52 798
Assignments Done
Successfully Done
In October 2017
Your physics homework can be a real challenge, and the due date can be really close — feel free to use our assistance and get the desired result.
Be sure that math assignments completed by our experts will be error-free and done according to your instructions specified in the submitted order form.
Our experts will gladly share their knowledge and help you with programming homework. Keep up with the world’s newest programming trends.

Answer on Algebra Question 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) ⇒ (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 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.

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!


No comments. Be first!

Leave a comment

Ask Your question