proof of the weak Nullstellensatz
Let be an algebraically closed field, let , and let be
an ideal in the polynomial ring . Suppose is
strictly smaller than . Then is contained in a
maximal ideal
of (note that we don’t have to
accept Zorn’s lemma to find such an , since is
Noetherian
by Hilbert’s basis theorem), and the quotient ring
is a field. We view as a subfield of via the natural
homomorphism
, and we denote the images of
in by . Let
be a transcendence basis of over ; it is
finite since is finitely generated
as a -algebra
. Now is
an algebraic extension
of . By multiplying the
minimal polynomial of over by a
suitable element of for each , we obtain
non-zero polynomials
with the
property that in :
for certain integers and polynomials with . Since is algebraically
closed (hence infinite), we can choose such that
for all . We define a homomorphism
by taking to be the identity on and sending to .
Let be the kernel of this homomorphism. Then can be
extended to the localization
of
. Since for all , the
are integral over this ring. Since is algebraically
closed, the extension theorem for ring homomorphisms implies that
can be extended to a homomorphism
Because is an extension field of and is the identity on
, we see that is actually an isomorphism
, that , and
that is the zero ideal
of . Now let . Then for all polynomials in the
ideal we started with, the fact that implies
We conclude that the zero set of is not empty.
Title | proof of the weak Nullstellensatz |
---|---|
Canonical name | ProofOfTheWeakNullstellensatz |
Date of creation | 2013-03-22 15:27:43 |
Last modified on | 2013-03-22 15:27:43 |
Owner | pbruin (1001) |
Last modified by | pbruin (1001) |
Numerical id | 4 |
Author | pbruin (1001) |
Entry type | Proof |
Classification | msc 13A10 |
Classification | msc 13A15 |