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Prime ideals in normalizing extensions
JOURNAL
OF ALGEBRA
73, 556-572
Prime
Ideals
(1981)
in Normalizing
Extensions*
D. S. PASSMAN University
of Wisconsin-Madison,
Madison,
Communicated Received
Wisconsin
53706
by I. N. Hernstein September
15, 1980
In this paper we offer a new approach to the study of prime ideals in finite normalizing extensions of rings. We introduce a “primitivity machine” which translates certain problems on prime ideals to analogous questions on primitive ideals. We then use the well-known Clifford theory of irreducible modules to study these primitive ideals.
Let S 2 R be an extension of rings with the same 1. We say that S is a finite normalizing extension of R if there exists x,, x2,..., x, E S with S = C; Rxi and with Rxj = XjR for all j. Note that the normalizing set {Xl, X2,“‘, x,} need not consist of units and need not be independent over R. Examples of such extensions are the finite centralizing extensions (also called liberal extensions) where each xi centralizes R (see [ 131) and crossed products S = RG, where G is a finite group (see [8]). Furthermore there is the commutative situation where S is a finite integral extension of R. Prime ideals in the commutative extensions have, of course, long been studied and in particular one is aware of the classical Krull relations of Lying Over, Going Up, Incomparability, etc. Furthermore analogs of these relations also hold in finite centralizing extensions [ 131 and in crossed products of finite groups [8, 9, 111. Recently, prime ideals in the general case of finite normalizing extensions have been studied by Bit-David and Robson [2], Lorenz [7] and Heinicke and Robson [6]. From this work it is apparent that the really deep and difficult relation is Incomparability. This was proved in part by Lorenz and finally obtained in the definitive paper of Heinicke and Robson. In ‘this manuscript we offer a completely different approach to this subject. For a ring R, we show that there is a suitable power series-polynomial ring R * over R in which the prime ideals of R extend to primitive ideals of R *. Furthermore, if S 2 R is a finite normalizing extension, then so is S * 2 R *, and using the classical Clifford theory we study the relationship between the * Research
supported
in part by NSF
Grant
No. MCS
556 0021.8693/81/120556-17$02.00/O Copyright All rights
0 198 1 by Academic Press, Inc. of reproduction in any form reserved.
77-01775.
PRIME IDEALS INNORMALIZINGEXTENSIONS
551
irreducible modules of S* and of R *. This then yields information comparing the primitive ideals of S* and of R* and then the prime ideals of S and of R. In this way, with the help of a result from [2], we are able to obtain all the basic Krull relations on primes with the exception of the difficult Incomparability. We also obtain certain structure theorems which hold for crossed products and normalizing extensions with a free basis. This paper is written in a reasonably self-contained manner. Because of this,, many of the lemmas and most of the theorems are known. We take credit only for the approach to the problem. 1. POWER SERIES-POLYNOMIALS In this section we develop the “primitivity machine” which translates prime ideals to primitives. If Y is a set of variables, let R((Y)) denote the formal power series ring over the ring R in the noncommuting variables Y. Since any word in the semigroup generated by Y has only finitely many initial and final segments, it is clear that multiplication in R((Y)) is defined. LEMMA 1.1. Let R be a prime ring and assume that 1Y( > IR 1, an equation of cardinalities. Then there exists a regular element f E R(( Y)) such that fj is regular for all nonzero g E R( (Y)).
ProoJ {y,jrER}.
Since ] Y 1> 1R 1, we can label some of the variables in Y as Let
f = & ry, E R(V)). We claim that f is the required element. To this end, by symmetry, it suffices to show that g, h # 0 implies fgF # 0. Let a be a monomial of shortest length in the support of g and let p be one of shortest length in h. Say g=sa+ ..* and h = t,LI+ .. - with s, t E R\O. Since R is prime, there exists r E R with srt # 0. It is then easy to see that
fgF = (lsrt) y, cry,/? + ... and that y,a~~,.p is a uniquely represented element. Hence fgfh # 0. If Z is a set of variables, let R(Z) denote the polynomial ring over R in the noncommuting variables Z. The following is based on ideas of Formanek (41 and is a special case of a result of [5]. LEMMA 1.2. Let R be a ring and supposethat R has a regular element r such that rsr is regular for all s E R \O. Let Z be an ifinite set of variables with IZI > (R (. Then R(Z) is primitive.
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D. S.PASSMAN
Proof: Clearly IZI = /R(Z)1 so to each fE R(Z)\0 distinct variable .z,E Z. We claim that the right ideal
N=
we can associate a
2 (1 +zrrfr>R(Z) f#O
is proper. If not, we have a finite sum 1 =
c f#O
(1
+zfrfrkf
and we can choosefoccurring in the above so that degf+ deg gf is maximal. Let a be a monomial of largest degree inf, /3 one of largest degree in gf and sayf=sa+... andgf=tP+.... Then, by properties of r, zrrfirg, contains the nonzerp term (rsrt) zraj3 and zfap is clearly a uniquely represented element in the whole right-hand expression C (1 + z&) gr. Thus N # R(Z). Now choose A4 a maximal right ideal of R(Z) with A4 2 N. We claim that M is comaximal with all nonzero ideals I. Indeed if fE 1\0, then 1 + zfrfr E M so M + Z = R(Z). It therefore follows that R(Z) is primitive. Let R(Y, Z) = (R(( Y)))(Z). have
Then by combining
the above two results we
LEMMA 1.3. Let R be a prime ring. Zf Y and Z are suflciently large, as functions of 1R 1,then R( Y, Z) is primitive.
In the remainder of this paper, when we deal with rings S 2 R we will choose Y and Z sufficiently large for S. They will then be sufficiently large for R and for all homomorphic images. With this understanding we will just write R*=R(Y,Z)
so that R* is a suitable power series-polynomial ring over R. Basic properties of this overring are listed in the next three lemmas. If A is a subset of R, we let A * = A( Y, Z) denote the set of all power series-polynomials with coefficients in A. LEMMA
1.4.
With the above notation we have
(i) Zf Z is an ideal of R, then I* is an ideal of R * with Z* n R = Z and (R/Z)* N R */I*. (ii) ZfAi are subsetsof R, then ((J Ai)* = n Ai*. (iii) If A and B are ideals of R, then A*B* c (AB)*. Hence A” = 0 implies (A *)” = 0.
PRIME IDEALS IN NORMALIZING
EXTENSIONS
559
(iv)
If P is a prime ideal of R, then P* is a primitive ideal of R”.
(v)
If P is a semiprimeideal of R, then P* is a semiprimitive ideal of
R*. Prooj
Parts (i). (ii) and (iii) are clear.
(iv) We have R*/P* E (R/P)* Lemma 1.3.
and this latter ring is primitive by
(v) Since P is a semiprime ideal, we have P = 0 Ai, an intersection of primes. Then P* = (-)A? and each AT is primitive so P* is semiprimitive. If Q is a prime ideal of R *, then it is not necessarily true that Qf7 R is prime in R. An example of this nature will be given in Section 4. However the result does hold if Q is also an annihilator ideal and indeed we have LEMMA 1.5. Let I be an ideal of R * with Z = l,*(J) for some subset J E R *. Then (I C7R )* 5 I. Furthermore, if I is prime, then In R is a prime ideal of R.
Proof. If r E In R, then rJ = 0. It then follows from the uniquenessof representation of elements in R * as power series-polynomials that r annihilates all coefficients of all elements of J. From this we deduce immediately that (I r7 R)* . J = 0 and hence that (In R)* c ZR*(J)= I. Suppose in addition that I is prime and let A, B be ideals of R with AB&InR. Then A*B* g (AB)” Y&(InR)*
c_z
by the above and hence either A* G I or B* E I. Thus either A g Zn R or BcInR. Finally we relate this idea to normalizing extensions. The following result is clear. LEMMA
1.6. Let S 2 R be an extension of rings.
(i) If S/R is a Jinite normalizing extension with normalizing set 1x13x2 ,.-.> xn}, then S*/R * is a jinite normalizing extension with the same normalizing set. (ii) In (i), if {x,, x2 ,..., x,} is a basisfor S/R (on the right or left or both), then {x, , x2 ,..., x, ] is a similar basisfor S*/R *. (iii) If S = RG is a crossedproduct with Gfinite, then S* = R *G. (iv)
IfIisanidealofSwithInR=O,thenZ*nR*=O.
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D.S.PASSMAN
2. IRREDUCIBLE
MODULES
In this section we assume that S = 5 Rxi = i xiR I 1 is a finite normalizing extension and we develop the usual Clifford relations between the irreducible right modules of S and of R. LEMMA 2.1. If V is an irreducible S-module, then V, = V, @ V, @ . . . @ Vk is a direct sum of k < n irreducible R-modules. Furthermore, if V, is faithful, then k > the number of minimal primes of R.
