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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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In your work, you show that $ae^2 = 0$. And $ae^2 = ae = r$. –  Dylan Moreland Jun 28 '12 at 19:18
    
You might consider dropping or swapping the commutative algebra tag... it's completely true for noncommutative rings with unity, also. –  rschwieb Jun 28 '12 at 19:31
    
Thank you for this comment, I've changed the tag to abstract algebra. –  PhillyPhil Jun 28 '12 at 20:51
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up vote 2 down vote accepted

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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Thank you! So $m \in M, 1-m \in N \Rightarrow (1-m)m=m(1-m) \in M \cap N=0 \Rightarrow m=m^2$. Is it important, that we look at $R$, $M$ and $N$ as $R$-modules? These arguments work also if we treat $R$, $M$ and $N$ as rings, right? So is it true, that $_RR=_RN \oplus_RM \iff R=N \oplus M$? –  PhillyPhil Jun 28 '12 at 22:17
    
@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. –  rschwieb Jun 28 '12 at 23:49
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