Today we’ll classify finitely generated projective modules over Dedekind domains which will finish off a very similar structure theorem to the one over a PID. First, we need an approximation theorem. Fix a Dedekind domain . If we specify an order of vanishing of at finitely many primes , then we can find an element with the property that and non-negative otherwise.

We can reduce to the case of finiding an element where all the orders of vanishing are positve, since we can divide in to get the negative terms. By the Chinese Remainder Theorem the map is surjective. Now take and let be the preimage of .

A quick consequence of this is that for any non-zero fractional ideals of there are elements such that is coprime to . The proof is to specify the appropriate orders of vanishing of and to cancel out and make for all prime ideals . By primary decomposition this shows they are coprime.

Now we need to prove that for any fractional ideals we have . By induction it suffices to prove that . If and are coprime, then and , so since the standard exact sequence splits we see that .

The statement follows for arbitrary because we can find such that and are coprime, but and .

Our structure theorem says that if is a finitely generated projective -module of rank , then . Recall that rank here is defined to be the dimension of , or in other words since is locally free from last time, the rank of the free module after localizing. This is NOT the number of generators of .

To prove the statement we’ll prove by induction that must be a direct sum of fractional ideals which by the rest of the post proves it. The base case just follows from flatness since . Now suppose has rank . Let be a rank submodule of and form . Tensoring with again shows that is rank and hence projective. Thus the sequence splits and by the inductive hypothesis is a sum of fractional ideals.

Now for the uniqueness of this representation we only have to worry about which can appear in . It turns out that this is unique up to choice of representative of the class . This follows easily because if , then they differ by a principal ideal and hence as -modules . If , then taking the determinant of both sides (top exterior product) kills off the as follows:

Lastly, let’s consider one special case of the above theory that will be used in our application next time. If is a rank one projective -module, then . It is well known in a Dedekind domain that is generated by at most elements. Thus either is generated by element, in which case it is free, or is generated by exactly elements. Thus an invertible module (locally free of rank ) over a Dedekind domain is not free if and only if it is generated by elements. This has a rather bizarre but “intuitively obvious” consequence for elliptic curves.

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May 6, 2013 at 9:15 pm

What’s the bizarre but “intuitively obvious” consequence for elliptic curves?

May 7, 2013 at 8:10 am

Ah. This is the next post.

May 10, 2013 at 1:31 pm

I don’t understand a step in your proof of the structure theorem. It doesn’t seem to be true in general that a rank module is necessarily projective, since e.g. it need not be torsion-free. So how do you know that is torsion-free?

May 10, 2013 at 2:48 pm

Ah yes. Good catch. We definitely can’t just pick some arbitrary , but we are allowed to pick it however we want. I think I was following J. P. May’s notes on this and he just says choose elements of that span an -dimensional subspace of . Again, that doesn’t seem good enough because by seems to satisfy that but still leaves torsion.

I feel like I had something in mind when I wrote this, but I can’t remember now.

May 10, 2013 at 3:48 pm

I’m not sure how to deal with this either. I posted a math.SE question.