# Adèlic points and algebraic closure

Consider $$\mathcal{X}$$ a projective and flat scheme over $$\text{Spec}(\mathcal{O}_K)$$, with $$\mathcal{O}_K$$ the ring of integers of a number field $$K$$.

Let $$F/K$$ vary over all finite Galois number field extensions and define $$\mathbf{A} := \mathbf{A}_{\overline{K}}$$ as the direct limit of the topological rings $$\mathbf{A}_F$$ of the adèles of each $$F$$.

Question 1 Is there a good and intrinsic definition of $$\mathcal{X}(\mathbf{A})$$ and does it agree with the direct limit topological space $$\varinjlim_{F/K}\mathcal{X}(\mathbf{A}_F)$$? (or is this latter the definition usually given?)

Question 2 Is $$\mathcal{X}(\mathbf{A})$$ compact?

The ring $$\mathbf{A}$$ must probably be replaced with a restricted product of countably many copies of $$\mathbf{C}$$, $$\mathbf{R}$$, $$(\mathbf{C}_p,\mathcal{O}_{\mathbf{C}_p})$$ for infinitely many $$p$$. Even so, $$\mathbf{C}_p$$ is not locally compact, so I expect the answer to question 2 is "no", no matter how we put it.

Since $$\mathcal X$$ is projective, a section is given by finitely many coordinates. If the $$i$$'th coordinate lies in $$\mathbf A_{F_i}$$ for some extension $$F_i$$ of $$K$$, then all the coordinates lie in $$\mathbf A_{F}$$ for $$F$$ the composition of the $$F_i$$.

So $$\mathcal X(\mathbf A_{\overline {K}})$$ agrees with the direct limit of $$\mathcal X(\mathbf A_{\overline {F}})$$ - at least as sets: I didn't check the topology but I imagine it's fine.

I'm not sure what you mean by an intrinsic definition. Both these definitions seem fine to me.

$$\mathcal X(\mathbf A_{\overline {K}})$$ is not compact. Choose a $$p$$-adic valuation on $$\overline{K}$$, We have maps $$\mathcal X ( \mathbf A_{\overline{K}}) \to \mathcal X (\overline{K}_p) = \mathcal X( \overline{\mathbb Q}_p) = \mathcal X( \overline{\mathbb Z}_p) \to \mathcal X ( \overline{\mathbb F}_p)$$ and the composition is continuous with the discrete topology on $$\mathcal X ( \overline{\mathbb F}_p)$$, but $$\mathcal X ( \overline{\mathbb F}_p)$$ is infinite.

Restricted to each individual $$\mathcal X( \mathbf A_{F})$$, we're restricting the valuation to $$F$$, mapping to the appropriate completion, clearing denominators (because we are working in a projective variety), and reducing modulo the uniformizer.

• Thanks! I've also asked a short follow-up question here.
– user335418
Commented Feb 10, 2022 at 17:32