Mathematische Zeitschrift

Math. Z. 176,447-484 (1981)

9 Springer-Verlag 1981

Finite Normalizing Extensions of Rings Martin Lorenz Fachbereich Mathematik, Universitgt Essen, D-4300 Essen 1, Federal Republik of Germany

Introduction Let S be a ring and let R be a subring with the same 1. The extension R c S is called a finite normalizing extension if there exist finitely many elements

x l , x 2.... , x n in S such that S = ~ R x i and Rxi=x~R for all i. i=1

An important special case of such an extension is given by K [ N ] ~ K [ H ] , where H is a group, N is a normal subgroup of H of finite index and K [H] and K IN] denote the group algebras of H, resp. N, over the field K. Indeed, K [H] can be written as a crossed product of the finite group G = H / N over the ring R = K [N], in short K [H] = g [ g ] * (H/N) = R* G (see 1-7] for definitions). In a recent series of papers ([7, 8, 9]), D.S. Passman and the author have studied the prime ideals in R , G , where R is any ring (with 1) and G is an arbitrary finite group. In particular, we gave a description of the primes in R*G that are disjoint from R in terms of a certain finite-dimensional algebra. As a by-product of this description, it follows that if P ~ I are ideals of R*G with P prime, then P ~ R c , I ~ R ([7], Theorem 1.2). In other words, the extension R c R * G satisfies Incomparability. Combining these results with the work of Roseblade in [17], Passman and the author could construct all prime ideals in group algebras of arbitrary polycyclic-by-finite groups ([10]). Finite normalizing extensions of rings also are of great importance in the study of PI-algebras ([15, 18]). Actually, the extensions considered there usually satisfy the stronger condition x~r=rx~ for all r~R and i = 1, 2 .... , n instead of x ~ R = R x i, and hence are called finite centralizing extensions (sometimes finite extensions or liberal extensions, [16]). Such extensions occur quite frequently. Some standard examples are S = R [ G ] with G a finite group, S=Mn(R ), the n x n-matrix ring over R, and S = R @ E , where R and E are K-algebras with E K

finite-dimensional over K. Extending work of Schelter on PI-rings ([18]), G.M.

0025-5874/81/0176/0447/$0760

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Bergman established Lying Over and Incomparability theorems for finite centralizing extensions (1,2,3]). A simplified exposition of Bergman's results has recently been given by Robson and Small in [16], where one can find a lot of additional interesting information on finite centralizing extensions. The purpose of this paper is to give a unified treatment of both types of ring extensions within the framework of finite normalizing extensions, with particular emphasis on prime ideals. So let

R c S = ~ Rxi, xiR=Rx~, be a finite normalizi=1

ing extension of rings. In I-9], Passman and the author have shown that S is integral over R in the sense of Schelter (1,-18]). As a consequence, we derived a Lying Over theorem for R c S . These results will be extended in the first section of the present paper. Indeed, we show in Theorem 1.3 that there exists an integer t > 1, depending only upon n, such that for any right ideal A of R any monomial sls2...s m with sjeAS and m>t can be expressed as a finite sum of elements of the form al sila2 siz...sizal+ 1,

where O
Pc~R= ~ (2~ for suitable prime ideals (21, Q2,

-.-, Q,n

i=1

of R such that the factor rings R/Q~ are pairwise isomorphic. Moreover, if P is primitive or maximal then so are the Q[s. Most of the results of this section have been independently obtained by J.C. Robson and J. Bit-David. In the remaining sections of this article the so-called Martindale ring of quotients will play an essential role. In order to get it to work properly, we require the following to be true: Suppose S is prime and A is an ideal of R having zero annihilator in R. Does AS contain a nonzero ideal of S? Unfortunately, this is false in general. At the end of Sect. 2 we will give an example, due to Allan Heinicke, demonstating this (Example 2.3). However, the answer is easily seen to be positive if instead of xiR=Rx~ we assume that there exist suitable automorphisms a~ of R such that xir~'=rx~ holds for all r~R. Conversely, if R is also assumed to be prime then a positive answer to the above question does actually imply the existence of such automorphisms cq. Note further that the above situation certainly covers crossed products and finite centralizing extensions. Therefore, the remaining sections are mainly concerned with extensions R c S = ~ g x i such that xirai=rxi (reR) for certain automorphisms cri of R. ~= 1 Section 3 is devoted to studying such extensions with S a prime ring. In particular, R is semiprime, by Theorem 2.2, and so the Martindale ring of

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quotients Qo(R) of R is at our disposal. We define a certain normal closure R(G) of R inside Qo(R) and embed the extension R c S in a suitable extension R(G) c S(G) which has a particularly nice structure. In fact, R(G) is a finite direct product of pairwise isomorphic prime rings corresponding to the finitely many minimal primes of R. Moreover, if el, e 2 .... , em~R(G) denote the idempotents corresponding to this decomposition, then each eiS(G)e j is a free left module over eiR(G). In particular, for each i, eiR(G)ceiS(G)e i is a free (on both sides) normalizing extension of rings (Proposition 3.7). We remark that R(G) is a subring of the normal closure of R that has been defined by Cohen and Montgomery in [4 I. Section 4 contains a detailed discussion of the latter type of extension, that is we consider extensions R c S with R a prime ring and with S = ~ Rx~, x~R i=1

=Rxi, such that the elements x~eS are left and right R-independent. In this case, the elements x~ are said to form a normalizing basis for R cS. Besides the extension R(G)mS(G) constructed in Sect. 3, any crossed product S=R*G, with G a finite group, has a finite normalizing basis over R. In fact, the results of Sect 4 generalize work of Passman and the author on crossed products of finite groups in [7, w However, the present situation is much more complicated. We give a short outline of the main techniques used in Sect. 4. First, we adjoin the Martindale ring of quotients of R to S and thus obtain an extension RocSo, with R o --Qo(R), which still has a finite normalizing basis. We then consider the normalizer T=Nso(Ro) of R o in So, that is T-{sESoJsRo=Ros }. Note that this is not a subring of S O in general, since T will in general not be closed under addition. However, even with this defect, Thas a fairly nice structure and many standard ring theoretic arguments still do apply in this setting. Thus we introduce an obvious notion of prime ideal in T and show that T has only finitely many such primes and, moreover, these are all maximal and their intersection is nilpotent (Proposition 4.9). Finally, we define operators " and e which yield an explicit one-to-one correspondence between the primes of T a n d the primes P of S satisfying P~R=O (Theorem4.12). In certain special cases, including finite centralizing extensions and crossed products, one can use the centralizer E = ~?so(Ro) of R o in S o instead of Tand thus work within a somewhat more familiar setting, since E is a finite-dimensional algebra. In general, this is unfortunately not possible, as we will show by an easy example. However, even in the general case, the subalgebra E can be used to a good deal to study primes in T. Finally, we give an example showing that the main results of Sect. 4 all break down if the extension R c X does not have a normalizing basis. Section 5 contains an Incomparability theorem for finite normalizing extensions of the form R c S with S = ~ Rx i and x~r~=rxi(reR) for certain autoi=1

morphisms a~ of R. Indeed, we show in Theorem 5.2 that if S is prime then any nonzero ideal of S intersects R nontrivially. This generalizes the results previously proved for crossed products ([7, Theorem 1.2]) and finite centralizing extensions ([3, Corollary 7], [16, Theorem 4.51). Theorem 5.2 is an outcome of the more precise results of Sect. 4, combined with those of Sect. 3.

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Except for a few isolated places in Sect. 1 (which will be pointed out explicitly), we will assume throughout that rings have identity elements which are inherited by subrings. Ideals are understood to be two-sided. Left R-module homomorphisms will usually be written on the right, ring homomorphisms in the exponent. It is a pleasure to thank Don Passman for the interest he has shown in this paper. The majority of our results have their origin in previous joint work on crossed products.

w Integrality In this section we consider arbitrary finite normalizing extensions R c S = ~ Rx~, x ~ R = R x i , and we will show that S is integral over R in the sense of i=l

Schelter [18]. In fact, we shall prove a somewhat stronger version of [9, Corollary 2] which works for arbitrary right ideals of R instead of just normal ideals, that is ideals A of R which satisfy x~A =Ax~ for all i. The main technique here, as in [13] and [9], is to reduce the integrality problem for R ~ S to a question about polynomials satisfied by matrices. The first general result in this direction is due to Par6 and Schelter ([13]), and our Proposition 1.2 is proved by a careful adaption of their argument to the present situation. The difference between our approach and the one in [13] is that we have to use a fairly drastically generalized concept of scalar matrices. The need for such a generalization first become apparent in [9], where the usual scalars were replaced by "transversals" in M,,(R), the ring of n xn-matrices over R. We widen the scope of this notion by the following definition. Definition. Let n > 1 be a positive integer and let d = ( A 1 , A2, ..., An) be a fixed sequence of n right ideals A~ of R. A matrix ~ M , ( R ) is called an d - m a t r i x if and only if all entries in the i-th row of c~ belong to A,. Note that the set of d matrices is a right ideal of M,(R). Let Dd denote the subring (in general without 1) of Mn(R ) consisting of those d - m a t r i c e s which are diagonal, so that D o v ~ _ A I | 1 7 4 .. A subring T e D d is called an d-transveraI if and only if T projects onto each component A i9

An example that will become important later on is as follows. Let R c S = ~, Rx~, x i R = R x i, be a finite normalizing extension and let A be a right ideal i=i

of R. For i = 1, 2, ..., n define A i= {reRIr xTxiA}.

It is easily seen that A i x i = x ~ A and that Ar is a right ideal of R. Let d = ( A 1 , A 2 , . . . , A n ) and let TA denote the set of all diagonal matrices d = diag(q, r2, ..,, r ~ ) e M d R ) such that there exists an element a e A with r~ x i = x i a for i = 1, 2, ..., n. Then the elements of TA are clearly d-matrices, and Ta is even an d-transversal. For, if t E A i is given then we can choose a E A with r x i = x i a ,

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and then suitable r j s A j with r j x j = x j a ( j # : i ) . Clearly, d = d i a g ( r l , . . . , r ~ _ l , r, ri+ 1, ..., r,) belongs to TA, and d maps to r under the projection map of TA to Ai. Thus TA maps onto each A~, and it is easily checked that TA is a subring of D~. Definition. Let d = ( A ~ , A2, ..., A,) be a fixed n-tuple of right ideals of JR and let T c D d be an J-transversal. Let cq, c%, ..., % be d-matrices. By a T-monomial in cq, ~2,-.., % we will mean a product of the form dl ~11d2 c~i2-- 9~i~ d~+ 1, where the dSs belong to T a n d where some of the dSs may be missing but at least one dj does occur. Thus, for example, if d 1 and d E belong to T t h e n dlcqc~~ decq and d 1 are T-monomials in el and c~2 but elc~2 is not (unless I~T). The degree of such a monomial will be the total degree in the e~'s. Thus d I cq c~2 d 2 cq has degree 4 and d~ has degree 0. Any sum of T-monomials will be called a T-polynomial, and its degree will be the highest degree of the monomials occuring in the sum. Note that any matrix that can be written as a T-polynomial in 7~,c~2, ..., % surely is an ~r The following lemma contains some notations and a few elementary observations that will be needed in the proof of Proposition 1.2. Lemma 1.1. Let R be a ring and let Mn(R) be the ring of n x n-matrices over R. Each matrix z E M , ( R ) can be partitioned as

where z' is a block of size (n - 1) x ( n - 1), z" has size (n - 1) x 1, and the *-entries stand for blocks of the appropriate sizes. We set

and for any z, p c M , ( R ) we write z - p if and only if z " = p ' . N o w let sC=(A1,Ae, ...,A~) be a f i x e d n-tuple of right ideals of R and let T c D ~ be an J-transversal. I f zl, Ze, ..., Zh, p E M , ( R ) are ~r with h > 1, then ?h'~h_ l ... Zl p=--'Ch'Ch_ I ... zl p + ~, where cp~M,(R) has the form ----%dh+ % % - idh- 1 + .-- +'~h~h- 1 "" ~1dl for suitable d l , d 2 , ..., d, e T. Proof First note that for any two z and p e M n ( R ) we have ( { p ) " = z ' p " . Hence, if Pl and p2~M,(R) are such that Pl =-Pz then {Pl =?P2. Now let z and p ~ M , ( R ) be d-matrices. Then we can find a suitable d E T such that {p=zp-zd.

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Indeed, since the (n, n)-entry of p belongs to A, and since Tprojects onto A,, we

[:

can find d e T s u c h that p - d has the form p - d =

. We obtain that

Thus ? p - z p - - c d, and since - z d = z ( - d ) has the required form we have proved the case h = 1 of the lemma. N o w let h > l and assume that ~ h _ l " ~ h _ 2 . . . { l p = - - ' C h _ 1 " C h _ 2 . . . ' C l p 4 7 @ for a suitable O=Zh_l dh_ 1 -t-72h_l'Ch_2dh_2-t-... ~-"Ch_lZ h 2,.."Cl dl, with di~T. Then it follows from the remark in the first paragraph of the proof that Th'~h_l...~lp~'~h(TYh_lTh

2 . . . "Cl p -}- I]/),

Since Zh_ 1 and 0 are ~r we see that the matrix Pl =TJh-I'Ch-2""TlP-}-I/I is an d-matrix, and hence as in the proof of the case h = 1, we can find a suitable d e T s u c h that T'hPl =---'ChPl - - Z h d , that is ~hP l ~ T,h ~2h_ l . . . "cl p q- "ChO -- "Chd

~ "ChTh_ 1 . . . z l

p Ar- ~O.

