8
$\begingroup$

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...

  • 6
    Hint: pick somewhere for $1$ to go under your homomorphism. Now where does an arbitrary $r$ have to go?2012-04-28
  • 2
    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 $\varphi(1)$.2012-04-28
  • 0
    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?2012-04-28
  • 0
    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).2012-04-28
  • 0
    @BrettFrankel Thanks. Looks like I was trying to make things much too complicated.2012-04-28
  • 0
    @BrettFrankel Please consider converting your comment into an answer, so that this question gets removed from the [unanswered tab](http://meta.math.stackexchange.com/q/3138). If you do so, it is helpful to post it to [this chat room](http://chat.stackexchange.com/rooms/9141) to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see [here](http://meta.stackexchange.com/q/143113), [here](http://meta.math.stackexchange.com/q/1148) or [here](http://meta.math.stackexchange.com/a/9868).2013-06-09

1 Answers 1