Q:

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

Background
Recall that an integral scheme $X$ is a scheme which is both irreducible and reduced; equivalently, its ring of functions is an integral domain on every open subset.
Given any point $p$, there is a local ring $R_p$ at $p$, which is given by localizing the ring of functions $R_U$ on any affine neighborhood $U$ of $p$ at the prime corresponding to $p$.  This local ring then has a residue field $K_p/p$.  The characteristic of $p$ is then the characteristic of the residue field at $p$.
Question
Its not too hard to come up with integral schemes where the characteristic of points jumps around.  The standard example of $Spec(\mathbb{Z})$ has a closed point of every positive characteristic, and the unique non-closed point (the generic point) has characteristic 0 (since the residue field is $\mathbb{Q}$).
If we don't require connectedness, then the disjoint union of $Spec(\mathbb{Q})$ and $Spec(\mathbb{Z}/2)$ has closed points of order 0 and 2, respectively.
However, I have been playing with examples, and I can't seem to come up with an example of an integral scheme that has closed points of both kinds.  Can an integral scheme have closed points of both positive and zero characteristic?
I would be curious to see such an example, since there is a wide gap in my intuition between schemes whose closed points have positive characteristic (which are inherently arithmetic) and schemes whose closed points have characteristic zero (which are either geometric, or certain arithmetic localizations).
Algebraic Version of Question (No Schemes Needed)
I should mention, though my motivation and curiosity are geometric in origin, the problem has an algebraic version.  Is there an integral domain $R$, and two maximal ideals $m$ and $n$, such that $R/m$ has positive characteristic, and $R/n$ has zero characteristic?

A:

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 $\text{Spec } \mathbb{Z}[x]$.

