Torsion-free module over a PID is flat Suppose a ring of integers $S$ is an extension of a ring of integers $R$ with $\mathfrak{q}$ a prime ideal in $S$ and $\mathfrak{p}=\mathfrak{q}^c$ in $R$. Is there a straightforward way of showing that $S_\mathfrak{q}$ is flat over $R_\mathfrak{p}$ from first principles? i.e. without just quoting a the theorem such as the one in the title (the proof of which in Bourbaki is somewhat obscure). I am happy that this then shows that $S$ is flat over $R$ which is the result I am actually after.
A later thought (in the bath!): $S$ is f.g. over $R$ and $S-\mathfrak{q}\subset S-\mathfrak{p}$, so $S_\mathfrak{q}$ is f.g. over $R_\mathfrak{p}$. Moreover $S_\mathfrak{q}$ is torsion-free, hence free hence flat. Does this work?
 A: Yes, the structure theorem for f.g. modules over a PID gives a quick resolution: such a module is a direct sum of a free module and a torsion module. 
Another method: $M$ is flat iff $\text{Tor}_1^R(M, R/I) = 0$ for every $R$-ideal $I$. If $R$ is a PID, every (nonzero) ideal is principal and generated by a nonzerodivisor, say $I = (a)$. Then using the long exact sequence for 
$$0 \to R \xrightarrow{a} R \to R/(a) \to 0$$
to compute $\text{Tor}$, we find that $\text{Tor}_1^R(M, R/(a)) \cong 0 :_M (a) = \{m \in M \mid am = 0\}$, which vanishes iff $a$ is a nonzerodivisor on $M$. Thus torsionfree $\implies$ flat over a PID, even without finite generation.
Also, for future reference: the following implications hold for any module, over any ring: 
$$\text{free} \implies \text{projective} \implies \text{locally free} \implies \text{flat} \implies \text{torsionfree}$$
A: This has been open long enough. The clearest demonstration I have found is in Qing Liu "Algebraic Geometry and Arithmetic Curves" Theorem 2.4 combined with Corollary 2.5. This just relies on part of the Snake Lemma which is not too heavy, Homologically speaking. It also relies on $R$ being a Dedekind domain and thus $R_\mathfrak{p}$ is a PID which I personally am happy with.
