Essential extensions and intersection theorems

If R is right and left noetherian, primitive factor rings are artinian, and R is right fully bounded, then a simple proof is given to show that finitely generated essential extensions of right artinian modules are artinian. An immediate corollary is that n 1jn = 0 for such a ring. Let R be a ring with 1 and J its Jacobson radical. Recall that a ring is right fully bounded if essential right ideals of prime factor rings contain nonzero two-sided ideals. That rings satisfying a polynomial identity are right fully bounded is shown in [1]; indeed essential right ideals contain central elements. Jategaonkar [3] has recently proved that if R is a right and left noetherian, right and left fully bounded ring, then finitely generated essential extensions of artinian modules are artinian. His proof relies heavily on some very complicated and difficult results also in [3]. Our purpose is to present a short elementary proof of the following theorem, which has Jategaonkar's result as an immediate consequence. We note, however, all known examples of right and left noetherian rings which are right fully bounded, are also left fully bounded and so our Theorem may not be more general. Implications of the Theorem for the intersections of the powers of ideals other than the radical are given in [4]. Theorem. If R is right and left noetherian, left1 primitive factor rings are artinian9 and R is right fully bounded then f.g. essential extensions of right artinian modules are right artinian. Before proving the Theorem we note that the sum of the injective hulls of simple right R-modules is faithful (indeed an injective cogenerator for Mod R), yet every f.g. submodule of this sum will be artinian, and so annihilated by Jn for some n. As Jategaonkar notes, this yields the following Corollary. If R is as above, n ljn = 0. Received by the editors May 17, 1974 and, in revised form, August 6, 1974 and October 7, 1974. AMS (MOS) subject classifications (1970). Primary 16A46, 13E05; Secondary 16A26, 16A38. 1 I would like to thank T. Lenagan for pointing out to me that we may drop the assumption on left primitive factor rings, since the sequence of ideals Q. in our proof may be chosen to be right primitive by using the ideas of [4]. Copyright ? 1975, American Mathematical Society 328 This content downloaded from 157.55.39.215 on Wed, 31 Aug 2016 04:44:19 UTC All use subject to http://about.jstor.org/terms ESSENTIAL EXTENSIONS AND INTERSECTION THEOREMS 329 We require the following generalization of Nakayama's lemma. Lemma. If R is any ring, RM a finitely generated R-module, then there exists a left primitive ideal Q such that QM C; M. Proof. Take In Rm. = M such that m I ( Y. Rm.= X, Then M z--l I i z2 I M/Rm 1 r) X ? RR 1 (D3 S where S is a simple homomorphic image of R-m. If g: M Al5 S as above, then g(annRS . M) = annRSg(M) = 0, so (annRS) . M M. O Proof of the Theorem. Let MR be f.g., and an essential extension of an artinian module AR. Let XR be a maximal artinian extension of AR in MR, and suppose X MA M. Take M'l;? XR in M such that annRAl/X is maximal among such annihilator ideals, and write annRM /X = P since it is clearly prime. P/annRMlI is a f.g. right R/annXR module, and the latter is artinian since it contains a finite product of primitives, namely the annihilators of the simple factor modules in a composition series for XR, Take a sequence of left primitive ideals Q1l Q2.... by the Lemma such that Q i it '' l( P/annR M# !r Qj_ I.'.' ' I t(P/ann eM) Thus QkQk 1-I Q1P C annRMA If Qi C P for some i we are done, since R/P is then artinian, and if not then 0 MQk.. . Q1 + X/X, and it has anni- hilator P, and so it is a faithful R/P module (and hence not singular since the intersection of the annihilators of a finite set of generators would be essential and so contain a nonzero two-sided ideal). But M Qk*Q1 has essential submodule X Al M Qk ... Ql so M'Qk... Q/X n MQ Qk.. Ql is a singular R/P module. E We remark in closing that it is well known that left fully bounded to- gether with left noetherian imply that left primitive factor rings are artinian. (See [2, Theorem] for a much more general and stronger statement or, in our case, simply argue as follows: if RS is a simple faithful R-module, and 0 g s c S, then annRs is not essential, so annRs n X =0 for some left ideal X, and thus X contains an isomorphic copy of S as a minimal left ideal. But prime left noetherian rings with a left minimal ideal are easily seen to be artinian.) BIBLIOGRAPHY 1. S. A. Amitsur, Prime rings having polynomial identities with arbitrary coef- ficients, Proc. London Math. Soc. (3) 17 (1967), 470-486. MAR 36 #209. 2. R. Gordon and J. Robson, Krull dimension, Mem. Amer. Math. Soc. No. 133 (1973), p. 58. 3. A. Jategaonkar, Jacobson's conjecture and modules over fully bounded noe? therian rings, J. Algebra 30 (1974), 103-121. This content downloaded from 157.55.39.215 on Wed, 31 Aug 2016 04:44:19 UTC All use subject to http://about.jstor.org/terms