How does this step in the proof of the structure theorem for f.g. modules over a Dedekind domain work?

Question

I am trying to show that every finitely generated projective module $P$ over a Dedekind domain $D$ is a direct sum of (fractional) ideals. May's notes on Dedekind domains claim the result can be proven by induction as follows: if $P$ has rank $n$, let $Q$ be a rank $n-1$ submodule of $P$ (constructed for example by taking elements spanning a subspace of dimension $n-1$ in $P \otimes \text{Frac}(D)$), and consider the short exact sequence

$$0 \to Q \to P \to P/Q \to 0.$$

By tensoring with $\text{Frac}(D)$ we see that $P/Q$ has rank $1$, and here May's notes claim this implies that $P/Q$ is projective, which is the crucial part of the inductive step.

I don't see how this works. It seems like you need to show that you can choose $Q$ so that you can guarantee that $P/Q$ is torsion-free. Otherwise consider, for example, the inclusion

$$\mathbb{Z} \ni 1 \mapsto (0, 2) \in \mathbb{Z} \oplus \mathbb{Z}.$$

Here the quotient is rank $1$ but has torsion.

So is this a genuine hole in the proof and, if so, how can it be repaired?

Answer 1

As indicated in the comments, you should saturate $Q$ in $P$ first, i.e. replace it by the preimage of the torsion in $P/Q,$ so that (after changing $Q$ in this way) we get that $P/Q$ is torsion-free.

The basic fact you need is that the saturation is again f.g., but this will follows from the fact that the torsion in $Q/P$ is f.g., being a submodule of the f.g. module $P/Q$.

The process of saturation is a common one in these sorts of situations, where we are playing off the relationship between modules over $D$ and its fraction field, and we want to avoid torsion appearing on the integral level. One way to proceed, which is a bit cleaner than your construction of $Q$, is to choose $V$ of dim'n $n-1$ in $P\otimes Frac(D)$, and set $Q = P \cap V$. Then $Q$ is f.g. (being a submodule of $P$) and is of rank $n-1$ (it spans $V$), and is automatically saturated in $P$ (because $P/Q$ embeds into $P\otimes D/V,$ which is a $Frac(D)$ vector space, and hence torsion free). (The difference b/w your construction and mine is that you took $Q$ to be the span of some specific vectors, hence free, whereas I didn't try to pin down the structure of $Q$ so precisely; I let it be whatever it had to be to be saturated.)

This latter way of proceeding is especially useful in iterated contexts (like if you had a filtration on $P\otimes Frac(D)$, and you want to obtain an underlying filtration on $P$ by some $Q$'s which are saturated, and which induces the original filtration after extending scalars back to $Frac(D)$: just intersect the vector spaces in the filtration with $P$!).

Answer 2

A few remarks:

1. You don't need to say "fractional" in your first sentence. By definition a fractional ideal is an $R$-submodule of $K$ of the form $\frac{1}{a} I$ for an integral ideal $I$ and $a \in R^{\bullet}$. As an $R$-module, this is visibly isomorphic to the integral ideal $I$.

2. Yes, I think you're right that this is a genuine hole in the proof. It can certainly be fixed, as in Matt E's answer. But the argument as it stands looks incomplete, and it may be worth contacting Professor May about it.

3. Did you look at my commutative algebra notes? I certainly don't mean to oversell them -- they are somewhat idiosyncratic in what they cover and the author is not one of the great experts in the subject, but a lot of effort has gone into them already, and they do cover some things well: one of those is Dedekind domains. In particular see $\S 20.5$ for my treatment of the result you're asking about. I honestly think it's better than May's approach. In fact it is essentially what Matt E says in his answer: if you set things up the right way, you will not have to worry about saturations.