Integral extension implies surjective map on spectra

This article gives the statement, and possibly proof, of a fact about how a property of a homomorphism of commutative unital rings, forces a property for the induced map on spectra
View other facts about induced maps 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.

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 $R_P$. 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.