Let $X$ be a smooth projective $k$-scheme, $k$ being a number field. Let $\mathcal{O}\_k$ be the ring of integers of $k$.


Fix a large enough category of schemes $\text{Sch}/k$ containing $X$, and consider the big étale site $(\text{Sch}/k)\_{\rm Ét}$. The functor of points $h\_X$ of $X$ is a sheaf on $(\text{Sch}/k)\_{\rm Ét}$.


Let $j : \text{Spec}(k)\to\text{Spec}(\mathcal{O}\_k)$ be the inclusion of the generic point. It induces a big push forward $(j\_{\rm Ét})\_\*$ on sheaves on $(\text{Sch}/k)\_{\rm Ét}$ to $(\text{Sch}/\mathcal{O}\_k)\_{\rm Ét}$, and we may consider the étale sheaf $\mathcal{X} := (j\_{\rm Ét})\_\*h\_X$.


**When is $\mathcal{X}$ representable?**


Does there exist a possibly singular integral model $\mathscr{X}/\mathcal{O}\_k$ scheme/algebraic space for $X$ that represents $\mathcal{X}$?



Néron models vs integral models
Note that sheaf pushforward and preheaf pushforward agree, so this a question about categories, not sites.



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> > **Lemma.** If $X$ is a finite type $k$-scheme with $\dim X > 0$, then $j\_\*h\_X$ is not representable by a finite type algebraic space over $\mathcal O\_k$.
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*Proof (sketch).* Suppose $Y$ is a finite type algebraic space over $\mathcal O\_k$ representing $j\_\*h\_X$. This means that
$$\operatorname{Hom}\_{\mathcal O\_k}(T,Y) = \operatorname{Hom}\_k(T \times\_{\mathcal O\_k} k, X)$$
for all $\mathcal O\_k$-schemes $T$. If $T$ is any scheme of positive characteristic with a map $T \to \operatorname{Spec} \mathcal O\_k$, then $T \times\_{\mathcal O\_k} k = \varnothing$. Thus, $\operatorname{Hom}\_{\mathcal O\_k}(T,Y)$ has exactly one element. Taking $T = \operatorname{Spec} \overline{\mathcal O\_k/\mathfrak p}$ for $\mathfrak p \subseteq \mathcal O\_k$ prime, we conclude that the special fibre $Y\_\mathfrak p$ is zero-dimensional, as it is a finite type algebraic space over $\mathcal O\_k/\mathfrak p$ with only one $\overline{\mathcal O\_k/\mathfrak p}$-point. (For example, you can use [Tag 06LZ](https://stacks.math.columbia.edu/tag/06LZ) and [Tag 0ADC](https://stacks.math.columbia.edu/tag/0ADC) to prove this, plus a bit of work.) This actually shows that $Y\_\mathfrak p = \operatorname{Spec} \mathcal O\_k/\mathfrak p$.


But a finite type algebraic space over $\mathcal O\_k$ whose closed fibres are zero-dimensional has all fibres zero-dimensional; for this you use that finite type schemes over $\mathcal O\_k$ are Jacobson (hence the points over finite fields are everywhere dense). But now the generic fibre doesn't have enough points: if $T = \operatorname{Spec} \bar k$, then $\operatorname{Hom}\_{\mathcal O\_k}(T, Y) = \operatorname{Hom}\_k(T, X)$ has infinitely many elements, as $\dim X > 0$. $\square$


There may very well be a way to construct $Y$ as a non-finite-type object. For example, if $X = \mathbb A^1$, then $j\_\* h\_X$ is the quasicoherent sheaf $\tilde{k}$ on $\operatorname{Spec} \mathcal O\_k$. You might be able to represent this as an algebraic space by writing down a resolution $\mathcal O\_k^J \to \mathcal O\_k^I \to k \to 0$ of $k$ as $\mathcal O\_k$-module, and constructing $Y$ as the quotient of $(\mathbb A^1)^I$ by the equivalence relation defined by the presentation above. It might even be possible to glue this construction to treat the case of arbitrary (finite type) $k$-schemes $X$. (I didn't work this out because it sounds painful, and I'm not sure if this would be of any use to you.)