Here we have set ~0= % 0 - "Ohd, and this clearly has the required form, with d h = - d. This proves the induction step, and the assertion of the l e m m a follows. We now prove our integrality result for matrices. Proposition 1.2. Let R be a ring and let M , ( R ) be the ring of n x n-matrices over R. Then there exists an integer t = t ( n ) > 1 such that the following holds: Let d = ( A 1 , A2, ...,A,) be a fixed n-tuple of right ideals of R and let TcD~r be an d-transversal. I f % , cq, ..., cq~M,(R) are arbitrary d-matrices then cq ~2 . . . ~t = q~(cq , %,

...,

c~,)

for a suitable T-polynomial ~ ( ~ ) in cq, % , ..., a t of degree at most t - 1 . Proof. We argue by induction on n, the case n = 1 being obvious with t(1)= 1. Now suppose that n > l and that the result holds for M , _ I ( R ). We view M , _ I(R) as being embedded in M , ( R ) as the matrices

~

, with ~' of size (n

-1) x(n-1), and we use L e m m a 1.1 and its notation. Let = (A1, A 2 .... , A,_ 1) and let T~ = {~1 de T} c M , _ 1(R). Then T1 is an ~r sal in M , _ 1(R).

.d 1

We isolate part of the proof of Proposition 1.2 in the following sublemma. Sublemma. Set u = t ( n - 1 ) and let %, ~2 .... , ~,+ I ~M,(R) be arbitrary ag-matrices. Then there exists an alt-matrix fi(cq, c~2, ..., %+ 1)eM,(R) of the form .

fl(0{1, 0(2, . . . , 0(uq_ 1) = 0~i ~2 9 .- 0{u+ 1 -}- X,

where X is a T-polynomial in ~ , ~2, ..., ~,+ ~ of degree u, such that fi(~l, ~2 . . . . , ~ , + 1 ) - [0].

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Proof Let cq, ~2, .--, % be d-matrices in M,(R). Then the matrices ~eM,_I(R ) are sgl-matrices. Thus, by assumption on M,_ 1(R), we know that ~1(~2 ...~u = ~[X((~i)

for a suitable Tl-polynomial 7"(~i) in ~1, ~2,..., ~, of degree at most u - 1 . We now apply Lemma 1.1 to each of the Ti-monomials occuring in 7'(~i), and to ~i ~2... ~,. If # = d i ~il d2 ~i2... ~irCTr+1 occurs in 7'(~i), then choose for each d~e T1, a corresponding d~eT. Then for any d - m a t r i x %+,eM,(R) we have #O~u+ i =-dl O:i dzO~iz...o~irdr + l O~u+l + ~o

where (p is a sum of matrices of the form #1 d, with de T and #1 = dl ~ d2 giz... ~ an initial piece of dlczid2Oqa...o~iJr+ 1. Of course, each such matrix is a Tmonomial in % , ~ , . . . , % of degree at most r
[0] =(~ ~2...a.--Z#) ~.+, = ~10~2 "'" ~u+ 1 "Jr-Z~

where Z = 0 - a%+ ~ - ~z is a T-polynomial in %, e2 .... , %+ 1 of degree u. Thus we may take/3(%, % , . . . , %+ 1) = el e2..- %+ 1 + X. Note that fi(~l, % , . - . , %+ ~) is an ~'-matrix since any T-polynomial in the e~'s is an sr This proves the sublemma. We now continue the proof of Proposition l.2. Set t = ( u + l ) ~ and let e~, % , . . . , e~ be arbitrary s~-matrices. Form

for i = 1, 2 .... , u + 1 as in the sublemma so that, in particular, each fl~ is an sOmatrix with fl~~-[0]. If ~a and T, are defined as in the first paragraph of the proof then we know, by assumption on M,_ I(R) again, that/?2fla.., ft,+ 1 can be written as a Ta-polynomial i n / ~ , fia, ..., fi,
~=~...~.+1-r

has the form

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M. Lorenz

_r~'/~'0,]

This follows easily from the fact that each/~i = [0], if we observe that for any ~, fl~m,(R) with c~,/~- [0] one has ~/~ = [ . -

. Because of the special form

of the/~'s, any T-polynomial in the/~'s is a T-polynomial in the ~'s, and the ~degree of such a polynomial is at most (u + 1) x/~-degree. Therefore, '~ : O~u+2~u+ 3 " " ~ t - - ~

for a suitable T-polynomial ~ in c~,+2, c~,+,, ..., cq of degree less than u(u + 1). Finally,/~ is an d-matrix, and hence we can find a suitable dEThaving the same (n,n)-entry as /?l. Since /~l=[0l, we see that / ~ l - d = ] :

~.

Thus we

obtain that [o] = (/~1 - d) 3'

=(~1~2-..~.+ ~+ z - d ) ( ~ . + 2 . . - " t - ~) = O~l O~2 . . . O~t "~ O ,

where O=Zo~u+Z...o:t-dO:u+Z...o~t-)~-k-d~ is a T-polynomial in el,~2, ...,~t of degree less than (u+ 1)2 =t. This proves the result for M,(R), with t ( n ) = ( t ( n - 1 ) + 1)2. We now apply Proposition 1.2 to prove the promised integrality result for finite normalizing extensions of rings. Let R ~ S =

~, R x i be given with xiR i=1

=Rx~, let A be a right ideal of R and let Sa,S2, ...,s,,~S. In analogy to the notion of T-monomials defined above, we define an A-monomial in sa, s2, ..., Sm to be a product in some order of the s~'s, each occuring at most finitely often, and of elements of A with at least one element of A occuring. Thus for example if al, a2EA , then a 1 and s~a 1 s 2 a 2 s a are A-monomials in s I and s 2 but s~ s 2 s 1 is not (unless A = R). By the degree of any such monomial we will mean the total degree in the si's. Any sum of A-monomials will be called an A-polynomial, and its degree will be the highest degree of the monomials occuring in the sum. The proof of the following theorem uses the example discussed at the beginning of this section. Theorem 1.3. Let R ~ S = ~ Rxi, x ~ R = R x i , be a finite normalizing extension of i=1

rings. Then there exists an integer t > 1, depending only upon n, such that for any right ideal A of R and any sl,s 2, ...,st~AS we have

sl s 2 . . . s , =

~ ( s l , s~ . . . . ,

s3,

where ~b(si) is an A-polynomial in sa, s2, ..., s t of degree less than t. Proof. Set F = R | 1 7 4 O R (n times) so that F is a free left R-module of rank n, and let ~:F-~S be the left R-module homomorphism given by (r D r 2, ...,r,) ~ = ~ rix v If K = F denotes the kernel of ~, then E = { c p ~ E n d R F [ K ~ = K } is a i=1

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455

subring of EndRF = M,(R). Furthermore, there is a natural ring h o m o m o r p h i s m

^:E-~EndRS. Indeed, if q)eE and s = ~, rixieS , then (~eEndRS is given by i=I S~ = ( r l ,

rn) ~~

r2,...,

Routine verifications show that gb is well-defined and that the m a p * thus obtained is a ring homomorphism. N o w let t=t(n) be the integer given by Proposition 1.2 and let

Sa,S2,...,steAS be given, with s L= ~ a}~

aj
j=l

able elements b}yeA i = {reRlr xlexiA}, and hence

x~s,= ~ xia}')xJ= ~ blOxixj= ~ riJa)xa j=l

j=l

j=l

for suitable r~j (~)eR which are, of course, in general not uniquely determined but can be chosen so that rlgeA ~for all j, l, since 0i;" (~)eA,. and A i is a right ideal of R. Let oh= [r}~)]eM,(R) for l = 1, 2, ..., t and set ~r A2, ..., An). Then each gl is an ~4-matrix. Moreover, for any f = ( r a , r 2 .... , r,)eF we have f ~l~ ~

Fir i=

i=1

----

ri r ij(l) j=l

ri -ri(o n

r r (1) . . . , -i-i2~

,

i=1 Xj =

ri

i

i=l

rij(t) x j j=l

r ixi sl =f=. s r

=

i=1

Hence we see that azeE and that the endomorphism a t of S is right multiplication by s z. Let TA denote the set of all diagonal matrices d=diag(rl, r2, ...,r~)eM,(R) such that r~xi=x~a for a suitable aeA. As we have seen earlier, TA is an sr transversal in M,(R). Moreover, as in the preceding paragraph, we see that if deT A then for a l l f e F we have fd~ = f ~ . a for a suitable aeA. Thus dee and ff is right multiplication by aeA. Since each a~ is an d - m a t r i x , we conclude from Proposition 1.2 that a l a 2 . . . a , can be expressed as a Ta-polynomial ~ ( a l , a2, ..., at) of degree less than t. Note that each factor in al~2...a,=q0(ai) belongs to E, since aidE and TACE. Therefore, we may apply the h o m o m o r p h i s m ^ to this expression, and letting the resulting h o m o m o r p h i s m of RS act on l e S we see that 1.~t~2...~,=sas2...s , can be expressed as a sum of monomials of the form l'dl~i~de...~.i

ffr+ l = a l

sila2...sl,

a r + l,

where O
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M. Lorenz

In particular, taking s 1 ~-s 2 . . . . . s e A S satisfies a "monic" equation

s t = s E A S in the above, we see that any

s ~= ~ (s)

over A. Part (i) of the following corollary is very useful in dealing with modules. Part (ii) generalizes the main content of the Lying Over theorem [9, Theorem 4].

Corollary 1.4. Let R ~ S = ~ R x i, x~R = R x i, be a finite normalizing extension of rings,

i= 1

i) I f A is a proper right ideal of R, then AS =~S. ii) Let A be an ideal of R. I f t denotes the integer given by Theorem 1.3, then (ASc~R) t ~ A c ASc~R. In particular, for semiprime A we have A = ASc~R. Proof (i) Assume that leAS. By our above remarks, we have 1 = 1t = ~(1), where 9 (1) is an A-polynomial in 1, and hence is a sum of monomials in suitable elements of A. Consequently, ~b(1)eA and leA. Part (i) follows. (ii) Note that ASc~R is clearly an ideal of R containing A. Thus the last assertion of (ii) will follow if we can show that (ASc~R)tcA. So let sl,Sz,...,st~ASc~R be given. Then, by Theorem 1.3, S1S2.,.St=~(Si) for a suitable A-polynomial ~b(sl) in sl,s2, ...,s t. Since at least one A-factor occurs in each monomial in ~b(s~) and since s ~ R for all i, we see that q~(si)eA. Thus s 1 s 2 ... steA , and part (ii) is proved. Note that part (i) above implies in particular that if R = S = ~ Rxi, x~R i=l

= R x i , and S is a division ring then R is a division ring. Another consequence of Corollary 1.4(i) is that any irreducible right R-module V can be embedded in a suitable irreducible right S-module W. Namely, just write V = R / A with A a maximal right ideal of R and choose a maximal right ideal B of S with A S c B . This is possible by the above corollary. Then W= S/B is irreducible and contains V, since Bc~R = A. This observation will be used in the proof of the following result. Here, J(.) denotes the Jacobson radical.

Theorem 1.5. Let R ~ S = i R x l , x i R = R x i , be a finite normalizing extension of rings. Then J(R) =J(S)c~R. Proof Let rEJ(S)c~R and let V be an irreducible right R-module. Then, as we have observed above, we can find an irreducible right S-module W with V ~ WR. Since Wr = 0, it follows that Vr = 0. Thus r annihilates each irreducible right Rmodule and hence belongs to J(R). Conversely, let r~J(R) and let W be an irreducible right S-module. Then, by a result of Formanek and Jategaonkar [5, Theorem 41, the restricted module WR is completely reducible. Since r annihilates the irreducible components of W~,we conclude that W r = 0. Thus rEJ(S), and the theorem follows.

