By A.A. Tuganbaev
Preface All jewelry are assumed to be associative and (except for nilrings and a few stipulated instances) to have nonzero id components. a hoop A is expounded to be common if for each point a E A, there exists a component b E A with a = aba. commonplace earrings are good studied. for instance, [163] and [350] are dedicated to typical jewelry. a hoop A is related to be tr-regular if for each aspect a E A, there's a component n b E A such that an = anba for a few optimistic integer n. a hoop A is expounded to be strongly tr-regular if for each a E A, there's a confident integer n with n 1 n an E a + An Aa +1. it really is proved in [128] is a strongly tr-regular ring if and provided that for each aspect a E A, there's a optimistic integer m with m 1 am E a + A. each strongly tr-regular ring is tr-regular [38]. If F is a department ring and M is a correct vector F-space with endless foundation {ei}~l' then End(MF) is a typical (and tr-regular) ring that isn't strongly tr-regular. The issue ring of the hoop of integers with recognize to the best generated through the integer four is a strongly tr-regular ring that's not regular.
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3) A is a semiprimitive right Artinian ring. (4) A is a semiprimitive left Artinian ring. (5) A is a right semisimple ring. (6) A is a left semisimple ring. (7) A is a finite direct product of simple Artinian rings. (8) There are division rings D l , ... , Dn and positive integers ml, ... , mn such that A ~ Al X ... X An, where Ai is isomorphic to the ring of all (mi X mi)matrices over Di (i = 1, ... , n). Proof. It is sufficient to prove the equivalence of conditions (1), (3), (5), (7), and (8).
10) Let M be a submodule of M, and let 7 E Hom(M, M) ~ Hom(M, N). Since N is a quasiinjective module, 7 can be extended to an endomorphism 9 of N. Let I be the restriction of 9 to M. Since M is a fully invariant submodule in N, we have I E End(M). Since 11M = 7, the module M is quasiinjective. (11) Since LfEHom(X,N) I(X) is a fully invariant submodule in N, the proof follows from (10). (12) Since the identity automorphism of X belongs to Hom(X, N), the proof follows from (11). 0 For a module N, a module M is said to be projective with respect to N (or N -projective) if for every epimorphism h : N -t N and each homomorphism M -t N, there is a homomorphism I : M -t N with = hi.
Therefore, the series f is right invertible in R. f is left invertible in R. (9) By (5), '\(F) is a right ideal of A and there is a series f E F with the lowest term 1. By 6, f is invertible in R. Therefore F = A. 3. Laurent series rings that are division rings or domains. Let A be a ring, and let