TITLE: Modules which are not free
QUESTION [5 upvotes]: I am trying to figure out the following: Given a principal ideal domain $R$ which is not a field, does there necessarily exist a module over $A$ which is not free?
$A$ is a PID, so taking submodules won't work. The only approach I could come up with is taking some maximal ideal $m$ and considering the quotient field $A/m$. If $A/m$ were free, then its rank would be at most one, since we have an epimorphism $A\to A/m$. Hence the problem amounts to showing that $A$ and $A/m$ are not isomorphic as $A$-modules. Is this true? If so, does anyone know how to prove this? If not, can someone please provide a counterexample?
Thanks a lot,
Roy

REPLY [4 votes]: Examples, examples, examples! Take your favorite non-field PID (this had better be $\mathbb{Z}$ unless you have a really good reason for using another) and investigate. You know all sorts of $\mathbb{Z}$-modules, because these are just the abelian groups. And you should know, or else prove for yourself, that a $\mathbb{Z}$-module is free if and only if it’s a direct sum of copies of $\mathbb{Z}$.