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w2. The Cutting Down Theorem Let R c S = ~ Rxi, xi R =Rxz, be a finite normalizing extension of rings. In this i=l

very short section we study the behaviour of semiprime, prime, primitive and maximal ideals of S under intersection with R. We need one preliminary lemma whose content is almost entirely notational. Lemma 2.1. Let R c S = ~ R x i , x i R = R x i , be a finite normalizing extension of i=l

rings and let A be an ideal of R. For i=1,2, ...,n we set AX'= {rERlx i r~Axi}, AXCl={r~RlrxicxiA}. Furthermore, we let ~ = d R(xi) and ~ i = *R(x~) denote the left, resp. right, annihilators of x z in R. Then A x', Axc 1, ~Lfl and ~ i are ideals of R, and we have (AX')xcl = A + ~ i and (A ~c ~)~ = A + Ni. Moreover, R / A ~ ~- R/(A + di) , R/A ~ ~~- R/(A + ~ ) . Proof It is routine to check that S~, ~ , A x~ and A ~cl are ideals of R and that the latter two satisfy x~A~=Ax~ and A ' x~=xiA. Combining these equalities we X" obtain that A x l -_ x i A x l -_ ( A X i )X i - 1 x i and x i A = A X '~- 1 xi=xi(A X y~ 1 )'. We deduce that A + 5~ -=-(AX9xc~ + 5~ and A + N~ = (A ~c i)~ + ~ . But, surely, (A~) ~ct contains 5a~ and (A~ ,)x, contains Ni so that A + 5eg= (A~')~c' and A + Ni = (A~ ~)~'. Finally, let fi: R ~ R / A ~ denote the map given by f i ( r ) = r ' + A ~, where r'eR is any element satisfying x i r ' = r x ~. Since YI~cA ~', it is easily seen that f~ is a well-defined ring homomorphism which is in fact epi. Moreover, r~Kerfi says that rxi~x~A ~, that is r ~ ( A ~ ) x C l = A + ~ i. Thus K e r f / = A + 5 ~ , and the first isomorphism follows. The second isomorphism is proved analogously. The notation of Lemma 2.1 will be kept henceforth. For any right (or left) ideal Y of S we let J d ( Y ) denote the unique largest ideal of S contained in Y More explicitely, J d ( Y ) = { s e S l t s ~ Y for all t~S}. If Y has the form Y = A S for some ideal A of R then the condition t s e Y for all t~S reduces to xisEY

for i = 1,2,...,n.

For, in this case we certainly have

for arbitrary r ~ R . This observation will be used freely in the sequel. Part (i) of the following theorem is well-known and, in addition, is an immediate consequence of part (ii). Nevertheless, we offer a short independent argument which has been shown to us by D.S. Passman. Parts (i) and (ii) have been independently proved by J. Bit-David and J.C. Robson.

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M. Lorenz

Theorem 2.2 (Cutting Down). Let R c S =

~ Rxi, xiR=Rxi, be a finite nori=1 realizing extension of rings and let 1 be an ideal of S. i) l f I is semiprime then so is lc~R. ii) I f I is prime then R has only finitely many minimal primes over lc~R. In fact, if these are Q1,Qa, ...,Qt then t < n and R/Qi~-R/Ql for all i. iii) If I is primitive then the Qg's are primitive. iv) If I is maximal then the Qi's are maximal. Proof. Since R/(Ic~R)cS/1 is a finite normalizing extension of rings, we may assume that 1 = 0 throughout and we have to show that the zero ideal of R has the appropriate properties. (i) We assume that S is a semiprime ring and prove that R is likewise. Suppose, by way of contradiction, that there exists a nonzero ideal A of R with A2=0. For convenience, let Xo= 1 and let Ao=A. For all i=O, 1,2, ..., n we will construct nonzero ideals A i of R with A i c A and AixjA~=O for all j
and O:JSai_ 1 xiai_ 1 c A x i A i _ 1 =xiAXiAi_l c x i A i.

Thus A ~ O , and A~ has all the properties we need. Finally, A, is nonzero and satisfies A,x~A,=O for all i, and hence

a contradiction. Therefore, A does not exist and R is shown to be semiprime. (ii) Assume that S is a prime ring. We first show that there exists an ideal Q of R which is maximal with respect to the property that Jd(QS) = 0. Indeed, let {Q~}~A be a chain of ideals of R with Jd(Q~S)=O for all )~eA and let Q = U Q~. Then we have QS= (.9 Q~S and hence 2eA 2~A

~r

(~ {seSlx~seO_.S}

i=1

= U

{seSlx, s Q s}

,~eA i= 1

= U :d(Q.~S)=0. 2cA

Thus Zorn's lemma applies. We claim that Q is prime. Indeed, if B 1 and B a are ideals of R strictly containing Q then, by the choice of Q, we have Jd(B~S)+ 0 for j = 1, 2 and hence

Finite Normalizing Extensions of Rings

459

J d ( B 1 S ) . J d ( B 2 S ) @ O , since S is prime. Now, surely, J d ( B 1 S ) . J d ( B 2 S ) c B 1 S . J d ( B z S ) c B a B E S so that ~ r ). Thus Jd(BaBzS)@O and B~B 2 CQ. This shows that Q is prime. Now assume that for i = 1, 2, ..., rn we have 5P~c Q, whereas 5~ r Q for i > m, and set A = Q + (~ ~ . (As usual, if m = n then A = R.) Then A strictly contains i>m

Q, and hence we conclude that J d ( A S ) + O. Therefore, the right S-module

W=AS/QS is faithful. For, if T denotes the right annihilator of W in S then A S T c QS and hence J d (AS) T ~ J d (QS)= 0. Since J d (AS)tO and S is prime, we conclude that T=0. Observe that for i = 1 , 2 .... ,n, W~=(Ax~+QS)/QS is a right Rsubmodule of W, and W= ~ Wi. Moreover, for i>m we have W i ~ ( ~ i x i i=1

+ QS)/QS = 0 and thus

i=1

Since Wi QX, = (A x~Q~' + QS)/QS = (AQ x i + QS)/QS = 0, we obtain that m

0=T~R~ 0 Qx~. i=1

Finally, by Lemma 2.1, each QX~ with i<__m is prime, and R/QX~R/Q. Since these primes intersect in 0, we conclude that the minimal primes of R are precisely the minimal members of {QXl, Qx~, ..., QX,,}. In particular, the corresponding factor rings are all isomorphic to R/Q. Part (ii) is thus proved. (iii) Assume that S is primitive and let W be a faithful irreducible right Smodule. Then, by a result of Formanek and Jategaonkar [5, Theorem 4], the restricted module WR is completely reducible of finite length. 'Thus WR = W1 | Wz @... 9 Wt for suitable irreducible right R-modules ~ . If Pi= annR(Wi), then each Pi is a primitive ideal of R, and (~ Pi=annR(W)=0. Hence the i=1

minimal primes of R all belong to {P1, P2,-.., Pt} and thus are primitive. (iv) Suppose S is simple and let A be any proper ideal of R. Then Jd(AS) = 0, for otherwise J d ( A S ) = S and hence AS = S. Corollary 1.4(i) would yield A =R, a contradiction. This shows that the ideal Q constructed in the proof of part (ii) is maximal in the present situation. We deduce that the factor rings R/Q~ corresponding to the minimal primes Qi of R are simple, since R/Qi~-R/Q. This proves (iv), and the proof of the theorem is complete. It would be nice if the prime ideal Q of R constructed in the proof of part (ii) above would automatically be minimal or, in other words, if S is prime and A is an ideal of R with J d ( A S ) - - 0 then A has a nonzero annihilator in R. However, the following example that was pointed out to me by Allan Heinicke shows that this is false in general.

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M. Lorenz

Example 2.3. Let

T be a prime ring with a surjective endomorphism o- such that Ker a + 0 . For example, we may take T=K[tl,t2,...], the polynomial ring in countably many indeterminates t i over the field K, and define a to be the endomorphism of T given by t~ = 0, t~ = t i_ l(i > 1). Let S = M2(T), the ring of 2

:~][aET~. Since

x2-matrices over T, and let R = t [ :

a is epi, one easily

LL--

checks that the matrix units e~1,e12,e21,ez2 each normalize R and that S z = ~ Re~j. Hence S is a finite normalizing extension of R, and S is surely i,j=l

prime. Now let A =

{I: ~

}

laEKera Then A is a nonzero ideal of R, since 0 K e r a + 0 , and hence A has zero annihilator in R, since R is prime. Yet AS ={[:

bo]la,b~Kera} c~

n~ n~176 ideal ~ S.

However, if there exist automorphisms o-i, o-a, ... , a,, of R such that xi r ~ r xi holds for all r~R then for any ideal A of R with zero annihilator in R we have Jd(AS)+0.

For, D = (~ A~ is nonzero and is contained in

Jd(AS),

since

i=1

xiDcxiA~=AxicAS

for all i. Similarly, Jd(SA) is nonzero. Assume further that S is prime and let Q denote the prime ideal of R constructed in the proof of Theorem 2.2(ii). Then J d ( Q S ) - - 0 and so we conclude that Q has a nonzero annihilator in R and thus is minimal. We also know that the minimal primes of R form a subset of {Q% Qx~, ..., Q~,}. Since, clearly, ( 2 ~ QX~for all i, we deduce that the minimal primes of R are precisely the conjugates of Q under G =
w3. The Martindale Ring of Quotients This section essentially splits into two parts. The first part contains the definition and some basic facts concerning the Martindale ring of quotients Qo(R) of a semiprime ring R. We also introduce the notion of G-closure of R, where G is a group acting on R. The ring Qo(R) was defined for prime rings by Martindale ( [ l l l ) and extended to semiprime rings by Amitsur ([-1]). Many of the preliminary results on Qo(R) are already in the literature, and we have included them here for the sake of having a convenient reference. In the second half of this section we will use Qo(R) to study finite normalizing extensions of the form R c S --- S

Rxi with xlr~ =rxi(r~R ) for suitable

automorphisms a i of R

i=1

and with S a prime ring. Let R be a semiprime ring. Then for any ideal A of R we have ~R(A)-~r Let ~ - - ~ ( R ) denote the filter of all ideals of R with zero annihilator in R. Then

Finite Normalizing Extensions of Rings

461

Qo(R) is defined to be the ring of (left) quotients with respect to ~ , that is Qo(e) = limmHom(RA, RR), where Hom(RA , RR) denotes the set of left R-module homomorphisms from A to R. A more explicit description of Qo(R) is as follows. The elements of Qo(R) are equivalence classes in 0 H~ RR) under the following equivalence reA~j~

lation: Let c~:RA--*RR and fi:RB~RR be left R-module homomorphisms, with A,B~@; then c~ and fi are equivalent if and only if there exists DcAc~B, De@, such that dc~=dfi holds for all d~D. It is easily seen that in this case e and fl do in fact agree on Ac~B. The equivalence class of e under this relation will be denoted by ~. The arithmetic in Qo(R) is defined in the following manner. Suppose ~: RA-~RR and fi: RB~RR, with A, B e ~ . Then c~+fi is the equivalence class of a + f i : A c ~ B ~ R , and ~fl is the class of the composite function aft: BA ~ R . It is easily checked that, with these definitions, Qo(R) becomes an associative ring with 1. Finally, if Pr denotes right multiplication by r6R, then the map r - * r r is an injective ring homomorphism from R to Q0(R). Hence we can view R as a subring of Qo(R) with the same 1. Under this identification, the m a p a : RA~RR translates in Qo(R) to right multiplication by ~. More explicitely, we have a c~=~3,~=~3, ~ = a ~ for all a~A. The center C = C(R) of Qo(R) is usually called the extended centroid of R. The following lemma contains some basic properties of 00(R) and C(R). Lemma 3.1. Let R be a semiprime ring, let Qo(R) be its Martindale ring of

quotients and let C(R) be the extended centroid of R. i) For every q~Qo(R) there exists an ideal A E ~ with A q c R. ii) If qeQo(R ) satisfies Aq=O for some A E ~ then q = 0 . iii) Each automorphism of R has a unique extension to Qo(R). iv) The extended centroid C(R) is the centralizer of R in Qo(R). Moreover, R is prime if and only if C(R) is a field. Proof Assertions (i) and (ii) follow directly from the definition of Qo(R). For details see [1]. For (iii), let a be an automorphism of R and let q = ~ Q o ( R ) with c~: RA--~RR, A E Y . Then one defines a map on A ~ f f by a~c~r r for aEA. Clearly, e~ is left R-linear and hence defines an element q'=K~Qo(R). The map q~q' is the desired extension of a to Qo(R), and it is easily seen to be unique. Finally, the first assertion in (iv) is an easy consequence of parts (i) and (ii) and the second assertion is [1, Theorem 5]. Let a be an automorphism of R, and hence of Qo(R). Following Kharchenko ([6~) we set

q5 = {qeQo(R)lq r r = rq for all reQo(R)}. As a consequence of Lemma 3.1(i), (ii), we see that q~qSo if and only if qr~ holds for all r~R. Note that each 4~o is a module over C = C(R). The automorphism r is said to be X-inner if ~o4:0. Part (i) of the following lemma, due to

462

M. Lorenz

Kharchenko [6], implies in particular that, in case R is a prime ring, each ~bo is at most one-dimensional over C. For a quick proof of part (ii) see [12, Lemma 2.2]. Lemma 3.2. Let R be a prime ring. i) l f q is a nonzero element of Qo(R) such that R q = q R then q is invertible in

Qo(R). ii) Let a be an automorphism of R and let a,b~R be elements with a4=O and arb=br~ a~ for all r~R. Then there exists n ~

with b=an.

