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

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.

• It seems you missed the following: if $P$ is a prime ideal of a commutative ring $R$, $S\subset R$ a multiplicative set, and $P\cap S=\emptyset$, then $S^{-1}P\cap R=P$. This is very easy to prove and I suggest you to try it. (Btw, this has nothing to do with $R$ being an UFD.) – user26857 Aug 31 '14 at 20:50
• Ok, i'm really thankful. I've proved what you suggested me, now I want to apply it to my original problem, I think this is what I need to do: Your $R$ is my $R[x]$, your $P$ is my $Q$ and $S$ is my original $R$ minus $\{0\}$, I'd only need to show that $QF[x]=S^{-1}Q$ and that's all. Am i right? – Daniel Aug 31 '14 at 21:15
• Yes, you are right. – user26857 Aug 31 '14 at 21:23

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 $S^{-1}Q\ne F[X]$.
• Oh, I'm sorry, I didn't even realized it was you the one in the comment under my question. Thanks again, you took me out from a jam. I only have one last question if you don't mind: $S^{-1}Q$ is defined as $\{q/s:q\in Q, s\in S\}$ and $QF[x]$ is the set of all finite sums of elements of the form $qf(x)$ with $q\in Q$ and $f(x)\in F[x]$, isn't it? Can I define $S^{-1}Q$ as the set of all finite sums of elements of the form $q/s$ with $q\in Q$ and $s\in S$? – Daniel Aug 31 '14 at 21:38
• The answer to both questions is yes, but why define the last ideal in that way? If you add all that fractions you get only one with numerator in $Q$ and denominator in $S$. – user26857 Aug 31 '14 at 21:40