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:

$f^{*} : S p e c (S) \to S p e c (R)$

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 $S p e c (K \otimes_{R} S)$ , which is the same as $S p e c (K \otimes_{R / P} S / P^{e})$ .

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).