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). 
 A: 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).
A: That's a good start. I am not sure which structure theorem you mean, but I do not think you need any. 
What you have to classify are all the modules up to isomorphy. 
Let $p \mid s$ be a prime/irreducible element. Then $(p)/(s) \subsetneq R/(s)$ is an ideal. It's image under the isomorphism is thus a proper submodule of $M$. 
For $M$ to be simple it needs to be $\{0\}$ that is $(p)/(s)$  is trivial and hence $(p)=(s)$ and $s$ is prime/irreducible itself. 
Conversely if $s$ is irreducible, it is also prime (as you are in a PID). Further since you are in a PID $(s)$ is not only prime but also maximal so that  $R/(s)$ is a field. Suppose $N \subset M$ is a submodule. Then its preimage under the isomorphism is an ideal of $R/(s)$. Since this is a field it is trivial or the full field, showing that $N = \{0\}$ or $N=M$. Thus $M$ is simple. 
To sum it up, the simple modules are up to isomorphy $R/(p)$ for $p$ irreducible/prime. 
A: Every simple module is isomorphic to the quotient of $R$ by a maximal ideal.  In a PID, an ideal is maximal iff it is generated by an irreducible element.  So simple modules (up to isomorphism) have the form $R/(p)$ for $p$ an irreducible element of $R$.  Also, since $R$ is commutative, the annihilator of $R/(p)$ is precisely $(p)$, so given irreducible elements $p,q \in R$, $R/(p) \cong R/(q)$ iff $(p) = (q)$.  Finally, $(p) = (q)$ iff $p$ and $q$ are "equivalent", meaning there is a unit $u \in R$ with $p = uq$.
Summary: The isomorphism classes of simple modules over a PID $R$ are bijective with the equivalence classes of irreducible elements of $R$ under the above equivalence relation.
