Integral extension implies surjective map on spectra

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 map:

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

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.

Related facts

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 $U = R ∖ P$ to get $S^{'}$ . Then $R_{P}$ is a local ring with unique maximal ideal $P^{'} = P R_{P}$ , 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 $P^{'} S^{'} \neq S^{'}$ (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 $R_{P}$ . For full proof, refer: finitely generated and integral implies finite
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.