DEMONSTRATIO MAT IJ KM ATI CA Vol. XXVII No 1 1994

Andrei V. Kelarev

FINITENESS CONDITIONS FOR SPECIAL BAND GRADED RINGS

A number of chain (initeness conditions arc well-known classical concepts of ring theory. Properlies of this soi l have been investigated by many authors for various ring constructions, in particular, for semigroup rings (cf. [25]). The aim of the present paper is to carry out analogous investigations for special band graded rings, which arc closely related to semigroup rings. Namely, we shall deal with right Artinian, right Noetherian, semilocal, local, scalarly local, right perfect and scmiprimary rings. Let D be a band, i.e. a semigroup consisting of idempotents. An associa- tive ring R is said to be /¿-graded if R = ©¡,eö Rb-, the direct sum of additive groups, and RaR>, C Ra^ for any a, b £ B. A survey of papers concerning band graded rings was given in [10]. Here we mention only more recent ar- ticles [11]—[13], [19]. If each ring /?,(, has an identity 11,, and 101(, = la(, for any a,b then the ring II = Rb is called a special band graded ring, or a special X?-graded ring. This concept was introduced in [19]. In the case of commutative B special band graded rings were considered earlier in [1], [6], [27], [30] and other papers. Special band graded rings have found effective applications to the study of distributive and semiprimitive semigroup rings ([29], [9]), and to the investigations of the radicals of semigroup rings of CliiTord semigroups ([27]), orlhogroups ([28]), and commutative semigroups ([7], [14])- First, let us turn to semilocal special band graded rings. A ring R is said to be semilocal if the quotient ring R/.J(R) is right Artinian, where J(R) is the Jacobson radical of R. Semilocal semigroup rings were considered in [20], [22] and [31]. A band is called a semilattice (a rectangular band; a left zero band; a right zero band) if it satisfies the identity xy = yx (xyx = x; xy = x; XV = V)• Let be a semilattice. A band B is said to be a semilattice S of 172 A. V. Kclarcv

rectangular bands II,, s G S, if B = Us(=s IIs, IIs fl //< = 0 whenever s ^ i, and //,//< C If st for any .s,t £ S. Each band is uniquely represented as a semilattice of rectangular bands (cf. [2], §4.2). This representation coincides with the splitting of JJ into classes by the equivalence relation a ~ b & aba = a and bah = b (cf. [19], §2).

THEOREM 1. Let a band B be a semilallice S of rectangular bands II s. A special band graded ring 11 = lib is semilocal if and only if S is finite and all Rb are semilocal. A ring is said to be local (scalary local) if factoring out its Jacobson radical one gets a simple right Artinian ring (a division ring). Local and scalarly local semigroup rings were considered in [17], [20], [22] and [26].

Theorem 2. A spccial band gnu led ring R — ©¡,6B Rb is local (scalarly local) if and only if IS is a rectangular band and all Rb are local (scalary local). A ring K is said to be right 71-nilpotcnl if for any sequence xi, X2,... € K there exists a positive integer n such that ,rn.rn_1 ...a-j = 0. A ring R is right perfect (scmiprimary) if R/J(R) is right Artinian and J(R) is right T-nilpotent (nilpotcnt). Kiglit, perfect and scmiprimary semigroup rings were considered in [3], [-1], [23] and [16].

THEOREM 3. Let, a band B be a semilatlice S of rectangular bands Hs, and let R = Rb be a spccial band graded ring. Then R is right per- fect (semiprimary) if and. only if S is finite and all Rb are right perfect (semiprimary). Right Artinian and Noetlieriau semigroup rings were considered in [18], [32], [21], [15], [22], [5] and [2-1]. In particular, it is proved in [32] that if a semigroup ring RS is right Artinian, then S is finite and R is right Artinian. The converse implication is not valid. In [5] right Artinian commutative semigroup rings are described completely. We shall obtain analogous results for special band graded rings.

Theorem 4. Let R = Rb be a special band graded ring. If R is right Artinian (right Noelhcrian), then B is finite and all Rb are right Artinian (right Noetherian). The following example shows that the converse is not true.

