Is the axiom of choice necessary to prove that closed points in the Zariski topology are maximal ideals?

Solution 1:

The following example shows that the use of the axiom of choice cannot be avoided. Let $A$ be the ring of entire functions, and let $B$ be the localization of $A$ obtained by inverting $z-a$ for each $a\in\mathbb{C}$ (equivalently, let $B$ be the ring of meromorphic functions on $\mathbb{C}$ with only finitely many poles). The ring $A$ is a Bezout domain, and hence so is $B$. In particular, $0$ is a prime ideal of $B$.

Now suppose $P\subset B$ is a prime ideal with a nonzero element $f\in P$. Since $f$ is not a unit in $B$, it must have infinitely many zeroes. Let $Z$ be the set of zeroes of $f$. Since $Z$ is a closed discrete subset of $\mathbb{C}$, it is in bijection with $\mathbb{N}$ (exercise: you can prove this without choice). Now let $U$ be the set of subsets $S\subseteq Z$ such that there exists $g\in P$ whose vanishing set is $S$. Then $U$ must be an ultrafilter on $Z$ (the fact that $U$ is closed under binary intersections comes from the fact that $B$ is a Bezout domain and intersections of zero sets correspond to gcds of functions; the fact that if $S\cup T=Z$ then either $S\in U$ or $T\in U$ is because you can find a function $g$ vanishing on $S$ and a function $h$ vanishing on $T$ so that $gh$ is divisible by $f$, so by primeness either $g\in P$ or $h\in P$). Moreover, since any function with finitely many zeroes is a unit in $B$, $U$ must be nonprincipal.

Thus if there exists a nonzero prime ideal in $B$, there exists a nonprincipal ultrafilter on $\mathbb{N}$. Now work in a model of ZF where there exist no nonprincipal ultrafilters on $\mathbb{N}$. In this model, $0$ is the only prime of $B$, so $0$ is a closed point of $\operatorname{Spec} B$. However, $0$ is not a maximal ideal of $B$.