Rings with Every Proper Image a Principal Ideal Ring

The main result of this paper states that if R is a right Noetherian right bounded prime ring such that nonzero prime ideals are maximal and such that every proper homomorphic image of R is a principal right ideal ring then R is right hereditary. In [10, Theorem 8] it is proved that if R is a right bounded prime ring of finite right Goldie dimension such that every proper homomorphic image is a right Artinian principal right ideal ring then R is right hereditary. It is not difficult to see that such rings are in fact right Noetherian and so [10, Theorem 8] is a consequence of the main result of this present paper. In fact, it is shown here that if R is a right Noetherian right bounded prime ring, such that nonzero prime ideals are maximal and such that for all nonzero prime ideals P, Q (not necessarily distinct) the ring R/PQ is a principal right ideal ring, then R is right hereditary (compare [10, Theorem 9]). In a recent paper, Hajarnavis and Norton [5, Theorem 6.4] proved that if R is a (right and left) Noetherian right bounded prime ring whose proper homomorphic images are ipri-rings then R is a Dedekind prime ring. The proof involves localization at every nonzero prime ideal of the ring R. We show how to deduce this result and in so doing completely remove localization techniques from the proof, making it much more elementary. If / is any ideal of a ring R then 6(7) will denote the set of elements c in R such that c + I is a regular element of the ring R/I. The ring R is an ipri-ring if every (two-sided) ideal is principal as a right ideal. Lemma 1. Let R be a right Noetherian prime ring such that every proper homomorphic image is an ipri-ring. If Mx and M2 are distinct maximal ideals of R then MXM2 = M2MX = Mxn M2. Proof. Since R is prime, but not simple, it follows that MXM2 n M2MX ̂ 0. Let R = R/(MXM2 n M2MX). Then R is a right Noetherian ipri-ring and Mx and M2 are distinct maximal ideals of R, where " denotes images in R. There exist elements cx, c2 in R such that Mx = cxR and M2 = c2R. By [2, Theorem 3.9], c, G G(M2) and c2 e C(A/,). It follows that Af, n M2 C cxM2 = MXM2 and hence Mx n M2 Ç MXM2. Thus Mx n M2 = MXM2 and similarly Mx n M2 = M2MX. Received by the editors October 30, 1979. AMS (MOS) subject classifications (1970). Primary 16A46, 16A60; Secondary 16A04, 16A12. © 1981 American Mathematical Society 0002-9939/81/0000-0100/$02.50 347 License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use