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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.

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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 $\beta\alpha$.

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  • $\begingroup$ Thanks a lot. Very helpful and clear! $\endgroup$ – Francesco Gemma Feb 9 '16 at 18:02

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