# Projective Modules over Dedekind Domains

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 ${R}$. If we specify an order of vanishing of ${n_i}$ at finitely many primes ${\frak{p}_i}$, then we can find an element ${x\in Frac(R)}$ with the property that ${ord_\frak{p_i}(x)=n_i}$ 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 ${Frac(R)}$ to get the negative terms. By the Chinese Remainder Theorem the map ${R\rightarrow R/\frak{p}_1^{n_1+1}\times \cdots \times R/\frak{p}_m^{n_m+1}}$ is surjective. Now take ${x_i\in \frak{p_i}^{n_i}\setminus \frak{p}_i^{n_i+1}}$ and let ${x}$ be the preimage of ${(x_1, \ldots, x_m)}$.

A quick consequence of this is that for any non-zero fractional ideals ${I, J}$ of ${R}$ there are elements ${x,y\in Frac(R)}$ such that ${yI}$ is coprime to ${xJ}$. The proof is to specify the appropriate orders of vanishing of ${x}$ and ${y}$ to cancel out and make ${ord_P(yI)\cdot ord_P(xJ)=0}$ for all prime ideals ${P}$. By primary decomposition this shows they are coprime.

Now we need to prove that for any fractional ideals ${I_1, \ldots, I_n}$ we have ${I_1\bigoplus\cdots \bigoplus I_n\simeq R^{n-1}\bigoplus I_1\cdots I_n}$. By induction it suffices to prove that ${I\bigoplus J\simeq R\bigoplus IJ}$. If ${I}$ and ${J}$ are coprime, then ${I\cap J=IJ}$ and ${I+J\simeq R}$, so since the standard exact sequence ${0\rightarrow I\cap J\rightarrow I\bigoplus J\rightarrow I+J\rightarrow 0}$ splits we see that ${I\bigoplus J\simeq R\bigoplus IJ}$.

The statement follows for arbitrary ${I, J}$ because we can find ${x,y\in Frac(R)}$ such that ${xI}$ and ${yJ}$ are coprime, but ${xI\simeq I}$ and ${yJ\simeq J}$.

Our structure theorem says that if ${P}$ is a finitely generated projective ${R}$-module of rank ${n}$, then ${P\simeq R^{n-1}\bigoplus I}$. Recall that rank here is defined to be the dimension of ${P\otimes_R Frac(R)}$, or in other words since ${P}$ is locally free from last time, the rank of the free module after localizing. This is NOT the number of generators of ${P}$.

To prove the statement we’ll prove by induction that ${P}$ must be a direct sum of ${n}$ fractional ideals which by the rest of the post proves it. The base case just follows from flatness since ${P\subset P\otimes_R Frac(R)\simeq Frac(R)}$. Now suppose ${P}$ has rank ${n}$. Let ${Q}$ be a rank ${n-1}$ submodule of ${P}$ and form ${0\rightarrow Q\rightarrow P \rightarrow P/Q\rightarrow 0}$. Tensoring with ${Frac(R)}$ again shows that ${P/Q}$ is rank ${1}$ and hence projective. Thus the sequence splits and by the inductive hypothesis ${P\simeq Q\bigoplus P/Q}$ is a sum of fractional ideals.

Now for the uniqueness of this representation we only have to worry about which ${I}$ can appear in ${P\simeq R^{n-1}\bigoplus I}$. It turns out that this is unique up to choice of representative of the class ${[I]\in Cl(Frac(R))}$. This follows easily because if ${[I]=[J]}$, then they differ by a principal ideal and hence as ${R}$-modules ${I\simeq J}$. If ${R^{n-1}\bigoplus I\simeq R^{n-1}\bigoplus J}$, then taking the determinant of both sides (top exterior product) kills off the ${R^{n-1}}$ as follows:

${\displaystyle \begin{matrix} \bigwedge^n (R^{n-1}\bigoplus I) & \simeq & \displaystyle \bigoplus_{p+q=n} \bigwedge^p(R^{n-1})\otimes \bigwedge^q(I) \\ & \simeq & \bigwedge^{n-1}R^{n-1}\otimes \bigwedge^1(I) \\ & \simeq & R\otimes I \\ & \simeq & I \end{matrix}}$

Lastly, let’s consider one special case of the above theory that will be used in our application next time. If ${P}$ is a rank one projective ${R}$-module, then ${P\simeq I}$. It is well known in a Dedekind domain that ${I}$ is generated by at most ${2}$ elements. Thus either ${P}$ is generated by ${1}$ element, in which case it is free, or ${P}$ is generated by exactly ${2}$ elements. Thus an invertible module (locally free of rank ${1}$) over a Dedekind domain is not free if and only if it is generated by ${2}$ elements. This has a rather bizarre but “intuitively obvious” consequence for elliptic curves.

## 7 thoughts on “Projective Modules over Dedekind Domains”

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

2. hilbertthm90 says:

Ah. This is the next post.

3. 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 $1$ module is necessarily projective, since e.g. it need not be torsion-free. So how do you know that $P/Q$ is torsion-free?

4. hilbertthm90 says:

Ah yes. Good catch. We definitely can’t just pick some arbitrary $0\to Q\to P$, 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 $n-1$ elements of $P$ that span an $(n-1)$-dimensional subspace of $P\otimes Frac(R)$. Again, that doesn’t seem good enough because $0 \to \mathbb{Z}\to \mathbb{Z}\oplus \mathbb{Z}$ by $1 \mapsto (0,2)$ 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.