Proof. If U, E V\O, then V= v,S = C; u,x,R; so VR is finitely generated. Hence we can choose a maximal R-submodule W and for each i we set WX,~‘={vE VIvXiE W}. Since xiR = Rxi, it follows easily that Wx,:’ is an R-submodule of V. Next suppose that u E n Wx,: I. Then uxi E W for all i, SO US = C ux,R E WR < V and hence u = 0. Thus n Wx; I = 0. Observe that v --f vxi yields into map V/Wxi’ + V/W which is easily seen to respect R-submodules. Hence V/ Wx;’ is either zero or irreducible and reducible. Indeed V, = n Wx; ’ = 0 implies that VR is completely v, @ Vz@ ..- @ Vk with k < n. Finally suppose VR is faithful and let Qi = ann, Vi. Then Qi is a prime ideal of R and 0; Qi = 0. Thus the minimal primes of R are the minimal members of {Q, , Q*,..., Qk) and hence k 2 the number of minimal primes. There is actually more information in the above which we have not listed. Indeed the irreducible constituents Vi are “conjugate” R-modules and the annihilators Qi are “conjugate” ideals (see [2, 71). We now consider certain induced modules. If M is a right ideal of R, then MS is a right ideal of S and hence S/MS is an S-module induced from the R-module RIM. LEMMA 2.2. Let M be a maximal right ideal of R and let V be the Smodule V = S/MS. Then VR is completely reducible with composition length < n. Hence V has composition length 6 It.
Proof.
We have V = S/MS =
a sum of the n R-submodules
= c (Rxi + MS)/MS, Wi = (Rx, + MS)/MS.
Observe that r -+ rxi
PRIME
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yields an onto map (R + MS)/MS + @xi + MS)/MS which respects Rsubmodules. Since (R + MS)/MS N R/(R n MS) is either zero or R/M, we conclude that each Wi is either zero or irreducible. Thus V, = CT Wi is completely reducible of length ,< n. In the next few results we obtain conditions faithfully on S/MS. LEMMA 2.3. Suppose ann, R/M = J, then
which guarantee that S acts
{x,, x2 ,..., x,,] is a free left basis for S over R. If
annSS/MS c @ 5 Jxi. 1
In particular, if R acts faithfully faithfully on @ Cj S/MjS. Proof.
on the direct sum @ cj RIMi,
For the first part, let I = annSS/MS
IcMS=~MRx,=~Mx,. I
then S acts
so that
I
For each k, note that
I, =
I
r E R 1x rixi E I, r = rk I I
is a two-sided ideal of R and that Ik c M by the left freeness of the generators. Thus I, g J, since J is the largest two-sided ideal of R contained in M, and hence IZJS=@CJxi. Finally supposeR acts faithfully on @ z R/Mj SO that nj Jj = 0, where Jj = annRR/Mj. If I is the annihilator in S of @ xj SIMjS, then we have ICI @ CiJjxi for all j and the left freenessyields I = 0. For crossed products we have a slightly stronger result. LEMMA
2.4.
Let
S = RG
be a crossed product
with G finite.
If
annRR/M = J, then
Proo$ As in the previous proof, if I = annSS/MS, Moreover, if x E G, then since I = X-‘IX = I” we have
481/73/2-I9
then I c JS.
562
D.
S. PASSMAN
We conclude from the direct sum that ZC (n,,G J”) S = L. Conversely L is easily seen to be a two-sided ideal of S with L c JS G MS. Thus L G annS S/MS = I. The following proof is based on an argument shown to me by M. Lorenz. As will be apparent in the next section, the assumption on the existence of the ideal H is equivalent to Q being a minimal prime of R. LEMMA 2.5. Let S be prime and let Q be a prime ideal of R with Q = arm, R/M for some right ideal M of R. Suppose that there exists a nonzero two-sided ideal H of R (possibly R itself) with Q n H = 0. Then S
acts faithfully
on SIMS.
ProoJ Let x,, = 1 and set X = (x0, x, ,..., xn}. Choose Y s X of maximal size with respect to the property that Y is right free over H. That is, C, yh, = 0 with h, E H implies all h, = 0. Note that Y # 0 since (x0} is such a subset. Let YH= C,,,yH and observe that this is an (R, R)bimodule. For y E Y set T, = H so that certainly Ty is a nonzero two-sided ideal of R contained in H with yT, c YH. For x E X\Y note that YU {x} is not right free over H so there exists a relation xh + C,,, yh, = 0 with all coefficients in H and not all coefficients zero. Now h = 0 implies CyEy yh, = 0 and hence all h, = 0, a contradiction. Thus h # 0, xh = - JJ yh, E YH and hence T,={hEHlxhE
YH}
is a nonzero two-sided ideal of R contained in H. Since Q is prime and TX n Q = 0 for all x, we deduce that
T=
n T,#O. XEX
Moreover XT G YH for all x E X and hence
ST=xRxTcRYH= x
YH.
If Z = annSS/MS, then Z is a two-sided ideal of S contained in MS. The goal, of course, is to show that Z = 0. Observe that Z c MS implies that
ZTgMSTcMYHc
YH.
For any x E X, note that L, = l,(x) is a two-sided ideal of R. We set Y’ = { y E Y ( L, G Q} and Y” = Y\Y’. Fix z E Y’ and let
A=A,=
I
aEH
I
c YEY
yh,EZT,allh,EH,a=h,
PRIME
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IN NORMALIZING
Since IT and YH are (R, R)-bimodules, of R contained in H. Furthermore, let
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it is clear that A is a two-sided
ideal
B = B, = {b E R 1bz = za for some a E A }. Again B is a two-sided ideal of R and Bz 2 zA. We show below that B G Q. Suppose b E B. Then bz = zu for some a E A and hence there exists C yh, E IT with h, E H and h, = a. Furthermore we know that IT G MYH; so we can write x yh, = c myj yLyj Y Y..i a suitable finite sum with myj E M, h; E H. Since myjy = yr, ryj E R and since ryjKyj E H, we deduce from 1 Yh, = x myj Y YJ
yt;yj
for some
= x yr,jj;yj YJ
and the right freeness of Y over H that h, = cj ryj h;?i. In particular we have
for y = z
bz = za = zh, = x zrLjh; = 2 m,jzlizj = C rnzjFzjz, j j where zKzj = Fzjz and fzj E R. Thus since z E Y’, this yields
and hence we see that b E M. In other words, we have shown that B, = B c M and therefore B, G Q since Q is the largest two-sided ideal of R contained in M. Since IT& YH, we see that IT&
c
zA, + c
ZCY'
Moreover
YH.
YEY"
for z E Y’ we have, by the above, zA, E B, z G Qz.
Thus ITE
QS + c
YEY”
YR.
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D. S.PASSMAN
Now let L = Hf? n,,,,, L,. Since H & Q and each L, g Q for y E Y”, we have L !-?2Q and hence L # 0. Furthermore the above inclusion implies that LIT = 0 and hence
(SLS) I(STS) = 0. Finally
since S is prime and SLS, STS # 0 we conclude that Z = 0.
We close this section with a lovely result and proof of a somewhat different nature due to Bit-David and Robson. Observe that the hypothesis on the existence of H in the above lemma really asserts that Q is not essential as an (R, R)-subbimodule of R. The next lemma handles the essential case. LEMMA 2.6 [2]. Let Q be an ideal of R and suppose that Q intersects nontrivially with all nonzero ideals of R. Then there exists an ideal I of S with O#InRGQ.
ProoJ Here all modules considered are (R, R)-bimodules so the hypothesis on Q asserts that Q ess R. Let K be a maximal complement for R in S so that R 0 K ess S. It then follows that L = Q @ K ess S. For each i, j observe that x; ‘Lx,: ’ = {SESIXiSXjEL) is also an (R, R)-bimodule. Furthermore we see easily that xi ‘Lx,: I ess S. Indeed let U be a nonzero (R, R)-subbimodule of S. If xi Uxj = 0, then U G xi ‘LxJ: ’ and x; ‘Lx,: ’ n U # 0. On the other hand, if xi Uxj # 0, then since L ess S, we have xi Uxj n L # 0 and again x,: ‘Lxj ’ n U # 0. It now follows that
Z = f-) x; ‘Lx,: ’ ess S. i.i
Since L is an (R, R)-bimodule Z = (s E S 1SsS c L}. Thus Z is contained in L. Next, since Z E L Finally since I ess S, we have In
and S = 2 xiR = C Rxi, it is clear that clearly the largest two-sided ideal of S = Q @ K G R @ K, we see that Z n R 5 Q.
R # 0.
3. FINITE NORMALIZING EXTENSIONS We now combine the ideas of the previous sections to obtain the main results. Again we assume that S = c; Rxi = 2: xiR is a finite normalizing extension. If I is an ideal of S, let -: S -+ S/I denote the natural map. Then S= 2: Rzi and ki = fiR; so S is a finite normalizing extension of the ring
PRIME IDEALS IN NORMALIZING
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E = (R + Z)/Z N R/(R n Z). We use the notation of Section 1. In particular, Lemma 1.6(i) implies that S* = 2: R*xi is also a finite normalizing extension. The first result was obtained independently by Lorenz and Bit-David and Robson.
