Finite morphism implies finite on spectra

Statement

Suppose $f:R\to S$ is a homomorphism of commutative unital rings that is finite; in other words, $S$ is a finitely generated module over $R$. Then, the induced map on spectra:

$f^{*}:Spec(S)\to Spec(R)$

that sends a prime ideal of $S$ to its contraction in $R$, has finite fibers: in other words, the inverse image of any point is a finite set.

Unlike the case of lying over, we do not assume injectivity of $f$.

Proof

Although the result is true for a general map, it actually suffices to prove it in the case where $f$ is injective (this may make things conceptually simpler, though the proof details do not change).

We start with a prime ideal $P$. The proof relies on the following key result:

If $K$ is a field of fractions of the integral domain $R/P$, then the prime ideals of $S$ that contract to $P$ in $R$, are in bijective correspondence with the elements of $Spec(K\otimes _{R}S)$, which is the same as $Spec(K\otimes _{R/P}S/P^{e})$.

With this result, the proof reduces to the observation that since $f$ is finite, $K\otimes _{R}S$ is finite-dimensional as a $K$-vector space, and hence is a zero-dimensional ring (true for any algebra that is finite over a field).