# Endomorphisms of a ring $R$ considered as $R$-module

Let $R$ be a ring. Determine all $R$-module homomorphisms $\varphi:R\rightarrow R$.

For any $\varphi$, $\ker\varphi$ and im $\varphi$ both have to be submodules of $R$. In this case, that makes them ideals of $R$. So every $\varphi$ is a surjective map from $R$ to an ideal of $R$ 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? –  user29743 Apr 28 '12 at 21:38
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

This is essentially an expansion of Brett Frankel's comment. I'm used to work with left actions, so I'll post an answer in that "language". If you work with right actions instead, you have to reverse everything that I say in the remainder (I'll explain the possibilities after the calculation).

So we are looking for all left module homomorphisms $\varphi:R\to R$. I claim that they are all given by multiplications from the right with a particular element $x$, call this map $\rho_x$. First, we have to check that these are indeed left module homomorphisms: $$\rho_x(rm)=(rm)x=r(mx)=r\rho_x(m)$$ by associativity in $R$ and furthermore $$\rho_x(m+n)=(m+n)x=mx+nx=\rho_x(m)+\rho_x(n)$$ by distributivity in $R$, so $\rho_x$ is a left module homomorphism.

Now we have to check that two elements $x$ and $y$ give different $\rho_x$ and $\rho_y$. So assume that they give the same map. Then $$x=1x=\rho_x(1)=\rho_y(1)=1y=y.$$

Thirdly we have to check that each $\varphi$ is in fact a $\rho_x$ for some $x$. For this define $x:=\varphi(1)$. Then we have: $$\varphi(r)=\varphi(r\cdot 1)=r\varphi(1)=rx=\rho_x(r)$$ Thus $\varphi=\rho_x$.

What we have proven up to now is that $\operatorname{End}(R)\cong R^{op}$ as sets, where $R^{op}$ is the opposite ring of $R$ given by the same underlying abelian group with new multiplication $r*s:=sr$. In fact this is an isomorphism of rings (if you use left notation as I will) which we will check now: $$\rho_x\rho_y(r)=\rho_x(ry)=ryx=\rho_{yx}(r)=\rho_{x*y}(r)$$ This we have that the map $R^{op}\to \operatorname{End}(R)$, $x\mapsto \rho_x$ is compatible with multiplication. I'll leave it to the reader to check the simpler verification that it is also compatible with addition.

One comment on how you can reverse things: If you use left notation of maps and right action of modules then you would have $\operatorname{End}(R)\cong R$. (without any op's). Similarly if you use right notation and left actions. If you use right notation and right action you would of course have $\operatorname{End}(R)\cong R^{op}$ again.

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