Zf P is (I prime ideal of S, then P n R = fit Qi is an intersection of k < n primes of R. THEOREM
Prooj
3.1 (Cutting Down [2, 71).
We can assumethat P = 0, so that S is prime.
Case 1. S primitive. Say S acts faithfully on the irreducible module V. Then R acts faithfully on V, = V, @ V, @ . .. 0 I’,, a direct sum of k < n irreducible modules by Lemma 2.1. If Qi = ann, Vi, then each Qi is prime and n Qi = annR V = 0.
Case 2. S prime. Since S is prime we have S* 3 R* and S* is primitive by Lemma 1.4(iv). Thus, by the above, there are k < n minimal primes Qi of R * with n Qi = 0. If N, = njti Qj, it follows easily that Qi = l&Ni) so that Qi is also an annihilator ideal. Thus, by Lemma 1.5, each Qin R is prime in R. Since 0: (Qin R) = 0, the theorem is proved. We remark that there is actually more to say in the above. Namely, the primes Qi are in fact all “conjugate” and the rings R/Qi are all isomorphic. This can be found in [2, 71 and can be shown here by way of a strengthening of Lemma 2.1. Observe that Cutting Down implies that if P is a prime ideal of S, then there are at most n primes of R minimal over Pn R. Conversely, Lying Over asserts that each prime Q of R is minimal over some Pn R. This result, due to Bit-David and Robson, appears as the lower bound in the next theorem and the proof given here is precisely theirs. On the other hand, the upper bounds in the next two theorems can be found in the work of Heinicke and Robson and were also obtained at the sametime in this paper. THEOREM 3.2 (Lying Over [ 2,6]). Let Q be a prime ideal of R and let K be the number of primes P of S with Q minimal over P n R. Then 1 < K < n.
ProoJ: We first obtain the lower bound 1 < K following [2]. To this end, let P be an ideal of S maximal with respect to the property that P n R c Q. Since Q is prime, it is easy to verify that P is prime. The goal is to show that Q is minimal over Pn R and for this it clearly sufftces to study S/P. In other words we may assumethat P = 0 and that every nonzero ideal Z of S satisfies Zn R G Q. It now follows from Lemma 2.6 that there exists an ideal H#OofRwithQ~H=O.Finally,byCuttingDown,O=P~R=~~Q,., a finite intersection of minimal primes of R. Since H# 0 there exists i with
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D. S. PASSMAN
Qi~H.ButthenQi2O=QnH,soQi~QandQisaminimalprimeofR. Thus 1 < K and we now consider the upper bound. Case 1. Q primitive. Let M be a maximal right ideal of R with Q = ann, R/M and consider the induced S-module V= S/MS. By Lemma 2.2, V has composition length < n. Let P be a prime ideal of S with Q minimal over P--n R and let : S + S/P denote - -- the natural map. Then S + S and MS -+ MS are onto and hence SIMS is a--homomorphic image of S/MS. Moreover since Pn R G Q c M, we have R/Q-R/Q and hence M is a maximal right ideal of R with -Q = ann,-R/M. By Cutting Down, it follows that there exists a nonzero ideal Z? of R with en E? = 0. Hence since - -- S is prime, we conclude - -- from Lemma 2.5 that S acts faithfully on S/MS. Furthermore since S/MS has finite composition length and S is prime, S acts faithfully on a composition - -factor of S/MS and hence on a composition factor of S/MS. Thus P is the annihilator in S of a composition factor of V= S/MS and since there are at most il such factors, this case follows. Case 2. Q prime. We have S* 2 R * and Q* is a primitive ideal by Lemmas 1.4(iv) and 1.6. Suppose Q is minimal over Pn R. By Cutting Down, Pn R = 0: Qi, a finite intersection of incomparable primes with say Q = Q,. Then clearly P* n R* = 0: QT, so we conclude easily that Q* is minimal over P* n R *. Since there are at most n primes of S* lying over the primitive ideal Q*, and since P* n S = P, the result follows. Two remarks are now in order. First, we have actually shown in the above two results that if Q is minimal over P n R, then Q is primitive if and only if P is primitive. Second, since homomorphic images of finite normalizing extensions are also finite normalizing extensions, Lying Over clearly yields a suitable Going Up theorem. THEOREM 3.3 (Lying Over Zero [6]). Let J be an ideal of R, let k denote the number of minimal primes of R/J and let K denote the number of primes P of S with P n R = J. Then krc < n.
Proof. We can clearly assume that K # 0, so there exists a prime P, with P, n R = J. Set Z = SJS so that Z n R c J. On the other hand, Z C P for any primePwithPnR=JandthusJ=P,nR~ZnR.HenceZnR=Jand we may now clearly work in S/Z. Therefore it suffices to assume that J= 0. Cutting Down now implies that 0 = P, n R = f-): Qi, an intersection of k < n minimal primes of R. Set Q = Q, and H = fl: Qi so that H# 0 and QnH=O.
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Case 1. Q primitive. Let M be a maximal right ideal of R with Q = annR R/M and consider the induced module V= S/MS. By Lemma 2.2, VR has composition length < n. Suppose V,, V, ,..., Vj are the composition factors of V such that (Vi)R is faithful. By Lemma 2.1, each (Vi& has composition length > k. Thus jk < n. If Pi = annS Vi, then Pi is a prime ideal of S and Pi n R = 0 since (Vi)R is faithful. It clearly suffices to show that these are precisely all the primes lying over zero. Let P be a prime of S with P n R = 0 and-- let ~: S + S/P be- the-- natural homomorphism. Then S-+ S and MS + MS are onto so S/MS is a homomorphic image of SIMS. - Since - P n R --= 0, we - have R N R and hence I%?is a maximal right ideal of R, -Q -= ann,R/M, H # 0 and Qn B = 0. By Lemma 2.5, S acts faithfully on S/MS. - --
Finally since S/MS has finite composition length - -- and S is prime, S acts faithfully on some composition factor of S/MS. Thus there exists a composition factor W of S/MS such that annS W = P. Moreover Pn R = 0, so W, is faithful and hence W = Vi for some i and P = arms Vi = Pi.
Case 2.
Q prime. have S* 2 R*, Q* is primitive, H* #O and Q*n H* =O. Furthermore each Ql is prime, n’; QT = 0 and the Qif: are clearly all incomparable. Thus R * has precisely k minimal primes. Observe that if P,, P,,... are primes of S with Pi f7 R = 0, then PI*, PT,... are distinct primes of S* with Pi” n R* = 0. It now follows from Case 1 that there are at most n/k such P,? and hence at most n/k such Pi. We
If X, ) x2 )...) x, are R-free on the left or right, then more can be said about S. We start a result of Lorenz and Passman on nilpotent ideals. THEOREM 3.4 (Nilpotent Ideals [lo]). Let R be a semiprime ring and assume that x,, x2 ,..., x, are R-free on the left (or right). Then S has a unique maximal nilpotent ideal T and T” = 0.
Proof T” = 0.
It clearly suffices to show that if T is a nilpotent ideal of S, then
Case 1. R semiprimitive. If (Mj} is the set of all maximal right ideals of R, then R acts faithfully on @ Cj RIMi. Thus, by Lemma 2.3, S acts faithfully on @ Cj S/M,S. By Lemma 2.2, each S/M,S has composition length < n and hence if JS is the Jacobson radical of S, then (JS)” kills each S/M,S. Thus (JS)” = 0 and since T E JS we have T" = 0. Case 2. R semiprime. We have S* GJR* and x, ,x2 ,..., x, are R *-free on the left, by Lemma 1.6(i, ii). Moreover R* is semiprimitive, by Lemma 1.4(v). Finally T
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D. S.PASSMAN
is nilpotent, so T* is nilpotent and hence (T*)” = 0 by the above. Since T E T* this yields T” = 0. Next we have a structure theorem due to Lorenz. THEOREM 3.5 (Wedderburn-type [7]). Let R be a prime ring and assume that x, ,x2 ,..., x, are R-free on the left. Then S has Jnitely many minimal primes P, , P, ,..., P, with k
Proof: Case 1. R primitive. Let M be a maximal right ideal of R such that R acts faithfully on R/M. Then by Lemmas 2.2 and 2.3, S acts faithfully on V= S/MS and V has composition length < n. Let P,, P, ,..., Pj with j < n be the annihilators of the composition factors of V. Then each Pi is prime and T = n’, Pj annihilates all such factors so T” = 0. Finally assumethat T = 0: Pi is an irredundant intersection with kR* and by Lemmas 1.6(ii) and 1.4(iv) we know that x,, x2,..., x, are R*-free on the left and that R* is primitive. Hence, by Case 1, S* has the appropriate structure. Now according to the previous theorem S has a unique maximal nilpotent ideal T with T” = 0 and certainly T is a semiprime ideal of S. Thus (T”)” = 0 and T* is a semiprimitive ideal of S*, by Lemma 1.4(v). Hence T* is clearly the unique largest nilpotent ideal of S*. This implies that S*/T* ‘v (S/T)* has finitely many minimal primes P,, P, ,..., P, with k
Proof: We know that S = RG has the set {X 1x E G} as a free basis. In particular S is a finite normalizing extension of R with n, the size of the normalizing set, equal to )G ]. In addition, we know that R is G-prime if and only if there exists a prime ideal Q of R with n,.G Q” = 0. Case 1. Q primitive. Let M be a maximal right ideal of R with Q = annRR/M. Then S acts
PRIME IDEALS IN NORMALIZING
EXTENSIONS
569
faithfully on V= S/MS by Lemma 2.4. The argument of Case 1 of the preceding theorem now yields the result. Case 2. Q prime. We have S* = R *G, by Lemma 1.6(iii), and Q* is primitive, by Lemma 1.4(iv). Furthermore it is clear that (Q*)” = (Q”)* and hence that flxEG (Q*)‘= 0. Th us we conclude from Case 1 that S* has the appropriate structure. Moreover R is semiprime, so S = RG has a unique maximal nilpotent ideal T by Theorem 3.4. The argument of Case 2 of the previous theorem now completes the proof. Finally we prove a theorem of Fisher and Montgomery on crossed products. THEOREM 3.1 (Maschke-type 131). Let RG be a crossedproduct of the finite group G over the semiprimering R. If R has no 1G j-torsion then RG is semiprime.