Example 1. Let 7?. be field of real numbers, B = {a, b} a right zero band. Then the semigroup ring RB is a special band graded ring, RB = Ra + Rb. However, IIIJ is neither right Artinian, nor right Noetherian. In fact, evidently R contains an infinite chain of subrings ... C R-1 C Ro C R\ C Then it is routine to chcck that R\(a — b) C ^(a — 6) C ... is an Finileness conditions 173 infinite asccnding chain of light ideals of RB, and Ro(a-b) C R-i(a-b) C iZ_2(n. — 6) C ... is an in finite descending chain of right ideals of RB. If B is commutative, then we get

COROLLARY 1. Let B be a scmilallice, R = Rb a special band graded ring. Then R is right Artinian (right Noetherian) if and only if B is finite and all Rb are right Artinian (right Noetherian).

LEMMA 1 [19]. Let II be a reetangidar band, R = @be/i Rb a special band graded ring, If, the identity of Rb. Then x\hV — xy for any x,y G R, h G H.

Proof. Let x = ^2beJIxb, y = X)6e//yb where xb,yb G Rb- Then x\hy = r x X X^a,6ef/' a lh b = Ea.be" XalalhlbXb = Za,f>e// a ^a 1&&6 = XIJ, because ahb = ab. The following key lemma is inspired by the result of [19].

LEMMA 2. Let a band B be a scmilallice S of reetangidar bands IIS, R = 06eB Rb a special band graded ring, U the identity of Rb. For each s G S choose an clement lis in II s and set Qs - Rb, Is = {® £ Qs | = 0}, / = Then I is an ideal of R contained in the Jacobson radical .1(11). The quotient ring R/I is a special S-graded ring,

R/I = Fs, where Fs = Rh,. The ideal I and rings Fs do not depend on the choice of the elements hs. Besides,T^ = 0 for every s G S. A large part of the proof of Lemma 2 follows from several technical lemmas of [19]. For the sake of completeness we shall give a-self-contained proof.

Proof. Take any .s G S and put II = //.,, h = hs. For any x,y G Qs, z G Is Lemma 1 and the definition of 7, yield xzy = xl^zl^y = 0. Therefore

QshQs = 0, and so Ps = 0. Now we take any g G IIs and set h's = g, I's = {a; G Qs | igxlg = 0}. By Lemma 1 for any x G we get l^sl/i = l^lgxlglh = 0, whence I'3 C I3. Similarly, Is C I's. Thus Is and I do not depend on the choice of the elements hs.

Let s,t e S, X G Is, y G Qt, h = hs, g - hst. Then ylg G QtQst Q Qst and 1 gX G QstQs C Qst• 1» view of Lemma 1 (lgx)(ylg) = lgxlhghy^g- Similarly, lg G Qsi, xlil!jhyln G Qst, and so by Lemma 1, lgxylg = lglhghxlhghylg- Given that \hgh = 1 h\g\h and x £ Is, we get lhgh^hgh = 0. Hence \gxylg = 0, that is xy £ Isl. Thus IsQt C Ist, which means that I is a right ideal in R. By duality, I is a left ideal of R, as well. Thus I is an ideal of R. The inclusion I C J(R) immediately follows from the description of the Jacobson radical of a special band graded ring which was obtained in [19]. 174 A. V. Kclarcv

Here we give a separate proof. Every ring Is is radical, since = 0; and I = ®s€5/s is a semiiaUice graded ring. Theorem 3.2 of [1] shows that the radical class of the .lacobsou radical is closed under formation of semilattice graded rings. Therefore / is radical. Sincc J{ R) contains all the radical ideals of R, we get J(R) D I.

