Let R be a semiprime Noetherian PI-ring and Q(R) the semisimple Artinian ring of fractions of R. We shall prove the following conditions are equivalent: (1) the Krull dimention of R is at most one, (2) Any ring between R and Q(R) is again right Noetherian, (3) Let a, b be central regular elements of Q(R). Then the subring R + aR[b] of Q(R) is right Noetherian.</p

Chiba, Katsuo

English

Okayama University Scientific Achievement Repository

Mathematical Journal of OkayamaUniversityVolume 42, Issue 1 2000 Article 7JANUARY 2000On Semiprime Noetherian PI-RingsKatsuo Chiba∗∗Niihama National College of TechonologyCopyright c©2000 by the authors. Mathematical Journal of Okayama University is produced byThe Berkeley Electronic Press (bepress). http://escholarship.lib.okayama-u.ac.jp/mjouOn Semiprime Noetherian PI-RingsKatsuo ChibaAbstractLet R be a semiprime Noetherian PI-ring and Q(R) the semisimple Artinian ring of fractionsof R. We shall prove the following conditions are equivalent: (1) the Krull dimention of R is atmost one, (2) Any ring between R and Q(R) is again right Noetherian, (3) Let a, b be centralregular elements of Q(R). Then the subring R + aR[b] of Q(R) is right Noetherian.Math. J. Okayama Univ. 42 (2000), 55-59ON SEMIPRIME NOETHERIAN PI-RINGSDedicated to Professor Yukio Tsushima on his 60th birthdayKATSUO CHIBAAbstract. Let R be a semiprime Noetherian PI-ring and Q(R) thesemisimple Artinian ring of fractions of R. We shall prove the followingconditions are equivalent: (1) the Krull dimention of R is at most one,(2) Any ring between R and Q(R) is again right Noetherian, (3) Leta; b be central regular elements of Q(R). Then the subring R + aR[b]of Q(R) is right Noetherian.Throughout this note all rings will have a unit element. Let R be aring. We denote by dim(R) the Krull dimention of R, i.e., the supremum ofthe lengths of chains of distinct prime ideals in R, and by Z(R) the centerof R. Cauchon showed that a semiprime PI-ring with the ascending chaincondition on two-sided ideals is left and right Noetherian (See [5, II, p.174]). Therefore we simply say semiprime Noetherian PI-rings for semiprimeleft and right Noetherian PI-rings. Let R be a semiprime Noetherian PI-ring. As is well known, the ring of fractions of R with respect to the setof central regular elements of R is a semisimple Artinian ring (See [5, II,p.174]). The main result of this note is the following:Theorem. Let R be a semiprime Noetherian PI-ring and Q = Q(R) thesemisimple Artinian ring of fractions of R. Then the following conditionsare equivalent:(1) dim(R) · 1.(2) Any ring between R and Q(R) is again right Noetherian.(3) Let a; b be central regular elements of Q(R). Then the subring R +aR[b] of Q(R) is right Noetherian.Remark. Let R and Q(R) be as in the above Theorem, and let T be a ringbetween R and Q(R). If T satis¯es the conditions of the above theorem,then T is semiprime, therefore T is left and right Noetherian.Let S be a ring and R a subring of S. We say that S is an extensionof R if S = RSR, where SR = fs 2 S ; sr = rs for all r 2 Rg and thatS is integral over R if each element s of S satis¯es an equation of the formsn + r1sn¡1 + r2sn¡2 + ¢ ¢ ¢ + rn = 0, where ri 2 R for all i = 1; 2; ¢ ¢ ¢ ; n.1991 Mathematics Subject Classi¯cation. 