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Let $R$ be a ring, and let $V$ denote the $R$-module $R$. Determine all homomorphisms $\varphi:V\rightarrow V$.

For any $\varphi$, ker $\varphi$ and im $\varphi$ both have to be submodules of $V$. In this case, that makes them ideals of $R$. So every $\varphi$ is a surjective map from $V$ to an ideal of $R$. $V$ consists of everything in $R$ so, if $I$ is some ideal of $R$ I'm really looking for every $\varphi:R \twoheadrightarrow I$.

That's about as far as I've managed to get. I'm not even sure what form the answer is supposed to take.

Thanks...

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Hint: pick somewhere for $1$ to go under your homomorphism. Now where does an arbitrary $r$ have to go? – countinghaus Apr 28 '12 at 21:38
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If you are working with right action, the homomorphisms are given by left multiplication, for precisely the reason that countinghaus mentioned, namely that $\varphi$ is determined by $\varphi(1)$. – Brett Frankel Apr 28 '12 at 21:41
So if $\varphi(1)=v$, then $\varphi(x)=vx$ for every $x\in V$. Then my answer is there is a homomorphism $\varphi_v$ that sends $x\leadsto xv$ for each $v\in V$. Is that right? – jobrien929 Apr 28 '12 at 22:13
Yes, assuming your ring action is on the right. Otherwise, just reverse the order of multiplication. (If you're working in a commutative ring, it's all the same of course). – Brett Frankel Apr 28 '12 at 22:16
@BrettFrankel Thanks. Looks like I was trying to make things much too complicated. – jobrien929 Apr 28 '12 at 22:50

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