Weak nullstellensatz for arbitrary fields

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 $k[x_{1},x_{2},\ldots ,x_{n}]$ 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 $k[x_{1},x_{2},\ldots ,x_{n}]$, and this yields a contradiction if $n\geq 1$.

Alternatively, we can observe that the injective map from $k[x_{1},x_{2},\ldots ,x_{n}]$ 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 $k$.