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. 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?

  2. 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.

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