Fibers of $\operatorname{Spec}(R)\to\operatorname{Spec}(S):\mathfrak{q}\mapsto \mathfrak{q}\cap S$ are discrete?

Question

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!

Answer 1

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.

Answer 2

Since the (topological) fiber of $\mathrm{Spec}(R) \to \mathrm{Spec}(S)$ at $\mathfrak{p} \in \mathrm{Spec}(S)$ is the underlying topological space of the scheme $\mathrm{Spec}(R \otimes_S Q(S/\mathfrak{p}))$ and $R \otimes_S Q(S/\mathfrak{p})$ is a finitely generated module over $Q(S/\mathfrak{p})$, it suffices to prove:

If $K$ is a field and $R$ is a commutative $K$-algebra whose underlying $K$-module (i.e. $K$-vector space) is finitely generated, then $\mathrm{Spec}(R)$ is discrete. This follows from the more general observation that the spectrum of an artinian ring is discrete. To prove this, notice that by the Chinese Remainder Theorem every artinian ring is a finite direct product of local artinian rings (for details, see the book by Atiyah and MacDonald), and that the spectrum of a local artinian ring has just one point.