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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?

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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 $K$.

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This makes perfect sense. Thanks! – Abel Castillo May 17 '12 at 8:16

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