Thus n = a(&f + &(I - f ) ) ~ - ' a in Tj, , a contradiction, therefo~e T is

regulai.

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Let X he a non-empty set. Let 5 be any well-order of X. Let rl. ..., r , E

X, ~vi th sl < ... < T~ and let s l . ..., s,. r l . ..., r,be non-negative integers. ITt

write

xi' ...2-; < x;' ... z:

to mean that either r , + ... + r , < sl + ... + s,. or r l + ... + r , = sl + ... + s , and

for some j such that 1 5 j 5 n we have 1.1 = HI, .... rJ-1 = sJ - l and r, < s,.

Let R he a ring and let p E R [ X ] . 1% say that p is monic (respect to 5 ) if

1 is the coefficient of the maximum monomial of the support of p. It is easy

to see that the above proof can he adapted to show that if R is a regular ring

whose primitive factor rings are artinian t,hen R[X] is right and left Ore and

Q,I (R[X]) is t,he right localization of R [ X ] at the set of monic polynomials.

The converse of t,his result is not true as the next example shows.

Example 2.7 Tlztrt exrste n srrnple non-artznzan regular rzng R such that

R[x] 2s rzght and left Ore.

PROOF. Let R be the ring constructed in the proof of [ 5 , Example '21.201.

It is easy to see that R[x] is right and left Ore. rn

In the next section we shall see that for some classes of regular rings the

converse of Theorern 2.6 is true.

In [$I, Goodearl and Menal present a construction which a~sociat~e to each

comnlutative ring R a commuta t i~e overring S, having stable range one. They

derive from this cancellation results for modules over arbitrary commutative

rings. \l:e generalize in some sense this results.

Recall that a ring R has stable range one provided that for any a , b E R

satisfying uR + bR = R. there exists c E R such that n + bc is invertible. By

[17, Theorem 21, t,his condition is left-right symmetric. Furthermore, by [17:

Theorem 31. R has stable range 1 if and only if M,(R) has stable range 1

for any positive integer. It is well-known that any ring with st,ahle range 1 is

direcly finit,e.

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POLYNOMIAL AND POWER SERIES RINGS 5141

Proposition 2.8 Let R be a ring such that R [ X ] is right Ore for an infinite

set X . Then QEl(RIX]) has stable range 1.

PROOF. Let p,q E Q i t ( R [ X ] ) . There exist a, b,c E R [ X ] , with c regular,

such that p = ac-' and q = bc-'. Since X is infinite, there exists x E X such

that a , b,c E R [ X \ { x } ] . We shall see that p - x and q - x-I are units in

Q ; ( R [ X ] ) . Since c is regular in R [ X \ { x ) ] , it is clear that a - xc, bx - c are

regular elements in R [ X ] . Thusp-x = (a-xc)c-' and q-x-l = ( b ~ - c ) c - ~ x - l

are units in QL1(R[X]) . By [7, Lemma 1.21, the result follows. m

If R is commutative, it is not difficult to show that Q,' , (R[x]) has stable

range 1.

Let R be a ring such that R [ x ] is right Ore. We do not know whether R [ X ]

is right Ore for all non-empty set X.

It is well-known that a ring R is directly finite if and only if so is Rix] .

We do not know whether R is right Ore for R [ x ] to be right Ore. The next

example shows us that this condition works wrong in power series rings. In

order to prove this we need the following lemma.

Lemma 2.9 Let R be a ring that has a subring isomorphic to RE, h ln , (S i ) ,

where n; < n ; + ~ and $7; is non-zero for all i. Then R [ x ] is neither right nor

left Ore.

PROOF. We may assume that S = nzl Mn, ( S , ) is a subring of R. Suppose

that R[a] is right Ore. Let e:: be the matrix units of Mn,(S , ) . Let a, =

n -1 (1) C3L1 e33+1, a = ( a , ) E S and b = (ek)n,) E S. Since R [ z ] is right Ore, there

exist P(x), Y ( X ) E &[XI , with q ( x ) = qo + qlx + ... + q,zT regular, such that

( 1 - a x ) p ( x ) = bq(x) . Thus there exists s 1 1 such that

Let n; > r + s and let e, be the central idempotent of S corresponding to the

'actor Mn, (5';). It is easy to see that

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CEDO AND HERBERA

for J = 0, ..., I-. Hence e,bq(x) = 0. but q ( x ) is regular and e,b # 0, a contra-

diction, thus R [ x ] is not right Ore. Similarly R [ x ] is not left Ore.

Example 2.10 There exzsts a rrng R such fhat zt zs nezther nght nor left Ore

and that Rux] 1s rlght and left Ore.

