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I'm reading the paper "How to use finite fields for problems concerning infinite fields" of Jean-Pierre Serre.

In pp. 2, Serre uses the fact that, if $\Lambda\subset\mathbb C$ is a ring finitely generated over $\mathbb Z$ and $\mathcal M$ is a maximal ideal of $\Lambda$, then $\Lambda/\mathcal M$ is a finite field.

Serre refers to p.68 of "Bourbaki, N., Algebre Commutative. Chapitre V. Entries, Hermann, Paris, 1964"

But in this book p. 68 I do not see the proof of this fact.

Can someone tell me where I can find a proof of this fact?

azimut
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zacarias
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1 Answers1

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It suffices to prove that a field $R$, which is of finite type over $\mathbb{Z}$, is already a finite field.

Let $P$ be the prime field of $R$. Then $R$ is of finite type over $P$, hence finite since both are fields; this is a well-known application of Noether normalization. Now applying Proposition 7.8 in Atiyah-Macdonald to $\mathbb{Z} \to P \to R$ shows that $\mathbb{Z} \to P$ is of finite type. Hence $P \not\cong \mathbb{Q}$ and therefore $P$ is a finite prime field. Hence also $R$ is finite.

Actually more is true: A commutative ring is finite iff it is artinian and of finite type over $\mathbb{Z}$. See here.

If you want to know a reference for the above proof, you should find one in the context of the Weil conjectures since there it is crucial that a finite type scheme over $\mathbb{Z}$ has finite residue fields.

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Martin Brandenburg
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  • Perhaps it's just a matter of terminology but I know homomorphisms of finite type, so when you say $,R,$ is of finite type over $,\Bbb Z,$ you mean that $,R\cong \Bbb Z[x_1,...,x_n],,,,,,x_1,...,x_n\in R,$ integral over $,\Bbb Z,$ ? – DonAntonio Mar 26 '13 at 18:51
  • @DonAntonio: No. An $R$-algebra (i.e. a homomorphism $R \to S$) is called of finite type if it is a quotient of a polynomial ring over $R$ in finitely many variables (i.e. it is finitely generated as an $R$-algebra). One can show that an $R$-algebra is finite (i.e. as a module) iff it is of finite type and integral. And this is the condition you have mentioned. – Martin Brandenburg Mar 26 '13 at 20:31