What is meant by the $\overline{\mathbb{F}_p}$-points of $\operatorname{Spec} \mathcal{O}_K$?

Question

Let $K$ be a number field and $\mathcal{O}_K$ be the ring of integers. Let $\overline{\mathbb{F}_{p}}$ be the algebraic closure of the finite field with $p$-elements. What are the points $\operatorname{Spec} \mathcal{O}_{K} (\overline{\mathbb{F}_p})$?

I have seen in the past the definition of the $R$-points of a scheme defined over $R$ as sections of the structure map, but in this case I don't see how $\operatorname{Spec} (\mathcal{O}_K)$ is naturally a scheme over $\overline{\mathbb{F}_p}$. My best guess is that we take the sections of the natural map $\operatorname{Spec}(\mathcal{O}_K) \times_{\operatorname{Spec} (\mathbb{Z})} \operatorname{Spec}(\overline{\mathbb{F}_p}) \to \operatorname{Spec} (\overline{\mathbb{F}_p})$, but I want to be sure.

Answer 1

You're right, $\operatorname{Spec}(\mathcal O_K)$ is not a scheme over $\overline{\mathbb F}_p$, if it was you'd have a morphism $\operatorname{Spec}(\mathcal O_K)\to\operatorname{Spec}(\overline{\mathbb F}_p)$ corresponding to a morphism $\overline{\mathbb F}_p\to\mathcal O_K$ which doesn't exist.

Here is a general definition: if $X$ and $T$ are schemes then an $T$-point of $X$ is just a morphism $T\to X$. If there is an understood base scheme $S$ you are working over then you take $S$-morphisms $T\to X$.

Note this coincides with the definition of $R$-point you have given, in the particular case where $T=\operatorname{Spec}(R)$ and you are working over the base $S=\operatorname{Spec}(R)$, as a $\operatorname{Spec}(R)$-point of $X$ (usually just called an $R$-point of $X$) is an $R$-morphism $\operatorname{Spec}(R)\to X$, which is the same as a section of the structure map.

With this general definition you can work out what an $\overline{\mathbb F}_p$-point is of $\operatorname{Spec}(\mathcal O_K)$.

Answer 2

To add to Alex's answer, here is some motivation for why we call a morphism $T \to X$ a $T$-point.

Let $k$ be a field, and let $X = \operatorname{Spec}k[x_1,\dots,x_n]/(f_1,\dots,f_m)$ be a closed subscheme of affine space. Then if $T = \operatorname{Spec}R$ is a $k$-scheme (equivalently, $R$ is a $k$-algebra), then a $T$-point of $X$ is given by a homomorphism $k[x_1,\dots,x_n]/(f_1,\dots,f_m) \to R$ of $k$-algebras. Since $R$ is a $k$-algebra, this is equivalent to picking an $n$-tuple of elements $\mathbf r = (r_1,\dots,r_n)$ of $R$ which satisfy $f_1(\mathbf r) = \dots = f_m(\mathbf r) = 0$, which is closer to what one usually thinks of as a point. When $R$ is also a field, the set-theoretic image of $\operatorname{Spec}R \to X$ is also a point, bringing the analogy closer.

One good reason for why we may consider this expanded definition of what constitutes a point is that we may often want to consider a continuous family of varieties/schemes, and wish to continuously pick a point on each element of that family. For instance, a genus one curve with a distinguished point is an elliptic curve, hence a family of elliptic curves should have a continuous way to pick a distinguished point.

Let $S$ be some variety over $k$, for which we may wish to consider a family of varieties $\{X_s\}$ indexed by the closed points $s \in S$. Then this is done by considering a morphism $X \to S$ for which the fibre at any $s \in S$ gives the variety $X_s$. A closed point of $X_s$ corresponds to a morphism $\operatorname{Spec} k \to X_s$, so if we wish to continuously choose a closed point on every $X_s$ at once, we may begin with a section $\sigma: S \to X$. Then for any closed $s$, since the fibre $X_s$ is the base change of $X \to S$ by the point $s: \operatorname{Spec}k \to S$, composing with the section $\sigma$ gives a morphism $\operatorname{Spec}k \to X$, and thus a morphism/point $\sigma_s: \operatorname{Spec}k \to X_s$ by the universal property.