It is a known fact that if $k$ is a field that is finitely generated as a ring, which is the same as having a surjective ring homomorphism $f:\mathbb{Z}[x_1,\dots,x_n]\to k$ for some $n\in \mathbb{N}$, then $k$ must be finite. Since finite generation as a ring implies finite generation over the prime field ($\mathbb Q$ or $\mathbb F_p$), by Noether normalization it follows that $k$ must be a finite extension of its prime field. In positive characteristic this finishes the job and in zero characteristic, should lead quickly to a contradiction, though I don't see immediately how. Is there an elementary/slick proof of this fact?
Proof That a Field Finitely Generated as a Ring is Finite
commutative-algebrafield-theory
Best Answer
One approach is via the theory of Hilbert-Jacobson rings. There are several equivalent definitions, including that every prime ideal be the intersection of the maximal ideals containing it. From this it is easy to see that a PID, for instance, is a Hilbert-Jacobson ring iff it has infinitely many maximal ideals, and that in particular $\mathbb{Z}$ is a Hilbert-Jacobson ring.
Now here is an important and useful result about Hilbert-Jacobson rings:
Theorem: Let $R$ be a Hilbert-Jacobson ring, and $S$ a finitely generated $R$-algebra. Then:
a) $S$ is a Hilbert-Jacobson ring.
b) For every maximal ideal $\mathfrak{P}$ of $S$, $\mathfrak{p} := \mathfrak{P} \cap R$ is a maximal ideal of $R$.
c) The degree $[S/\mathfrak{P} : R/\mathfrak{p}]$ is finite.
(This result and its proof can be found in these notes: see Theorem 283 in $\S 12.2$.)
In particular every field which is a quotient of $\mathbb{Z}[t_1,\ldots,t_n]$ has finite degree over $\mathbb{Z}/(p)$ so is finite.