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). 

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. 
