Answer on Algebra Question for Melvin Henriksen
(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.
(3) ⇒ (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) ⇒ (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
(∗) a1R ⊆ a1R + a2R⊆ a1R + a2R+ a3R ⊆ · · · (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∈ R. Since e belongs to some a1R+ · · · + anR, the chain (∗) stabilizes after n steps—acontradiction.
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