16R20, 16P40.551Chiba: On Semiprime Noetherian PI-RingsProduced by The Berkeley Electronic Press, 200056 KATSUO CHIBALet R be a prime PI-ring and Q(R) the central simple ring of fractions ofR. The ring obtained by adjoining to R all elements of Z(Q(R)) whichare integral over R is called centrally integral closure of R. The proof ofTheorem is similar to that in the commutative case as given by Kaplansky[2,Theorem 93 and Exercise 20, p.64]. For the proof of Theorem we needseveral lemmas.Lemma 1 ([6, Theorem 1]). Let S be a PI-ring and R a subring of S. If Sis an extension of R, i.e. S = RSR and integral over R, then the followinghold;(1) For any prime ideal P in R there exists a prime ideal Q in S withQ\R = P (lying over).(2) For any pair of prime ideals P ½ P1 in R and a prime ideal Q in Swith Q \ R = P , then there exists a prime ideal Q1 in S satisfyingQ ½ Q1 and Q1 \R = P1 (going up).(3) Two di®erent primes in S with the same contraction in R cannot becomparable (incomparability).Lemma 2 ([7, Theorem 3]). If R is a prime PI-ring and integral over anintegrally closed central subring A of R, then A ½ R has going down, i.e. ,given prime ideals P0 ½ P in A and a prime ideal Q in R with Q\A = Pthen there exists a prime ideal Q0 in R satisfying Q0 ½ Q and Q0\A = P0.Lemma 3 ([6, Theorem 2]). If R is a prime PI-ring with ascending chaincondition on centrally generated ideals, then the coe±cients of the reducedcharacteristic polynomial of any element of R are integral over R.As a corollary of Lemma 3 we shall prove the following lemma.Lemma 4. Let R be a Noetherian prime PI-ring with the central simplering of fractions Q(R) and let R¤ be its centrally integral closure, the ringobtained by adjoining to R all elements of Z(Q(R)) which are integral overR, then:(1) R¤ is integral over R.(2) R¤ is integral over Z(R¤).(3) Z(R¤) is integrally closed in its ¯eld of fractions.Proof. (1) Let x 2 R¤ then there are ¯nite elements t1; ¢ ¢ ¢ ; tk 2 Z(Q(R))which are integral over R and x 2 R[t1; ¢ ¢ ¢ ; tk]. Clearly, R[t1; ¢ ¢ ¢ ; tk] isa ¯nitely generated R-module. Since R is Noetherian, x is integral overR. (2) If x 2 R¤, then there are ¯nite elements t1; ¢ ¢ ¢ ; tk 2 Z(Q(R)) suchthat x 2 R[t1; ¢ ¢ ¢ ; tk] as in the proof of (1). Let µ be a coe±cient of thereduced characteristic polynomial of x. It is enough to show that µ is anelement of Z(R¤). Since R[t1; ¢ ¢ ¢ ; tk] is a Noetherian prime PI-ring, by2Mathematical Journal of Okayama University, Vol. 42 [2000], Iss. 1, Art. 7http://escholarship.lib.okayama-u.ac.jp/mjou/vol42/iss1/7ON SEMIPRIME NOETHERIAN PI-RINGS 57Lemma 3, µ is integral over R[t1; ¢ ¢ ¢ ; tk], hence R[t1; ¢ ¢ ¢ ; tk; µ] is a ¯niteR-module, therefore µ is integral over R. This shows that µ is an elementof Z(R¤). (3) Let t be an element of Z(Q(R)) and integral over Z(R¤).Then, as in (1), t is integral over R[t1; ¢ ¢ ¢ ; tk], where t1; ¢ ¢ ¢ ; tk 2 Z(Q(R))are integral over R, hence t is integral over R, therefore, by the de¯nitionof R¤, t 2 Z(R¤).Lemma 5. Let R be a PI-ring. Then R is a right Noetherian ring withdim(R) = 0 if and only if R is a right Artinian ring.Proof. Suppose R is a right Noetherian ring with dim(R) = 0. Then allits prime