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...

• Hint: pick somewhere for $1$ to go under your homomorphism. Now where does an arbitrary $r$ have to go?
– user29743
Commented Apr 28, 2012 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 $\varphi(1)$. Commented Apr 28, 2012 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? Commented Apr 28, 2012 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). Commented Apr 28, 2012 at 22:16
• @BrettFrankel Thanks. Looks like I was trying to make things much too complicated. Commented Apr 28, 2012 at 22:50

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.