TITLE: How does this step in the proof of the structure theorem for f.g. modules over a Dedekind domain work?
QUESTION [9 upvotes]: 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?

REPLY [5 votes]: 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.