ideals are both minimal and maximal prime ideals and thereare only ¯nitely many such prime idals, say M1;M2; ¢ ¢ ¢ ;Mn. We haveJ(R) =M1\M2\¢ ¢ ¢\Mn, where J(R) is the Jacobson radical of R. SinceJ(R) is nil, by Levitzki's theorem [1, Theorem 1.4.5], J(R) is nillpotentand hence (M1 ¢ ¢ ¢Mn)k = 0 for some k. Since R=Mi (i = 1; 2; ¢ ¢ ¢ ; n) aresimple Artinian rings, we can re¯ne the series of R-modules R ¾ M1 ¾M1M2 ¾ ¢ ¢ ¢ ¾M1 ¢ ¢ ¢Mn ¾ ¢ ¢ ¢ ¾ 0 and then we have a composition seriesof the right R-module R. Thus R is a right Artinian ring. The converse iswell known.Lemma 6. Let R be a semiprime Noetherian PI-ring with ¯nite Krulldimention dim(R). Let a be a non-unit central regular element of R. Thendim(R=aR) < dim(R).Proof. Since R is a semiprime Noetherian PI-ring, R has a ¯nite set ofminimal prime ideals, say P1; P2; ¢ ¢ ¢ ; Pn. We show that the canonicalimage of a in the factor ring R=Pi is regular for each i. Suppose ¹a is notregular in R=Pi for some i, where ¹a denote the canonical image of a inR=Pi, then there is an element b in R such that ¹a¹b = 0 and ¹b 6= 0 in R=Pi.SinceTj 6=i Pj 6½ Pi, there is a non-zero central element ¹c in R=Pi suchthat ¹c 2 Tj 6=i Pj ¡ Pi by [4, Theorem 2]. Then abc 2 TPi = 0 so bc = 0,implying ¹b = 0, a contradiction. Now, let Q0 ½ Q1 ½ ¢ ¢ ¢ ½ Qn be a chainof prime ideals of R such that a 2 Q0, then Q0 is not a minimal primeideal of R. This shows that dim(R=aR) < dim(R).Now we shall prove Theorem:Proof of Theorem. (1) implies (2). Let T be a ring between R and Q andlet X be a right ideal of T . Then XQ© Y = Q for some right ideal Y ofQ. Since Q is the central localization of R, we have xa¡1 + ya¡1 = 1, forsuitable elements a 2 Z(R); x 2 X \ R and y 2 Y \ R. Hence the rightideal X© (Y \T ) of T contains a regular central element a of R. It su±cesto show that T=aT is a ¯nite generated right R-module. Since the Krull3Chiba: On Semiprime Noetherian PI-RingsProduced by The Berkeley Electronic Press, 200058 KATSUO CHIBAdimention of R is at most 1, by Lemma 6 it follows that the Krull dimentionof R=aR is 0. By Lemma 5, R=aR is right Artinian. Thus the descendingchain of ideals famT \ R + aRjm = 1; 2; ¢ ¢ ¢ g in R becomes stable, sayat anT \ R + aR. For this n we assert that T=aT µ (a¡nR + aT )=aT .Let t = zc¡1 be an element of T , where z 2 R and c is a regular centralelement of R. Since R=cR is right Artinian, then akR µ ak+1R + cR, forsome k, so akt = ak+1rt+ cr1t (r; r1 2 R), whence t 2 a¡hR+ aT for someh. Let us suppose that the equation t 2 a¡hR + aT has been arrangedwith the smallest possible value of h. We shall prove that h · n. Supposeh > n. We write t = a¡hu+at1; u 2 R and t1 2 T . Then u = ah(t¡at1) 2ahT \R µ ah+1T \R+aR. So we can write u = ah+1t2+au1, where t2 2 Tand u1 2 R. Thus we have t = a¡(h¡1)u1+ a(t1+ t2). This contradicts theminimal choice of h.(2) implies (3). Trivial.