# 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?

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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 $\text{Spec } \mathbb{Z}[x]$.
One speedy way to do this is to let $R=\mathbb Z[\mathbb Q^\times]$ be the group ring of the multiplicative ring of $\mathbb Q$. – Mariano Suárez-Alvarez Nov 16 '10 at 0:50
Dear Qiaochu, I wouldn't call the closed points like $(px−1)$ random.It is instructive to think about this example inside $\mathbb P^1$ over $R'$ (where as in your answer $R'$ denotes the localization of $\mathbb Z$ at $p$;although taking $R'$ to be $\mathbb Z_p$ would be just as good). Then $\mathbb P^1_{R'}$ has a generic fibre (i.e. its fibre over the generic point of Spec $R'$) as well as a special fibre (its fibre over the closed point of Spec $R'$). Given a closed point of the generic fibre, one can form its closure in $\mathbb P^1_{R'}$, which will then also meet the special fibre ... – Matt E Nov 16 '10 at 2:30
... in a closed point. In this way one obtains the so-called specialization map from closed points of $\mathbb P^1$ over $\mathbb Q$ (the fraction field of $R'$) to closed points of $\mathbb P^1$ over $\mathbb F_p$ (the residue field of $R'$). Now Spec $R'[x]$ is open in $\mathbb P^1_{R'}$, but it is not closed. The closed points of Spec $R'[x]$ with residue char. $0$ are points lying in $\mathbb A^1_{\mathbb Q}$ (which is the intersection of Spec $R'[x]$ with the generic fibre of $\mathbb P^1_{R'}$) whose specialization is the point at infinity in $\mathbb P^1_{\mathbb F_p}$. – Matt E Nov 16 '10 at 2:31
Thus they are closed in Spec $R'[x]$ because when you try to close them up over $R'$, there is no room inside Spec $R'[x]$ for them to specialize into characteristic $p$. E.g. the point $(p x - 1)$ can be written more intuitively as $x = 1/p$, which specializes to infinity mod $p$. So although this is not closed in $\mathbb P^1_{R'}$, it is closed in Spec $R'[x]$. (Another way to think of this is that we can't always specialize from $\mathbb A^1_{\mathbb Q}$ to $\mathbb A^1_{\mathbb F_p}$, and the points that don't specialize are precisely the closed points of residue char. $0$. ... – Matt E Nov 16 '10 at 2:31
... . This fits in with the valuative criterion for properness; this guarantees that specialization is possible for $\mathbb P^1$ (which is proper over $R'$), but says nothing about $\mathbb A^1$ (which is not proper over $R'$).) – Matt E Nov 16 '10 at 2:37