$M$ is a simple module if and only if $M \cong R/I$ for some $I$ maximal ideal in $R$.

Question

I am trying to show the following statement (taken from Rotman's Advanced Modern Algebra):

Let $M$ be an $R$-module. Then $M$ is a simple module if and only if $M \cong R/I$ for some $I$ maximal ideal in $R$.

If $M$ is simple, then $0 \neq M$, so I can pick $x \in M, x \neq 0$. Since $M$ is simple, we have $Rx=M$. If I define $f: R \to M$ as $f(r)=rx$ then by the first isomorphism theorem, $R/Ker(f) \cong Im(f)=M$. How can I show that $Ker(f)$ is a maximal ideal?

I would also appreciate some help to prove the other implication.

Answer 1

Denote $I=\ker f$.

$R/I=M$ is simple, hence the only submodule (ideal) of $R/I$ is $0$ and itself. This show that $R/I$ is a field, so $I$ is maximal.

If $I$ is maximal, then $M=R/I$ is a field hence simple.

Answer 2

Assume not. Then because every ideal is contained in a maximal ideal. If $\ker f$ is not maximal, then $(\ker f)\subset M_0\subset R$ with $M_0$ maximal, and $M_0/(\ker f)$ is a non-trivial ideal of $R/(\ker f)$, contradicting simplicity of $M$.