# 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.

(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)

(3)

(4)

(

The union

*⇒**(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*with a fixed rank 1 free direct summand of_{R}*F*,then*A*is the image of some*R*-endomorphism of*F*. SinceEnd(*F*) is von Neumann regular,_{R}*A*is a direct summandof*F*, and hence of_{R}*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(

*∗*)*a*_{1}*R**⊆**a*_{1}*R*+*a*_{2}*R**⊆**a*_{1}*R*+*a*_{2}*R*+*a*_{3}*R**⊆**· · ·*(in*R*)*.*The union

*A*of this chain isa countably generated right ideal, and hence a direct summand of*R*by (4). But this means that_{R}*A*=*eR*for some*e*=*e*^{2}*∈**R*. Since*e*belongs to some*a*_{1}*R*+*· · ·*+*a*, the chain (_{n}R*∗*) stabilizes after*n*steps—acontradiction.Need a fast expert's response?

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