Proof: Let S = RG. By forming the extension S[ l/]G]] = R [ l/]G]] can clearly assumethat l/]G] E R.
G, we
Case 1. R semiprimitive. If (Mj} is the set of all maximal right ideals of R, then R acts faithfully on @ Cj R/kfj. Thus, by Lemma 2.3, S acts faithfully on V= @ cj S/M,S. Furthermore, VR is completely reducible, by Lemma 2.2, so we conclude from Maschke’s theorem (see for example [ 10, Lemma 11) that V is also completely reducible. Hence, if JS denotes the Jacobson radical of S, then V(JS) = 0; so JS = 0. In other words, S is semiprimitive and hence also semiprime. Case 2. R semiprime. We have S* = R*G, by Lemma 1.6(iii), and R* is semiprimitive, by Lemma 1.4(v). Hence by the above, S* is semiprime and, by Lemma 1.4(iii), so is S. 4. EXAMPLES In this section we discusssomeinteresting and relevant examples. The first concerns Lemma 1.5 and offers an example of a prime ideal P of R* with P 17R not prime and in fact not even semiprime. Observe that by setting all but one variable in R * = R(Y, Z) equal to zero, we can obtain an Rhomomorphism of R* onto the ordinary power series ring R[ [t;]]. Thus it sufftces to find a prime ideal P of R [[Cl] with P n R not semiprime. To this end, we have the following argument of G. Bergman. We thank Professor Bergman for allowing it to be included here.
D.S.PASSMAN
570
LEMMA 4.1 [l]. Let Q, s QZE -.a be an ascending chain of prime ideals in a ring R. Then P= U, Q,[ [Cl] is a prime ideal of R[ [[I].
ProoJ Here of course Q,[ [cl] is the ideal of all power serieswith coefficients in Q,. Now supposea and p are elements of R [[[I] not contained in P= U, Q,[Kll. We construct inductively a power series
f = 5 riccci) i=l
such that for all n > 1 (1)
a(Cr=, ricCi)) p has a term sncd(n)with s, 6$Q,,
(2)
c(n + 1) > d(n), c(n).
Suppose we have already found Cr=, rirti) for some n > 0 with the understanding that this expression is zero for n = 0. Choose any integer c(n + 1) larger than d(n) and c(n). There are now two casesto consider. If a(Cy=, ri@)P @ Q,+,[ [[I], take r,,+, = 0. Then certainly
has a term So+i cd@’ i) with s, + i 6ZQ, + i . On the other hand, supposethat a(Cy,, ri4@))/3E Q,+,[[[]]. Then for any r,,, E R we have
Since a, p 66Q,, , [ [[]I and Q, + i is prime, we can surely find r,+ , with r(“+‘)/?& Q,+,[[[]]. Indeed, we can do this by considering the terms in a and in /I of lowest degree not contained in Q,, , and use the primeness ofQnt,- With this choice of r,, +1E R, the induction step is proved. It now follows easily from (1) and (2) that the coefficient of cd(“) in afl is s, @Q,. Hence @P @Q,[[C]] and since this holds for all n we conclude that ajJ3@P. Therefore P is prime.
artIt
EXAMPLE
4.2 [l].
A prime ideal P of R [ [t;]] with P n R not semiprime.
Proof. Let R = k(x, y) be the free associative algebra over the field k in the noncommuting variables x and y. For each n > 1, let Q, be the ideal of R generated by xx, xyx, xy*x ,..., xy”-’ x. Then Q, is prime since a,P & Q, implies easily that ay”/3 6?Q,. Furthermore Q, + ,Z Q, and Q = (j Q, = (RxR)* n
PRIME IDEALS IN NORMALIZING
EXTENSIONS
571
so that Q is not semiprime. Finally, by Lemma 4.1, P = U, Q,[ [[I] is a prime ideal of R [ [[I]. Since P n R = U, Q, = Q, we conclude that P n R is not semiprime. The remaining examples all involve extensions constructed in the following simple manner. LEMMA 4.3. Let I and J be ideals of R with In J = 0 and set R’ = R/I, R” = R/J. Then S = R’ OR” is a finite centralizing extension of R, with R embedded diagonally. Furthermore, the minimal primes of S are given by P’ @RI’ and R’ 0 Q” where P’ and Q” are minimal primes of R’ and R”, respectively.
ProoJ Since Z n J = 0, the map r + (r’, r”) embedsR into S with (I’, 0) and (0, 1”) a centralizing set of generators. With J = 0 and varying R and 1, several examples are now easily given. EXAMPLE 4.4. (i) Let R be a primitive ring acting faithfully on R/M, with A4 a maximal right ideal and assumethat R is not simple. If I is any nonzero ideal of R, then M+I=R so M’=R’ and MS = R ’ ClJM” 1 R’ @ 0. Hence S does not act faithfully on S/MS (compare with Lemmas 2.3 and 2.5).
(ii) Let R be a commutative integral domain with ideal I such that R/Z has no maximal nilpotent ideal. Then clearly S has no maximal nilpotent ideal (compare with Theorem 3.4). (iii) Let R be prime and let I be a nonminimal prime ideal of R. Then 0 @ R ” is a minimal prime of S which lies over I. Moreover R ’ @ 0 is the only minimal prime of S lying over zero, so the intersection of all such is not nilpotent (compare with Theorem 3.5). (iv) Let R be a commutative integral domain such that R/I has infinitely many minimal primes. Then S has infinitely many minimal primes, while R has the unique minimal prime zero (compare with Theorem 3.5). We close this paper with a remark of a somewhat different nature. Let G be a group acting as automorphisms on a ring R. Then certainly G acts on R * and for the fixed rings we have clearly (R*)G = (RG)*. Thus the “primitivity machine” may also have applications to the study of rings with finite group actions. Indeed it can be used to obtain some of the basic relations between the primes of R and of RG, but these have already all been proved by Montgomery [ 121, from skew group ring results, in a quite satisfactory manner.
572
D. S. PASSMAN REFERENCES
1. G. M. BERGMAN, Chains of prime ideals in R and prime ideals in R[[t]], unpublished. 2. .I. BIT-DAVID AND J. C. ROBSON, Normalizing extensions, I, in “Proc. Ring Theory Conference 1980),” Springer Lecture Notes, Springer-Veilag, (Antwerp, Berlin/Heidelberg/New York, 1980. 3. J. FISHER AND S. MONTGOMERY, Semiprime skew group rings, J. Algebra 52 (1978), 241-247..
E. FORMANEK, Group rings of free products are primitive, J. Algebra 26 (1973), 508-5 11. 5. D. HANDELMAN AND J. LAWRENCE, Strongly prime rings, Trans. Amer. Math. Sot. 211 4.
(1975),
209-233.
6. A. G. HEINICKE AND J. C. ROBSON, Normalizing extensions: prime ideals and incomparability, to appear. 7. M. LORENZ, Finite normalizing extensions of rings, Math. Z. 176 (198 I), 447-484. 8. M. LORENZ AND D. S. PASSMAN, Prime ideals in crossed products of finite groups, Israel J. Math. 33 (1979), 89-132. 9. M. LORENZ AND D. S. PASSMAN, Integrality and normalizing extensions, J. Algebra 61 (1979),
289-297.
