Integral extension implies surjective map on spectra

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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.

Related facts

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.