Answer to Question #17364 in Algebra for Melvin Henriksen

Question

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.

Answer 1

(1) ⇒ (2) since MR is semisimple, and (2) ⇒ (3) is a tautology.
(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 R_R with a fixed rank 1 free direct summand of F,then A is the image of some R-endomorphism of F. SinceEnd(F_R) is von Neumann regular, A is a direct summandof F_R, 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 R_Rby (4). But this means that A = eR for some e = e²∈ R. Since e belongs to some a₁R+ · · · + a_nR, the chain (∗) stabilizes after n steps—acontradiction.