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PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 46, Number 3, December 74 CHAIN CONDITIONSON SYMMETRICELEMENTS1 SUSAN MONTGOMERY ABSTRACT. Recently orem for the Jordan ring Britten tion of characteristic not orem to the where only situation an ample show that subring if S has has proven S of symmetric 2. In this paper symmetric ring elements. on quadratic ideals, of Goldie's in a ring with we first R is an arbitrary of the ACC an analog elements extend Britten's and the Jordan We apply this then the- involu- the (Jordan) thering is result to nil radi- cal of S is nilpotent. Recently Jordan Britten ally, he has shown a prime Jordan quadratic Rj, [l] has proven ring of symmetric then has an analog S ^ Sj, ascending In this paper, we first extend which symmetric elements. the simplification The proof given comes for the More specificS of R are chain elements is *-simple Britten's ring and the Jordan Britten's; theorem elements or descending the symmetric a ring of quotients R is an arbitrary of Goldie's of a ring with involution. that if 2R = R and the symmetric ring with either ideals, and Rj elements condition of a *-prime ring Artinian. theorem to the situation ring is only an ample subring here shorter is considerably from using on an idea of Lanski where of the than [5] on di- rect sums of right ideals. As an application, condition on quadratic question remains Before associative open beginning, ring, of R of period and we show that ideals, then for general we need if S satisfies the nil radical Jordan some 2). We let An ideal the product / of R is called of two nonzero *-ideals a *-ideal is, = x\ denote ments of R, TR = \x + x \x £ R] the traces, norms. of S is nilpotent. R will always on R (that SR = \x £ R\x chain This algebras. definitions. * an involution the ascending the symmetric and NR = \xx ii I is nonzero, denote an an antiautomorphism = I; thus, ele- \x £ R\ the R is *-prime and ^-simple if there if are no Received by the editors August 13, 1973. AMS iMOS) subject classifications (1970). Primary 16A28, 16A68; Secondary 16A46. Key words and phrases. Involution, symmetric elements, Goldie ring, nil radical. 1 This research was supported in part by NSF Grant #GP 29119—X. Copyright © 1974, American Mathematical Society License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use 325 326 SUSAN MONTGOMERY proper *-ideals. it is easy We say to see that In what follows, / R is semiprime a *-prime / will be an additive if R contains no nilpotent ideals; Jordan that is, ring is semiprime. will denote a special quadratic ring; of R closed under the quadratic operator 2 xU = yxy and the binary composition x . A Jordan subring / of SR is y * called ample if it contains all norms and traces and if xjx C /, for all x £ R. An additive subgroup subgroup We abbreviate Recall that R has no infinite Risa We begin / C / isa the ascending right direct Goldie sums with an easy Lemma 1. Let quadratic ideal of / if JU. C /. (descending) chain ring if R has of right lemma. condition ACC on right by ACC (DCC). annihilators and ideals. Part R be a semiprime (ii) is part of Britten's ring with *, and Lemma J an ample A [l]. Jordan subring of S. Then: (i) If a £ J with a]a = 0, then a = 0. (ii) // / is a right {left) ideal of R with I (1 / = 0, then Q = \x + x*\ x £ 1} is a quadratic ideal of J. Proof, axax (i) Since a £ J and a = 0. On the other hand, axa = —ax a, and so axaxa is a nil right R has ideal a nilpotent (ii) Say that 0. Thus, ideal theorem Theorem 1. Let sums of quadratic index. (a contradiction), ideal, £ J and in /, a*Ja xax £ J for all unless and choose a*saja*sa Theorem R be a *-prime ideals, {ax) By a theorem = 0; that is, aR [3, p. l], a = 0. a £ I. Then a]a that JU C / n/ = 0. Since « Ç