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# Answer to Question #17364 in Algebra for Melvin Henriksen

Question #17364
For any ring R, show that the following statements are equivalent:
(1) R is semisimple.
(2) The endomorphism ring of every right R-module M is von Neumann regular.
(3) The endomorphism ring of some infinite-rank free right R-module F is von Neumann regular.
(4) Any countably generated right ideal A ⊆ R splits in R.
Expert's answer
(1) &rArr; (2) since MR is semisimple, and (2) &rArr; (3) is a tautology.
(3) &rArr; (4). Let F and A be as in (3) and (4). Certainly, A isthe image of some R-homomorphism from F to R. If weidentify RR with a fixed rank 1 free direct summand of F,then A is the image of some R-endomorphism of F. SinceEnd(FR) is von Neumann regular, A is a direct summandof FR, and hence of R.
(4) &rArr; (1). Under assumption (4), R is certainly von Neumann regular.(1) will follow if we can show that R is right noetherian. Assume,instead, that there exists a strictly ascending chain of right ideals in R.From this, we can construct a strictly ascending chain
(&lowast;) a1R &sube; a1R + a2R&sube; a1R + a2R+ a3R &sube; &middot; &middot; &middot; (in R).
The union A of this chain isa countably generated right ideal, and hence a direct summand of RRby (4). But this means that A = eR for some e = e2&isin; R. Since e belongs to some a1R+ &middot; &middot; &middot; + anR, the chain (&lowast;) stabilizes after n steps&mdash;acontradiction.

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