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?
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1Dear KotelKanim: Please see this answer. – Pierre-Yves Gaillard Jan 29 '12 at 09:31
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@Pierre-YvesGaillard: Per chance you can give one extracted answer, as this is not precisely a duplication, is it? – awllower Jan 29 '12 at 10:40
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Dear @awllower: It's kind of ironic, because the answer I gave here didn't answer the question in the exact form it was asked (because the answer was slightly stronger than needed), but it answered exactly the question asked here. I could make a copy and paste of the previous answer, but I thought things were clearer that way. – Pierre-Yves Gaillard Jan 29 '12 at 11:06
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Yes indeed, this can simplify things a lot. Also, this situation is not devoid of resemblances to many other great discoveries in the history of the science indeed. – awllower Jan 29 '12 at 11:10
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The answer in the reference is great. I just wanted to check if by chance there is an elementary argument that I am missing. It is still surprising for me that the zero characteristic is not trivial. thanks. – KotelKanim Jan 29 '12 at 17:28
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It's a 'little bit' late, but there's a short proof of this as Lemma 1.1 here: http://dept.math.lsa.umich.edu/~mityab/beilinson/SamREU07.pdf – Cocopuffs May 23 '13 at 15:55
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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.
Pete L. Clark
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1This is a nice approach. I should get more familiar with Hilbert rings. – KotelKanim Jan 29 '12 at 17:29