Q, fot all = s a £ I. 1 of Britten. ring and J ample. then gives of Levitzki = 0. By (i), a*sa = 0. It now follows extends x £ R. Thus, £ J so a{x + x )a = 0. This = 0, all x £ R. But then / is a right The first direct is ample, of R of bounded if s £ ], a*sa was arbitrary / x + x R has no infinite If J has no infinite direct sums of right or left ideals. Proof. Assume sums of quadratic ideals whose that ideals the result is false, that is, / has no infinite but, in R, M = \T. ! is an infinite direct collection of right tT¡ by an appropriate subset) sum is direct. We claim that we may assume that any finite sum of ideals (by replacing in % intersects / only in (0). For, \Tn, Tn ., • • • }. Now, if 111= tIÎj fails, there exists let %. = a finite sum Sj = T,1 +T, 2 + --- + T «1 —.1 such that S.1 n /■>^ (0). Then try• /Tln\ = \T nY ,•••!. If 1ÍÍ fails, there must be a finite sum Sn = T + — + T , with n j ' 2 n\ License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use 722-1 327 CHAIN CONDITIONS ON SYMMETRIC ELEMENTS Lfl/ 4 (0). Repeat the argument: if )1I fails, there is some Sk . 4 + ■•• + T , with S, ,nj4 (0). But if S, exists for all k, then T the set \S, C\J\ is an infinite direct sum, a contradiction. finite sum of right set of quadratic Thus, ideals for some intersects ideals of / which form a k, it must be that in IK / nontrivially. nk is the desired % , no sub- set. Now for each T. el, dratic ideal of/. Since/ consider Q . = \x + x*\x £ T. i. By Lemma 1, Q . is a qua- has no infinite direct set \Qj, • • • , Q I which is dependent: a. = x.l + x*l fot 1l nite direct x.l £ T.7 and let sum in {Q fö„+i» ••• ,Qml sums of quadratic ideals, there is a finite for some q. £ Q., not all zero, Sç. = 0. Write a = Sx. 7 £ £"_.T.. Z —1 ,, • • • Î, there must Since I be another finite there is no infi- dependent set Say that 1pj = 0 for not all p. £ Q. zero. Let p. = y. + y* and b = Ey/ ;. e" Sm /; n +1,T..7 Then a + a* = b + b* = 0:' thus a and b ate skew and are nonzero by independence of the T,. Now let r £ R. Then ore + br*a £ (ST. + ST ) O / = (0), and so are = -fJ7-*a. But then arb £ {1T{) n{2T ) = (0) by the directness of \T k\. Thus, izrt3= 0, all r £ R. But then {RaR){RbR) =0, and RaR and RbR ate *-ideals of R since a and a contradiction. b ate skew. Thus, Now if 17 is a subset annihilator ann.fi it, a properly B . = ann;A .. Then an infinite and that that R be *-prime. of right the ideals. of U. If annihilators in R, let B . are a properly descending A . = ann B .. 7 r The next lemma is also part of Britten's Lemma 2. Let sum of right the left annihilator chain to verify of R, a = 0, or b = 0, direct ann U = \x £ R\ Ux = 0], the right denotes ascending it is easy chain of left annihilators have of R, we write of U. Similarly, Aj C A2 C • • • By the *-primeness R cannot Then 7 Lemma A. if C is the right {left) annihilator of a nonzero ideal of R, C ft) C* = 0. Proof. Let / be the ideal of R, so that IC = 0. Then C7 is a right ideal of R with {CI) - 0, and so CI = 0 since R is semiprime. Applying have I*C* = 0 and thus (/ + 1*){C O C*) = 0. Since R{C nC*)R of R, we must have C DC We are now able plification of his =0 to extend since *, we is a *-ideal R is *-prime. Britten's Theorem 4. The argument is a sim- argument. Theorem 2. Let R be *-prime and J ample. If J has ACC or DCC on quadratic Proof. ideals, then If / has R is a Goldie ring. ACC or DCC on quadratic ideals, License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use then it is easy to see 328 SUSAN MONTGOMERY that / has no infinite it suffices We first treat not Goldie, right direct to show that let sums of quadratic R satisfies the case ideals. ACC on right when / has Thus, ACC on quadratic A. C iá, C • •'•'C A. C • •« be a proper annihilators in R, and let by Theorem annihilators. ideals. If