Inverse Image of Maximal Ideals

Solution 1:

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 $R$.

Solution 2:

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