10. M. LORENZ AND D. S. PASSMAN, Two applications of Maschke’s theorem, Comm. Algebra 8 (1980), 1853-1866. Il. M. LORENZ AND D. S. PASSMAN, Addendum-Prime ideals in crossed products of finite groups, Israel J. Math., in press. 12. S. MONTGOMERY, Prime ideals in fixed rings, Comm. Algebra 9 (1981), 423449. 13. J. C. ROBSON,Some results on ring extensions, Lecture Notes, University of Essen, 1979.
OF ALGEBRA
73, 556-572
Prime
Ideals
(1981)
in Normalizing
Extensions*
D. S. PASSMAN University
of Wisconsin-Madison,
Madison,
Communicated Received
Wisconsin
53706
by I. N. Hernstein September
15, 1980
In this paper we offer a new approach to the study of prime ideals in finite normalizing extensions of rings. We introduce a “primitivity machine” which translates certain problems on prime ideals to analogous questions on primitive ideals. We then use the well-known Clifford theory of irreducible modules to study these primitive ideals.
Let S 2 R be an extension of rings with the same 1. We say that S is a finite normalizing extension of R if there exists x,, x2,..., x, E S with S = C; Rxi and with Rxj = XjR for all j. Note that the normalizing set {Xl, X2,“‘, x,} need not consist of units and need not be independent over R. Examples of such extensions are the finite centralizing extensions (also called liberal extensions) where each xi centralizes R (see [ 131) and crossed products S = RG, where G is a finite group (see [8]). Furthermore there is the commutative situation where S is a finite integral extension of R. Prime ideals in the commutative extensions have, of course, long been studied and in particular one is aware of the classical Krull relations of Lying Over, Going Up, Incomparability, etc. Furthermore analogs of these relations also hold in finite centralizing extensions [ 131 and in crossed products of finite groups [8, 9, 111. Recently, prime ideals in the general case of finite normalizing extensions have been studied by Bit-David and Robson [2], Lorenz [7] and Heinicke and Robson [6]. From this work it is apparent that the really deep and difficult relation is Incomparability. This was proved in part by Lorenz and finally obtained in the definitive paper of Heinicke and Robson. In ‘this manuscript we offer a completely different approach to this subject. For a ring R, we show that there is a suitable power series-polynomial ring R * over R in which the prime ideals of R extend to primitive ideals of R *. Furthermore, if S 2 R is a finite normalizing extension, then so is S * 2 R *, and using the classical Clifford theory we study the relationship between the * Research
supported
in part by NSF
Grant
No. MCS
556 0021.8693/81/120556-17$02.00/O Copyright All rights
0 198 1 by Academic Press, Inc. of reproduction in any form reserved.
77-01775.
PRIME IDEALS INNORMALIZINGEXTENSIONS
551
irreducible modules of S* and of R *. This then yields information comparing the primitive ideals of S* and of R* and then the prime ideals of S and of R. In this way, with the help of a result from [2], we are able to obtain all the basic Krull relations on primes with the exception of the difficult Incomparability. We also obtain certain structure theorems which hold for crossed products and normalizing extensions with a free basis. This paper is written in a reasonably self-contained manner. Because of this,, many of the lemmas and most of the theorems are known. We take credit only for the approach to the problem. 1. POWER SERIES-POLYNOMIALS In this section we develop the “primitivity machine” which translates prime ideals to primitives. If Y is a set of variables, let R((Y)) denote the formal power series ring over the ring R in the noncommuting variables Y. Since any word in the semigroup generated by Y has only finitely many initial and final segments, it is clear that multiplication in R((Y)) is defined. LEMMA 1.1. Let R be a prime ring and assume that 1Y( > IR 1, an equation of cardinalities. Then there exists a regular element f E R(( Y)) such that fj is regular for all nonzero g E R( (Y)).
ProoJ {y,jrER}.
Since ] Y 1> 1R 1, we can label some of the variables in Y as Let
f = & ry, E R(V)). We claim that f is the required element. To this end, by symmetry, it suffices to show that g, h # 0 implies fgF # 0. Let a be a monomial of shortest length in the support of g and let p be one of shortest length in h. Say g=sa+ ..* and h = t,LI+ .. - with s, t E R\O. Since R is prime, there exists r E R with srt # 0. It is then easy to see that
fgF = (lsrt) y, cry,/? + ... and that y,a~~,.p is a uniquely represented element. Hence fgfh # 0. If Z is a set of variables, let R(Z) denote the polynomial ring over R in the noncommuting variables Z. The following is based on ideas of Formanek (41 and is a special case of a result of [5]. LEMMA 1.2. Let R be a ring and supposethat R has a regular element r such that rsr is regular for all s E R \O. Let Z be an ifinite set of variables with IZI > (R (. Then R(Z) is primitive.
558
D. S.PASSMAN
Proof: Clearly IZI = /R(Z)1 so to each fE R(Z)\0 distinct variable .z,E Z. We claim that the right ideal
N=
we can associate a
2 (1 +zrrfr>R(Z) f#O
is proper. If not, we have a finite sum 1 =
c f#O
(1
+zfrfrkf
and we can choosefoccurring in the above so that degf+ deg gf is maximal. Let a be a monomial of largest degree inf, /3 one of largest degree in gf and sayf=sa+... andgf=tP+.... Then, by properties of r, zrrfirg, contains the nonzerp term (rsrt) zraj3 and zfap is clearly a uniquely represented element in the whole right-hand expression C (1 + z&) gr. Thus N # R(Z). Now choose A4 a maximal right ideal of R(Z) with A4 2 N. We claim that M is comaximal with all nonzero ideals I. Indeed if fE 1\0, then 1 + zfrfr E M so M + Z = R(Z). It therefore follows that R(Z) is primitive. Let R(Y, Z) = (R(( Y)))(Z). have
Then by combining
the above two results we
LEMMA 1.3. Let R be a prime ring. Zf Y and Z are suflciently large, as functions of 1R 1,then R( Y, Z) is primitive.
In the remainder of this paper, when we deal with rings S 2 R we will choose Y and Z sufficiently large for S. They will then be sufficiently large for R and for all homomorphic images. With this understanding we will just write R*=R(Y,Z)
so that R* is a suitable power series-polynomial ring over R. Basic properties of this overring are listed in the next three lemmas. If A is a subset of R, we let A * = A( Y, Z) denote the set of all power series-polynomials with coefficients in A. LEMMA
1.4.
With the above notation we have
(i) Zf Z is an ideal of R, then I* is an ideal of R * with Z* n R = Z and (R/Z)* N R */I*. (ii) ZfAi are subsetsof R, then ((J Ai)* = n Ai*. (iii) If A and B are ideals of R, then A*B* c (AB)*. Hence A” = 0 implies (A *)” = 0.
PRIME IDEALS IN NORMALIZING
EXTENSIONS
559
(iv)
If P is a prime ideal of R, then P* is a primitive ideal of R”.
(v)
If P is a semiprimeideal of R, then P* is a semiprimitive ideal of
R*. Prooj
Parts (i). (ii) and (iii) are clear.
(iv) We have R*/P* E (R/P)* Lemma 1.3.
and this latter ring is primitive by
(v) Since P is a semiprime ideal, we have P = 0 Ai, an intersection of primes. Then P* = (-)A? and each AT is primitive so P* is semiprimitive. If Q is a prime ideal of R *, then it is not necessarily true that Qf7 R is prime in R. An example of this nature will be given in Section 4. However the result does hold if Q is also an annihilator ideal and indeed we have LEMMA 1.5. Let I be an ideal of R * with Z = l,*(J) for some subset J E R *. Then (I C7R )* 5 I. Furthermore, if I is prime, then In R is a prime ideal of R.
Proof. If r E In R, then rJ = 0. It then follows from the uniquenessof representation of elements in R * as power series-polynomials that r annihilates all coefficients of all elements of J. From this we deduce immediately that (I r7 R)* . J = 0 and hence that (In R)* c ZR*(J)= I. Suppose in addition that I is prime and let A, B be ideals of R with AB&InR. Then A*B* g (AB)” Y&(InR)*
c_z
by the above and hence either A* G I or B* E I. Thus either A g Zn R or BcInR. Finally we relate this idea to normalizing extensions. The following result is clear. LEMMA
1.6. Let S 2 R be an extension of rings.
(i) If S/R is a Jinite normalizing extension with normalizing set 1x13x2 ,.-.> xn}, then S*/R * is a jinite normalizing extension with the same normalizing set. (ii) In (i), if {x,, x2 ,..., x,} is a basisfor S/R (on the right or left or both), then {x, , x2 ,..., x, ] is a similar basisfor S*/R *. (iii) If S = RG is a crossedproduct with Gfinite, then S* = R *G. (iv)
IfIisanidealofSwithInR=O,thenZ*nR*=O.
560
D.S.PASSMAN
2. IRREDUCIBLE
MODULES
In this section we assume that S = 5 Rxi = i xiR I 1 is a finite normalizing extension and we develop the usual Clifford relations between the irreducible right modules of S and of R. LEMMA 2.1. If V is an irreducible S-module, then V, = V, @ V, @ . . . @ Vk is a direct sum of k < n irreducible R-modules. Furthermore, if V, is faithful, then k > the number of minimal primes of R.