R is ascending B . be the corresponding 1, ; chain descending of chain of left annihilators. First assume that we may as well assume an ascending n/ chain C5 A . DJ 4 0 for some that of quadratic A = • • • , We claim i. Since A . n / 4 0 for all ideals that this gives the i. Then A . ate ascending, \A . D J] of /, 7 so for some a contradiction. is m, A 777777 n J = A For, choose , + 1 a £ A O / and y' £ A772+ ,. Then y a + ay* £ A 777+1. D J = A772'D /, and so 0 = 1 J * B 777{ya + ay*) = B mJ ya, ' since J J S 777772 A , 7^0, and a £ J"_' C S, a = 0 byI Lemma 2. Thus Am n / = 0. ■ <2- be given that 1. Since a—a a + a* £ Q xm =-{a—a 777 that a is skew. is, ' (R Q. ate ascending, , = Q , so x m1 each any .A 772+ 1 y £ A But then for some Aa = 0. As above, 2. Then this ' contradicts m, the eives & Q = a £ A 772+1'., a mm £ A . So we may' assume ay* + ya* = ay* — ya £ B 777+ 1. ay* = 0 = B777+ ,ya ' 1-7 777+ / A ., let a + a* = a mm1+ a* , tot some + l Now choose A 777+ ,1 = A 777', which ' the )* £ A 777+7'., 4 A m' , and is skew. 777 A 777+ ,2 n / = 0,» so ay* = ya. J J Thus, A . n / = 0, for all ¿. For We claim that A777+ ,1 = A777. For,' if not,' choose a i A 777. Then ' assume as in Lemma Q x 777+ 1,=•••. That {B 777A 777+1,)a = 0. Since + 1' We may therefore Thus, a £ A 777 = ann r R 777. Thus since a £ A 777+ ,.1 o = 0,» a contradiction. A ,. being t> a proper r r chain. The proof for DCC is very similar; if B . n / = 0 for all i, get the Q.' s as above. again have assume chain Then we can choose y*a = ay, so a{B that a skew in B 772+1',, not in fi 777+¿,. A B . O / ^ 0, for some of quadratic 1 ideals, ' A = 0 and ¿. But the so for some may assume ß see that a(S A .) = 0,' so a = 0. 772 777+2 7 O / 4 0. But then, 772 We are now able Corollary 1. Let to extend ' Britten's J be an ample a = 0 as before. 1/5.0 /J m,' B 772 n/ choosing 6 main subring to an ample Proof. Let as [2, Lemma P{R) Jordan denote subring J. l], P{R) H / Ç P{J), the prime 777+ 1, »J n/ we may descending = • • • and we C\ J and y £ B , we . . ' m' theorem: of SR, where of a *-prime the prime radical Thus, are a proper = ß a £ B *. If J is prime and has ACC or DCC on quadratic morphic If y £ B m1 , we J ideals, Goldie of R. Then radical of /. R is a ring with then J is iso- ring Rj. by the same proof But since / is prime, P{J) = 0; thus, P{R) D / = 0. Let Rj = R/P{R); Rj is semiprime. Now /,, /j the image of / in Rj, is isomorphic to / since License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use P{R) HJf = 0. Also CHAIN CONDITIONSON SYMMETRICELEMENTS is ample since / is (R, It remains has an induced only to show that Rj involution is actually since 329 P{R) * = P{R)). *-prime (since then Rl Goldie follows from Theorem 2). Say AB =0, where A, B ate 4 0 *-ideals of R. Then A nj 4 0 (since if A O / = 0, a + a* = 0=o*a = a2, for all a £ A, and R would contain a nilpotent ideal). Similarly, B CiJ 4 0. But AH/ =0, and fin/ are ideals of /, and {A ^DUgf^j contradicting / being prime. Since R. of quotients in fact is a *-prime A which A is *-simple / has a Jordan Goldie ring, by Goldie's is semisimple Artinian. (see Theorem Britten's theorem Rj It is not difficult has a ring to see that 5). We can therefore show that ring of quotients. Corollary 2. Let J, R, J., and R. be as in Corollary 1. Then J has a Jordan ring of quotients Q(J). If A is the associative ring of quotients for Rj, then Q{J) ~ Q{JA is an ample subring of SA, and Q{SR ) = SA. When 2R = R, Q{J) is a simple Jordan ring. Proof. Using every regular Corollary element that rx - 0 for some x £ R,. -x*, so * = -*** 1, we may work entirely of / is regular £ J. of r is nil of index tradiction). Thus, x U = 0, x is ample ment of / is regular The fact that =0, whenever 4.5], which that and say so x + x* = 0. Then x = =0. But this xr = 0 implies R has We claim r £ J, regular, 2, and so R, contains x = 0. Similarly, We can now apply [7, Theorem and Q{J,) in R,. For let Then {x + x*)U Since annihilator S,, in R.. says that the right a nilpotent ideal (a con- x = 0. states exactly a ring of quotients that Q(JiR) = A and every ele- in R. when 2R = R, (?