Let $M$ be a module over a ring $A$ and $R=Hom_A(M,M)$ its endomorphism ring (with respect to the composition). I need to show these following conditions are equivalent:

  1. $\alpha = \alpha \beta \alpha$, for some $\beta \in R$
  2. $\ker(\alpha)$ and im$(\alpha)$ are direct summands of $A$
  3. the right ideal $\alpha R$ is a direct summand of $R_R$
  4. the left ideal $R \alpha$ is a direct summand of $_RR$

Searching on the web and on this site, I found out we are talking about Von Neumann regular rings. Also, I found a proof of $1\Leftrightarrow2$. However, I can't prove the other implications. Can someone help me or just give any suitable reference where I can find a proof? Thank you all.


1 Answer 1


Assuming 1, $\alpha\beta$ is idempotent, and $\alpha R=\alpha\beta\alpha\subseteq \alpha\beta R\subseteq \alpha R$, so there is equality. Clearly $\alpha\beta R$ is a summand.

In the other direction, if $aR=eR$ is a summand ($e$ idempotent) then $a=er$, and consequently $ea=er=a$. Also, $e=as$, but multiplying on the right by $a$ you get $asa=ea=a$. (Sorry, I got tired of Greek letters.)

We've shown $1\iff 3$ at this point.

You can prove $1\iff 4$ analogously with $\beta\alpha$.

  • $\begingroup$ Thanks a lot. Very helpful and clear! $\endgroup$ Feb 9, 2016 at 18:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.