To show module homomorphism being injective

Question

Suppose $R$ is a field, $M$ is a right $R$ module, and $f : R_R \rightarrow M$ is a non-zero homomorphism. Show that $f$ is injective.

My work: The homomorphism is uniquely determined by $f(1)$. If $f(1)=0$ then it is the zero homomorphism. So suppose $f(1)=m\ne0$. Then $f(r)=mr$. To show that it is injective, we need to show that $mr\ne0$ for all $r\ne0$.

So the question becomes showing that $mr\ne0$ for all $m\ne0$ and $r\ne0$.

Are my arguments correct? Then how to show $mr\ne0$ for all $m\ne0$ and $r\ne0$?

Answer 1

Hint: Consider $\ker f$: it's an $R$-submodule of $R_R$. But the $R$-submodules of $R$ are its ideals.

Solution:

$\ker f$ is an ideal of $R$. Since $R$ is a field, $\ker f$ is either $0$ or $R$. Since $f$ is a non-zero homomorphism, $\ker f$ cannot be $R$ and so must be $0$. This means that $f$ is injective.

Alternatively, we can use your argument:

If $mr=0$ with $r\ne0$, then $m=m1=m(r(1/r))=(mr)(1/r)=0$, which means that $f$ is the zero homomorphism.