Integral models and adelic points

Question

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?

Answer 1

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!