Fields finitely generated as $\mathbb Z$-algebras are finite?

Question

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?

Answer 1

Yes!
Consider the morphism $f:\mathbb Z\to k$ and the ideal $\mathfrak m=f^{-1}(0)\subset \mathbb Z$.
Since $\mathbb Z$ is a Jacobson ring and $(0)\subset k$ is maximal, $\mathfrak m$ is maximal too and we obtain a morphism $\bar f:\mathbb F_p\to k$.
Since $k$ is finitely generated over $\mathbb F_p$ and is a field, it is actually a finite extension ("Zariski's version of the Nullstellensatz") and thus $k$ is (set-theoretically!) finite.

Edit
Considering the comments below , I had better state explicitly the theorem I have used.
Theorem
Let $f:A\to B$ be a finitely generated $A$-algebra.
If $A$ is a Jacobson ring, , then $B$ is also Jacobson and for every maximal ideal $\mathfrak m\subset B$ the ideal $f^{-1}(\mathfrak m)\subset A$ is also maximal.

New Edit
Here is a proof in the style of algebraic geometry, due to Akaki Tikaradze.
If $k$ has characteristic $p$ we conclude as above, using Zariski.
If $\operatorname{char}k=0$, the morphism $f:\mathbb Z\to k \:$ is injective, so $A$ has no $\mathbb Z$-torsion and is thus $\mathbb Z$-flat.
But then $\operatorname{Spec}(k)\to\operatorname{Spec}(\mathbb Z)$ is open (flat+finite presentation $\implies$ open), i.e. $(0)\in\operatorname{Spec}(\mathbb Z)$ is open. Contradiction.
(This proof is absurdly sophisticated but it will probably appeal to scheme-theory addicts. Take it as some kind of joke...)

Answer 2

In case you are interested, here is a proof which also uses Zariski's lemma but no difficult theorems about Jacobson rings.

Write $R$ for the image of the unique ring homomorphism $\mathbb{Z} \to k$, so that $k$ is a finitely generated $R$-algebra and hence a finite extension of the fraction field of $R$ by Zariski's lemma. Thus it suffices to show that $R = \mathbb{F}_p$, which is to say $k$ has positive characteristic. If $R = \mathbb{Z}$, meaning $k$ has characteristic zero, then $k$ is a number field which is a finitely generated ring. But this is impossible: if we write $k = \mathbb{Z}[\alpha_1,\dots,\alpha_r]$, then one can choose $n \in \mathbb{Z}$ so that all the denominators of coefficients in the minimal polynomials over $\mathbb{Q}$ of $\alpha_1,\dots,\alpha_r$ divide $n$. This implies that $k$ is integral over $\mathbb{Z}[1/n]$. Then $\mathbb{Z}[1/n]$ must be a field, which is absurd.