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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,


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All $R$ modules are free if and only if $R$ is a division ring.

Go ahead and try it as an exercise! It's accessible to undergraduate abstract algebra students.

Let $I$ be any nontrivial ideal of a PID which isn't a division ring.

Then $I$ is certainly not a direct summand of $R$. Now if all $R$ modules were free, then $R/I$ would be free, hence projective. As such, $I$ would be a summand of $R$, but as we have just noticed that is impossible. So, $R/I$ cannot be free (or even projective).

Or, more simply, as I just thought, you only need to produce an $R$ module with a nonzero annihilator (since all free modules are faithful.) That is a MUCH better way to see why $R/I$ isn't free.

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$A/m$ has a non-trivial annihilator, namely $m$, while $A$, being an integral domain, has no zero divisor, and hence, in particular, no non-trivial annihilator.

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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 $\mathbb{Z}$.

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