All simple modules over a PID. The exercise is to classify all simple modules over a PID. I tried the following: 
If $M$ is a simple module over $R$ (PID) then for a $m \in M$ with $m \neq 0$ we have $M = (m)$ then i can define $f:R \rightarrow M$ by $f(r) = rm$. The homomorphism theorem give that $R/\ker(f) \simeq M$ and $\ker(f) = (s)$ for some $s \in R$, because $R$ is a PID. Now come my doubt, what specifically is to classify?
I have to decompose $s$ in irreducible elements and apply the structure theorem? (the exercise is in the structure theorem section in the book). 

Let $S$ be a simple module over a commutative ring $R$. Then $S\ne\{0\}$ by definition and, if $x\in S$, $x\ne0$, we have $Rx=S$ because $S$ is simple.
Then the map $\varphi\colon R\to S$ defined by $r\mapsto rx$ is surjective and so
$$
S\cong R/\ker\varphi
$$
Since $S$ is simple, it follows from the homomorphism theorems that $I=\ker\varphi$ is a maximal ideal.
We can also note that $IS=\{0\}$, so $I$ is contained in the annihilator of $S$. On the other hand, $S=RS\ne\{0\}$ and, by maximality, $I$ is the annihilator of $S$.
Conversely, if $I$ is a maximal ideal of $R$, then $R/I$ is simple.
It is clear that isomorphic modules have the same annihilator, so we have a complete classification of the simple modules: a complete and irredundant set of representatives of the simple modules is given by the family of quotients $R/I$, where $I$ is a maximal ideal.
In the case of a PID, an ideal $(p)$ is maximal if and only if $p$ is irreducible. For two irreducible elements $p$ and $q$ we have
$$
R/(p)\cong R/(q)
$$
if and only if $(p)=(q)$ (by what has been proved above), which is equivalent to $p$ and $q$ being associate (that is $q=up$ where $u$ is invertible).
