# 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

## Contents

## Name

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

## Statement

Suppose is an integral extension of a ring , in other words is an injective homomorphism of commutative unital rings with the property that every element of is integral over the image of . Then, the following are true.

The map:

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

This result is sometimes termed the **lying over theorem**.

Note that injectivity of 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 of , we can find a prime ideal of such that .

We localize at , and localize at the image of to get . Then is a local ring with unique maximal ideal , and induces a map .

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

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

- Since is integral over , is integral over
- If , then the element can be written as a -linear combination of finitely many elements from
- Let be the -subalgebra generated by these finitely many elements. Then is finitely generated and integral over , hence it is finitely generated as a module over .
*For full proof, refer: finitely generated and integral implies finite* - We thus have (since ). Since is the Jacobson radical of , Nakayama's lemma tells us that , yielding a contradiction.