5
$\begingroup$

Suppose that you have a ring $\Lambda \subset \mathbf{C}$ such that $\Lambda$ is finitely generated over $\mathbf{Z}$. Let $\mathfrak{m}$ be a maximal ideal of $\Lambda$. Then the field $\Lambda / \mathfrak{m}$ must be finite.

Serre uses this fact in his paper How to use Finite Fields for Problems Concerning Infinite Fields, and cites Bourbaki1 for a proof. Since I lack a copy of Bourbaki, I figured I'd ask how to prove this here. I have a gut-feeling a proof shouldn't be too tough, but the fact that Serre decided to cite a source for this fact has me a bit worried. The only approach I've thought to try so far is to just write down an arbitrary finitely-generated ring over $\mathbf{Z}$ and figure out what a max ideal would have to look like in term of those generators, but maybe there's a slicker way.

  1. Bourbaki, N. Algèbre Commutative. Chapitre V. Entiers, Hermann, Paris, 1964. (p. 68, cor. 1)
$\endgroup$
3
$\begingroup$

This question is addressed in this MathOverflow post A finitely generated $\mathbf{Z}$-algebra that is a field has to be finite. Proving the statement comes down to knowing some basic properties of Jacobson rings, including a version of Hilbert's Nullstellensatz for Jacobson rings which can be found in these great notes by Matthew Emerton. I'll sketch out the proofs that appear in the answers to that MathOverflow post here.

Sketch of Proof: The ring $\mathbf{Z}$ is a Jacobson ring, and so is any finite extension $\Lambda$ of $\mathbf{Z}$. Moreover if you consider the natural inclusion map $\mathbf{Z} \hookrightarrow \Lambda$, by the Nullstellensatz a max ideal $\mathfrak{m}$ of $\Lambda$ will pull back to a max ideal $\mathfrak{n}$ of $\mathbf{Z}$. You can also see that the pullback will be a max ideal because the inclusion map is a finite map, so the preimage of the max ideal cannot be zero. Anyways, by another consequence of the general Nullstellensatz for Jacobson rings, $\Lambda/\mathfrak{m}$ will be a finite extension of $\mathbf{Z}/\mathfrak{n}$, so since $\mathbf{Z}/\mathfrak{n}$ is a finite field, $\Lambda/\mathfrak{m}$ will be too. $_\square$

This last proof is from a commutative algebra perspective. You could make the same sort of argument from the more geometric side of things: The inclusion map $\mathbf{Z} \hookrightarrow \Lambda$ induces a morphism $\operatorname{Spec}(\Lambda) \to \operatorname{Spec}(\mathbf{Z})$. This is a finite morphism of schemes, which must send constructable sets (finite unions of locally closed sets) to constructable sets. In particular it must send closed points to closed points, and not to the generic point, which would be the geometric analogue of the max ideal $\mathfrak{m}$ pulling back to $0$.

$\endgroup$
0
$\begingroup$

Note that fields of characteristic $0$ are not finitely generated over $\Bbb Z$. And infinite fields of positive characteristic are not finitely generated over the prime field.

$\endgroup$
  • 2
    $\begingroup$ So you're saying this because if $\Lambda / \mathfrak{m}$ were infinite you'd get a contradiction because it must be finitely-generated over either $\mathbf{Z}$ or over it's prime subfield. Must $\Lambda / \mathfrak{m}$ be finitely-generated over either $\mathbf{Z}$ or $\mathbf{Z}/p\mathbf{Z}$? Maybe this bit is obvious, and I'm just overthinking it. $\endgroup$ – Mike Pierce May 11 '18 at 15:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.