# Integral models and adelic points

Let $$k$$ be a number field and denote by $$\Omega _k$$ the set of places of $$k$$, by $$\Omega _\infty$$ the set of archimedean places of $$k$$, and by $$S$$ a nonempty finite subset of $$\Omega _k$$ such that $$\Omega _\infty \subset S$$.

Let $$X$$ be a variety over $$k$$. We define an integral model $$\mathcal{X}$$ of $$X$$ to be a faithfully flat of finite type scheme that is separated over the ring of $$S$$-integers $$\mathcal{O}_S$$ such that we have an isomorphism $$\mathcal{X} \times _{\mathcal{O}_S} \mathrm{Spec}\,k \cong X$$ over $$k$$. An integral point is defined to be an element of $$\mathcal{X}(\mathcal{O}_S)$$.

Question 1. Let $$(x_v)$$ be an adelic point of $$X$$, i.e., $$(x_v) \in X(\mathbb{A}_k)$$. By definition, we know that for all but finitely many places $$v \in \Omega _k$$, we have $$x_v \in \mathcal{O}_v$$, i.e., $$v(x_v) \geq 0$$. How do we prove that we can find a finite subset $$S$$ containing $$\Omega _\infty$$ and an integral model $$\mathcal{X}$$ over $$\mathcal{O}_S$$ such that $$(x_v) \in (\prod_{v \in S} X(k_v) \times \prod _{v \notin S} \mathcal{X}(\mathcal{O}_v))?$$ Intuitively, I believe this set $$S$$ has to be the set of finite places $$v$$ such that $$x_v$$ is not contained in $$\mathcal{O}_v$$. It follows that a point $$x_v$$ in $$X(\mathcal{O}_v)$$ would also be in $$\mathcal{X}(\mathcal{O}_v)$$ but I have no idea how to show this concretely.

Now we consider the map $$\mathcal{X}(\mathcal{O}_S) \rightarrow (\prod _{v \in S} X(k_v) \times \prod _{v \notin S} \mathcal{X}(\mathcal{O}_v)).$$

For $$v \notin S$$ and an integral point $$x \in \mathcal{X}(\mathcal{O}_S)$$, we have $$v(x) \geq 0$$ and so $$x$$ is contained in the valuation ring $$\mathcal{O}_v$$. Hence the map $$\mathcal{X}(\mathcal{O}_S) \rightarrow \prod _{v \notin S} \mathcal{X}(\mathcal{O}_v)$$ is simply the diagonal map.

Question 2. For $$v \in S$$, how do we define the map $$\mathcal{X}(\mathcal{O}_S) \rightarrow X(k_v)?$$

Lastly, let $$\Gamma_k$$ denote the absolute Galois group $$\mathrm{Gal}(\bar{k}/k)$$.

Question 3. Is it true that $$\mathcal{X}(\bar{k})^{\Gamma_k} = \mathcal{X}(\mathcal{O}_S)?$$ In other words, under the action of $$\Gamma_k$$, are the fixed $$\bar{k}$$-points of $$\mathcal{X}$$ simply its integral points?

• It will help you to answer your questions yourself if you assume that your variety $X$ is affine and is embedded into an affine space ${\Bbb A}^n$. Then your adelic point $(x_v)$ has $n$ coordinates $(x_v)_i$, $i=1,\dots,n$, and $S$ in Question 1 is the union of $\Omega_\infty$ and of the set of those $v\in\Omega_f$ for which at least one of the coordinates $(x_v)_i$ is not $v$-integral. Apr 4, 2021 at 8:46

for your first question, the idea is basically what MikhailBorovoi said but of course you have to be careful to get a model outside $$S$$: add all the prime divisors of dominators of the equations defining $$X$$ to $$S$$ to obtain a (faithfully flat)model for $$X$$ outside $$S$$. In general you want to glue several affine varieties so you have to also add all the dominators of equations and gluing data to $$S$$.(if I want to be more precise you have to be more careful for example you have to choose monic equations a more abstract way to say this is to use generic flatness which ensures that outside finitely many prime there is a flat model for $$X$$.)
for your second question you have obvious maps $$O_S\to k\to k_v$$ and by definition $$\mathcal{X}(k_v)=X(k_v)$$.
for the third why do you think $$\mathcal{X}(\bar{k})^{\Gamma_k} = \mathcal{X}(\mathcal{O}_S)$$ instead of $$\mathcal{X}(\bar{k})^{\Gamma_k} = \mathcal{X}(k)$$? Galios invariance does not imply integrality!
• About the third question, if $X$ (and therefore $\mathcal{X}$) is an affine group scheme, I am trying to apply the $\Gamma(k,-)$-functor on the Kummer sequence $0 \rightarrow \mathcal{X}[n] \rightarrow \mathcal{X} \xrightarrow{n} \mathcal{X} \rightarrow 0$. From the resulting long cohomology sequence, there should be $\mathcal{X}(\mathcal{O}_S) \rightarrow H^1(k,\mathcal{X}[n]) \rightarrow H^1(k,\mathcal{X})$, hence my question. Here $H^i(k,\mathcal{X})$ is short for $H^i(\mathrm{Gal}(\bar{k}/k),\mathcal{X}(\bar{k}))$. Apr 5, 2021 at 5:54
• Ok I think I got it, the Kummer sequence is supposed to be $0 \rightarrow \mathcal{X}[n](\bar{\mathcal{O}}_S) \rightarrow \mathcal{X}(\bar{\mathcal{O}}_S) \rightarrow \mathcal{X}(\bar{\mathcal{O}}_S) \rightarrow 0$. Applying the functor will give us $H^0(k,\mathcal{X}(\bar{\mathcal{O}}_S)) = \mathcal{X}(\mathcal{O}_S)$ as desired. Apr 5, 2021 at 11:57