Can an integral scheme have closed points of both positive and zero characteristic?

Solution 1:

Yes, but you're not going to like it. Let $R = \mathbb{Z}[x_1, x_2, ... ]$. Let $q_1, q_2, ... $ be an enumeration of the rationals; then the map $R \to \mathbb{Q}$ which sends $x_i$ to $q_i$ is surjective, so its kernel is a maximal ideal $n$ such that $R/n$ has characteristic zero. On the other hand, let $m = (p, x_1, x_2, ...)$. Then $R/m \simeq \mathbb{F}_p$.

Edit: Okay, so here is a Noetherian example. Let $R'$ denote the localization of $\mathbb{Z}$ at $p$ (since $\mathbb{Z}_p$ generally means something else) and let $R = R'[x]$. Then the map $R \to \mathbb{Q}$ which sends $x$ to $\frac{1}{p}$ is surjective, and so is the map $R \to \mathbb{F}_p$ which sends $x$ to, say, $1$. It seems interesting to try to picture $\text{Spec } R$; it might help to stare at Mumford's picture of $\text{Spec } \mathbb{Z}[x]$.