Proof. If U, E V\O, then V= v,S = C; u,x,R; so VR is finitely generated. Hence we can choose a maximal R-submodule W and for each i we set WX,~‘={vE VIvXiE W}. Since xiR = Rxi, it follows easily that Wx,:’ is an R-submodule of V. Next suppose that u E n Wx,: I. Then uxi E W for all i, SO US = C ux,R E WR < V and hence u = 0. Thus n Wx; I = 0. Observe that v --f vxi yields into map V/Wxi’ + V/W which is easily seen to respect R-submodules. Hence V/ Wx;’ is either zero or irreducible and reducible. Indeed V, = n Wx; ’ = 0 implies that VR is completely v, @ Vz@ ..- @ Vk with k < n. Finally suppose VR is faithful and let Qi = ann, Vi. Then Qi is a prime ideal of R and 0; Qi = 0. Thus the minimal primes of R are the minimal members of {Q, , Q*,..., Qk) and hence k 2 the number of minimal primes. There is actually more information in the above which we have not listed. Indeed the irreducible constituents Vi are “conjugate” R-modules and the annihilators Qi are “conjugate” ideals (see [2, 71). We now consider certain induced modules. If M is a right ideal of R, then MS is a right ideal of S and hence S/MS is an S-module induced from the R-module RIM. LEMMA 2.2. Let M be a maximal right ideal of R and let V be the Smodule V = S/MS. Then VR is completely reducible with composition length < n. Hence V has composition length 6 It.
Proof.
We have V = S/MS =
a sum of the n R-submodules
= c (Rxi + MS)/MS, Wi = (Rx, + MS)/MS.
Observe that r -+ rxi
PRIME
IDEALS
IN
NORMALIZING
561
EXTENSIONS
yields an onto map (R + MS)/MS + @xi + MS)/MS which respects Rsubmodules. Since (R + MS)/MS N R/(R n MS) is either zero or R/M, we conclude that each Wi is either zero or irreducible. Thus V, = CT Wi is completely reducible of length ,< n. In the next few results we obtain conditions faithfully on S/MS. LEMMA 2.3. Suppose ann, R/M = J, then
which guarantee that S acts
{x,, x2 ,..., x,,] is a free left basis for S over R. If
annSS/MS c @ 5 Jxi. 1
In particular, if R acts faithfully faithfully on @ Cj S/MjS. Proof.
on the direct sum @ cj RIMi,
For the first part, let I = annSS/MS
IcMS=~MRx,=~Mx,. I
then S acts
so that
I
For each k, note that
I, =
I
r E R 1x rixi E I, r = rk I I
is a two-sided ideal of R and that Ik c M by the left freeness of the generators. Thus I, g J, since J is the largest two-sided ideal of R contained in M, and hence IZJS=@CJxi. Finally supposeR acts faithfully on @ z R/Mj SO that nj Jj = 0, where Jj = annRR/Mj. If I is the annihilator in S of @ xj SIMjS, then we have ICI @ CiJjxi for all j and the left freenessyields I = 0. For crossed products we have a slightly stronger result. LEMMA
2.4.
Let
S = RG
be a crossed product
with G finite.
If
annRR/M = J, then
Proo$ As in the previous proof, if I = annSS/MS, Moreover, if x E G, then since I = X-‘IX = I” we have
481/73/2-I9
then I c JS.
562
D.
S. PASSMAN
We conclude from the direct sum that ZC (n,,G J”) S = L. Conversely L is easily seen to be a two-sided ideal of S with L c JS G MS. Thus L G annS S/MS = I. The following proof is based on an argument shown to me by M. Lorenz. As will be apparent in the next section, the assumption on the existence of the ideal H is equivalent to Q being a minimal prime of R. LEMMA 2.5. Let S be prime and let Q be a prime ideal of R with Q = arm, R/M for some right ideal M of R. Suppose that there exists a nonzero two-sided ideal H of R (possibly R itself) with Q n H = 0. Then S
acts faithfully
on SIMS.
ProoJ Let x,, = 1 and set X = (x0, x, ,..., xn}. Choose Y s X of maximal size with respect to the property that Y is right free over H. That is, C, yh, = 0 with h, E H implies all h, = 0. Note that Y # 0 since (x0} is such a subset. Let YH= C,,,yH and observe that this is an (R, R)bimodule. For y E Y set T, = H so that certainly Ty is a nonzero two-sided ideal of R contained in H with yT, c YH. For x E X\Y note that YU {x} is not right free over H so there exists a relation xh + C,,, yh, = 0 with all coefficients in H and not all coefficients zero. Now h = 0 implies CyEy yh, = 0 and hence all h, = 0, a contradiction. Thus h # 0, xh = - JJ yh, E YH and hence T,={hEHlxhE
YH}
is a nonzero two-sided ideal of R contained in H. Since Q is prime and TX n Q = 0 for all x, we deduce that
T=
n T,#O. XEX
Moreover XT G YH for all x E X and hence
ST=xRxTcRYH= x
YH.
If Z = annSS/MS, then Z is a two-sided ideal of S contained in MS. The goal, of course, is to show that Z = 0. Observe that Z c MS implies that
ZTgMSTcMYHc
YH.
For any x E X, note that L, = l,(x) is a two-sided ideal of R. We set Y’ = { y E Y ( L, G Q} and Y” = Y\Y’. Fix z E Y’ and let
A=A,=
I
aEH
I
c YEY
yh,EZT,allh,EH,a=h,
PRIME
IDEALS
IN NORMALIZING
Since IT and YH are (R, R)-bimodules, of R contained in H. Furthermore, let
563
EXTENSIONS
it is clear that A is a two-sided
ideal
B = B, = {b E R 1bz = za for some a E A }. Again B is a two-sided ideal of R and Bz 2 zA. We show below that B G Q. Suppose b E B. Then bz = zu for some a E A and hence there exists C yh, E IT with h, E H and h, = a. Furthermore we know that IT G MYH; so we can write x yh, = c myj yLyj Y Y..i a suitable finite sum with myj E M, h; E H. Since myjy = yr, ryj E R and since ryjKyj E H, we deduce from 1 Yh, = x myj Y YJ
yt;yj
for some
= x yr,jj;yj YJ
and the right freeness of Y over H that h, = cj ryj h;?i. In particular we have
for y = z
bz = za = zh, = x zrLjh; = 2 m,jzlizj = C rnzjFzjz, j j where zKzj = Fzjz and fzj E R. Thus since z E Y’, this yields
and hence we see that b E M. In other words, we have shown that B, = B c M and therefore B, G Q since Q is the largest two-sided ideal of R contained in M. Since IT& YH, we see that IT&
c
zA, + c
ZCY'
Moreover
YH.
YEY"
for z E Y’ we have, by the above, zA, E B, z G Qz.
Thus ITE
QS + c
YEY”
YR.
564
D. S.PASSMAN
Now let L = Hf? n,,,,, L,. Since H & Q and each L, g Q for y E Y”, we have L !-?2Q and hence L # 0. Furthermore the above inclusion implies that LIT = 0 and hence
(SLS) I(STS) = 0. Finally
since S is prime and SLS, STS # 0 we conclude that Z = 0.
We close this section with a lovely result and proof of a somewhat different nature due to Bit-David and Robson. Observe that the hypothesis on the existence of H in the above lemma really asserts that Q is not essential as an (R, R)-subbimodule of R. The next lemma handles the essential case. LEMMA 2.6 [2]. Let Q be an ideal of R and suppose that Q intersects nontrivially with all nonzero ideals of R. Then there exists an ideal I of S with O#InRGQ.
ProoJ Here all modules considered are (R, R)-bimodules so the hypothesis on Q asserts that Q ess R. Let K be a maximal complement for R in S so that R 0 K ess S. It then follows that L = Q @ K ess S. For each i, j observe that x; ‘Lx,: ’ = {SESIXiSXjEL) is also an (R, R)-bimodule. Furthermore we see easily that xi ‘Lx,: I ess S. Indeed let U be a nonzero (R, R)-subbimodule of S. If xi Uxj = 0, then U G xi ‘LxJ: ’ and x; ‘Lx,: ’ n U # 0. On the other hand, if xi Uxj # 0, then since L ess S, we have xi Uxj n L # 0 and again x,: ‘Lxj ’ n U # 0. It now follows that
Z = f-) x; ‘Lx,: ’ ess S. i.i
Since L is an (R, R)-bimodule Z = (s E S 1SsS c L}. Thus Z is contained in L. Next, since Z E L Finally since I ess S, we have In
and S = 2 xiR = C Rxi, it is clear that clearly the largest two-sided ideal of S = Q @ K G R @ K, we see that Z n R 5 Q.
R # 0.
