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

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 $$k(T)$$.
 * 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).

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 $$n \ge 1$$.

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 $$k$$.