This is from an old Ph.D Qualifying Exam of Algebra.
Let $R$ be a commutative Noetherian ring with unity and $M$ be a nonzero simple $R$-module. Prove that there exists a prime ideal $I\subset R$ such that $R/I \cong M$ as an $R$-module.
My solution: Since $M$ is simple, it is cyclic. i.e. it is generated by a single element, say $m$.
Then $\text{Ann}(M)=\text{Ann}(m)$, where $\text{Ann}$ means the annihilator of a given module(or an element). Now prove that $I=\text{Ann}(m)$.
Define a map $\phi: R\to M$ as $\phi(r)=rm$, $r\in R$, $m$ is a generator of $M$. Clearly $\phi$ is an $R$-module epimorphism and $\ker(\phi)=\text{Ann}(m)$. Therefore, $R/\text{Ann}(m)\cong M$.
Now it remains to show that $\text{Ann}(m)$ is a prime ideal of $R$.
Let $a,b\in R$ and $ab\in \text{Ann}(m)$. Suppose that $b\notin \text{Ann}(m)$. Then $bm\neq 0$, so the submodule $(bm)$ of $M$ is nonzero and thus $(bm)=M$ by simplicity of $M$.
$\therefore \text{Ann}(bm)=\text{Ann}(m)$, and $abm=0$ implies $a\in \text{Ann}(bm)$, so $a\in \text{Ann}(m)$. This completes the proof.
On the solution above, the Noetherian condition of $R$ is not used. Is my solution correct? If so, then the Noetherian condition is indeed unnecessary.
That seems okay to me. A slight generalization: one can take $I$ maximal.
As you said, given a nonzero $m \in M$ we have a module epimorphism,
$$ l : R \longrightarrow M \\ r \mapsto rm $$
and therefore if $I := \ker l$ by the first isomorphism theorem we get that $M \simeq R/I$. Now, this epimorphism induces a bijection from the $R$-submodules of $M$ to the $R$-submodules of $R/I$, which are in correspondence with the $R$-submodules (that is, the ideals) of $R$ that contain $I$. In particular, since $M$ has only two submodules, there are only two submodules that contain $I$: either $R$ or itself. This proves that $I$ is maximal, and in particular, prime.