# Néron models vs integral models

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 $\mathcal{X}$?

• I don't think the Neron model represents this functor on the site of schemes over $\mathcal O_K$. For instance, we could take the scheme to be a different model of $X$, and you would have to say that every model maps to the Neron model, which isn't true. The two functors agree only on the etale sets. – Will Sawin Jan 14 '18 at 10:28
• Yes I agree. I realized it only a few minutes ago – user92332 Jan 14 '18 at 10:32
• While that functor is not representable, let me suggest that there may be hope for what you're trying to do. Whatever geometric object you want should morally be the limit over all models of this $k$-scheme. The rigid analytic space / Berkovich space associated with this variety sort of functions like this, as does the valuative space of Kato if you prefer log geometry. – Joe Berner Jan 14 '18 at 23:24
• At some point I realized that I misread your comment as being about a local field. For global fields you could still use the "relative rigid spaces" (see Bosch's book), and there is at least one person (Jerome Poineau) trying to establish some of the foundational results of Berkovich for spaces over Banach rings that aren't fields. These are difficult issues though! – Joe Berner Jan 15 '18 at 13:20

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 $\mathcal O_k$.
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 and 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.)