Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question

3 Answers 3

Yes.

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.

share|improve this answer

$A/m$ 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.

share|improve this answer

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 $\mathbb{Z}$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.