Now let G be a group of automorphisms of R, and hence of Qo(R). Then for any two a, ~ G we have ~ q ~ , . Thus B(G)= y' 4~ is a C-subalgebra of aeG

Qo(R), the so-called algebra of the group G. Since Rq~o=q~R for all craG, we see that R(G)= ~ Rq~=RB(G) a~G

also is a C-subalgebra of Qo(R) which we call the G-closure of R. The terminology is justified by part (i) of the following lemma. Lemma 3.3. Let R be a semiprime ring and let G be a group acting on R. i) The G-closure R(G) of R is a semiprime ring and G acts on R(G) (via its

action on Qo(R)). Moreover, the G-closure of R(G) equals R(G). ii) Let 0 q=e = e 2 be a primitive idempotent of Z(R), the center of R, and let H denote the stabilizer subgroup of e in G. Then H acts on the semiprime ring e R, and there is a canonical isomorphism eR(G)~-(eR)(H). Proof. (i) The semiprimeness of R(G) follows from the fact that any nonzero ideal of R(G) meets R nontrivially, by Lemma 3.1(i), (ii). Furthermore, if a, ~ G then

Therefore, G acts on R(G). Now let neQo(R(G)) be given with n r ~ = r n for all reR(G), where craG is fixed. Let n be represented by V:R(G)I-+R(G)R(G), with Ie~(R(G)). The equation n r~= r n translates to

qrv=qvr ~ for all q~l and r~R(G). Set I~ = {r~lc~Rlr v~R}. Then, using the above identity, we see that 11 is an ideal of R. We claim that I I ~ ( R ). Indeed, suppose that Ilr = 0 for some r~R and let q~I be arbitrary. By Lemma 3.1(i), there exist suitable ideals A,B~@(R) with A q c R and B q v c R . Thus ( A B q ) v = A B ( q v ) c R and A B q c l n R so that A B q c I ~ . Therefore, ABqr=O, and since A B l Y ( R ) we conclude from Lemma 3.1(ii) that qr=O. Thus lr=O, and hence r = 0 . This proves that Ii~o~(R ). Consequently, if v~ denotes the restriction of v to I~, then v~ defines an element n~ of Qo(R) which clearly belongs to ~ , since

irvl =ivl r~

Finite Normalizing Extensions of Rings

463

holds for all ieI a and r~R, that is rn~=n~r ~. As we have r e m a r k e d above, the latter equality does even hold for all reQo(R ). N o w let I z = R ( G ) I ~ R ( G ) so that 12 is an ideal of R(G) with I1 ~ I 2 ~ I . We claim that I2eo~(R(G)). For, if q / 2 = 0 for some q~R(G) then qI, = 0 . Choosing A e ~ ( R ) with A q c R , by L e r n m a 3.1(i), we conclude that AqI, = 0 and hence Aq=O, since I~e~(R). L e m m a 3.1(ii) n o w implies that q = 0 . Thus we have in fact I2e~(R(G)). To finish the p r o o f of (i), it suffices to show that v acts on 12 as right multiplication by nleR(G). So let n~i~m~eI 2 be given, with n~,m~eR(G) and i~eI~. Then using the equations i~m~ v I

= i~ v m~ and m~ n~ = n 1 m~ we obtain

1

l

1

= 2 n~(i~n~)rng = ( 2 n~ i~ra~)n~. l

l

Thus we see that v=p,, on I2, and hence heR(G). This completes the p r o o f of part (i). (ii) We first establish an i s o m o r p h i s m ~o:eQo(R)--+Qo(eR). So let q = ~eQo(R ) be given with a: RA--+RR, A ~ ~(R). Then e A e ~(eR) and the restriction of a to eA defines an eR-linear m a p a 1 : eA--+eR. Let 0(q) denote the class of a 1 in Qo(eR). Then 0 is easily seen to be a well-defined ring h o m o m o r p h i s m from Qo(R) to Qo(eR) with K e r O = ( 1 - e ) Q o ( R ). To see the latter, just note that O ( q ) = 0 for q=~eQo(R ) says that ( e A ) a = 0 or, equivalently, ( a ( 1 - e ) ) a = a a for all aeA, that is (1-e)q=q. Thus the restriction of ~ to eQo(R ) yields a m o n o m o r p h i s m q0: eQo(R)~Qo(eR ). Conversely, let B e ~ ( e R ) and let a 1 " B--+eR be left eR-linear. Then A = B | and the m a p a: A--+R defined by ( b + ( 1 - e ) r ) c ~ = b a 1 for beB, rER is left R-linear and satisfies a=pea. It is readily verified that the m a p sending the class of a t in Qo(eR) to ~ e Q o ( R ) is the desired inverse for ~0. Thus ~0 is in fact an isomorphism. N o t e that for reeR we have ~o(r) = reQo(eR ). N o w let aeH. Then a acts on eR and hence on Qo(eR). We claim that for all qeeQo(R ) we have (p(q~) = q)(q)~. Indeed, ifq =02 with a: RA~RR, Ae~(R), then q~ = a ~, where a~:A~--+R is defined by a ~ a ~ = ( a a ) ~ for all a~A. If a~eeA ~ then a ~ = e a ~ = (e a) ~ and a ~ a ~ = (e a a) ~. Thus we see that a~[ eA~ = (a[ eA)~ Taking equivalence classes in Qo(eR) we obtain that ~o(q~)= q)(q)~, as claimed. It follows that

(p(edb~)={ql~Qo(eR)Iq lr~=rlq ~ for all

r~eQo(eR)}

=r Finally, i f a e G \ H then e ~ e = 0 , since e is primitive in Z(R). Since e is central in Qo(R), we have e ~ =(b e and, on the other hand, e ~ = ~ e ~. Thus e ( b = 0 for a e G \ H , and we obtain

cp(eR(G))=cp(eR. ~, ecb~)=cp(eR). Z qo(e~b~) er~G

aEH

=eR. ~ q~(eR)=(eR)(H). a~H

This finishes the p r o o f of the lemma.

464

M. L o r e n z

Note that part (i) above implies in particular that the extended centroid of R(G) is contained in R(G) and hence is equal to the extended centroid of R. Furthermore, taking o-= 1 in the proof of part (ii), we obtain a canonical isomorphism

e C (R)=eq~ I ~ i ( e R ) =

C(e R).

This isomorphism will be used in the proof of the following lemma which is just a combination of several results in [1]. It refers to the situation R c S = ~ Rx~, i=l

xiR =Rx~, with S a prime ring. By Theorem 2.2 (ii), we know that in this case R is semiprime with finitely many minimal primes. Lemma 3.4. Let R be a semiprime ring having only finitely many minimal primes Q1, Qa, ..., Q,. Then there exist orthogonal primitive idempotents el, e2, ..., e,~ C = C(R) such that i) e l + e 2 + . . . + e t = l . ii) Q~=(1-e)Qo(R)cvR. iii) If rz is an automorphism of R and Q~ = Qj, then e~ = ej. iv) If G is a group acting on R, then each e~R(G) is a prime ring. Proof. The existence of primitive orthogonal idempotents e ~ C with (i) and (ii) is part of [1, Theorem 11]. (iii) Using (ii), we can write our assumption Q~=Q; as (1-e[)Qo(R)c~R = (1 - ei) Qo(R) c~R. Since both 1 - e7 and 1 - ej are idempotents in C, we can apply [1, Corollary 9] to conclude that e~=ej. (iv) By [1, Theorem 8 (iii)], we have C(R/Q)~-eiC. Thus eiC is a field, by Lemma 3.1 (iv). Moreover, as we have remarked above, Lemma 3.3 (i) implies that C(R (G)) = C, and this yields e~C ~ C (eiR (G)). Thus, by Lemma 3.1 (iv) again, eiR(G)) is prime. This finishes the proof of the lemma. We now return to finite normalizing extensions of rings. Let R c S = ~, Rx~ i=1

be given with x~rr = r x~(reR) for suitable automorphisms r;~ of R and suppose that S is prime. Let G denote the subgroup of Aut(R) generated by rrl, or2,..., r~. Our next goal is to embed R c S in a suitable finite normalizing extension R(G)cS(G), where R(G) is the G-closure of the semiprime ring R. This will be done in Lemma 3.6. The following observation will be needed in the proof. Lemma 3.5. Let R ~ S = ~ Rxi, xi:~=r xi(r~R), be given with S a prime ring. If A e o ~ = ~ ( R ) then i=1 i) Jcd(AS)~Rao~ and JM(SA)c~Ra~ ii) doo(s)(A) = ~eots)(A) = O.

Proof. (i) As we have observed in Sect. 2, the ideal D = (~ A ~i is contained in i=l

Jrd(AS) and, Jd(SA)c~RE~.

clearly,

D~

Hence,

Jd(AS) c~Re~

and,

similarly,

Finite Normalizing Extensions of Rings

465

(ii) Suppose that sA=O for some seQo(S ). By Lemma 3.1 (i), there exists a nonzero ideal I of S with I s c S . Since IsA=O, we have Is.Jd(AS)=O. But J d ( A S ) is nonzero, by part (i), and S is prime so we deduce that Is=O. Lemma 3.1 (ii) yields s = 0 , and A has zero left annihilator in Qo(S). If As=O for some s~Qo(S ) then Jd(SA)s=O. Since Jd(SA) is nonzero, by part (i), Lemma 3.1 (ii) yields s = 0 . Thus A has zero right annihilator in Qo(S), and the lemma is proved. Lemma 3.6. Let R ~ S = ~ Rxi, x ir ~ =r xi(reR), be given and assume that S is i=1

prime. Let G = ( G 1 , aa, ..., a , ) c A u t ( R ) and let R(G) denote the G-closure of R. Then there exists an embedding # of R(G) into Qo(S) such that # is the identity on R and xig(q ~') = #(q) xi holds for all qER(G) and i = 1 , 2 , . . . , n . (Note that R(G)~=R(G) for all i, by Lemma 3.3 (i).) Proof. Let q = ~ ron~R(G), a finite sum with r ~ R , n~bo, and write n ~ = ~ , aEG

where c~: RA~--*RR, A ~ e ~ Restricting to A = (~ A~e~- if necessary we may ff

assume that A~=A for all a. Then q=c2 where ~=~proc% is defined on A. We ff

extend a from A to a map a' defined on the left ideal SA of S by setting

(2 sj a~) c(= 2 sj(a~c~)= Z sj(aj r~c~) j

j

j,,r

for sj~S, ayA. To prove that ~' is well-defined we show that ~ s j a j = O implies J

~,sj(ajr~)=O for all ~. Indeed, if ~ sjaj=O then for all b~A we have J

J

0 = ( ~ sj aj) (r, b ~,) = ~ sj(aj r, b ~,) J

J

= F~ sj(aj ro ~o) b ~ J

since ~ , = n , ~ # ,

implies (b~ b ) a , = ( b 1%)b * for all b, bteA. Hence we see that and hence Lemma3.5(ii) implies that

(~sj(ajr, c~o))A*=O. But A ~ , J

~sj(ajr, a,)=O for all o-. This proves that c( is well-defined. Clearly, ct' is left SJ

linear, and if q=r~R then c~' is just right multiplication by r. Since Jd(SA) is nonzero, by Lemma 3.5 (i), the restriction of e' to Jd(SA) defines an element q' of Qo(S). It is easily checked that the map # sending q to q' is a well-defined ring homorphism of R(G) into Qo(S) which is the identity on R. Since any nonzero ideal of R(G) intersects R nontrivially, we see that Ker # = 0 so that # is injective. Finally, note that if q = ~ r ~ n ~ R ( G ) is as above then q~ is the t7

equivalence class of the map c~~

defined by a~'cd~=(acQ ~ for a~A.

466

M. Lorenz

Thus p~, .(e~)' is defined on SA, and for ~ s j a j ~ S A we have J (2 Sj

J

a j) Pxi " (c~ai)' = ( 2 S j X i a~') (~~ = ~ S j X i (a~~c~~

J

J = 2 sj x i(aj ~)~' = ~, sj (aj ~) x i J

J

= ( ~ sjaj) c( p~ . J Thus p~.(~~ and taking equivalence classes we obtain that x~#(q ~) =#(q) x i. This proves the lemma. Identifying R(G) with #(R(G)) in the above situation, we may form S(G) = { ~ qlsllqleR(G), sl~S } = ~, R(G)x i inside Qo(S), and this is a subring of Qo(S) l

i=1

containing R(G). More precisely, S(G) is a finite normalizing extension of R(G) with

x i q~ = q xi for all i and all qsR(G). Moreover, S(G) is prime, for it follows from Lemma 3.1 (i), (ii) that any nonzero ideal of S(G) intersects S nontrivially. The following proposition further describes the structure of the extension R(G) c S(G). It implies in particular that if R is also assumed to be prime, then S(G) is free as a left and right R(G)-module.