PROOF. Let K be a field. Let S = n,,, - .%(I{). Let R = S [ x ] . B y

Lemma 2.9, R is neither right nor left Ore. Let en be the central idempotent

of S corresponding to the factor M,(li). Since

e n R [ y ] is right and left Ore. Let a , b E R [ y ] , with b regular. Then eib is

regular in e,R[yJl. Thus there exist c; ,d; E e ; R [ y ] such that c; is regular in

eiR[yJl and eiac; = e;bd;. Hence

It is easy to see that Cr==, cnyn is regular in R [ y ] . Thus Ray] is right Ore.

Similarly vie see that R[y] is left Ore. rn

Recall that a module is directly finite if it is not isomorphic to any proper

direct summand of itself.

Lemma 2.11 Let R be a rzng such that for all a , b E R there exzst p ( x ) , q ( x ) E

R [ z ] , u ~ t h q ( z ) regular such that ( 1 - a x ) p ( z ) = bq(x) . Let I , J be rzght deals

o,f R such that R/ I 2 R / I 9 R / J . Then J contazns the coeficients of a regular

polynomcal.

PROOF. Let a . b E R such that f: R / I @ R I J + R I I defined b y f (c. d ) =

(LC + bd is an isomorphism. There exist p ( x ) , q ( x ) E R [ x ] , with q ( x ) = qo + q l s $ ... $ q,xr regular such that (1 - a x ) ~ ( x ) = bq(x). Thus there exists s 2 1

such that

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POLYNOMIAL AND POWER SERIES RINGS 5143

By induction. it is easy to see that q,, ql. ..., q, E J .

Corollary 2.12 Let R be a regular rzng such that for all a . b E R there exz.st

p(x),q(x) E R[x], with q(x) regular such that ( 1 - ax)p(x) = bq(x). Then the

cyclzc rzght R-modules are dzrectly jnzte.

PROOF. Let I. J be right ideals of R such that R /1 2 R / I c;j R I J . By

Lemma 2.1 1. there exist qo, ql, .. ., q, E J such that qo + qlx + . .. + q,xT is regular

in R[x]. Since R is regular, qoR + qlR $ ... + q,R = R. Thus R/J = 0 and

R I I is directly finite.

Corollary 2.13 Let R be a rzng such that R[x] zs rzght Ore. Then the non-

szngular cyclic rzght R-modules are dzrectly jnzte.

PROOF. Let I be a right ideal of R such that R / I is nonsingular. Let J be

a right ideal of R such that R I I F & / I $ R I J . Since R I I is nonsingular, so is

R I J . By Lemma 2.11, there exist qo. ql, ..., q, E J such that qo+qlx + ...+q, xT

is regular in R[x]. Since R[x] is right Ore, qoR + qlR + ... + qrR <, R. Thus

J Se R and, since R I J is nonsingular, R I J = 0. 8

Corollary 2.14 Let R be a regular ring such that R[x] zs rlght Ore. Then the

j h t t l y generated rzght R-modules are dzrectly jnl te .

PROOF. Let AI be a finitely generated right R-module. Thus there exist

ml, .... m, E ill such that .%f = mlR + ... + m,R. Consider the right Mn(R)-

module M n = (ml, ... m,),%In(R). By Lemma 2.4 and Corollary 2.12, M n is

directly finite. Suppose that M 2 M $ A for some right R-module A. Then

Mn E llMn @ An as right A4n(R)-modules. Thus An = 0 and so A = 0.

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3 Monic localization and repetitive rings

?Jote that the difficultv in the study of the Ore condition in polynomial and

poxver series rings proceeds from the fact t,hat we do not know very well the

regular elelllents of this rings. Kow we shall study when the set of monic

polynomials is a right Ore set'.

Following [(i]. we say t,llat a ring R is right repet,itive if for all a , b E R the

right ideal I,,, - u 1 6 R is finitely generated. In [ 6 ] , Goodearl shows that ,II,(R)

is right repetitive for all n 2 1 if and only if all surjective endomorphisms of

finitely generated right R-modules are automorphisms. The next result shows

us tha t this is equivalent t o the rnonic localization of R [ x ] .

Theorem 3.1 Lei R be a rlng. T h c n the followzng condzttons ure e q u ~ r a l e n f .

(1) 31n(R) 15 rzyht rfpetrtzre for all 11 > 1 .

