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.

Let $R$ be a Dedekind domain, $K$ its field of fractions, and $U$ a finite-dimensional vector space over $K$. Let $M$ be an $R$-submodule of $U$ that contains a basis of $U$ (so $M$ "spans" $U$).

Question: How does one produce an example of a module $M$ as described above that is not a free module? Or, how could one determine the existence of such a (non-free) module?

The setting comes from Cassels and Froelich's "Algebraic Number Theory", Chapter 1, Section 3. On the top half of page 10, the author refers to "remov[ing] the restriction that $M$ and $N$ are free modules", which suggests the existence of non-free $R$-modules embedded in a finite dimensional vector space over the field of fractions of a Dedekind domain $R$.

I suspect that this should not happen for the ring of integers of a number field, but have not tried to write up a proof of this. I have considered looking at an arbitrary Dedekind domain that is not a PID, and constructing $M$ using non-principal ideals somehow, but am not sure exactly how to go about this.

Is there some very simple example I'm not seeing? Or will finding such a module require looking much deeper into the structure of Dedekind domains?

share|improve this question
add comment

1 Answer

up vote 3 down vote accepted

Your suggestion of using non-principle ideals is right on the money. Let $\mathcal{O}$ be the ring of integers of a number field $K$ that is not a PID. It is Dedekind, and if $I$ is a non-principal ideal, then it is certainly not free over $\mathcal{O}$ (if it was, it would have to be of rank 1, and then it would be generated by one element over $\mathcal{O}$, i.e. would be principal after all). Moreover, it is a full rank lattice in $\mathcal{O}$, so it spans $K$.

share|improve this answer
    
This makes perfect sense. Thanks! –  Abel Castillo May 17 '12 at 8:16
add comment

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.