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 $k$ is a field and $K$ is a field extension of $k$ , such that $K$ is finitely generated as a $k$ -algebra. Then, $K$ is algebraic over $k$ , and in fact, is a finite field extension of $k$ .
Suppose $M$ is a maximal ideal in a polynomial ring in finitely many variables over $k$ . Then the quotient field for $M$ is a finite field extension of $k$

Applications

Weak nullstellensatz for algebraically closed fields

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 $k (T)$ of $K$ , which is the field of fractions of a subset $T$ of $k$ , such that $K$ is algebraic over $k (T)$ . In our case, since $K$ is finitely generated over $k$ , it is also finitely generated over $k (T)$ , so in fact $K$ is a finite field extension of $k (T)$ . Further information: Finitely generated and integral implies finite
We use Artin-Tate lemma and the fact that fields are Noetherian, to deduce that $k (T)$ is finitely generated as a $k$ -algebra (here $A = k, B = k (T), C = K$ )
We now use the fact that if $T$ is nonempty, $k (T)$ can never be finitely generated over $k$ . Thus, $T$ is empty, forcing $K$ to be a finite field extension of $k$ (This uses the fact that the polynomial ring is a unique factorization domain with infinitely many irreducibles).

Proof using Noether normalization theorem

Facts used

Noether normalization theorem

Proof outline

Suppose $K$ is a finitely generated algebra over $k$ , that happens to be a field. Then, by the Noether normalization theorem, there exists a polynomial algebra $k [x_{1}, x_{2}, \dots, x_{n}]$ inside $K$ such that $K$ is finite over this polynomial algebra.

But $K$ being a field, must at any rate contain the field of fractions of $k [x_{1}, x_{2}, \dots, x_{n}]$ , and this yields a contradiction if $n \geq 1$ .

Alternatively, we can observe that the injective map from $k [x_{1}, x_{2}, \dots, x_{n}]$ to $K$ being a finite morphism, must give a surjective map on spectra, but the spectrum of $K$ has one element, and so can surject to the spectrum of a polynomial ring only if $n = 0$ .

This yields that $K$ is finite-dimensional over $k$ .