Evidently, R/I = ®4gs Fs where Fs = (Qs + /)//. Furthermore, Fa Si Qa/(Qs n /) = Qs/Is. Hence the Fs do not depend on the choice of ha, s € S. Let s 6 S, g 6 //.,, x € Rg, h = ha. Then 1/tXl/i € Rh and it is clear that x —l/f-cl/i € h- Therefore Rg C Rh + Ia implying Rh + I3 = Qs- Besides i?/,il/, = 0, because Ihylh = y for every y € Rh. Hence Fa = Qa/Ia = (Rh + Is)/is = R)J(Rh n ls) s Rk. Let / be the natural liomomorphism of R onto R/I. Since — \g G Ia for every g,li £ IIs, then /(l/J = /(and it is easily seen that f(lg) coincides with the identity of Fs. Now take any s,t € S, g € II3, h € Ilf Then f(lg)f(lh) = /(l.,U) = /(lflA). This shows that f{lg)f{lh) is equal to the identity of Fsi- Tlicrefore R/I = (&S£SFs is a special band graded ring.

LEMMA 3 ([31], Proposition 1.3). Let S be a semilattice, F = ©a£5 Fs a semilattice graded ring, Fs / J(FS) for every s 6 S. Then F is semilocal if and only if F is finite and all Fs are semilocal. Proof of Theorem 1. Define in R the ideal I as in Lemma 2. Since I C J(R), then R is semilocal if and only if F = R/I is semilocal. By

Lemma 2 F =-®s€S Fs is a special semilattice graded ring, and Fa — Rh,- Since hs was taken arbitrarily, Fs = R(, for every 6 £ IIs. Besides Fa / J(Fa) because Fa has an identity. Therefore Lemma 3 yields that R is semilocal if and only if S is finite and all Rb are semilocal.

LEMMA 4 ([6]). Let S be a finite semilattice and R = ©sg5 Rs a special semilattice graded ring. Then R is isomorphic to the direct product Ra. Proof of Theorem 2. Let a band B be a semilattice S of rectan- gular bands IIa. Define in R the ideal I as in Lemma 2. If R is local, then Theorem 1 implies that S is finite. By Lemma 2 the quotient ring F = R/I is a special semilatticc graded ring, F = rises^»' whcre Fa — Rh, for ha € IIa- Lemma 4 tells us that F = f^es Fa. Since all Fa have identities, F can be local only if 5' is a singleton, i.e. B is a rectangular band. Now let B be a rectangular band. Then R is local if and only if F = R/I is local. Lemma 2 tells us that F = Rh for every h € B. Hence R is local if and only if all Rb arc local. For the scalarly localness the proof is analogous.

LEMMA 5. Let S be a finite semilattice, I = 0aes Is a semilattice graded ring, where all Is are nilpolent. Then I is nilpotent, too. Finileness conditions 175

Proof. We shall use the induction by n = |S|. If n = 1, then S = {s},

I = Is and the claim is obvious. Let n > 1. Choose in 5 a maximal element m and put T = S\ {m}, A' = h• Then T is a semilattice, |T| = n — 1, and by the induction assumption K is nilpotent. It is routine to verify that T is an ideal of S. Hence I is an extension of its ideal I( by the nilpotent

Im. Therefore I is nilpotent. The following lemma is proved in [19], the proof of Theorem 3.1, steps (i) and (ii).

s a LEMMA G (19]). If It. = ©¡,e/j Rb i- special band graded ring and b £ B, then Rb is a liornoinorphic image of II.

Proof of Theorem 3. The 'only if' part follows from Theorem 1 and Lemma 6. Now let S be finite, and let all Rb be right perfect. Consider in R the ideal I = ©s65/s defined as in Lemma 2. Lemma 2 says that F = R/I = ©ae5 Fs is a special scniilattice graded ring, where Fs = Rhs for any hs € Hs. By Lemma 4 F = fLes R-h,- Hence F is right perfect. Since = 0 for every s € 5, Lemma 5 shows that I is a nilpotent ideal of R. Therefore R is right perfect, too. For seinipiimary rings the proof is analogous. r w cre T Let R = ©(,6W Rb, r e R, r = Y^beB i>' h b 6 Rb- Denote by supp(r) the set of b e B such that rb ^ 0. If M C 72, then put supp(M) = UrgAf SUpp(r).

Lemma 7 ([8], Lemma 2.1). Let S be a semilattice, R = ©se5-Rs a special semilattice graded ring. If R is right Noetherian, then S is finite.

Proof of Theorem 4. Suppose that R is cither right Artinian or right Noetherian. Then Lemma 6 implies that all Rb are Artinian or Noethe- rian, respectively.

