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I'm trying to prove that the following statements are equivalent for a commutative ring $R$

A: $_RR=_RN \oplus_RM$ for some submodules $_RN$, $_RM\subseteq_RR$

B: There exists an element $e=e^2\in R$ such that $N=Re$ and $M=R(1-e)$

I have no idea how to show $A \implies B$, but $B\implies A$ seems easy: every $r \in R$ can be written as $re+r(1-e)$ and for $r\in N\cap M$ $r=ae=b(1-e)$ for some $a,b\in R$, so $re=ae^2=b(1-e)e=b(e-e^2)=0$. Now if $R$ is a domain, $e=0$ and $M=R$ or $r=0$ and $N\cap M=\emptyset$. But what if it's not a domain?

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    Thank you for this comment, I've changed the tag to abstract algebra.2012-06-28

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For $B\implies A$, since $(1-e)e=0$, you see that $ae\in R(1-e)$ implies $ae=ae^2=0$. Thus the intersection is zero.

For $A\implies B$, look at $1=m+n\in M\oplus N$. Clearly n=1-m. So $m$ and $n$ are pretty good candidates for idempotents: check!

(Hint: one might start by multiplying $1=m+n$ on the left and right by $n$ to prove something about $mn$ and $nm$.)

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    @PhillyPhil Nice, your method is even better! About your generalization: not exactly. A ring decomposition $R=S\oplus T$ corresponds to a pair of *central* idempotents. The proper generalization of your question is this: for any module $_RM$, $M=N\oplus P$ iff there is an idempotent $e\in End(_RM)$ such that $e(M)=N$ and $(1-e)(M)=P$. Since $End(_RR)\cong R$ acting by right multiplication, you retrieve your question.2012-06-28