Proposition 3.7. Let R c S = ~

R x i , xlr~ = r x i (r~R), be given with S prime and

i=1

let R ( G ) c S ( G ) be as above Furthermore, let el, e2, . . . , e r i C = C(R) denote the idempotents corresponding to the minimal prime ideals of R (cf. Lemma 3.4). Then i)

~ ejR(G), a finite direct sum of prime rings ejR(G), with j=t

R(G)=|

e jR(G) ~- e 1R(G). ii) S(G)=@ ~

ejS(G)el, and each ejS(G)e l is a free left module over the

j,l= 1

prime ring ej R(G). iii) In particular, for each j, e j R ( G ) c e j S ( G ) e j is a free normalizing extension of rings. More precisely, there exists a subset Tjc {ile~ ~= e j} such that ejS(G) ej= | ~ ejR(G)yi, i~ T j

where the elements Yi = e j x i ejeejS(G)e~ are left and right ejR(G)-independent and satisfy r Yi = Yl r~' for all r eejR (G). Proof Clearly, R ( G ) = |

ejR(G) and S(G)=@ ~ j=l

ejS(G)el, since the ej's are

j,l=l

pairwise orthogonal and satisfy e l + e 2 + . . . + e ~ = l . By Lemma 3.4(iv), each ejR(G) is a prime ring. Moreover, as we have seen in Sect. 2, the group G = @1, ~2 ..... a , ) permutes the minimal primes of R transitively. Lemma 3.4 (iii)

Finite Normalizing Extensions of Rings

467

implies that G acts transitively on {e~,ez, ...,e,}. Thus for each j there exists a e G with e j - e t ,~- and hence we have ejR(G)=e~R(G)=(e~R(G))L Therefore, the rings e jR(G) are pairwise isomorphic, a n d (i) follows. Consider Sjz = ejS(G)e l. W e set Mjt = {ile~ = e~}. If i~Mj~ then e~ e~~c~-- 0 and hence e jR ( G) x i e I = e jR ( G) et~"~ *x~=0. O n the other hand, if ieM3~ then ej e~~ --- ej and we obtain eaR(G ) x~ e~ = e j R ( G ) x i. Thus we see that

Sa,= ~ eaR(G)x~e,= Z i= 1

e~R(G)xi

ieMjl

= ~. ejR(G)Yi, i ~ M jl

we have set y i = e j x i = x i e t = e a x i e t E S j l . In particular, Sja=eaS(G)Q = ~. QR(G)y~, and this is a ring containing ejR(G) as a subring. Moreover, for

where

i~Mjj

each ieMja , a i induces an a u t o m o r p h i s m of ejR(G) and if r e e j R ( G ) then ry~ = r ej x i = e jr x i = e a x i r '*~= y~ r% Thus ejR (G) c Sjj is a finite normalizing extension of rings, with Sjj and e~R(G) prime. N o w choose M ~ Mj~ m a x i m a l such that F = ~ e jR (G)yi is a free left Q R (G)ieM

m o d u l e with basis {Y,IieM}. We show that F = Sit. Otherwise there exists rneMi, with y,,,~E The maximality of M implies that there is a relation

s = r m y m + ~, riyi=O i~M

with rm, r~eejR(G), rm+O. N o w for any reejR(G) we have

O=rmrs-sr

Gm

r,.~ m = ~ ( r , . r r l - r i f f

-~'-* r,,~-~c* )Yl,

iEM

since the y,,-coefficients cancel. Since ej = e~' = e j, by definition of Mjz, we see that a,,a7 1 is an a u t o m o r p h i s m of eaR(G ) for each iEM. Therefore, the elements r , ~ r r i - r l r ~ c ' r m~'~c~ belong to ejR(G), and the freeness of {yilieM} over e jR(G) implies that rrn r r i = r i r

-i cr~,(Yi

ffmffr - i

rm

for all t e e jR(G). Since e jR(G) is a prime ring, L e m m a 3.2 (ii) yields elements ni~Qo(QR(G)) satisfying nir~'~cl= r n i for all r~eiR(G), and r i = r,, n i. Thus, if tlj denotes the stabilizer of ej in G, then nj~(ejR(G))(Hj), since a m a i - ' e H j. L e m m a 3.3 implies th at (ei R (G)) (Hj) = ej R (G). Thus n i ~ ei R (G) and we ,obtain that 0 = s = r , , t with t = y ~ + ~ niyieejS(G ). Moreover, for any rEejR(G) we have ieM

r t = r ( Y m + ~ nlYi)=Ymff'~+ ~, niy~r ~ ' ~ c ~ i~M

ieM

=(Y,.+ ~, n l y l ) r ~ = t r ~'~. ieM

It follows that 0 = A a t, where A a = e jR(G)r m e jR(G) is a nonzero ideal of e jR(G). Since ejR(G) is prime, we see that A = T A j @ R ( G ) ( 1 - Q ) is an ideal of R(G) with

468

M. Lorenz

zero annihilator in R(G) which satisfies A t = R(G) (1 - e j) t = R(G) (1 - Q) ej t = 0, since t = ej t. Since S(G) is prime, we conclude from L e m m a 3.5 (ii) that t = 0, that is ym= ~ (-nl)yi~ E a contradiction. Thus we must have Sit=F, and it follows iaM

that Sj~ is free as a left module over ejR(G). This completes the proof of the proposition. We remark that if the x:s are units in S(G) or, more generally, if there exist units titS(G ) with t7 1 r t i = : ~ for all raR(G), then we have

S(G) ~- M,(el S(G) e~), a t x t - m a t r i x ring over elS(G)el=@ ~ elR(G)y i. This follows from [14, iaT1

L e m m a 6.1.6], since G = @1, 0"2,..., an) permutes {el, % , . . . , 6} transitively.

w

Extensions with a Finite Normalizing Basis

The goal of this section is to study the situation described in Proposition 3.7 (iii). More precisely, we will consider extensions R c S

such that S = ~ Rxi for i=1

suitable elements xg~S which satisfy the following conditions: (1) The set {xl, x2, ..., xn} is left R-independent. (2) There exist automorphisms o-i of R such that x i : ' = r x i holds for all rsR and i = 1,2, ..., n. In this case, the elements x~ are said to form a normalizing basis for the extension R c S . In other words, {x~, x 2 .... , xn} is a normalizing basis for R c S if and only if S = ~, Rxi, x i R = R x ~ and the x~'s are both left and right Ri=1

independent. Throughout this section we will further assume that (3) R is a prime ring. Our goal is to describe the prime ideals of S that intersect R in 0, by extending the methods of [7, w to the present situation. Let Ro=Qo(R ) denote the Martindale ring of quotients of R and let S O = R o @ S . Then S O contains S via s--*l| and, with this identification, S o is a R

free left Ro-module with basis {xl, x2, ... , xn}. The multiplication of S extends to a multiplication of S o as follows. By L e m m a 3.1 (iii), there exist unique extensions of the automorphisms ai from R to R o which will also be denoted by ag. Let rijt~R be the structure constants of S, defined by X i Xj = ~ r itf x l. l=l

Then if qi xi and qj x: are elements of S o, with qi, qj~Ro, we let

(qi xi) (qj x j) = ~ qi q~:~ rift Xt" l=l

Finite Normalizing Extensions of Rings

469

L e m m a 4.1. The above multiplication, extended distributively, makes S o an asso-

ciative ring. Proof For any raR we have ~ rrljzx,=r xixa= xixjr~'a'= 2 rljzx, r~'a'= Z rij, ra'a'~ 1

l

l

and hence by (1),

r rij~_- r~j~r aiajal-

1

As we have noted in w this identity extends to all reR o. N o w let q~, qj and qkCRo be given. Then, in So, we have

[(ql xl) (qj x j)] (% Xk) = ~(~, ql q j~ m

r u t qko,-1 rlkm) Xm

l

and a-1

(qixl)[(qjxj)(qkXk)]=2(2qiqj

a f lo/- I a i - t

'

m

qk

rjkt

ri,m)x~.

l

Thus, in order to prove associativity we have to show that for m = 1, 2, ..., n and all qeR o we have 2 rljl l

tTi-1

q

rlkm =

=

~1

~ qa~ ~,

a~l rj~z rilm"

l

Since S is associative, this relation surely holds for q = 1 so that the right hand side equals ~ qa; laClr~jz r~km" Thus it suffices to show that l =la~l

qaj

if-1

~ rift = rift q

holds for all i,j, l and all qaR o. But, after replacing q by q~J, the above identity becomes q rii1=rij lq~'~jac' which has been established in the first paragraph of the proof. Thus the lemma is proved. It is clear that the above multiplication extends the multiplication of S and is in fact the unique such extension. We also remark that Lemma4.1, with the same proof, works for semiprime R. Note that, as an immediate consequence of L e m m a 3.1 (i), (ii), there exists to any s~S o a nonzero ideal A of R with A s c S , and A s = O implies s = 0 . In the following, we identify R o with its image R o 91 = R o | 1 inside S o. Thus R 0 c S o is a finite normalizing extension with normalizing basis {xl,x2, ...,xn}. Let s = ~ rlxieSo be given, with rieR o. Then the r~'s i=l

are uniquely determined and the set Supp(s) = {ilrl 4=0} will be called the support of N o w let

s.

T=Nso(R) = {saS o fRs = s R }

470

M. Lorenz

and for any automorphism a of R set

T={seSolsr~

for all

reR}.

In particular, T~ is the centralizer of R in S O which will be denoted by E in the sequel. E is a subring of S o which contains the extended centroid C = C(R) in its center. Furthermore, each Tr is a left and right E-submodule of S o. More generally, for any two o, zeAut(R) we have T~- T c T~. Tis at least closed under multiplication but it is not necessarily true that t~ + t 2 ~ T i f tl, t2~T. Recall that T j ~ R o has been denoted by ~ in w and that o- is said to be X-inner if and only if ~ is nonzero. Lemma 4.2. With the above notations we have

i) T=

U r~. ~eAut(R)

ii) T~= ~ cb~clx ~. In particular, T~ is at most n-dimensional over C= C(R). i--1

Moreover, if t~To then we have q t = t q ~ for all q~R o. iii) For each o-tAut(R) we have RoT~=T~R o. If o-,z~Aut(R) are such that T~4=O, T~4=O then RoT~=RoT~ if and only if za -1 is X-inner. iv) As left Ro-modules we have Ro T~- Ro@ T~. C

Proof. (i) Let 0 ~ t ~ T a n d assume that t r = 0 for some r~R. Then it follows that 0 = R t r = t R r . Write t in the form t =

r~x~= ~ x~r~~ w i t h r i ~ R o . Sincethexi's i=1

i=1

are right R0-independent it follows that rT'Rr=O. Now t is nonzero, and hence at least one r~ is nonzero. Thus, by Lemma 3.1 (i), (ii), we have 04=ART' c R for some ideal A of R. Hence (ArT')Rr=0, and the primeness of R implies that r = 0 . Similarly, r t = 0 implies r = 0 . Thus to any r e R there exists a unique r~ER with r t = t r ~ and the map r-~r ~ is easily seen to be an automorphism of R. This proves that te T~. (ii) Let t = ~,

rixiETr

with riER o. Then for all reR we have

i=1

rrlxi=rt i=1

= t r Y = ~, rS~C~x~. Since the x~'s are left R0-independent, we conclude that ~=~ r r i = r S ~ for all reR, that is r i ~ - ~ . Conversely, each ~ o _ l x i is clearly contained in T~, and hence we have shown that To= ~ ~o~, ~x~. Since each i=1

qs~c ~ is at most one-dimensional over C, by Lemma 3.2, we conclude that T~ is at most n-dimensional over C. Now let teT~ and qeR o. Then, by Lemma 3.1, there exists a nonzero ideal A of R with A q c R . Thus for all aEA we have

(a q) t = t(a q)r = t a~q~ =a t q~. It follows that A ( q t - t q ~ ) = O and hence q t = t q ~ by our above remark. This proves (ii).

Finite Normalizing Extensions of Rings

471

(iii) The equality RoT~= ToR o follows immediately from the fact that any element of T~ normalizes R0, by part (ii). Now let a, TeAut(R) be given such that zo--1 is X-inner. Then there exists a unit u ~ _ ~ , and since uTo~T~_~T~cT~ we obtain

RoTr162

~.

Similarly, using the fact that az -1 is also X-inner, one shows that R o T e c R o T ~ . Thus equality holds. Conversely, assume that R o T~=R o T~. By part (ii), we have Supp(R o T~) = U Supp(s) = {ilo-a/- 1 is X-inner}, ~RoT~

and similarly for R 0 T~. Thus it follows that {i] ao-F 1 is X-inner} = {i]za~-1 is Xinner}, and if these sets are nonempty then o-a~-~ and za~- ~ are X-inner for some i. It follows that (z a/-1) (o-a/-1) -1 = z a -1 is X-inner, and part (iii) is proved. (iv) By part (ii), we know that T ~ = ~ ~ c ~xi, where the sum runs over those i i

with ~o~c~ :F0. For each such i choose O ~ u ~ c ~ and set 2~=u~x~T~. Then it follows from Lemma 3.2 that u~ is a unit in Tc~Ro and that ~ = Cuv Hence T~= ~, C2~, and hence the 2i's form a C-basis for To which is simultaneously R oi

independent since the x~'s are. Thus the 2~'s form an R0-basis for R 0 T~, and the assertion follows. An immediate consequence of Lemma4.2 is as follows. For any two i,j~{1, 2, ..., n} write i ~ j if and only if aiaf 1 is X-inner. Then part (iii) above shows that R o To, = R o T~j if and only if i,-@ Let M denote a complete system of representatives for {1, 2, ..., n}/~. Then since xi~T~ we have

i

i

i-- 1

i= 1

z

maM

Moreover, the elements of R o T,, have their support in [i], the equivalence class o f i under ~ . Since [i]c~[j] = Z for i,j~M, i@j, we see that the sum ~ R o T ~ is m~M

in fact direct. Thus

So = | Z R0 tom = | mEM

Y, Yoao. mEM

Although Tis not a subring of S o in general, we define a left ideal of T to be a subset L of T w i t h 0~L such that for all leL and t E T w e have tl~L and, k

k

moreover, if ll,12,...,lkeL then ~ lFL, provided ~ l~eT. Right ideals and i=l

j=l

ideals of T are defined analogously. We give a simple example to illustrate these notions.

