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Given a map of commutative rings with unit, it is often the case that the inverse image of a maximal ideal is not maximal. For example, consider the inclusion $\mathbb{Z} \subseteq \mathbb{Q}$.

However, it is well-known that the inverse image of a maximal ideal under a map of finitely generated algebras over an algebraically closed field is maximal.

Are there other examples where we see this same behavior? For example,

Is the inverse image of a maximal ideal under a map of finitely generated $\mathbb{Z}$-algebras maximal?

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2 Answers 2

In the following argument $\mathbb{F}_p$ denotes either a finite field or the field $\mathbb{F}_0 = \mathbb{Q}$.

Yes. Let $\phi : R \to S$ be a morphism of finitely-generated $\mathbb{Z}$-algebras and let $m$ be a maximal ideal of $S$. Then $S/m$ is a finitely generated (as a ring) field.

Lemma: Finitely generated fields are finite fields.

Proof. Any finitely generated field $F$ is finitely generated over $\mathbb{F}_p$ for some $p$. By Noether normalization $F$ is a finite integral extension of $\mathbb{F}_p[x_1, ... x_n]$ for some $n$, and since it is a field we must have $n = 0$. Hence $F$ is either a finite field or a number field, but the latter is impossible as rings of integers in number fields have infinitely many primes.

It follows that the image of $R$ in $S/m$ is a subring of a finite field, hence also a finite field. Hence $m$ is sent to a maximal ideal of $R$.

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Yes, because $\mathbb{Z}$ is a Hilbert-Jacobson ring. See e.g. $\S 12.2$ of these notes.

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I was just about to mention: for any morphism $f\colon A\to B$ of locally finite type from a Jacobson ring $A$, then maximal ideals in $B$ map to maximal ideals in $A$. Also, $B$ will automatically be Jacobson, so all finitely generated algebras over Jacobson rings are themselves Jacobson. –  George Lowther Apr 9 '11 at 1:50
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...which I see is captured by Theorem 269 of your notes. –  George Lowther Apr 9 '11 at 1:54

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