Prime Ideals in a Localised Ring

Let $R$ be a commutative ring, and let $S \subseteq R$ be a multiplicatively closed subset (not containing $0$). Then we construct the localised ring $R [ S^{-1} ]$. I understand that prime ideals in $R [ S^{-1} ]$ correspond to prime ideals in $R$ which do not intersect $S$. Is it always true that if $\mathfrak{q}$ is prime in $R$, then $\mathfrak{q}R[S^{-1}]$ is prime in $R[S^{-1}]$?

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Either $\mathfrak q$ does not intersect $S$, and then you know that the extension to $S^{-1}R$ is prime, or it intersects $S$ in which case $\mathfrak q(S^{-1}R) = S^{-1}R$ is a very specific non-prime.
Here's an elementary way to prove the first assertion: show that $a/s$ is in $\mathfrak q(S^{-1}R)$ if and only if $a \in \mathfrak q$. You'll need to use that $\mathfrak q$ is prime and that $S \cap \mathfrak q=\varnothing$.
OK, my proof is a bit abstract so I wasn't actually sure what prime in $S^{-1} R$ got mapped to $\mathfrak q$. – Paul Slevin May 20 '12 at 16:25
@Paul It's precisely $\mathfrak q(S^{-1}R)$! I'll add a proof of that when I get some time later, so that the answer has some meat to it. – Dylan Moreland May 20 '12 at 16:29
The way we proved that the map was surjective was to use the universal property of localised rings using the maps $R \to R / \mathfrak q \to \operatorname{Frac}(R / \mathfrak q)$ and $R \to R[S^{-1}]$ to get a map $h:R[S^{-1}] \to \operatorname{Frac}(R / \mathfrak q)$. the kernel of $h$ is the prime that gets mapped to $\mathfrak q$. I would really appreciate it if I could see your proof. – Paul Slevin May 20 '12 at 16:36
It's easy to see elements of $q(S^{-1} R)$ have that form though right? Every element looks like $\displaystyle \frac{b_1}{1}\frac{a_1}{s_1} + \cdots + \frac{b_n}{1}\frac{a_n}{s_n} = \frac{b_1a_1s_2\cdots s_n + \cdots + s_1\cdots s_{n-1}b_n a_n}{s_1 \cdots s_n}$, I think, and the top is in $\mathfrak q$ and the bottom is in $S$. – Paul Slevin May 21 '12 at 14:34