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). 

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.