(3) implies (1). Assume ¯rst that R is prime. Let R¤ be the centrallyintegral closure of R, then dim(R¤) = dim(R) by Lemma 4 (1) and [6,Corollary 1, p.247]. It su±ce to show that dim(R¤) = 1. By Lemma 4(3), R¤ is integral over Z(R¤) and Z(R¤) is integrally closed. Therefore,we may assume that R is integral over Z(R) and Z(R) is integrally closed.If dim(R) > 1 then there exist prime ideals 0 6= Q ½ P in R. By Lemma1 (3), Z(R) \ Q ½ Z(R) \ P are distinct primes in Z(R). Take x 2Z(R) \ Q; x 6= 0. Since R is Noetherian, there are only ¯nitely manyprime ideals minimal over xR, say P1; ¢ ¢ ¢ ; Pn by [3, Corollary 2.4, p.108].By Lemma 2, Pk \ Z(R) is a minimal prime ideal over x in Z(R) andthus P \ Z(R) 6½ Pk \ Z(R) for any k. Hence we have P \ Z(R) 6½(P1 \ Z(R)) [ ¢ ¢ ¢ [ (Pn \ Z(R)) by [2, Theorem 81]. Take y 2 P \ Z(R)with y 62 (P1 \ Z(R)) [ ¢ ¢ ¢ [ (Pn \ Z(R)) . Let T = R + xR[y¡1] andI = xR[y¡1]. We assert that I is not a ¯nitely generated ideal in T . If I isa ¯nitely generated ideal then I is generated by xy¡i for some i. Then wehave xR[y¡1] = xy¡iT , and so R[y¡1] = T . Let y¡1 = a + xby¡j ; a; b 2R; j ¸ 1. Then yj¡1(1 ¡ ay) = xb 2 xR ½ Pk for all k. Therefore1 ¡ ay 2 Pk for all k, and then, we have (1 ¡ ay)m 2 xR for some m.Expanding (1¡ ay)m = xc; c 2 R, we have the relation 1 = yd+ xc whered 2 R, which leads a contradiction,1 2 P .Suppose R is semiprime. Since R is Noetherian, there are ¯nitelymany minimal prime ideals, say Q1; ¢ ¢ ¢ ;Qr. Put Ri = R=Qi. We showthat the Krull dimention of Ri is at most 1 for any i. Let Q(Ri) (i =1; 2; ¢ ¢ ¢ ; r) be the ring of fractions of Ri. Suppose dim(R=Qk) > 1 forsome k. By the above argument, there are regular central elements ¹x; ¹yof Rk such that the subring Rk + ¹xRk[¹y¡1] in Q(Rk) is non-Noetherian,where ¹x; ¹y are the canonical image of x; y (2 R) in Rk. Using the injections4Mathematical Journal of Okayama University, Vol. 42 [2000], Iss. 1, Art. 7http://escholarship.lib.okayama-u.ac.jp/mjou/vol42/iss1/7ON SEMIPRIME NOETHERIAN PI-RINGS 59R!Qri=1Ri µQri=1Q(Ri); the ring Qri=1Q(Ri) is considered as the ringof fractions of R (See [5, I, Theorem 3.2.27 and II, p.174]). Let xk =(1+Q1; ¢ ¢ ¢ ; x+Qk; ¢ ¢ ¢ ; 1+Qr); yk = (1+Q1; ¢ ¢ ¢ ; y+Qk; ¢ ¢ ¢ ; 1+Qr) bethe elements ofQri=1Q(Ri). By the hypotheses, the subring R+ xkR[y¡1k ]ofQri=1Q(Ri) is Noetherian and any homomorphic image of R+xkR[y¡1k ]is Noetherian, hence the subring Rk + ¹xR[¹y¡1] of Q(Rk) is Noetherian,which is a contradiction. This completes the proof.AcknowledgementsThe author thanks Professor Y. Hirano of Okayama University forhis helpful suggestions.References[1] I. N. Herstein: Non-commutative Rings, Carus Math. Monographs 15, Math.Assn. of America, J. Wiley and Sons, 1968.[2] I. Kaplansky: Commutative Rings, Allyn and Bacon, Boston, Mass., 1970.[3] C. Procesi: Rings with Polynomial Identities, Marcel Dekker, Inc., New York,1973.[4] L. Rowen: Some results on the center of a ring with polynomial identity, Bull.Amer. Math. Soc. 79 (1973), 219-223.[5] L. Rowen: Ring Theory I, II, Academic Press, New York 1988.[6] W. Schelter: Integral extensions of rings satisfying a polynomial identity, J. Al-gebra 40 (1976), 245-257.[7] W. Schelter: Non-commutative a±ne P.I. rings are catenary, J. Algebra 51 (1978),12-18.KATSUO CHIBANiihama National College of Techonology7-1 Yagumo-cho Niihama 792-8580Japan(Received November 15, 2000)5Chiba: On Semiprime Noetherian PI-RingsProduced by The Berkeley Electronic Press, 2000

On Semiprime Noetherian PI-Rings

Abstract

Similar works

Full text

Available Versions

Okayama University Scientific Achievement Repository