This is a follow-up to [this][1] question.  So that you don't have to flick back and forwards I'll briefly summarize:

My original question was on how to prove that a polynomial $g(x)$ obtained from $f(t,x)$ by specializing $t$ to some rational value has Galois group over $\mathbb Q$ a subgroup of that of $f$ over $\mathbb{Q}(t)$.  Keith Conrad wrote out a nice proof using the factorization of prime ideals in extensions of Dedekind domains.

I have been trying to extend this result to the multivariate case, to show that eg. given some polynomial $f(s,t,x)$ in $\mathbb{Q}[s,t,x]$, a rational specialization of $s$ will result in a polynomial with Galois group over $\mathbb{Q}(t)$ which is a subgroup of that of $f$ over $\mathbb{Q}(s,t)$.  

Unfortunately it is not as simple as replacing $\mathbb{Q}[t]$ with $\mathbb{Q}[s,t]$ in Keith's proof and considering the splitting of the prime ideal $(s-s_0)$ (where $s_0$ is the rational value being substituted in for $s$).  $\mathbb{Q}[s,t]$ is not a Dedekind domain, and so the proof would fall through in various places, not least because $(s-s_0)$ is not a maximal ideal.

Does anyone know if this result still holds for Galois groups over function fields in more than one variable?  And more generally, does unique factorization of prime ideals hold for any non-Dedekind domains?


  [1]: http://mathoverflow.net/questions/34521/proof-of-the-result-that-the-galois-group-of-a-specialization-is-a-subgroup-of-th