((0 9 = 1 + rR[a.] is n rzght Ore set zn R[r ]

( l tz) The sc t of rnonzc yolynorn~als o j R [ x ] 1.5 a u g h f Ore se t ~n R [ x ]

PROOF. ( i ) =+ ( i i ) . Let a j x ) = 1 - a lx - ... - a,xn E S and b ( x ) =

bo + bl;c + ... + 6,,xr" E R[r] . \Ye may assume that n > m. Let C,? ,nz i x i =

a ( x ) - I E R [ x ] and c(;c.) = Ci,,c,.i' - = a ( x ) - ' b ( x ) E R [ x ] . We shall see that

there exists q ( s ) E S such that c ( ; c )q (x ) E R[.L.].

Consider the following nlatrices of M, ( R ) :

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POLYNOMIAL AND POWER SERIES RINGS

Since m o = 1, M is invertible, in fact

From the relation 1 = (C+,, m z x z ) a ( x ) we have -

MA' =

for 1 _< J < n and n 5 I . Thus

and

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5146 CEDO AND HERBERA

Since the last row of AAII-' is (O.1,0 , ..., 0 ) and A"'B = i l , lI- 'MA", the

last ron of A"lB is ( c , + ~ , 0 , .... 0 ) for all a >_ 0.

Since *lI,(R) is right repetitive, there exist t > 1 and Q1. ....Qt E M n ( R )

such that

Att 'B = A B Q t + A 2 B Q t - l + ... $ AtBQ1.

Let q, be the elenlent of R in the first column and the first row of Q,. Then

by the component (n. 1) of the above matrix relation,

Let q ( z ) = 1 - qlx - ... - qtxt . Then p ( x ) = C ( ~ ) ~ ( X ) is a polynomial of degree

5 t and a ( z ) p ( x ) = h ( x ) ~ ( x ) .

(ii) + ( i ) . Suppose that ,S is a right Ore set in R[.r]. It is easy to see that

then R is right repetitive. So it is sufficient to show that S t = 1 $ x , I&(R)[z]

is a right Ore set in M 2 ( R ) [ x ] .

Let .4(x) E '5'' and B ( r ) E M , ( R ) [ x ] . We shall see that A ( x ) M 2 ( R ) [ x ] n B ( x ) S t f 0. There exist a l ( x ) , a 4 ( z ) E S and a 2 ( x ) , a s ( x ) E sR[;c] such that

Since S is a right Ore set i n R [ x ] , there exist r r ( x ) , ? ( z ) E R [ x ] and J ( x ) . 6 ( x ) E

S such that a l ( x ) a ( x ) = a 2 ( ~ ) a l ( x ) 3 ( . r ) and a 4 ( x ) - y ( x ) = a 3 ( x ) a 4 ( x ) S ( x ) . Let

s1[.r) = nl( .r)a4(n.)S(z) - a 2 ( r ) ? ( x ) and s 2 ( x ) = a 4 ( x ) a l ( z ) f ( x ) - a 3 ( x ) r r ( x ) .

Then

Thus we may assume that a z ( x ) = n 3 ( x ) = 0. Let

Now there exist P 1 J s ) . p 2 ( x ) , p 3 ( x ) , p 4 ( x ) E R [ x ] and q ( x ) E S such that

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POLYNOMIAL AND POWER SERIES RINGS 5 147

Hence

The proof of (ii) (iii) is an easy exercise.

Now we shall see that the converse of Theorem 2.6 is true for some classes

of regular rings.

Recall that a pseudo-rank function on a regular ring R is a map P: R --+

[O, 11 such that

(a) P(1) = 1.

( b ) P(xy) 5 P ( x ) . P ( y ) for all a , y E R,

(c) P(e + f ) = P ( e ) + P( f ) for all orthogonal idernpotents e. f E R.

Denote by P(R) the set of all pseudo-rank functions on R, and let

for all x E R. Then the rule 6(x. y ) = Ar"(x - Y) defines a pseudo-matric on

R; we say R is 11'"-complete if S is a metric and R is complete with respect to

it.

We say that a regular ring R is right No-continuous if the lattice of principal

right ideals L(R,q) is upper No-continuous, that is, every countable subset of

L(R,q) has a supremum in L ( R R ) and

for all A and all countable linearly ordered subsets {B,) in L(R,q).

Recall that a ring R is called right No-injective provided every homomor-

phism from a countably generated right ideal into R is given by left multipli-

cation by an element of R.

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Parts of the following theorem are due to Armendariz. Fisher and Snider

j.L]. and Goodrarl ant1 Lloncasi [6].

Theorem 3.2 Lct R be n r ~ g u l a r ring ~oh lch 1s .\"-complete, o r rlght o r left

Ro -con t~nuous , or l ~ f t E O - i n ~ e c t i ~ e . Then th t Jo l lowny condz t~ons ale tquirn-

lent:

(ij R is right r epc t i t i~e .

