If $Q$ is a prime ideal of $R[x]$ then $QF[x]\cap R[x]=Q$

Question

I'm filling the gaps in a proof and I'm stuck in this part:

Suppose $R$ is a UFD and $Q$ is a prime ideal of $R[x]$, if $F$ is the quotient field of $R$ and $R\cap Q=\{0\}$, then $QF[x]\cap R[x]=Q$.

I've been dealing with this problem for some days and every idea I have never works. I consider now is the time to come here, so I hope that some of you could give me a hand with this.

I don't post what I've done in order to solve the problem because, as I have told, anything of it worked.

Hints will suffice. Thanks a lot.

Answer 1

Your claim is wrong!

Take $R=\mathbb Z$ and $Q=2\mathbb Z[X]$. Then $Q\mathbb Q[X]=\mathbb Q[X]$.

However, if you add to the hypothesis $Q\cap R=(0)$, then the equality holds since then the extension of $Q$ to the ring of fractions $F[X]=S^{-1}R[X]$, where $S=R-\{0\}$, is $S^{-1}Q\ne F[X]$.