Closed points of a finitely generated $\mathbb{Z}$-algebra

Question

Let $A$ be a finitely generated $\mathbb Z$-algebra of characteristic 0. Is there any reason to believe the set of closed points of $\textrm{Spec } A$ is infinite?

Answer 1

Recall that the closed points of $\operatorname{Spec}A$ are precisely the maximal ideals of $A$, so suppose for contradiction that $A$ has finitely many maximal ideals. Since $\operatorname{char}A=0$, $A$ contains an isomorphic copy of $\mathbb{Z}$. We thus have $\langle p\rangle+\langle q\rangle=A$ for each pair of non-associate primes $p,q\in\mathbb{Z}$ (why?), and so – since every non-unit of a commutative ring is contained in a maximal ideal – at most finitely many of the primes of $\mathbb{Z}$ are non-units of $A$. Denoting the primes of $\mathbb{Z}$ that do not lie in $A^\times$ by $p_1,\dots,p_n$, we then have that $$B:=\frac{A[t_1,\dots,t_n]}{\langle p_1t_1-1,\dots,p_nt_n-1\rangle}$$ is a finitely generated $\mathbb{Z}$-algebra containing an isomorphic copy of $\mathbb{Q}$. ($B$ is the localization of $A$ at the elements $p_1,\dots,p_n$.) Then, letting $M$ be any maximal ideal of $B$, the ring $C:=B\big/M$ is a field of characteristic $0$ that is finitely generated as a $\mathbb{Z}$-algebra. In particular, $C$ is finitely generated as a $\mathbb{Q}$-algebra, and so by Zariski's lemma $C$ is actually finite over $\mathbb{Q}$. Thus by the Artin-Tate lemma applied to the ring extensions $\mathbb{Z}\subset\mathbb{Q}\subseteq C$, we have that $\mathbb{Q}$ is finitely generated as a $\mathbb{Z}$-algebra. But this is impossible; indeed, if $D$ is any subring of $\mathbb{Q}$ that is finitely generated as a $\mathbb{Z}$-algebra, say by elements $a_1/b_1,\dots,a_k/b_k$, let $p$ be any prime of $\mathbb{Z}$ that does not divide any of the $b_i$. Can you show that $p$ is not a unit in $D$?

Answer 2

Suppose that $A$ has finitely many maximal ideals. Then their product to some power is $(0)$; see Jacobson radical of a finitely generated $\mathbb Z$-algebra is nilpotent. It follows that $A$ is isomorphic to a finite direct product of rings of the form $A/\mathfrak m^k$ with $\mathfrak m$ a maximal ideal. But $A/\mathfrak m$ is a finite field (see Fields finitely generated as $\mathbb Z$-algebras are finite?), so there is a prime number $p$ such that $p(A/\mathfrak m)=0$. It follows that $p^k(A/\mathfrak m^k)=0$ and therefore $A$ is a ring of positive characteristic, a contradiction.