(11) l? h a s boundtd 1ndc.r of ndpot t t zc t .

( Z I L ) ,112 p r i m r t m factor rings of H a r t artlnznn.

PROOF. ( i i ) =+ ( i i i ) [5. C'orollary i .10].

( z i i ) =$ ( 2 2 ' ) [2. Corollary 2.61.

( i i i ) =$ i i ) By [2 , Theorem 2.31 and [6, Theorem ' i j .

( i j 5 ( i i ) By 16. Proposition 41. all surjective endornorphisnls of cyclic

rigtit Fi-modules are auto~norphisms. In particular all cyclic right R-modules

are clirect,ly finite. By the proof of [d: Theoren1 6 ( ( t ) =+ ( a ) ) ] : H has hounded

index of nilpotence.

t i ? ) + ( i i ) By the proof of [2, Theorem 1.1 ( ( a ) + ( b ) ) ] , all injective

eridonlorphisrns of cyclic right R-modules are auto~norpl~isms. In particular

all cyclic right R - n ~ o t h ~ l ~ s are directly finite. By the proof of [8: Theorem 6

((t.) j ( a ) ) ] R has bounded index of nilpotence.

( i i z ) =+ ( z > ) Theorem 2.6.

(2.) + ( i i i ) By Corollary 2.11, all finitely generated right R- n~odules are

directly finit,e. By [d, Theorem 61. the result follows. rn

Note that this result gives a partial positive answer t o a question of Good-

earl [6] , i . e.

Let R be n right r e p ~ t i t i v t rcgular ring. I s R strogly a-regular3?

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POLYNOMIAL AND POWER SERIES RINGS 5 149

4 The Jacobson radical and semilocal rings

Let R be a ring we denote by J ( R ) the Jacobson radical of R. Recall that R is

semilocal provided R / J ( R ) is semisimple artinian. If in addition J ( R ) is right

(left) T-nilpotent then R is said to be right (left) perfect. If R is semilocal

and J (R) is nilpotent then R is semiprimary.

In [3, Theorem 11 Camps and Dicks prove that a ring R is semilocal if and

only if there exists a homomorphism from R to a semisimple artinian ring S

taking non-units of R to non-units of S, that is the image of R \ U ( R ) lies

in S \ I.'(S). Assume that Qi1(R[x]) exists and is semilocal, then if QE,(R)

exists the natural homomorphism Q:{(R) --+ Q,',(R[s])/J(QE1(R[x]) carries

non-units to non-units and thus we have,

Lemma 4.1 Let R be a rzght Ore rzng such that Q : l ( R [ ~ ] ) exzsts and zs semllo-

cal. Then Q:l(R) zs semzlocal.

In this section we shall show that whenever J(QE1(R[x])) n R[s] = I[.r] for

some ideal I of R, if QLI(R[x]) exists and is semilocal then Q',[(R) exists and,

by Lemma 4.1, is also semilocal.

First we shall study when the condition J(Q:l(R[x]))n Rix] = I[x] for some

ideal I of R holds.

Lemma 4.2 Let R be a ring such that Q:,(R[x]) exzsts. Then the followzng

statements are equzoalent:

(i) J(Qb(R[x]) ) n R[x] = I[x] for some ideal I of R.

(ZLZ) l f p ( x ) E J(QT,l(R[s])) n R[x] . then there exzsts k > 1 such that

PROOF. We only need to prove that (iii) j ( i ) . Consider p ( x ) = a0 + . . . $ a,xn a polynomial in J(QIl(R[z])) n R[x]. We m a y assume that a. and

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5150 CEDO AND HERBERA

a, are different from zero. By hypothesis there exist k l , . . . , I;, > 1 such tha t

p(.~"). . . . .p(rk1'.-"" are elements in J(Qzi(R[.r])) n R[x]. Consider the matrix

\;lc h a r e that *\I [ a : ] E nJ(Q:l(R[r])). Slnce >tI is an invertible matrix we

a n

can conclude that a o . . . . , a , E .J(Q,Pi(R[r])) n R.

Proposition 4.3 L f t R be a c o m m u t a t ~ v e u n g . Then J(Q,!(R[x])) n R [ x ] =

I [ s ] for sonre d f n l I of 12.

PROOF. The re5ult will follorv from Lemma 4.2 if we show that whenever

p(z) is in b(Q,l(R[n.])), then so is For that we only need t o prove tha t

1 - JI(T')# 1s an invertible element ot Qc, iR[~] ) . Since $$$$ = % me may assume that sf(x) = s(x2) and me must show that s(x2) - p(x2)r(x)

IS regular In R[x]. Obserxe that smce R [ x ] is isomorphic to R[x2], s(x2) -

p(x2)r(r2) is regular in R[z2].

