All simple cyclic module over $k[x]$ is of the form $k[x]/(f)$ with $f$ prime. Let $M$ a (left) $k[x]$ module. We suppose $M$ cyclic. Prove that $M$ is of the form $k[x]/(f)$ with $f$ prime.
I tried as follow : We have that $M=k[x]m$ for some $m$. Now, if $\varphi:k[x]\longrightarrow M$ (where $k[x]$ is seen as a $k[x]$ module) is s.t. $\varphi(1)=m$, we have that $$\varphi(p(x))=0\iff p(x)m=0$$
and thus $$M\cong k[x]/\text{Ann}(m).$$
Now, why $\text{Ann}(m)=(f)$ for $f$ a prime element of $M$. (By the way, what is a prime element of a module ?) 
 A: Let $R=k[x]$.
You forgot to mention in the question that $M$ is simple, as $R$-module (you wrote it in the title, though). Otherwise this is wrong, since $R$ is cyclic but not of the form $R/(f)$ with $f$ prime ($0$ is not considered as prime), because $R/(f)$ has torsion as $R$-module in this case (while $R$ is a free $R$-module, so it has no torsion).

You're right about $M\cong R/\mathrm{Ann}_{R}(m)$ (as $R$-modules). Since $R=k[x]$ is a PID (because $k$ is a field), the ideal $\mathrm{Ann}_{R}(m)$ is $(f)$  for some $f \in R$. Notice that $f$ belongs to the ring $R$, not to the module $M$, so talking about "prime elements" (in the ring) has some meaning.
Now, $M \cong R/(f)$ has no proper submodules. The $R$-submodules of $R/(f)$ are exactly the $R/(f)$-submodules of $R/(f)$, which are the ideals of $R/(f)$. An $R$-modules isomorphism gives a one-to-one correspondance between the submodules of $M$ and the ideals of $R/(f)$.
I let you conclude that $f$ must be prime in $R$ (equivalently, irreducible, because $R$ is a UFD).
