A Decomposition Theorem for Noetherian orders in Artinian Rings

Ginn, S. M.; Moss, P. B.

doi:10.1112/blms/9.2.177

Cited by 15 publications

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“…We do not know whether in case A = B, A is a direct summand of R. Assume in addition that R is right and left noetherian, then A (=B) is a direct summand of R (see [8]). …”

Section: A Ring R Is a Right Order In A Hereditarily Artinian Ring Ifmentioning

confidence: 99%

A characterization of artinian rings

Huynh

Dũng

1988

Glasgow Math. J.

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Throughout this paper we consider associative rings with identity and assume that all modules are unitary. As is well known, cyclic modules play an important role in ring theory. Many nice properties of rings can be characterized by their cyclic modules, even by their simple modules. See, for example, [2], [3], [6], [7], [13], [14], [15], [16], [18], [21]. One of the most important results in this direction is the result of Osofsky [14, Theorem] which says: a ring R is semisimple (i.e. right artinian with zero Jacobson radical) if and only if every cyclic right R-module is injective. The other one is due to Vamos [18]: a ring R is right artinian if and only if every cyclic right R-module is finitely embedded.

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“…We do not know whether in case A = B, A is a direct summand of R. Assume in addition that R is right and left noetherian, then A (=B) is a direct summand of R (see [8]). …”

Section: A Ring R Is a Right Order In A Hereditarily Artinian Ring Ifmentioning

confidence: 99%

A characterization of artinian rings

Huynh

Dũng

1988

Glasgow Math. J.

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“…Let R be a left and right Noetherian ring which has an Artinian (classical) quotient ring. It was shown by S. M. Ginn and P. B. Moss [2,Theorem 10] that there is a central idempotent element e of R such that eR is the largest Artinian ideal of R. We shall extend this result, using a different method of proof, to show that the idempotent e is also related to the socle of R/N (where N, throughout, denotes the largest nilpotent ideal of R) and to the intersection of all the principal right (or left) ideals of R generated by regular elements (i.e. by elements which are not zero-divisors).…”

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confidence: 87%

“…We have hyf = 0, so that hf = hg. Set / = hf; then j'g = j = j 2 and / e fRf. But g is primitive, and gj and g -gj are orthogonal idempotents in fRf.…”

Section: Lemma (B) With the Above Notation Eehmentioning

confidence: 99%

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A note on Noetherian orders in Artinian rings

Chatters¹

1979

Glasgow Math. J.

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Throughout this note, rings are associative with identity element but are not necessarily commutative. Let R be a left and right Noetherian ring which has an Artinian (classical) quotient ring. It was shown by S. M. Ginn and P. B. Moss [2, Theorem 10] that there is a central idempotent element e of R such that eR is the largest Artinian ideal of R. We shall extend this result, using a different method of proof, to show that the idempotent e is also related to the socle of R/N (where N, throughout, denotes the largest nilpotent ideal of R) and to the intersection of all the principal right (or left) ideals of R generated by regular elements (i.e. by elements which are not zero-divisors). Before proving the main theorem we shall give for the reader's convenience the proof of a basic result which is a consequence of Goldie's theorem but which does not seem to be readily available in the literature. Let R be a semi-prime right Noetherian ring and let S be the right socle of R; we shall show that S is generated by a central idempotent element of R (in fact this result is true for any semi-prime right Goldie ring). There is a right ideal K of R such that S n K = 0 and S + K is an essential right ideal of R. We have KS = 0 because S is a two-sided ideal of R. By Goldie's theorem there is a regular element c of R such that ceS + K. Thus c = s + k for some s e S and k e K. Because S and cS have the same length, as right R -modules, we have S = cS. Hence s = ct for some teS. We have ct = (s + k)t = ct 2 + kt = ct 2 so that t = t 2 . Also if ueS then cu = (ct + k)u = ctu so that u = tu. Therefore S = tR. Finally, we have (S(l -i)) 2 = S(l -1 ) . fS(l -1) = 0 so that S(1 -1) = 0 from which it follows easily that S = Rt and that t is central.

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