Assume t h a t (s(r2) - p(x2)r(x))q(x) = O . Smce r(x) = rl(r2) + x ~ ~ ( 1 ~ )

and q(n.) = ql(r2) + ,rQ2(x2), n e have that

(s(.rL) - P(t2)r1(x2))ql(x2) - x2p(x2)~2(x2)q2(x2) = 0

(s(x2) - p(x2)r1(x2))q2(x2) - p(x2)r2(x2)q1(x2) = 0.

But u = A(") - p(x2)rl(.r2) is regular in R[r2], thus from the first equal-

lity we grt y , ( ~ 2 ) = U - ~ X ~ ~ ( ~ ~ ) T ~ ( . T ~ ) ~ ~ ( X ~ ) . By substituting to the second

equallit: me h a w

71q2jxZ) - p ( x 2 ) r 2 ( x 2 ) ~ - 1 x 2 p ( x 2 ) r 2 ( . ~ 2 ) q Z ( 2 2 ) =

(U - p(x2)r2(x2)~-1x2~(x2)r2(x2))q~(~2) = uq2(x2) = 0,

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POLYNOMIAL AND POWER SERIES RINGS 5151

where z. = ~ - ~ ( x ~ ) r ~ ( x ~ ) ~ - ~ x ~ p ( x ~ ) r ~ ( x ~ ) is a unit in Q C l ( R [ z 2 ] ) , thus 42(z2) =

0 and so ql(x2) = 0. Hence s ( x Z ) - p(x2)r(x j is regular in R [ x ] . m

For non necessarily commutative ring we can obtain a similar result, but

assuming 2-I 6 QE,(R[x]) .

Proposition 4.4 Let R be a rzng such that Qrl (R[x ] ) exzsts and 2 zs mvertzble

zn Q,' , (R[x]) . Then J(Q,T/(R[z] ) ) n R[x] = I [x ] for some dea l I o f R.

PROOF. Let f ( x ) be an element of J ( Q i , ( R [ x ] ) ) fl R[x] of minimal degree

satisfying f ( x ) $' (J(QE,(R[x]) ) n R ) [ x ] . Then

Thus f ( r c ) + f ( - x ) = 2 f l ( x 2 ) and since 2 is inveltible, f i ( x ) E J ( Q : , ( R [ x ] ) ) .

On the other hand f ( x ) - f ( - x ) = 2f2(x2), thus f 2 ( x ) E J(Q: , (R[x ] j ) . Since

f i ( x ) or f2 (x ) are polynomials of lower degree than f ( x ) , they must be elements

of (J (Q , ' , (R[x ] ) ) f l R ) [ x ] . But the coefficents off ( x ) are the coefficients of ~ f l ( x )

and . f 2 ( x ) , thus f ( x ) E (J (Ql , (R[x ] j ) n R ) [ x ] , which contradicts the choice of

f ( x ) .

Anlitsur in [I] proved that J ( R [ x ] ) = N [ x ] where N is a nilideal of R,

Bergman cf. [15, pg.1941 gave a new proof of this result. The argument in

the proof of Proposition 4.4 is very similar to Bergman's one. By looking at

Bergman's proof ~t is not difficult to see that the stament of Proposition 4.4

is still true if we assume that R contains a primitive odd root of unity in its

centre instead of the condition 2-' E Q,TI(R[x]).

The next result is a consequence of this kind of condition on the ,Jacobson

radical of Q: , (R[x ] ) .

Proposition 4.5 Let R bc a ring such that R[x] is righf Ore and

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CEDO AND HERBERA

for some ideal 1 of R. Then

( I ) If Q',,(R[x]) 2s semzlocal, then Q,P1(R) exzsts and 1s sem~locul

j z z ) If Qi ,(R[x]) 1s rrght (left) perfect, then QZl(R) zs rzght (leftj p~rfect .

j i i i ) If Q:)(R[x]) is ~einiprimary, then QE,(R) is semiprimary. I

PROOF. Let S = {p(r) E R[x] / l [x] / p(r) is regular in R[x]) . It is easy t o

see that S is a multiplicatively closed subset of regular elements of R[x] / l [x]

and that Q : l ( R [ r ] ) / J ( Q ~ , ( R [ r ] ) ) is t,he right localization of R[x]/ I[x] at S.

