On Hilbert's Nullstellensatz Theorem

Question

I was reading Ravi Vakil's notes on his website and he states the Hilbert Nullstellensatz (3.2.5.): If $k$ is any field, every maximal ideal of $k[x_1, ..., x_n]$ has residue field a finite extension of $k$. Translation: any field extension of $k$ that is finitely generated as a ring is necessarily also finitely generated as a module (i.e., is a finite field extension).

I understand (at least I think I do) the statement of the theorem, but I just don't understand why this statement of the theorem translates to what he wrote in "Translation". Could someone please explain me how this works? Thanks!

Answer 1

The "translation" amounts to noting that "residue fields of maximal ideals" and "field extensions that are finitely generated rings" are the same.

More precisely, a field extension $K$ of $k$ is a finitely generated $k$-algebra if and only if it can be written as the residue field of a maximal ideal in some polynomial ring over $k$. Indeed, if $K$ is a finitely generated $k$-algebra, then $K=k[x_1,\dots,x_n]/J$ for some ideal $J$. The quotient is a field, therefore $J$ is maximal. For the converse, note that any quotient of a polynomial ring over $k$ is a finitely generated $k$-algebra.

Answer 2

If $B$ is a finitely generated algebra over $k$, then it's a quotient of $A= k[x_1,\ldots,x_n]$ for some $n$. Say then that $B\cong A/I$.

If $M$ is a maximal ideal of $B$, then $B/M\cong (A/I)/(M'/I)\cong A/M'$ for some maximal ideal $M'$ in $A$. So, having proven this for $A$, we've also proven it for $B$.