3. FINITE NORMALIZING EXTENSIONS We now combine the ideas of the previous sections to obtain the main results. Again we assume that S = c; Rxi = 2: xiR is a finite normalizing extension. If I is an ideal of S, let -: S -+ S/I denote the natural map. Then S= 2: Rzi and ki = fiR; so S is a finite normalizing extension of the ring
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E = (R + Z)/Z N R/(R n Z). We use the notation of Section 1. In particular, Lemma 1.6(i) implies that S* = 2: R*xi is also a finite normalizing extension. The first result was obtained independently by Lorenz and Bit-David and Robson.
Zf P is (I prime ideal of S, then P n R = fit Qi is an intersection of k < n primes of R. THEOREM
Prooj
3.1 (Cutting Down [2, 71).
We can assumethat P = 0, so that S is prime.
Case 1. S primitive. Say S acts faithfully on the irreducible module V. Then R acts faithfully on V, = V, @ V, @ . .. 0 I’,, a direct sum of k < n irreducible modules by Lemma 2.1. If Qi = ann, Vi, then each Qi is prime and n Qi = annR V = 0.
Case 2. S prime. Since S is prime we have S* 3 R* and S* is primitive by Lemma 1.4(iv). Thus, by the above, there are k < n minimal primes Qi of R * with n Qi = 0. If N, = njti Qj, it follows easily that Qi = l&Ni) so that Qi is also an annihilator ideal. Thus, by Lemma 1.5, each Qin R is prime in R. Since 0: (Qin R) = 0, the theorem is proved. We remark that there is actually more to say in the above. Namely, the primes Qi are in fact all “conjugate” and the rings R/Qi are all isomorphic. This can be found in [2, 71 and can be shown here by way of a strengthening of Lemma 2.1. Observe that Cutting Down implies that if P is a prime ideal of S, then there are at most n primes of R minimal over Pn R. Conversely, Lying Over asserts that each prime Q of R is minimal over some Pn R. This result, due to Bit-David and Robson, appears as the lower bound in the next theorem and the proof given here is precisely theirs. On the other hand, the upper bounds in the next two theorems can be found in the work of Heinicke and Robson and were also obtained at the sametime in this paper. THEOREM 3.2 (Lying Over [ 2,6]). Let Q be a prime ideal of R and let K be the number of primes P of S with Q minimal over P n R. Then 1 < K < n.
ProoJ: We first obtain the lower bound 1 < K following [2]. To this end, let P be an ideal of S maximal with respect to the property that P n R c Q. Since Q is prime, it is easy to verify that P is prime. The goal is to show that Q is minimal over Pn R and for this it clearly sufftces to study S/P. In other words we may assumethat P = 0 and that every nonzero ideal Z of S satisfies Zn R G Q. It now follows from Lemma 2.6 that there exists an ideal H#OofRwithQ~H=O.Finally,byCuttingDown,O=P~R=~~Q,., a finite intersection of minimal primes of R. Since H# 0 there exists i with
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D. S. PASSMAN
Qi~H.ButthenQi2O=QnH,soQi~QandQisaminimalprimeofR. Thus 1 < K and we now consider the upper bound. Case 1. Q primitive. Let M be a maximal right ideal of R with Q = ann, R/M and consider the induced S-module V= S/MS. By Lemma 2.2, V has composition length < n. Let P be a prime ideal of S with Q minimal over P--n R and let : S + S/P denote - -- the natural map. Then S + S and MS -+ MS are onto and hence SIMS is a--homomorphic image of S/MS. Moreover since Pn R G Q c M, we have R/Q-R/Q and hence M is a maximal right ideal of R with -Q = ann,-R/M. By Cutting Down, it follows that there exists a nonzero ideal Z? of R with en E? = 0. Hence since - -- S is prime, we conclude - -- from Lemma 2.5 that S acts faithfully on S/MS. Furthermore since S/MS has finite composition length and S is prime, S acts faithfully on a composition - -factor of S/MS and hence on a composition factor of S/MS. Thus P is the annihilator in S of a composition factor of V= S/MS and since there are at most il such factors, this case follows. Case 2. Q prime. We have S* 2 R * and Q* is a primitive ideal by Lemmas 1.4(iv) and 1.6. Suppose Q is minimal over Pn R. By Cutting Down, Pn R = 0: Qi, a finite intersection of incomparable primes with say Q = Q,. Then clearly P* n R* = 0: QT, so we conclude easily that Q* is minimal over P* n R *. Since there are at most n primes of S* lying over the primitive ideal Q*, and since P* n S = P, the result follows. Two remarks are now in order. First, we have actually shown in the above two results that if Q is minimal over P n R, then Q is primitive if and only if P is primitive. Second, since homomorphic images of finite normalizing extensions are also finite normalizing extensions, Lying Over clearly yields a suitable Going Up theorem. THEOREM 3.3 (Lying Over Zero [6]). Let J be an ideal of R, let k denote the number of minimal primes of R/J and let K denote the number of primes P of S with P n R = J. Then krc < n.
Proof. We can clearly assume that K # 0, so there exists a prime P, with P, n R = J. Set Z = SJS so that Z n R c J. On the other hand, Z C P for any primePwithPnR=JandthusJ=P,nR~ZnR.HenceZnR=Jand we may now clearly work in S/Z. Therefore it suffices to assume that J= 0. Cutting Down now implies that 0 = P, n R = f-): Qi, an intersection of k < n minimal primes of R. Set Q = Q, and H = fl: Qi so that H# 0 and QnH=O.
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IDEALS
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Case 1. Q primitive. Let M be a maximal right ideal of R with Q = annR R/M and consider the induced module V= S/MS. By Lemma 2.2, VR has composition length < n. Suppose V,, V, ,..., Vj are the composition factors of V such that (Vi)R is faithful. By Lemma 2.1, each (Vi& has composition length > k. Thus jk < n. If Pi = annS Vi, then Pi is a prime ideal of S and Pi n R = 0 since (Vi)R is faithful. It clearly suffices to show that these are precisely all the primes lying over zero. Let P be a prime of S with P n R = 0 and-- let ~: S + S/P be- the-- natural homomorphism. Then S-+ S and MS + MS are onto so S/MS is a homomorphic image of SIMS. - Since - P n R --= 0, we - have R N R and hence I%?is a maximal right ideal of R, -Q -= ann,R/M, H # 0 and Qn B = 0. By Lemma 2.5, S acts faithfully on S/MS. - --
Finally since S/MS has finite composition length - -- and S is prime, S acts faithfully on some composition factor of S/MS. Thus there exists a composition factor W of S/MS such that annS W = P. Moreover Pn R = 0, so W, is faithful and hence W = Vi for some i and P = arms Vi = Pi.
Case 2.
Q prime. have S* 2 R*, Q* is primitive, H* #O and Q*n H* =O. Furthermore each Ql is prime, n’; QT = 0 and the Qif: are clearly all incomparable. Thus R * has precisely k minimal primes. Observe that if P,, P,,... are primes of S with Pi f7 R = 0, then PI*, PT,... are distinct primes of S* with Pi” n R* = 0. It now follows from Case 1 that there are at most n/k such P,? and hence at most n/k such Pi. We
If X, ) x2 )...) x, are R-free on the left or right, then more can be said about S. We start a result of Lorenz and Passman on nilpotent ideals. THEOREM 3.4 (Nilpotent Ideals [lo]). Let R be a semiprime ring and assume that x,, x2 ,..., x, are R-free on the left (or right). Then S has a unique maximal nilpotent ideal T and T” = 0.
Proof T” = 0.
It clearly suffices to show that if T is a nilpotent ideal of S, then
Case 1. R semiprimitive. If (Mj} is the set of all maximal right ideals of R, then R acts faithfully on @ Cj RIMi. Thus, by Lemma 2.3, S acts faithfully on @ Cj S/M,S. By Lemma 2.2, each S/M,S has composition length < n and hence if JS is the Jacobson radical of S, then (JS)” kills each S/M,S. Thus (JS)” = 0 and since T E JS we have T" = 0. Case 2. R semiprime. We have S* GJR* and x, ,x2 ,..., x, are R *-free on the left, by Lemma 1.6(i, ii). Moreover R* is semiprimitive, by Lemma 1.4(v). Finally T
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is nilpotent, so T* is nilpotent and hence (T*)” = 0 by the above. Since T E T* this yields T” = 0. Next we have a structure theorem due to Lorenz. THEOREM 3.5 (Wedderburn-type [7]). Let R be a prime ring and assume that x, ,x2 ,..., x, are R-free on the left. Then S has Jnitely many minimal primes P, , P, ,..., P, with k
Proof: Case 1. R primitive. Let M be a maximal right ideal of R such that R acts faithfully on R/M. Then by Lemmas 2.2 and 2.3, S acts faithfully on V= S/MS and V has composition length < n. Let P,, P, ,..., Pj with j < n be the annihilators of the composition factors of V. Then each Pi is prime and T = n’, Pj annihilates all such factors so T” = 0. Finally assumethat T = 0: Pi is an irredundant intersection with k
Proof: We know that S = RG has the set {X 1x E G} as a free basis. In particular S is a finite normalizing extension of R with n, the size of the normalizing set, equal to )G ]. In addition, we know that R is G-prime if and only if there exists a prime ideal Q of R with n,.G Q” = 0. Case 1. Q primitive. Let M be a maximal right ideal of R with Q = annRR/M. Then S acts
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faithfully on V= S/MS by Lemma 2.4. The argument of Case 1 of the preceding theorem now yields the result. Case 2. Q prime. We have S* = R *G, by Lemma 1.6(iii), and Q* is primitive, by Lemma 1.4(iv). Furthermore it is clear that (Q*)” = (Q”)* and hence that flxEG (Q*)‘= 0. Th us we conclude from Case 1 that S* has the appropriate structure. Moreover R is semiprime, so S = RG has a unique maximal nilpotent ideal T by Theorem 3.4. The argument of Case 2 of the previous theorem now completes the proof. Finally we prove a theorem of Fisher and Montgomery on crossed products. THEOREM 3.1 (Maschke-type 131). Let RG be a crossedproduct of the finite group G over the semiprimering R. If R has no 1G j-torsion then RG is semiprime.