(/) — S . is simple follows from a theorem of Herstein [3, Theorem 2.6].. Having extended apply them to study the Jordan radical ring homomorphic As in Corollary P{J) S, since the (Jordan) the symmetric to arbitrary of S. Let 1, P{R) prime elements N{J) characteristic, denote we now the nil radical will be the (associative) radical of /. of prime We work with./, might not be preserved under images. Lemma 3. Let R be a *-prime on quadratic ideals. Proof. results the nil radical /. of R, and rather than Britten's Since ring in which J has either ACC or DCC Then N{J) =0. / is ample and N{J) is an ideal of /, N{J) is a core ideal License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use 330 SUSAN MONTGOMERY of S {tot details associative see [6, p. 387]). *-ideal Thus, B of R such if N{J) 4 0, there that the core exists KAB) Ç N{J), a nonzero where KAB) = \b+ b* + Se. r3*|è, b. £ B}[6, Corollary, p 387]. We claim that B must contain a nonzero 0, or b* = -h, is nil. Then nil right ideal. If KAB) = 0, this is trivial; tot all b £ B. But then also bb* = -h We may therefore x" = 0, x"~ assume 4 0 for some that KAB) 4 0. Choose 72, so letting for b + b* = =0, and so B itself x £ KAB), y = xn~ we have x 4 0. y =0, y 4 0. Now for any b £ B, yb + b*y £ XAB) Ç N{J), and so for some k, {yb + b*y) = 0. Multiplying on the right by yb, we see {yb) + = 0; that is, yB is nil. By Theorem subring 2, R is a Goldie of R is nilpotent. an ideal in a semiprime Thus, ring, ring. By Lanski's B contains so is itself theorem a nilpotent [A], any nil ideal. a semiprime ring, ring with * such But B is a contradiction. Thus, N{J) = 0. Theorem 3. Let R be an associative on quadratic ideals. Then on quadratic ideals, then Proof. We will first this by showing For, consider prime. the nil radical of S is nilpotent. the nil radical show that of S equals that if P is any proper prime ideal S has ACC If S has the prime N{S) = P{S) in either the ring R = R/P OP* (possibly that situation. of R, then DCC radical. We do P 2 N{S). P DP* = P = P*); R is *- Let / = S , the image of S in R. Then / is ample, has ACC or DCC on quadratic ideals, and so N{J) = 0 by the lemma. But N{S) C N{J); thus, N{S)ÇP DP* Ç P. Now P{R) is the intersection of all prime ideals of R, and so N{S) Ç P{R). By [2, Theorem 3], P{R) n S = P{S). But N{S)Ç P{R) n S and N{S) D P{S). Thus, N{S) = P{S). In the case ideal, when S has ACC, P{S) will simply be the maximal nilpotent so N{S) is nilpotent. REFERENCES 1. Daniel Britten, On prime Jordan rings HiR) with chain condition, J. Algebra 27(1973), 414-421. 2. T. S. Erickson and S. Montgomery, The prime radical in special Jordan rings, Chicago, 111., Trans. Amer. Math. Soc. 156 (1971), 155-164. MR 43 #306. 3. I. N. Herstein, Topics in ring theory, Univ. of Chicago Press, 1969. MR 42 #6018. 4. C. Lanski, Nil subrings of Goldie rings are nilpotent, (1969), 904-907. MR 40 #1428. License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use Cañad. J. Math. 21 331 CHAIN CONDITIONS ON SYMMETRIC ELEMENTS 5. C. Lanski, 6. Kevin Chain McCrimmon, conditions in rings On Herstein with involution s theorems (to appear). relating Jordan and associative algebras, J. Algebra 13 (1969), 382-392. MR 40 #2721. 7. Susan Montgomery, Rings of quotients fora class of special Jordan rings, J. Algebra 31 (1974), 154-165. DEPARTMENT OF MATHEMATICS, UNIVERSITY OF SOUTHERN CALIFORNIA, LOS ANGELES, CALIFORNIA 90007 License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use