Fibers of $\operatorname{Spec}(R)\to\operatorname{Spec}(S):\mathfrak{q}\mapsto \mathfrak{q}\cap S$ are discrete? Suppose $S$ is a subring of a commutative ring $R$, such that $R$ is finitely generated as an $S$-module. I"m curious about a property of the map $\operatorname{Spec}(R)\to\operatorname{Spec}(S):\mathfrak{q}\mapsto\mathfrak{q}\cap S$, which shows up frequently in the discussion around the Going Up Theorem. 
The fibers of the map are precisely the sets of primes lying above a prime in $S$. I know that these fibers are finite. However, can anybody explain why are these fibers also discrete? Since $\operatorname{Spec}(R)$ is generally not Hausdorff, I'm having trouble separating each prime from the others, so that each point would be open in the fiber in the induced topology.
I know $\operatorname{Spec}(R)$ is compact, but I don't think it's necessarily true that a finite subset of a compact space need be discrete in general. Thanks!

Since $R$ is finitely generated as an $S$-module, it has relative dimension zero. That implies if $\mathfrak{q}_1$ and $\mathfrak{q}_2$ are primes of $R$ lying over $\mathfrak{p}\in\operatorname{Spec} S$, then neither is contained in the other. 
Suppose $\mathfrak{q}$ is some prime over $\mathfrak{p}$ and $\mathfrak{q}_1,...,\mathfrak{q}_n$ are all the other ones. Pick $f_i \in \mathfrak{q}_i - \mathfrak{q}$ and let $f$ be their product. Then $D(f)$ separates $\mathfrak{q}$ from $\mathfrak{q}_i$ in $\operatorname{Spec} R$, therefore its restriction to the fibre over $\mathfrak{p}$ does the same in the induced topology.