Assunle that Qz,(R[x]) is semilocal. Then Q ~ , ( R [ Z ] ) / J ( Q ~ ~ ( R [ X ] ) ) is ar-

tinian and therefore is classical. Thus

By [13. Theorem 2.11 RII is right Ore and QI,(R/I) is a semisimple artinian

ring. Let r ( z ) be a regular element of R[r]. Then r(;c) is regular in R[.r]/ l[z]

and, by [16, Lemma 21, there exists q(r) E R[r] such that one of the coefficients

of r(r)q js ) is regular in R. By [9: Lemma 2.11. R is right Ore. By Lemma 4.1,

QE,(R) is semilocal.

In order t o prove ( i i ) and (iii) by ( i ) wr may assume that R is classical

and so also is RJI . By using t,he same argument in ( i ) we can conclude that

RII is semisinlplp a r h i a n . Since J(Q:I(R[x])) n R[x] = I [ x ] . I inherits the

nilpotency conditions of J(Q:l(R[.r])) and thus ( i i ) and ( i i i ) follow. rn

We have now the following result for co~nrnutative rings.

Theorem 4.6 L f t R bc a commutative rmg.

( 1 ) If Q,,(R[.c]) 1s sernilocal then Q,,(R) zs semzlocal.

( 1 2 ) Qcl (R[x]) 1s perject zf and only zf QC1(R) 2.3 also perject.

( z ~ z ) Q,I (R[x] ) 1s semlpr1mary zf and only ~f QcI(R) zs semzprzmary.

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POLYNOMIAL AND POWER SERIES RINGS 5153

PROOF. (i) and the .'only if "part of ( i i) and ( i i i ) follow from Proposition

4.5 and Proposition 4.3. It only renlains to s h o ~ that if Q , ! ( R ) is serrrilocal

and the Jacobson radical is nilpotent or T-nilpotent. then these properties are

inherited by t'he classical ring of quotients of the polynomial ring. IVithout

lost of generality we may assunle that R = Q,,(R). If J ( R ) is a nilideal we can

apply Kobson's characterization of orders in sernilocal rings, e. g. [13, pg.:36s]

and conclude R[x] is an order in a semilocal ring. Since J ( R ) is a nilideal by

[I] J ( K [ . r ] ) = J ( R ) [ s ] , thus R [ x ] is an order in a senlilocal ring whose Jacobson

radical satisfies the same nilpotency conditions that J ( R ) satisfies. m

The converse of Theorem 4.6ji) is riot true. It is inlplicit in the proof of

Proposition i4.3 that if R[.r] has a serriilocal right classical ring of quotients

then the tight ideal of R generated by the coetficients of a regular element of

R[s] contains a regular element of R. The following esamplc shows that tliis

can fail when R is a srmilocal commutative ring.

Example 4.7 Therc m s i s a corrzmutatrt~c sem2locnl rrng R such tha t Q L c ( H [ r ] )

15 not semilocal.

PROOF. Consider the power series ring Ii[.s.iJJ over a field K. Let ,S be

the set of irreducible elements of Ii '[s. i j and set

Let R be the trivial estcnsion K [ s . t j ;x -LI. It is clrar that R is scmilocal.

Consider the polynomial p ( a ) = (5 .0 ) + ( t , O ) z E R[s] . Since I = (,c.O)R + (t. O)R is faithful. p(x) is regular in R[:u]. But I clops not contain any regular

elenrent of R. Thns Q C I ( R [ s ] ) is not sernilocal.

Theorem 4.8 Lct K b f a r lng such tha t Qel(R[r]) C . Z I S ~ a n d 2-' E QTI(R[.r])

( i j I f Q ; , ( R [ a ] ) is semilocal. thrn QI: I (R) exists and is .stmilocal.

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5154 CEDO AND HERBERA

(ii) I f Q: ,(R[s]) is right ( l r f t ) pcrfict, then Q;!(H) crists and is right ( l ~ f t )

P ~ l f e ~ t .

( i i i ) If Q : , ( R [ r ] ) i.5 .st rrziprinary, i h t n Q:!(R) e.rists and ifi s t r r ~ i p ~ i m a r y .

Goodearl in [Ei. Esarnple '31 gives an esample of a semiprimary ring R which

is riot riglit repetitive and it is easy to see that R[s] is neither a right Ore ring.

5 K-algebras of small dimension

In [,his section we s tudy the Ore cor~tlition on the polyilonlial ring R[r] over an

algel~ra R ovcr a field Ii such that d i ~ n ~ ; R < jlij. F i r ~ t IVC g i v ~ a characteri-

zation of the regular polyilomial over such algc111.a~ mherr I< is uncountalde.

A11 the rcsults ill this section a1.r ~ i t h the collaboration of Professor L.

Romen.

