Weak nullstellensatz for arbitrary fields
This fact is an application of the following pivotal fact/result/idea: Artin-Tate lemma
View other applications of Artin-Tate lemma OR Read a survey article on applying Artin-Tate lemma
This fact is an application of the following pivotal fact/result/idea: Noether normalization theorem
View other applications of Noether normalization theorem OR Read a survey article on applying Noether normalization theorem
Statement
Here are two equivalent formulations:
- Suppose is a field and is a field extension of , such that is finitely generated as a -algebra. Then, is algebraic over , and in fact, is a finite field extension of .
- Suppose is a maximal ideal in a polynomial ring in finitely many variables over . Then the quotient field for is a finite field extension of
Applications
Proof using Artin-Tate lemma
Facts used
- Steinitz theorem: This states that any field extension can be expressed as an algebraic extension of a purely transcendental extension
- Artin-Tate lemma
- The fact that a purely transcendental field extension cannot be finitely generated as an algebra over the field
Proof outline
- We use Steinitz theorem to show that we can find a subfield of , which is the field of fractions of a subset of , such that is algebraic over . In our case, since is finitely generated over , it is also finitely generated over , so in fact is a finite field extension of . Further information: Finitely generated and integral implies finite
- We use Artin-Tate lemma and the fact that fields are Noetherian, to deduce that is finitely generated as a -algebra (here )
- We now use the fact that if is nonempty, can never be finitely generated over . Thus, is empty, forcing to be a finite field extension of (This uses the fact that the polynomial ring is a unique factorization domain with infinitely many irreducibles).
Proof using Noether normalization theorem
Facts used
Proof outline
Suppose is a finitely generated algebra over , that happens to be a field. Then, by the Noether normalization theorem, there exists a polynomial algebra inside such that is finite over this polynomial algebra.
But being a field, must at any rate contain the field of fractions of , and this yields a contradiction if .
Alternatively, we can observe that the injective map from to being a finite morphism, must give a surjective map on spectra, but the spectrum of has one element, and so can surject to the spectrum of a polynomial ring only if .
This yields that is finite-dimensional over .