Proof: Let S = RG. By forming the extension S[ l/]G]] = R [ l/]G]] can clearly assumethat l/]G] E R.
G, we
Case 1. R semiprimitive. If (Mj} is the set of all maximal right ideals of R, then R acts faithfully on @ Cj R/kfj. Thus, by Lemma 2.3, S acts faithfully on V= @ cj S/M,S. Furthermore, VR is completely reducible, by Lemma 2.2, so we conclude from Maschke’s theorem (see for example [ 10, Lemma 11) that V is also completely reducible. Hence, if JS denotes the Jacobson radical of S, then V(JS) = 0; so JS = 0. In other words, S is semiprimitive and hence also semiprime. Case 2. R semiprime. We have S* = R*G, by Lemma 1.6(iii), and R* is semiprimitive, by Lemma 1.4(v). Hence by the above, S* is semiprime and, by Lemma 1.4(iii), so is S. 4. EXAMPLES In this section we discusssomeinteresting and relevant examples. The first concerns Lemma 1.5 and offers an example of a prime ideal P of R* with P 17R not prime and in fact not even semiprime. Observe that by setting all but one variable in R * = R(Y, Z) equal to zero, we can obtain an Rhomomorphism of R* onto the ordinary power series ring R[ [t;]]. Thus it sufftces to find a prime ideal P of R [[Cl] with P n R not semiprime. To this end, we have the following argument of G. Bergman. We thank Professor Bergman for allowing it to be included here.
D.S.PASSMAN
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LEMMA 4.1 [l]. Let Q, s QZE -.a be an ascending chain of prime ideals in a ring R. Then P= U, Q,[ [Cl] is a prime ideal of R[ [[I].
ProoJ Here of course Q,[ [cl] is the ideal of all power serieswith coefficients in Q,. Now supposea and p are elements of R [[[I] not contained in P= U, Q,[Kll. We construct inductively a power series
f = 5 riccci) i=l
such that for all n > 1 (1)
a(Cr=, ricCi)) p has a term sncd(n)with s, 6$Q,,
(2)
c(n + 1) > d(n), c(n).
Suppose we have already found Cr=, rirti) for some n > 0 with the understanding that this expression is zero for n = 0. Choose any integer c(n + 1) larger than d(n) and c(n). There are now two casesto consider. If a(Cy=, ri@)P @ Q,+,[ [[I], take r,,+, = 0. Then certainly
has a term So+i cd@’ i) with s, + i 6ZQ, + i . On the other hand, supposethat a(Cy,, ri4@))/3E Q,+,[[[]]. Then for any r,,, E R we have
Since a, p 66Q,, , [ [[]I and Q, + i is prime, we can surely find r,+ , with r(“+‘)/?& Q,+,[[[]]. Indeed, we can do this by considering the terms in a and in /I of lowest degree not contained in Q,, , and use the primeness ofQnt,- With this choice of r,, +1E R, the induction step is proved. It now follows easily from (1) and (2) that the coefficient of cd(“) in afl is s, @Q,. Hence @P @Q,[[C]] and since this holds for all n we conclude that ajJ3@P. Therefore P is prime.
artIt
EXAMPLE
4.2 [l].
A prime ideal P of R [ [t;]] with P n R not semiprime.
Proof. Let R = k(x, y) be the free associative algebra over the field k in the noncommuting variables x and y. For each n > 1, let Q, be the ideal of R generated by xx, xyx, xy*x ,..., xy”-’ x. Then Q, is prime since a,P & Q, implies easily that ay”/3 6?Q,. Furthermore Q, + ,Z Q, and Q = (j Q, = (RxR)* n
PRIME IDEALS IN NORMALIZING
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571
so that Q is not semiprime. Finally, by Lemma 4.1, P = U, Q,[ [[I] is a prime ideal of R [ [[I]. Since P n R = U, Q, = Q, we conclude that P n R is not semiprime. The remaining examples all involve extensions constructed in the following simple manner. LEMMA 4.3. Let I and J be ideals of R with In J = 0 and set R’ = R/I, R” = R/J. Then S = R’ OR” is a finite centralizing extension of R, with R embedded diagonally. Furthermore, the minimal primes of S are given by P’ @RI’ and R’ 0 Q” where P’ and Q” are minimal primes of R’ and R”, respectively.
ProoJ Since Z n J = 0, the map r + (r’, r”) embedsR into S with (I’, 0) and (0, 1”) a centralizing set of generators. With J = 0 and varying R and 1, several examples are now easily given. EXAMPLE 4.4. (i) Let R be a primitive ring acting faithfully on R/M, with A4 a maximal right ideal and assumethat R is not simple. If I is any nonzero ideal of R, then M+I=R so M’=R’ and MS = R ’ ClJM” 1 R’ @ 0. Hence S does not act faithfully on S/MS (compare with Lemmas 2.3 and 2.5).
(ii) Let R be a commutative integral domain with ideal I such that R/Z has no maximal nilpotent ideal. Then clearly S has no maximal nilpotent ideal (compare with Theorem 3.4). (iii) Let R be prime and let I be a nonminimal prime ideal of R. Then 0 @ R ” is a minimal prime of S which lies over I. Moreover R ’ @ 0 is the only minimal prime of S lying over zero, so the intersection of all such is not nilpotent (compare with Theorem 3.5). (iv) Let R be a commutative integral domain such that R/I has infinitely many minimal primes. Then S has infinitely many minimal primes, while R has the unique minimal prime zero (compare with Theorem 3.5). We close this paper with a remark of a somewhat different nature. Let G be a group acting as automorphisms on a ring R. Then certainly G acts on R * and for the fixed rings we have clearly (R*)G = (RG)*. Thus the “primitivity machine” may also have applications to the study of rings with finite group actions. Indeed it can be used to obtain some of the basic relations between the primes of R and of RG, but these have already all been proved by Montgomery [ 121, from skew group ring results, in a quite satisfactory manner.
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D. S. PASSMAN REFERENCES
1. G. M. BERGMAN, Chains of prime ideals in R and prime ideals in R[[t]], unpublished. 2. .I. BIT-DAVID AND J. C. ROBSON, Normalizing extensions, I, in “Proc. Ring Theory Conference 1980),” Springer Lecture Notes, Springer-Veilag, (Antwerp, Berlin/Heidelberg/New York, 1980. 3. J. FISHER AND S. MONTGOMERY, Semiprime skew group rings, J. Algebra 52 (1978), 241-247..
E. FORMANEK, Group rings of free products are primitive, J. Algebra 26 (1973), 508-5 11. 5. D. HANDELMAN AND J. LAWRENCE, Strongly prime rings, Trans. Amer. Math. Sot. 211 4.
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6. A. G. HEINICKE AND J. C. ROBSON, Normalizing extensions: prime ideals and incomparability, to appear. 7. M. LORENZ, Finite normalizing extensions of rings, Math. Z. 176 (198 I), 447-484. 8. M. LORENZ AND D. S. PASSMAN, Prime ideals in crossed products of finite groups, Israel J. Math. 33 (1979), 89-132. 9. M. LORENZ AND D. S. PASSMAN, Integrality and normalizing extensions, J. Algebra 61 (1979),
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10. M. LORENZ AND D. S. PASSMAN, Two applications of Maschke’s theorem, Comm. Algebra 8 (1980), 1853-1866. Il. M. LORENZ AND D. S. PASSMAN, Addendum-Prime ideals in crossed products of finite groups, Israel J. Math., in press. 12. S. MONTGOMERY, Prime ideals in fixed rings, Comm. Algebra 9 (1981), 423449. 13. J. C. ROBSON,Some results on ring extensions, Lecture Notes, University of Essen, 1979.