PROOF. 1,et p j . r ) E R[.r]. Suppose that there esists X E I< sucll that p ( X )

is regular in R. Lct q ( . r ) = p(.r + X ) . It is clear that p( . r ) is regular if arid only

if q ! , r ) is rcgnlar. Since q ( 0 ) = p(X) is regular in R, q(.r) is rcyplar. Thus p ( . r )

is regular.

Con\-crscly. suppose t h a t p( . r ) = no -+ n,.r + ... + rr,,sn E R[r] is regular. Let

be a basis of R ovcr I<. Then

fol iorne X 3 k . o , , E h . Let r he t he subfield of Ii' generated b~ a,,.

1 . j . X E I Since 111 < /I\ / and I< 1s uncountable the tranicendence cleg~ec of

Ii over F is inhnite Lct t bt. an element of A' transcendental over E . \Ye

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POLYNOMIAL AND POWER SERIES RINGS 5155

shall see that p ( t ) is regular in R. Let S be the F-subalgebra of R generated

by e,. z E I. Let S1 be the F(t)-subalgebra of R generated by e, , z E I . Then

it is easy to see that

and that the nlap p: S[x] --+ S1 defined by p jq(x)) = q(t) is a ring monomor-

phism. Suppose that there exists r E R \ (0) such that p ( t ) r = 0. Since R is

a free left S1-module. we ma] assume that r E S1. Kow it is easy to see that

there ex~sts q( t ) E F[ t ] \ (0) such that rq(t) E p(S[s]) \ (0). Let s (x ) E S[s]

such that s ( t ) = rq(t). Then g(p(x)s(x)) = p( t ) ry( t ) = 0. so p ( x ) s ( r ) = 0 and

s (x) # 0. a contradiction, thus r ~ ( ~ ( t ) ) = 0. Similarly l ~ ( p ( t ) ) = 0. Hence

p ( t ) is regular in R.

Theorem 5.2 Let R be a classzcal algebra orer a j k l d K . S u p p o s e that

d2mKR < (I{(. Then R[x] zs an Ore rzng and Q:,(R[s]) = R[x]S-I whfre

S' = It'[r] \ (0) .

PROOF Suppose that jlil 5 Ro. Then R is a finite dimensional I<-algebra.

so R is Artinian and therefore R is classical. Furthermore, Ii'(x)lzA R is a finite

dimensional Ii(x)-algebra and thus it is classical. Since R [ x ] " I < [ X ] % ~ R, it is

clear that R[x] is an Ore ring and Q:,(Rjx]) = R[x]S-' where S = K[z] \ {O}.

Suppose that 11' is uncountable. Let { E , } , ~ ~ be a basis of R o\er K . Let

a (x ) = ag + alx + ... + a,xn and b(x) = bo + blx + ... + b,xn' be polynomials

over R. Suppose that a ( r ) is regular. There exists a , , k , al,, 3,, E Ii' (2.1. k E I.

0 5 1 5 n . 0 5 r 5 in) such that

Let F be the subfield of K generated by o,,n, ail, ,3,,. (z, J k E I. 0 5 I 5 n ,

0 5 r 5 m ) . Let S be the F-subalgebra of R generated by 6 , . (z 6 I ) . Sote

that (Si 5 max(No. dimh R) < JIr'l. Thus the transcendence degree of K over

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5156 CEDO AND HERBERA

F is infinite. Let T he a transcende~lce basis of li over F. Let t E T. By t'he

proof of t h e above Proposition. a ( t ) is regular in R. Since R is classical, a ( t )

is a rmit. By [7, Lemma 2.11 r R ( l - C L ( ~ ) C X ) , for Q E I<. are independent. Thus

therc esists JI 2 I< with 1 Jll = jI<j, such that r R ( l - -a ( t )u ) = 0 for all (1. E J1.

Similarly. there exists J2 C_ J1 v i t h 1521 = lKI. such that lR( l - a j t ) a ) = 0 for

all a E J2 . Since R is classical, 1 -a ( t )n is a unit for all a E J 2 . By t'he proof of

[12. Lemma 7.1.'). a ( t ) is algebraic over Ii'. so it is algebraic over F ( T ) . Thus

there exists p E F[T]\{O) such tha t a(t)-lp E ,?[TI t,he suhring of R generated

by .5' and T . It, is easy to see t<hat the map q: S[T \ {t)][x] + SIT] defined by

p(q( .c)) = q(t) is an isornorpllism of F-algebras. Let c(x) E S[T \ {t)][z] such

t>hat c ( t ) = a ( t ) - ' p . Let d ( r ) E F[T \ {t)][x] C Iijs] such that d ( t ) = p. S o w

we have

a ( s ) c ( x ) b ( x ) = b(x)d(x),

and since d ( x ) E I f [x ] \ (0). the result follows.