Example 4.3. Let R be a commutative domain with a nontrivial automorphism oand let U = R , [ y ] be the skew polynomial ring, with r y = y r ~ for all rsR. Then I =y2U is an ideal of U and S = U/I can be written as S = R O R x , with x = y + I . Since r x = x r ~ for all reR, we may take x ~ = l and x 2 = x in the above and see that the extension R c S satisfies (1), (2) and (3). Moreover, R o is the field of

472

M. L o r e n z

fractions of R and S o = R o | , with x 2 = 0 and r x = x r ~ for all r e R o. It is easily checked that T = N s o ( R o ) = R o w R o x = E w T ~. In this situation, the left ideals of T are 0, Rox and T They are all two-sided and are identical with the right ideals of T. Part (i) of the following technical lemma shows that Tis "left Artinian". The property described in the last assertion of part (ii) was called "R-cancelability" in [7, w Part (ii) further shows that the left ideals of Tare precisely the subsets of T o f the form L=Ic~T, where I is a suitable left ideal of S 0. The analogous assertions for right ideals are of course also valid. Lemma 4.4. i) Any chain of left ideals in T is finite and, in fact, has length at most n.

ii) Let L be a left ideal of T. Then S o L = { ~ s~Ij]sj~So, l;eL} has the form J

SoL= | Y eo(rnL~), rnaM

and this is an Ro-direct summand of S o. Moreover, S o L c ~ T = L and if seS o satisfies A s c S o L for some nonzero ideal A of R then saSoL. Proof. We begin with a preliminary remark. Let ~ denote the equivalence relation on {1,2, ...,n} defined above and let M be a complete set of representatives for {1, 2, ..., n } / ~ . We claim that for any left ideal L of T and any element l~L there exists a suitable m ~ M and a unit uaTc~R o with u l e L m =LcaTr

This if of course obvious if l = 0 so let I4=0 and write I= ~ rlxl, with i=l

ri~R o. If leT,, say, then r i a q ) ~ l , by Lemma4.2(ii). Since 14:0, some r i is nonzero and hence is a unit in TC~Ro, by Lemma 3.2. Choose m e M such that i ~ m . Then amO-~-1 is X-inner and there exists a unit ve~om~cl. Now let u = v r i -1. Then u is a unit in T c~Ro, and u 1= v r ~ 1 It ~ ~ c l ~,~- ~T~caL ~ T~ c~L = L m. This proves our claim. (i) Note that for any left ideal L of T, each L ~ = L c ~ T ~ is a left C-subspace of T r We define q)(L) to be the nonnegative integer (p(L)= ~ dimcL m. m~M

Since Tern= ~ ~r

x;, by Lemma 4.2 (ii), we see that dim c T ~ = ]l-re]I, the size

j~m

of the equivalence class of re. It follows that ~, d i m c T o = n , and hence co(L)<__n. mffM

Thus, in order to prove part (i), it suffices to show that the map (p is strictly increasing. Indeed, if L and L' are left ideals of T with L ~ E then choose l~L, 14E. By our preliminary remark, there exists a unit ueTc~R o such that u l e L m for some m and, clearly, u l~E. Consequently (p(L)>,(p(E), and part (i) follows. (ii) Let L be a left ideal of T a n d for each m e M choose a C-complement K~ for Lm=LC~T~ in T r Then it follows from Lemma 4.2 (iv) that R o T ~ = Ro LmGRo Km. Therefore, S o = @ ~ RoT~.,=@ ~, (RoLmQRoKm) m~M

m~M

=(@ ~ RoLm)QK, msM

Finite Normalizing Extensions of Rings

where K = |

473

~ R o K m is a left Ro-submodule o f S o. Clearly, G ~ RoLmcSo L. m~M

mEM

On the other hand, since xi~Tand L is a left ideal of T, it follows that x ~ L ~ L for all i, and hence SoL= ~ R o x i L = R o L . Moreover, as we have seen above, if l~L i=1

then there exists a unit u~Tc~R o such that u l~L,, for some m~M. It follows that R o l = R o u l c R o L , , and, consequently, R o L = O ~ RoL m. Therefore, we have me~M

equality, SOL=|

~ RoLm, and it also follows that K is the desired R 0mEM

complement for SoL in S O. Now let t~SoLc~Tand, as above, choose a unit u~Tc~R o with u t ~ T ~ for some m~M. If ut~L,~ then we can choose K,, above so that utEK,~. It follows that ut~KC~SoL=O , a contradiction. Thus we conclude that u t ~ L and so t = u - t ( u t ) ~ L . This shows that SoLc~T=L. Finally, let s~S o be given with As~SoL for some nonzero ideal A of R. Writing s=s 1 +s 2 with slESoL and s2~K, we obtain that

As 2 = A ( s - sl) ~ K c~SoL = O. Since A is nonzero, we conclude that s 2 = 0 so that s = s ~ S o L . This completes the proof of the lemma. We now prove a result of the type "nil implies nilpotent" for T. Let Ybe an ideal of T. A left ideal L of Tis called nil modulo Yif every element l~L satisfies lkEYfor some positive integer k=k(l). L is said to be nilpotent modulo Yif there exists a fixed k such that for any ll, 12, ..., lkeL we have l a 12 ... IkE Y. The smallest such k is called the nilpotence degree of L modulo Y The following lemma is proved by adapting the usual proof of Levitzki's theorem, keeping in mind that addition in T is only partially defined. If L~ and L 2 are left ideals of T t h e n we let L~ + L 2 denote the left ideal of T generated by L~ and L2, that is L 1 + L 2 is the intersection of all left ideals of T containing L~ and L 2. More explicitely, L~ + L 2 consists of all elements of the form ~ t y T w i t h t j ~ L t u L 2. J

Lemma 4.5. Let Y be an ideal of T and let L be a left ideal of T which is nil modulo Y Then L is nilpotent modulo Y

Proof. Suppose L~ and L 2 are left ideals of Twhich are nilpotent modulo Y, of nilpotence degree kl, resp. k 2. Then it is easily seen that L ~ + L 2 is nilpotent modulo Y o f degree at most k ~ + k 2 - 1 . Since any chain of left ideals in T i s finite, by Lemma 4.4 (i), we conclude that there exists a unique largest left ideal M of Twhich is nilpotent modulo Y M is in fact two-sided. For, if m e M and t~ T then Tm t = { ~ t~ m t~ rl tic T} is a left ideal of T which is nilpotent modulo J

Y, since M is nilpotent modulo Y Thus we have T m t c M and hence m t ~ M so that M is in fact two-sided. Now assume, by way of contradiction, that L is a left ideal of Twhich is nil modulo Ybut not nilpotent modulo Y. Then surely L is not contained in M. For any l s L we set r { t ~ T [ l t c M } . It is easily checked that this is a right

474

M. L o r e n z

ideal of T and, using the right sided version of Lemma 4.4(i), we can find l~L, l(~M such that ~T(I+M) is maximal. We claim that I t l E M for all teT. Indeed, suppose Itl(~M for some teT. Since t l ~ L is nilpotent modulo Y, and hence modulo M, there exists p > l with l(tI)P(~M, I(tl)P+lEM. We conclude that t le~r(l(t l)p + M), t lq~,r(I + M). But this contradicts the maximality of ~r(l + M), since ~T(I+M)C,T(I(tl)P+M ) and l ( t I ) P e L \ M . Thus we have in fact I t l ~ M for all teT. It follows that the left ideal Tl = { ~ tj l~ TI t ~ T } is nilpotent of degree at J

most 2 modulo M, and hence is nilpotent modulo Y. Therefore, Tl c M and leM, a contradiction. This proves that L is in fact nilpotent modulo Y. Our main goal in this section is to contruct all prime ideals P of S with Pc~R = 0 starting from certain ideals of T. These ideals will be called prime and are defined, in analogy to the usual notion of prime ideal, as follows.

Definition. An ideal X of T is said to be prime if X =# T and if for any two t 1 and t 2 E Z , t 1 tt2EX for all tETimplies that t l e X or t2eX. We note a few consequences of this definition which do of course come as no surprise. Indeed, the proofs are straightforward adaptions of the usual arguments to the present situation. If Y1 and II2 are ideals of T, then we let Y1 II2 denote the intersection of all ideals of T containing the products y~y2 with yl~Y1 and y2eY2. Thus I71 Y2 is an ideal of Twhich is contained in Ylc~Y2. Lemma 4.6. Let X be an ideal of T, with X :t: T i) X is prime if and only if Y1 Y2 c X for any two ideals I71 and I12 of T implies

that ~ c X for some l. ii) Any maximal ideal of Tis prime. iii) I f X is prime then any left ideal of T which is nil modulo X is contained in X. Note that part (iii) above also holds for a semiprime ideal X of T, with the obvious notion of semiprime ideal in T Note further that part (i) above implies in particular that if X is prime and Ylc~Ya c X for certain ideals I11, Y2 of Tthen Y~~ X for some I. Our next goal is to show that, in fact, every prime ideal of Tis maximal and, moreover, that there exist only finitely many primes in T and that their intersection is nilpotent. This will be done in Proposition 4.9. In order to be able to work in a more familiar setting we will use the C-algebra E = q2so(Ro)~ T t o a good deal. The following considerations may be of independent interest and hence contain more detail than is actually needed for the proof of Proposition 4.9. Let t e T b e given with teT~, say. Then, by Theorem 1.3, t satisfies a monic equation of the form t z= T(t), where T(t) is a sum of Ro-monomials in t of degree less than l, and l 0 minimal such that there exists a relation s=rm tm~_rm_l

t m - 1 _{_ . . . + r ~ = 0

with ri~Ro, rm~0. Using Lemma 3.10), (ii) we may even assume that ri~R for i=O, 1..... m. Since teT~ we have for any r~R

Finite Normalizing Extensions of Rings

475

qm O=rmrs--sr6ha rm =

m--1

2 (rmrri-r i=0

i r a m - i rm~m-i) ti,

By the minimality of m, we conclude that rm

r

ri~ rir

6m-i (Tm-i rm

for all r e R and i = 0 , 1, ..., m. By L e m m a 3.2, there exist elements n i e q ~ _ , with r~=rmn ~ for i = 0 , 1, ...,m, and n ~ = l . Then for all i and all r e R we have rnitl=niram-'tl~nit

9

l a m-i+I r

In particular, letting s o = W +n,~_ 1 t m- 1 + ... + n o we see that r s o = s o r ~m holds for all r e R so that s o e T ~ . Since s = r J o = O and r,,:t=0 we conclude, as in the p r o o f of L e m m a 4.2 (i), that s o = 0. Thus

t m + n ~ _ l t m-1 + . . . + n o = 0 , and this is clearly the unique m o n i c relation over R o of degree m satisfied by t. We will denote it by (o(t) and call it the minimal relation for t (over Ro). Any other relation for t over R o is a left multiple of ~o(t). In particular, since m--1

O=~t(t)=tq)(t)-q)(t)t=

2 (n~ Z - n i ) t i+ i, i=0

we conclude that (n~_i-nm_l)no=0 and

( n m _ l _ n m _ i ) n i + 1 -_- Hi~-1 - G-1

ni

~-1

a-1

for i = 0 , 1 , . . . , m - 1 . N o t e that n m _ l - n m _ l e ( , l , ~ ) +@~=@~. Thus if nm_ 1 - n .... 1 is nonzero then it is invertible, by L e m m a 3 . 2 , and using the above equalities we successively obtain n o = 0 , n l = 0 , . . . , n m _ l - - 0 , a contradiction. a-I (r-1 Therefore, n m_ 1 - n~_ 1 = 0 and it follows that n i - n~= 0 for i = 0, 1.... , m - 1. Thus all coefficients in q~(t) are a-invariant. L e m m a 4.7. Let t e T be given. i) I f t is left (or right) regular in T then it is invertible in T. ii) There exists a positive integer i with i
Proof. (i). Assume that teT~, say, and let (p(t)=t"+nm_ 1 t " - 1 + . . . + n o be the minimal relation for t over R o. Set s = t " - ~ + n , , _ 1 tm--2Ar - . . . - ~ - n 1. Then, by our above remarks, we have s t = t s and r s = s r ~m-~ for all r e R . Hence s e T and, moreover, s =t=0, by the minimality of m = deg (o(t). Thus if t is left (or right) regular in T then it follows from O = c p ( t ) = s t + n o = t S + n o that no=~0. By L e m m a 3.2, n o is invertible in T, and hence we have - n o i s t = 1 and - n o ~ seT. This proves (i). (ii). The assertion is of course obvious if tin=0 so assume that t ~ + 0 . Then some coefficient nj of q~(t) must be nonzero and hence is a unit in Ro, by L e m m a

