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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?

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A quick glance at another post makes me think you should be able to first show that $K$ is of prime characteristic. Perhaps once that's established, you can determine the degree of $K$ over its prime subfield is finite? – vgty6h7uij May 23 '12 at 13:04
Georges: It appears I do not have enough reputation to comment, so I am breaking the rules by posting an answer just to thank you. The point I was missing was that because $k$ is Jacobson, the inverse image of a maximal ideal must be maximal. Thanks! --- Willo – WillO May 23 '12 at 14:58
Dear Willo, I believe that you can always comment on your own question, and maybe (?) on answers to them as well, but the problem is that the account you wrote this answer from (which seems to be a registered account) is different from the account you posted the question from (which seems to be unregistered). Perhaps you can merge the two accounts; then you would be in good shape to accrue reputation and remove commenting and other problems. If you can't merge the two accounts yourself (I'm not sure if its possible), you can ask a moderator to do it for you, I believe. Regards, – Matt E May 23 '12 at 15:01
up vote 13 down vote accepted

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.

Considering the comments below , I had better state explicitly the theorem I have used.
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 $char.k=0$, the morphism $f:\mathbb Z\to k \:$ is injective, so $A$ has no $\mathbb Z$-torsion and is thus $\mathbb Z$-flat.
But then $Spec(k)\to Spec(\mathbb Z)$ is open (flat+finite presentation$\implies$ open), i.e. $(0)\in Spec( \mathbb Z)$ is open. Contradiction.
(This proof is absurdly sophisticated but it will probably appeal to scheme-theory addicts.
Take it as some kind of joke...)

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Georges: My apologies....I thought this made perfect sense this morning but now I suddenly don't see it anymore. ${\mathbb Q}$ is also a Jacobson ring, but the inverse image of (0) under the map ${\mathbb Z}\rightarrow {\mathbb Q}$ is not a maximal ideal. What additional property are you using? – WillO May 24 '12 at 3:21
Dear Will, I'm using that $k$ is finitely generated over $\mathbb Z$ (and that $\mathbb Z $ is Jacobson).This is crucial as your counterexample correctly shows. An online reference is De Jong and collaborators' Stack Project, [Chapter 7: Commutative Algebra, Proposition 32.18 ]( – Georges Elencwajg May 24 '12 at 5:58
Another reference is the generalization of the nullstellensatz stated here: . – Justin Young May 24 '12 at 12:12
Georges: Yes, as I said, I understood this the first time, but somehow when I reread it late at night I lost track of the main point. I remember why it made sense now! Thanks for being patient with me. – WillO May 24 '12 at 12:43
PS: You wrote "Since "k" is a Jacobson ring" but I think you meant "Since ${\mathbb Z}$ is a Jacobson ring"; I think you need this to apply the Nullstellensatz. – WillO May 24 '12 at 14:28

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.

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Justin, why is k integral over $\mathbb{Z}[1/n]$? – se0808 Nov 12 '14 at 20:09
Oops, now I remember realizing this answer was wrong and never fixing it. Thanks for the save. – Justin Campbell Nov 13 '14 at 0:02
I think I fixed it, is this OK? – Justin Campbell Nov 13 '14 at 0:06
Well, the proof was wrong until I fixed it just now because it didn't use the finite generation assumption in characteristic zero... – Justin Campbell Nov 13 '14 at 0:34
Thank you, Justin! – se0808 Nov 13 '14 at 9:41

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