Proposition 5.3 Lct R be an algebra over a field Ii' such that d1rnr, R < / I</ .

If R[x] I S a rrght Ore rrng t h e n R zs n ngh! Ore ring.

PROOF. Clearly we ma) assume that li is uncountable. Let a. b E R and

suppose that h is rrgular. Since R[s] is a right Ore ring there exist p(x), qjx) E

R[x] such that p(x) is irgular and

By Proposition 5.1, there exists X E Ii such that p(X) is regular. Since ap(X) =

bq(X). R is a right Ore ring.

\Ve do not know whether the converse of this result is true. The next result

is an easy generalization of [ i , Theorem 2.21 and give some information about

this problem (cf. Lemma 2.1).

As in [7] . we say that a ring R satisfies (right) unit 1-stable range if when-

ever a , h. c E R satisfy ctc + h = 1, there exists a unit u E R such that a + bu is

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POLYNOMIAL AND POWER SERIES RINGS 5157

a unit. It is clear that a ring with unit 1-stable range has stable range 1 and

thus it is directly finite.

Theorem 5.4 Let R be a rlght Ore algebra over a jkld Ii. Suppose that R

contalns no dzrect sums of lKl nonzero rzght or left zdeals. Then (&(R) has

1-stable range and thus zt as dzrectly finzte. Furthermore, i f Ir' zs znfinzte then

Q;(R) has unlt 1-stable range.

PROOF. Let Q = Q;[(R). If Ii is finite then R is Artinian, so R = Q and

it has 1-stable range.

Suppose that Ii is infinite. Let a , b E R, b regular and q = ab-' E Q.

By [i. Lemma 2.11, r Q ( q - a ) for a E lr' are independent right ideals of Q.

Since the map p: Q + Q defined by p ( s ) = 6-'s is an isomorphism of right

Q-modules. we have that r Q ( a - cub) for a E I( are independent. Hence there

exists J1 E Ii with 111' \ J I J < llil such that r ~ ( a - ah) = 0. \da E J1. B y

[ i , Lem~na 2.11. lQ(q - a ) for a E Ir' are independent left ideals of Q. Since

/ & ( Q - CY) = l Q ( a - ab), there exists Jz C J1 with IIi' \ \I2/ < IKj such that

lR(a - ah) = 0, V a E 1 2 . Thus q - cr is a unit in Q for all cu E Jz. Since

lIi \ J21 < JIi/, the result follo~vs by [7. Lemma 1.21.

ACKNOWLEDGMENTS

The results in section 5 are in collaboration with Professor L. Rowen. We

thank him for his kindness in allowing us to include these results in this paper.

The authors thank UBrren Dicks and Jaume Moncasi for their comments

and suggestions.

This work was partially supported by the grant PB92-0586 of the DGICYT.

References

[I] S. A. Amitsur, Radzcals of Polynornznl rzngs. Can. J . Math. 8 (1956), 353-

361.

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5158 CEDO AND HERBERA

[2] E. P Xrmendariz, J . \Y. Flsher and R. L. Snider. O n T ~ I J ~ C ~ ~ I ' E and s u r ~ e c -

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[3] Ro3a Camps and ITarien Dicks, O n ,iernzlocal r m p , Israel J . of hIatll.81

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[j] I< R . (;oodearl. 1 on .\~umnnn Rrqulnr R m g s , 2nd e d ~ t ~ o n , I<rlegel Pub

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[GI K. R. Goodrarl. S U ) J C C ~ ~ I ~ E ~ i ~ d o m o r p h ~ n z of finitely generat fd modules,

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[C)] D Herbcra and P. Pllla). l n j t c t z z '~ c lass~cal quo t l e t~ f 71ngs of polynornznl

~ z n q s arc qunsl-Frobfnlus. J o u ~ r ~ a l of Pure and App. Alg. 8 6 (1993). 51-63.

[lo] .Jeanne IVald Kcrr. Tht pouter ser l t s rlng ouer an Orc donla171 need not

be Orr . J . of Xlg. 75 (1983). 175-177.

[ l l ] P. I lenal . A\Ior~ta cquzrnlrnce artd quof lent rlngs, Results in itlath. 13

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[I.?] D. 5. Passrnan, The 4lgebra7c Structure of Group Rzngs, ll'lleq-

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[14] Richard Resco, Lance 51.. Small and J . T. Stafford, Iiru11 and global dz-

menszons of semzprzme A'oetherzan PI-rzngs, Trans. A.M.S. 274 (1982),

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[15] L. H. Rowen. Rzng Theory. z~olume I, Pure and .Applied Vathematics vol.

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Received: January 1994

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