476

M. L o r e n z

3.2, since n~-C~bom-j.Thus letting i = m - j and u = n f 1 we see that O
Proof. By Lemma 4.6(iii), it suffices to show that L is nil modulo X. So let leL be given. Then, by Lemma 4.700, there exists a positive integer i and a unit u~Tc~R o such that ul~sE. Clearly, u l ~ L and hence we deduce from our assumption on L ~ E that (u l~)k~X for some k >0. Since l~normalizes Ro, we can write (u Ii)k = V Iik for some unit v E T n R o. Thus v l i k ~ x and hence l l k ~ x . Therefore, L is nil modulo X, and the lemma follows. We now come to the promised result on prime ideals in T. The main part of the proof is based on the usual proof of Goldie's theorem. Proposition 4.9. Thas only finitely many prime ideals X1, X2,..., Xk, and these are k all maximal. Moreover, k
degree at most n. Proof. Let X be a prime ideal of T. If L is any left ideal of T with L r X then there exists an element leLc~E, lCX with EE(I+X)={eeEIelEX} =E~(IZ+X). Indeed, Lemma 4.8 shows that there exists an element t~k,c~E which is not nilpotent modulo Xc~E. Since E is finite-dimensional over C = C(R), a suitable power, l, of t satisfies ~ ( l + X) = ~E(I2 + X). Now suppose that Y is an ideal of T strictly containing X. Since X is prime, it follows that for any left ideal L of T, Y ~ L c X implies that L c X . By our above observation, we can find y~Y~E, yCX with ~ ( y + X ) = # e ( y 2 +X). It follows that the sum Ey+E~(y+X) is direct modulo Xc~E. For, if (e y) y e X for some eeE then e~f ~(y2 + X ) = f e(y + X), and hence e y ~ X c~E. Now suppose that we have already found elements Y=Yo, y l , . . . , y , ~ E with y~Y~ i--1

= Yc~ ~ #r(y~+X), y~(~X such that the sum j=0

Ey + Ey 1+... + Ey,, + (Ymc~E~(ym+ X)) is direct modulo Xc~E. Note that this surely holds for n = 0 , with yo=y and Yo =Y. If Y , , + I = Y s ~ f r ( y ~ + X ) r then, by the first paragraph of this proof, we can find ym+~Ym+xC~E, ym+l(~X with ~(ym+~+X) =~'~(y~+ 2 1 + x ) . As above, we see that the sum Eym+~+(u is direct modulo X mE. Moreover, E y,,+ ~+ ( Y,,+ ~m# ~(y,,+ ~+ X)) is contained in Y,,c~#~(y,,+ X). To see the latter, just note that y,,+aeY,,+~mE=Y,,c~#~(y,,+X). Therefore, the sum

Ey+ Ey~ +... + Eym+ ~+ (Ym+a ~[E(Ym+ 1 + X))

Finite Normalizing Extensions of Rings

477

is direct modulo X ~ E . Since E/(X~E) is finite-dimensional over C = C(R), the process must stop eventually. Thus for some m we must have Y~+~

= gc~ ~ f r ( y j + X ) c X . As we have remarked above, this implies (~ ~r(Yj j o j=o + X ) c X . Set e = y + y ~ + . . . + y m e E . Then f e e X for f e E implies that fy, fy~,...,fy,~ all belong to X, since the sum E y + E y ~ + . . . + E y m is direct modulo Xc~E. Thus we see that EE(e+X)= (~ ~E(yj+X)~Xc~E. In other j=o words, e is left regular modulo X n E in E. Since E/(Xc~E) is finite-dimensional, it follows that e is invertible modulo Xc~E in E. Finally, e e Y and we deduce that 1E Y and Y= T. This proves that X is maximal. To prove that T has only finitely many primes, let X~, X2, ..., X k be distinct prime ideals of T Then

X I c~E ~ X I c~X2~E ~ ... ~ X I ~X2c~...~Xg~E i

i+1

is a strictly descending chain of ideals in T. For, if ~ X / ~ E = ~ X j ~ E then j=l

j=l

i

Lemma 4.8, with L= (~ Xj and X=XI+I, implies that L c X i + 1. Lemma 4.6(i) j=i yields X j c X i + ~ for j<=i and since Xj is maximal we obtain that Xj=Xi+I, a contradiction. Thus the above chain is strictly decreasing, and it follows that k =
Finally, let X1,X2, . . . , X k be all the prime ideals of T and set J = ~ Xj. It j=l remains to show that J is nilpotent, of degree at most n. A standard argument shows that J is a nil ideal of T, and Lemma 4.5, with Y= 0, implies that J is in fact nilpotent. If p denotes the nilpotence degree of J then the chain

j ~ j 2 = j j ~ . . . ~ji+ ~= j j i ~ . . . ~JP=O is strictly descending and hence has at most n terms, by Lemma 4.4(i). This finishes the proof. We remark that dimcE=l{ila i is X-inner}l, by Lemma 4.2(ii), and hence surely dimcE __
yu = So Yc~S = { ~ sj y je SIsF So, y j~ Y }. J

it) If I is an ideal of S then we let Id= {t~TIAt c I for some nonzero ideal A of R}. The maps u and a are clearly inclusion preserving. The following lemma, in particular part (it), is the crux of the matter.

478

M. Lorenz

L e m m a 4.10. i) I f Y is an ideal of T then Y " = YRoc~S=RoYC~S, and this is an ideal of S with Y = y,d. ii) I f I is an ideal of S then I a is an ideal of T, and I c I a". Proof (i) Let Y be an ideal of T. Then, by L e m m a 4,4(ii), we have S o Y = R o Y and, analogously, YS o = Y R o. Therefore, S OY = YS o and this is dearly an ideal of S 0. It follows that Y" is an ideal of S. In order to prove the inclusion Y c y , a we have to show that for any y e Y there exists a nonzero ideal A of R with A y c Y". But there exists a nonzero ideal A of R with A y e s and so we have A y c S o Yc~S = Y". Thus Y c Y % Conversely, assume that y e T satisfies A y e Y " c S o Y for some nonzero ideal A of R. Then, by L e m m a 4.4 (ii), we conclude that y ~ S o Yc~ T = Y Thus we have shown that Y= yud, and part (i) is proved. (ii) Let I be an ideal orS. Let tffI d be given with A t c I , say, and let q e T b e arbitrary. Then te Tr and q ~T~ for suitable automorphisms a and z of R, and we can find a nonzero ideal B of R with B t i c S . We obtain that AB ~ '(ttl) = ( A t ) ( B t l ) ~ I S = I and B A ~ - ' ( t l t ) = ( B t l ) ( A t ) c S I = I . Since AB ~-~ and BA ~-1 are nonzero ideals of R, we conclude that tt> t~ t e I a. Finally, let t~, t2, ..., tkeI a k

be given with A ~ t ~ I

for suitable nonzero ideals Ai of R. Then D = ~ A~ is

nonzero and satisfies D(t~ + t 2 +... + tk) ~ A 1 t 1 q - A 2 t 2 J r . . . q - A k t k ~ I . Thus t x + t 2 +... + t k e I d, provided t~ + t 2 +... + tkeT. Since surely 0 e I d, we have shown that I d is an ideal of T. N o w suppose / is nonzero. For any subset M of {1,2 ..... n} we set I M =Ic~ ~ R x i , and we denote by ~{ the set of all subsets M of {1,2 ..... n} such ieM

that I~t + 0 but I v, = 0 for all M ' ~ M. Then ~ is a finite nonempty set of subsets of {1,2 .... ,n}. Consider a fixed M e J [ and let m s M . We set C M , , , = { r e R l t h e r e exists r~xiE1 M with r~=r}. Then CM, ~ is a nonzero ideal of R, and to each rEC%,m ieM

there exists a unique s~= ~ r~x~el M with r,,=r. Let Oq=reCv,,, be given. Then ieM

for each veR we have I~rvsr-s~v~

r

~ ( r v r i - r l v .... i-* r ..... 1)xi. i~M

Since the xm-coefficient of the element on the right is zero, this element belongs to Iu..{m} and hence equals zero. We conclude that for all i e M and all vER we have r v r i = r i v ~'~r

1 r~,~r

1.

L e m m a 3.2 yields units nlecb .... , , with r~=r ni, n m= 1. It follows that s~ =r t, with t = ~ n~x~. Note that t belongs to T~,. and that the xm-coefficient of t equals 1. In ieM

particular, we have I ~ R s ~ R = R r t R = (RrR)t so that t e I a. Furthermore, t only depends upon m and M, not on reCM, ~. For, if t, and t 2 a r e two elements of Idc~ ~ R o x i with Xm-COefficients equal to 1 then there exist nonzero ideals A, and

ieM A 2

of R such that A , t,, A 2 t 2 c I. Thus for any b e B = A ,

c~ A 2

we have b(t~

Finite Normalizing Extensions of Rings

479

-t2)~IM~{,~} =0, and we deduce that t I = t 2. Therefore, the above element t only depends upon m and M and will henceforth be denoted by tM,,n. Thus we have IM = CM, m tin, m"

We show, by induction on the length l(s)= ISupp(s)l of s, that for any s e I we have s e l a". The case l(s)=0, that is s =0, is of course trivial so we may ~tssume that l(s)=~O and that the result holds for all s'el with l(s')
has xm-coefficient 0. Since Supp(s')cSupp(s), we see that l(s')
Proof. (i) We have Y ~ Y I R o = R o Y z for l = 1 , 2 , and hence u Y ~ c Y~R o 9Y2Ro = t11 Y2 "Roc(Y~ Y2)Ro 9 Thus Y?. Y ; c ( Y 1 Y 2 ) R o ~ S = ( u I12)~(ii) Clearly, (Y1 c~ I12)"~ Y,"c~ Yf since " preserves inclusions. Conversely, let D = Y~ c~ Y~. Then, by L e m m a 4.10 (i), we have D e = (Y~ c~ y~)a ~ y~a c~ y~a = I11c~ Y2. By L e m m a 4.10(ii), we conclude that DcDduc(Ylc~Y2)" which is the reverse inclusion. Thus equality holds, and the lemma is proved. We now come to the main result of this section. Theorem

4.12.

Let

R c S = ~ R x i be an

extension

of

rings such

that

i=1

{xl, x 2.... , x~} is a normalizing basis for S over R : Assume that R is prime. Then the maps ~ and a yield a one-to-one correspondence between the prime ideals P of S with P?~R = 0 and the prime ideals of T = N s o ( R ). More precisely: i) I f P is a prime ideal of S with Pc~R = 0 then pa is a prime ideal of T, and P = pdu.

ii) t f X is a prime ideal of T then X ~ is a prime ideal of S with X ~ R = 0 , X = X ~d.

and

480

M. Lorenz

We prove this theorem simultaneously with the following l e m m a which contains a special case of the more general Incomparability theorem that will be proved in the next section. L e m m a 4.13. Let R ~ S be as in the theorem and let P be a prime ideal of S with P c~R = O. Then for any ideal I of S we have I a ~ pd if and only if I ~ P. I f I strictly contains P then I c~R ~ O.

Proof (i) Let P be a prime ideal of S with P ~ R = 0. We first show that P = pa,. By L e m m a 4.10(ii). we know already that P c P a". N o w let s s P a". Then s can be written as a finite sum s = ~ tjsj. with tjeP a and sjeS o. Choose nonzero ideals J Aj of R with Aj t j c P and set A = ~ Aj. Furthermore. choose nonzero ideals Bj J of R with B j s j c S . Each tjBj can be written as t j B j = B ) t j for a suitable nonzero

ABs=2A B t s =2(Ajt )

ideal Bj. of R, since tj~T. We set B = ( ~ B } . Then J j J (Bjsj)cP, and hence J d ( S A B ) . s c P . But ~d(SAB)c~R contains the nonzero

ideal (~ (AB) ~ of R and thus we have d;d(SAB)egP, by assumption on P. i=1 The primeness of P implies that s~P. This proves the reverse inclusion and so we have shown that P = pe,. To show that pe is prime in T. let I11 and I12 be ideals of T with I11Y2 c pd. Then. by L e m m a 4.11 (i). we have Y~- Y2~~(Yx Y2)"~pa~ = p and hence Y;' ~ P for some l. It follows from L e m m a 4.10(i) that Y1 = t~"a=Pd so that pa is prime in T For L e m m a 4.13. let I be an ideal of S. Clearly. if I c P then I a c P a. Conversely, if I a ~ pa then L e m m a 4.10 (ii) implies that I ~ I d~c pa, = p. Finally, if I strictly contains P then we deduce that I a strictly contains W and since pa is prime and hence maximal in T. by Proposition 4.9. we obtain that l e I a. The definition of I a shows that I ~ R + O . and L e m m a 4.13 is proved. (ii) Let X be a prime ideal of T. Then, by L e m m a 4.10(i), we have X = X "e, If A = X ~ c ~ R is nonzero, then L e m m a 4.4(ii) implies that l e S o X , since A. 1 E X " ~ S o X. Thus. by L e m m a 4.4(ii) again, we obtain that I ~ S o X ~ T = X . a contradiction. Therefore. we conclude that X" c~R = 0. N o w suppose that I is any ideal of S properly containing X ~. Then Ia~ X ua = X . But Ia~=X since otherwise we would have I ~ I e " = X ~. by L e m m a 4.10(ii). Thus the maximality of X implies that 1 ~ I d from which we deduce that I c~R=~O. We conclude from this that X" is prime. Indeed. if I~. 1 2 ~ X ~ then IlcvR=~O and I2cvR=~O so O~(I~cvR)(I2cvR)~I~I2c~R. Since X~cvR=O. this yields I~I 2 C X " so that X ~ is in fact prime. The theorem is proved. Corollary

4.14.

