Prove that if A is simple every R-module endomorphism is either the zero map or an isomorphism. How can I prove that: "Prove that if R has an identity and A is a nonzero unitary R-module and if A is simple every R-module endomorphism is either the zero map or an isomorphism" **My answer is:** Let $f: A \rightarrow A$ be the R-module endomorphism, since A is simple then Kerf is either 0 or A because kerf is a submodule of A, Imf is either 0 or A because Imf is a submodule of A.if f is not the zero map, then Kerf is 0 and Imgf is A, and f induces an isomorphism A/Kerf isomorphic to Imf by first isomorphism theorem. Is my answer right? Thanks!

It's right, but it can be made a bit simpler. First, a simple module $A$ is nonzero. If $f\colon A\to A$ is a homomorphism and $x\in A$, $x\
e0$, then there are two cases:

1. $x\in \ker f$
2. $x\
otin\ker f$



In case 1, $\ker f\
e\\{0\\}$, so $\ker f=A$ and $f$ is the zero map.

In case 2, $\ker f\
e A$, so $\ker f=\\{0\\}$ and $f$ is injective; moreover $f(x)\
e0$ and $f(x)\in\operatorname{im}f$, so $\operatorname{im}f=A$ and $f$ is surjective.

Of course, the key fact is that $\ker f$ and $\operatorname{im}f$ are submodules of $A$.