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.