TITLE: Fields finitely generated as $\mathbb Z$-algebras are finite?
QUESTION [22 upvotes]: Suppose $k$ is a field that is finitely generated as a ${\mathbb Z}$-algebra.  (That is, $k$ is a quotient of ${\mathbb Z}[X_1,\dots,X_n]$ for some $n$). Does it follow that $k$ is finite?

REPLY [3 votes]: This is a mild generalization of Justin Campbell's answer, and the proof is almost the same as that of Proposition 7.8 in Atiyah and MacDonald's Introduction to Commutative Algebra. (This proposition is sometimes called the Artin-Tate Lemma.) The notation has been designed to make the analogy as clear as possible.

Let $A\subset B\subset C$ be a tower of domains such that $B$ is the field of fractions of $A$, and $C$ is a field which is finitely generated as an $A$-algebra. Then (a) $C$ is a finite extension of $B$, and (b) $B$ is a finitely generated $A$-algebra.

As (a) results immediately from Zariski's Lemma, it suffices to prove (b). The following statement is ironically stronger and easier to prove than (b):

Assume that $C=A[x_1,\dots,x_m]$; that $y_1,\dots,y_n$ is a $B$-basis of $C$ with $y_1=1$; and that we have $$x_i=\sum_jb_{ij}y_j,\quad y_iy_j=\sum_kb_{ijk}y_k$$ with $b_{ij}$, $b_{ijk}$ in $B$. Then the $A$-algebra $B$ is generated by the $b_{ij}$ and the $b_{ijk}$.

The proof is straightforward, and left to the reader.