Let

R c S = ~ R x i be an extension

of rings such

that

i=1

{x 1. x 2..... x,} is a normalizing basis for S over R. Assume that R is prime. i) A prime ideal P of S is minimal if and only if P ~ R = O. ii) There are only finitely many such primes, say P1, P2..... Pk. and in fact k ~ l { i l a i is X-inner}INn. k

iii) N = ("] Pj satisfies N " = O. and hence N is the unique largest nilpotent ideal of S. j-1

Finite Normalizing Extensions of Rings

Proof Let X 1. X

2 .....

X k

48l

be all the prime ideals of T. Then, by Proposition 4.9, k

k
Theorem 4.12, we know that PI, P2..... Pk are precisely the prime ideals of S k

which intersect R in 0. Set N = ~ Pj. Then, by Lemma 4.11 (ii), we have N = J " j=l

and so, by Lemma 4.11(i), N"=(Ju)"c(J")u=Ou=O. Therefore, N is clearly the unique largest nilpotent ideal of S and, moreover, each prime ideal of S contains some P~. Thus the minimal primes of S are precisely the minimal members of {P~, P2..... Pk}. But since Pid=Xi, Lemma 4.13 shows that there are no/inclusions among the P[s so that P1, P2..... Pk are precisely the minimal primes of S. This finishes the proof. The following example shows that Corollary 4.14 fails miserably if the extension R c S is not free. The example was obtained in a conversation with D.S. Passman.

Example. Let R be a prime ring and let I be a nonzero ideal of R. Then the ring direct sum S=RO(R/I) contains R as a subring via the diagonal map r--*(nO, w h e r e - : R ~ R / I is the natural map. Observe that 1 and e =(0, i) are normalizing (in fact centralizing) generators for S over R which are of course not R-independent, since Ie=O. Let Q be a minimal covering prime o f / i n R. Then P = R | is a prime ideal of S, since S/P~_R/Q, and P is in fact minimal. On the other hand, P intersects R nontrivially, since 0 4 : I | Thus part (i) of Corollary 4.14 fails in this situation. Part (ii) also fails, because we may choose R and I so that I has infinitely many minimal covering primes Q1,Q2 .... in R to obtain infinitely many minimal primes P~=R@(~i in S. Finally, if J is any ideal of R then 0 Q J is an ideal of S which is nilpotent if J is. Hence if/~ = R/I has no largest nilpotent ideal then neither does S, and part (iii) fails. To obtain concrete examples, take R = K [ t 1, t 2 .... ], the polynomial ring in countably many commuting variables over the field K, and let Ii=(t2~-t~, t~@i, j = l,2 .... ; i # j ) and I : = ( t i + lli= l,2 .... ). Then R/I 1 ~_K (~'~ has infinitely many (minimal) primes and R/I 2 has no largest nilpotent ideal. We close this section with a few remarks on a certain special case which does at least cover crossed products and free centralizing extensions. Namely, assume that in addition to (1), (2) and (3) we have (4)

There exist units

tl, t 2 .....

t,~S such that t F l r t i =r ~' holds for all r~R.

Note that this assumption is surely satisfied if the xi's are units, or induce inner automorphisms on R. The elements t~ can be used to extend the automorphisms O-i from R to S and to So, and we denote the extended automorphisms by the same letter. If G = (or 1, 02 O'n) c Aut (So) then any O-~ G certainly stabilizes E=ff2so(R ), and hence we can talk about G-invariant, G-prime and G-maximal ideals of E, in the usual sense. Moreover, since E is finite-dimensional over C = C(R), the G-prime ideals of E coincide with the G-maximal ideals. . . . . .

482

M. Lorenz If I is an ideal of S, then I d is an ideal of T and hence

I,~ = Id c~E is an ideal, in the usual sense, of E which is easily seen to be G-invariant. Furthermore, the argument in the first paragraph of the proof of Lemma 4.4 shows that for any ideal Y of T we have Y= ~ T(Yc~ T~m). Since T ( Y ~ T~,,) meM

= rt2~(Yc~ T~,,) and t,~ ~(Yc~ T~,,)c Y c~E, we obtain that Y= T(Yc~ E). In particular, Y " = S o ( Y n E ) ~ S . For any G-invariant ideal H of E set H~ =SorterS. Then one shows without difficulty, using the known properties of the maps " and d together with our above remarks, that the maps 1" and $ yield a one-to-one correspondence between the G-maximal ideals of E and the prime ideals P of S with Pc~R=O. Thus if we assume (4), then we obtain a somewhat nicer description of the prime ideals of S that intersect R trivially, just as in [-7, Theorem 2.5]. However, without (4) the map T does not necessarily map Gmaximal ideals of E to prime ideals of S, as can be seen from Example 4.3.

w 5. Incomparability In this final section, we combine the results of Sections 3 and 4 to prove an Incomparability theorem for extensions R c S = ~ R x~ with x i r ~ = r x i (rER) for i=1

suitable automorphisms a~ of R. It would of course be interesting to drop the latter assumption and merely assume x~R=Rx~ instead. However, the present techniques seem to require that, for S prime, any ideal A of R with zero annihilator in R satisfies J d ( A S ) + O. As we have pointed out in Lemma 3.5, this is trivially true if we assume the existence of the above automorphisms ai, and thus we could form the extension R(G)cS(G) in Sect. 3. We remark that if one was interested in proving Incomparability alone, then Sect. 4 could of course be substantially shortened. We need one preliminary lemma.

Lemma 5.1. Let R c S =

~, Rx~, xir ~ =rx~ (r~R), be given with S a prime ring. i=1

Set G = @1, a2,-.-, % ) c A u t ( R ) and let R(G)~S(G) be the extension constructed in Lemma 3.6. I f I is an ideal of S such that S(G)IS(G)~R(G)+O then I n R 4=O. Proof Note that S(G)IS(G)= ~ ~br

Thus if O=#qES(G)IS(G)nR(G) then q

0-~ ~EG

can be written as q = ~ n zilm ~ with n I and m~ R-normalizing, say mt6q~,, and z it~I. By Lemma 3.1 (i), there exist ideals A~, B~, D s j ~ = ~ ( R ) such that D q c'R, A z n l c R and Blml=mlB~'cR. Thus, if we let H = ~ (AI~B;')~D then H e ~ l

Finite Normalizing Extensions of Rings

483

a n d H q H ~ I c ~ R . F i n a l l y , HqH:#O, for o t h e r w i s e Hq w o u l d be a n i l p o t e n t left ideal o f R a n d h e n c e w o u l d be zero, since R is s e m i p r i m e , by T h e o r e m 2.2 (ii). B u t t h e n L e m m a 3.1 (ii) w o u l d y i e l d q = 0 , a c o n t r a d i c t i o n . T h u s we o b t a i n t h a t I c~ R + 0, as c l a i m e d . T h e o r e m 5.2 ( I n c o m p a r a b i l i t y ) . Let R c S= ~ R xi, xir~i=rxi (reR), be given

i=1 with S a prime ring. 7hen each nonzero ideal of S intersects R nontrivially.

Proof F o r m R(G)cS(G) as in L e m m a 3.6 a n d let e~C=C(R) be o n e of t h e p r i m i t i v e i d e m p o t e n t s in C c o r r e s p o n d i n g to t h e m i n i m a l p r i m e s of R (see L e m m a 3.4). T h e n , by P r o p o s i t i o n 3.7, e R ( G ) c e S ( G ) e is an e x t e n s i o n w i t h a finite n o r m a l i z i n g basis, a n d b o t h e R(G) a n d e S(G)e are p r i m e rings. Now e S(G) e, so that, Lemma

let I be a n o n z e r o i d e a l of S. T h e n eS(G)IS(G)e is a n o n z e r o i d e a l of since S (G) is p r i m e . L e m m a 4.13 i m p l i e s t h a t e S(G) IS(G) e c~e R (G) + 0 in p a r t i c u l a r , S(G)IS(G)c~R(G)+O. T h e t h e o r e m n o w f o l l o w s f r o m 5.1.

Note Added in Proof (Dec. 1, 1980). A.G. Heinicke and J.C. Robson have recently succeeded to extend Theorem 5.2 (Incomparability) to arbitrary finite normalizing extensions. Their results will appear in an article entitled "Normalizing extensions: prime ideals and incomparability" in the Proc. London Math. Soc.. The work of J. Bit-David and J.C. Robson has appeared under the title "Normalizing extensions I, II" in "Ring Theory: Proc. of the 1980 Antwerp conference", Springer Lecture Notes 824. D.S. Passman has come up with an interesting alternative approach to many of the results presented here. In "Prime ideals in normalizing extensions" (preprint) he describes a mechanism that enables him to pass from prime rings to primitive rings where he can consider irreducible modules and use Clifford type arguments. S. Montgomery has pointed out to me that the normal closure of R as defined in [4] does indeed have a closure property similar to the one described in Lemma 3.3 (i). More specifically, let R be a semiprime ring, let Qm(R) denote its (1eft) maximal ring of quotients and set N = {xeQm(R)lxR =Rx}. Moreover, let S be a ring with RcScQm(R ) and let Ss~ denote the ring of(left) quotients of S with respect to the filter ~ of essential two-sided ideals of S. Then any xeS~ with xR=Rx is contained in N. Finally, R. Resco's proof of Theorem 1.5 will appear in "Radicals of finite normalizing extensions" in Commun. in Algebra, and J. Park's is contained in "Perfectness of normalizing extensions of rings" (preprint). Both their proofs are shorter than the one presented here. On the other hand, our proof of Theorem 1.5 is an easy application of integrality which we feel is a basic property of finite normalizing extensions.

References 1. Amitsur, S.A.: On rings of quotients. Symposia Mathematica 8, 149-164. New York-London: Academic Press 1972 2. Bergman, G.M.: Lifting prime ideals to extensions by centralizing elements, (unpublished) 3. Bergman, G.M.: More on extensions by centralizing elements, (unpublished) 4. Cohen, M., Montgomery, S.: The normal closure of a semiprime ring. In: Ring Theory, ed. F. Van Oystaeyen. New York: Marcel Dekker 1979 5. Formanek, E., Jategaonkar, A.V.: Subrings of Noetherian rings. Proc. Amer. Math. So c. 46, 181186 (1974) 6. Kharchenko, V.K.: Generalized identities with automorphisms. Algebra and Logic 14, 132-148 (1975)

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7. Lorenz, M., Passman, D.S.: Prime ideals in crossed products of finite groups. Israel J. Math. 33, 89-132 (1979) 8. Lorenz, M., Passman, D.S.: Addendum - Prime ideals in crossed products of finite groups. Israel J. Math. 35, 311-322 (1980) 9. Lorenz, M., Passman, D.S.: Integrality and normalizing extensions of rings. J. Algebra 61, 289297 (1979) 10. Lorenz, M., Passman, D.S.: Prime ideals in group algebras of polycyclic-by-finite groups. Proc. London Math. Soc. (to appear) 11. Martindale, W.S.: Prime rings satisfying a generalized polynomial identity. J. Algebra 12, 576584 (1969) 12. Montgomery, S., Passman, D.S.: Crossed products over prime rings. Israel J. Math. 31, 224-256 (1978) 13. Par+, R., Schelter, W.: Finite extensions are integral. J. Algebra 53, 477-479 (1978) 14. Passman, D.S.: The Algebraic Structure of Group Rings. New York: Wiley-Interscience 1977 15. Procesi, C.: Rings with Polynomial Identities. New York: Marcel Dekker 1973 16. Robson, J.C., Small, L.W.: Liberal extensions. Proc. London Math. Soc., (to appear) 17. Roseblade, J.E.: Prime ideals in group rings of polycyclic groups. Proc. London Math. Soc. 36, 385-447 (1978) 18. Schelter, W.: Non-commutative affine PI-rings are catenary. J. Algebra 51, 12-18 (1978) Received February 25, 1980