Represent B as a scniilattice S of rectangular bands IIs. If R is right Artinian, then the finileness of S follows from Theorem 1. Now consider the case where R is right Noetherian. Define the ideal I and rings Fs as in Lemma 2. Then F •— 11/1 is right Noetherian, and F = Fs is a special semilattice graded ring, and S is finite by Lemma 7. Now we claim that B is finite. Suppose the contrary, and consider the set M of s € S such that IIs is infinite. Since B is infinite, M ^ 0. Choose in M a maximal element m. Put U = Us>m Hsi D = B \ U. A standard verification shows that IIm is an ideal in U, and D is an ideal in B. Factoring out the ideal ©deD Rd of 11, we may assume that from the very beginning D = 0, that is m is least element of 5. Then B = U and IIm is an ideal of R. It is well-known and easily proven that IIm = Y X Z for a left zero band 176 A. V. Kclarev

Y and a right zero band Z. Since IIm is infinite, at least one of Y and Z is infinite. First we consider tlic case wlicn Y is infinite. Then Y contains infinte chains of subsets D] D VJ> D ... and A\ C An C Set Pi = Di X Z, Qi = A{ x Z, where i = 1,2, Obviously, every subset of a left zero semigroup Y is a right ideal of Y. Hence all Pi and Q, are right ideals in

IIm. Given that Hm is an ideal of U and IIm consists of idempotents, it is easy to show that every right ideal of IIm is also a right ideal of U. Put

Li = Pu Mi = Qi, where ¿=1,2,.... Let Vj = ®66i/. Rb, W{ = ®6eA/. Rb for i = 1,2,..Then V'i D V2 D •• • and C I'^ C ... are infinite chains of right ideals of R. This gives a contradiction, as R is right Artinian or Noetherian. Now consider the case where Z is infinite. Fix any y in Y and put II = {(y,z) | z £ Z}. Clearly 11 is a right zero band. Besides II is a right ideal in

Hm, because y is a left zero of Y. Put T = U \ IIm. By the choice of m the set T is a union of a finite number of finite sets IIS, where s > in. Ilence T is finite. Let c = |T|. Adjoining an identity e to 13, we put Te = T U {e}. For h 6 II one gets hT = h(hT) C hllm C II. Hence II = (J/ie// hTe. By induction define elements /i,- 6 II and infinite subsets Gi C H for i = 1,2, If h € If, then |/i71e| < c+1. Take any pairwise distinct elements 9i,92,----,9c+'i of II. Each set liTe docs not contain at least one of them. Hence there exists j such that 1 < j < c + 2 and the set M\ = {h € II \ gj £ hTe} is infinite. Put hi — gj, G\ = M\ \ h\Te. Suppose that hi and

Gi are already defined for i = l,...,7i. Take elements gi, • • •, gc+2 in Gn.

Each set hTe docs not contain at least one of them. Ilence Mn+1 = {h G

Gn | gj hTe} is infinite for some j, 1 < j < c + 2. Then we put hn+\ = gj,

Gn+1 = Mn+1 \ hn+iTe We claim that hi ^ hjTe for any positive integers i £ j. li i < j, then hj € Gj C Gj, because G\ D G2 D Then the choice of Gi implies hi <£ hjTe. On the other hand, if i > j, then hjTe f\G{ C hjTe n Gj = 0. Therefore again hi £ hjTe, because hi G G,.

Consider the elements c,- = 1 iH - 1^. for ¿ = 2,3, Given that II is a right zero band, we get ej = 1/,, - 1 hlhi - l^.^ + U, = 0. Let Ii = e.il1.

It is easy to see that supp(/l) C h\Te U hiTe, because B = T U IIm and eiRb = 0 for any b e JIm. Since hi $ U2

/2C/2+/3C/2I/3I I\ C ... and an infinite descending chain ®2

Rcfcrcnces

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DEPARTMENT OF1 MATHEMATICS AND MECHANICS URAL STATE UNIVERSITY Lenina 51 EKATERINBURG 620151, RUSSIA

Received March 23, 1992.