In the paper "Kronecker function rings and flat $D[X]$-modules" by Arnold and Brewer it's proved the following result:
Lemma: Let $D$ be an integral domain. If $Q$ is a prime ideal of $D[X]$ such that $D[X]_Q$ is a valuation ring and if $(Q\cap D)D[X]\subset Q$, then $Q\cap D=\{0\}$.
I was trying to prove the above lemma, but I couldn't. Then when I read the proof given in the paper I was unable to fully understand the proof. In order to make this post self contained, let me include a sketch of such proof.
Using the fact that $D_{Q\cap D}=D[X]_Q\cap K$ (where $K$ is the quotient field of $D$), the authors claim that it's enough to assume that $D$ is a valuation ring and $Q\cap D$ is its maximal ideal. I don't understand why.
On the other hand, they find that $Q$ contains a monic polynomial $f$ such that $f\notin (Q\cap D)D[X]_Q$. I guess they are using that $(Q\cap D)D[X]\subset Q$.
Then they take $y\in Q\cap D$, by the hypothesis ($D[X]_Q$ is a valuation ring) they conclude that $f\mid y$, i.e., there are $g\in D[X]$, $h\in D[X]\setminus Q$ such that $y=fg/h$. Then, if we suppose that $y\neq 0$, we get $h=f(g/y)$, thus $f\mid h$ in $K[X]$. After this, they conclude that $f\mid h$ in $D[X]$ by using theorem 8.4 of Gilmer's book "Multiplicative Ideal Theory".
I looked for the theorem, but I didn't find anything because I have the edition of 1996 and in such book section 8 is an "Allerlei" where the author makes additional remarks, gives further references, etc.
Basically, what I wrote above is their proof. I think we can add the following in order to show the contradiction by assuming that $y\neq 0$.
As $D[X]\setminus Q$ is a saturated multiplicatively closed subset of $D[X]$, then $f\in D[X]\setminus Q$, contradicting the fact that $f\in Q$. Hence, it must be $y=0$, i.e., $Q\cap D=\{0\}$.
Finally, I have two questions regarding the above proof.
1) Why can we assume that $D$ is a valuation ring and $Q\cap D$ is its maximal ideal?
2) Which theorem the authors are using in the last part of their proof, when they conclude that $f\mid h$ in $K[X]$ implies $f\mid h$ in $D[X]$?