Integral extension implies surjective map on spectra

Name
This result is sometimes termed lying over, and is a precursor to the going up theorem.

Statement
Suppose $$S$$ is an integral extension of a ring $$R$$, in other words $$f:R \to S$$ is an injective homomorphism of commutative unital rings with the property that every element of $$S$$ is integral over the image of $$R$$. Then, the following are true.

The map:

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

from the spectrum of $$S$$ to that of $$R$$, that sends a prime ideal of $$S$$ to its contraction in $$R$$, is surjective. In other words, every prime ideal of $$R$$ occurs as the contraction of a prime ideal of $$S$$.

This result is sometimes termed the lying over theorem.

Note that injectivity of $$f$$ is crucial for surjectivity of the map on spectra; this is analogous to the fact that surjective ring homomorphisms induce injective maps on spectra.

Related facts

 * Integral extension implies inverse image of max-spectrum is max-spectrum
 * Going up theorem: The going up theorem is a somewhat stronger version of this result, and also follows as a corollary of this result.

Proof
The goal is to prove that starting with a prime ideal $$P$$ of $$R$$, we can find a prime ideal $$Q$$ of $$S$$ such that $$f^{-1}(Q) = P$$.

We localize $$R$$ at $$P$$, and localize $$S$$ at the image of $$U = R \setminus P$$ to get $$S'$$. Then $$R_P$$ is a local ring with unique maximal ideal $$P' = PR_P$$, and $$f$$ induces a map $$R_P \to S' = S[U^{-1}]$$.

We thus have an inclusion $$f:R_P \to S'$$. Consider the image $$P'S'$$. This is an ideal of $$S'$$. If $$P'S'$$ is a proper ideal, it is contained in some maximal ideal $$M$$, and the contraction of that maximal ideal to $$R_P$$ is precisely $$P'$$. Contracting back along the localization, we find a prime ideal of $$S'$$, whose contraction is exactly $$P$$. (we are using the fact that contracting a maximal ideal of $$S'$$ yields a prime, though not necessarily maximal, ideal of $$S$$).

Thus, the main goal is to show that $$P'S' \ne S'$$ (this is where we need to use integrality). The idea is to construct a $$R$$-subalgebra of $$S'$$, called $$S$$, that is finite'' over $$R_P$$, and use Nakayama's lemma to derive a contradiction. Here are the steps:


 * Since $$S$$ is integral over $$R$$, $$S'$$ is integral over $$R_P$$
 * If $$P'S' = S'$$, then the element $$1 \in S'$$ can be written as a $$P'$$-linear combination of finitely many elements from $$S'$$
 * Let $$S$$ be the $$R_P$$-subalgebra generated by these finitely many elements. Then $$S$$ is finitely generated and integral over $$R_P$$, hence it is finitely generated as a module over $$R_P$$.
 * We thus have $$P'S = S$$ (since $$1 \in P'S$$). Since $$P'$$ is the Jacobson radical of $$R_P$$, Nakayama's lemma tells us that $$S = 0$$, yielding a contradiction.