# Do abelian varieties have Neron models over arbitrary valuation rings?

Let $\mathcal{O}_K$ be a valuation ring with fraction field $K$. Let $A$ be an abelian variety over $K$. Does $A$ have a Neron model?

If $\mathcal{O}_K$ is a discrete valuation ring, then this is proven in the book of Bosch-Lutkebohmert-Raynaud on Neron models.

• An easy way to convince yourself that it may not exist is the following example. Consider an elliptic curve $E$ over a field $K:=\operatorname{Frac}(R)$ (where $R$ is a dvr) with a bad multiplicative reduction. And assume that a special fiber of the Neron model $\mathcal E$ has $n$ connected components. Now note that for any integer m we have $E[m](K)=\mathcal E[m](\mathcal O_K)$ and $\mathcal E[m]$ is etale whenever $m$ is coprime with char. $k$. Thus $|\mathcal E[m](\mathcal O_K)| \leq |\mathcal E_0[m](k)|\leq nm$ where $E_0$ is the special fibre of $\mathcal E$. – gdb Aug 5 '18 at 21:45
• This implies that a semi-stable elliptic curve over $K$ with at most $n$ connected components in the special fibre of the Neron model cannot have more than $nm$ $m$-torsion points (for $m$ coprime with char of $k$). So, consider any valuation ring with alg. closed fraction field (for example, $\mathcal O_{\mathbf C_p}$), any semi-stable elliptic curve $E$ over $K$ has full $m^2$ rational $m$-torsion points for any $m$ coprime with $K$. Hence, the arg.above shows that if the Neron model of $E$ exists its special fibre must have infinitely many connected components. – gdb Aug 5 '18 at 22:02
• In particular, it can't be a finite type scheme (it is a part of the definition of Neron model to be a finite type $R$-scheme) – gdb Aug 5 '18 at 22:07
• @gdb Thank you. This is very convincing. – Kriss Aug 5 '18 at 22:49
• There is a notion of "lft-Néron model" in Bosch-Lütkebohmert-Raynaud (lft=locally of finite type). Such models might exist, at least in the semistable case. – Laurent Moret-Bailly Aug 6 '18 at 7:05

As a partial answer to your question, let me sketch a proof of the following result found in 2015 in collaboration with David Holmes.

Proposition. For a valuation ring $V$, every abelian $V$-scheme $A$ is the Néron model of its generic fiber in the sense that for every smooth $V$-scheme $S$, we have $A(S) = A(S_K)$ via pullback, where $K := \mathrm{Frac}(V)$.

Proof. Firstly, the result should hold: by weak uniformization (which is still conjectural in positive and mixed characteristic), every valuation ring is a filtered direct limit of regular local rings, and over a regular ring every abelian scheme is its own Néron model: the claim basically amounts to extending line bundles on the dual abelian scheme, and line bundles over regular bases do extend. As far as the actual proof goes, the idea is to deduce the claim via a limit argument from the extension result of Weil about rational maps into group schemes. Now for the details.

Since $A$ is separated over $V$ and $S_K$ is schematically dense in $S$, the injectivity of $A(S) \rightarrow A(S_K)$ is clear (it follows from EGA I, 9.5.6, as usual). Thus, we need to show that any $V$-map $S_K \rightarrow A$ extends to a $V$-map $S \rightarrow A$. For this, since the extensions are unique by the previous step, we may work Zariski or even étale locally on $S$, so we fix a point $s \in S$ around which we want to extend and may assume that $S$ is affine, connected, and admits an étale $V$-map $S \rightarrow \mathbb{A}^n_V$ for some $n \ge 0$ (we use EGA IV, 17.11.4).

Let $v \in \mathrm{Spec}(V)$ be the image of $s$. By replacing $V$ by its localization $\mathcal{O}_{V, v}$, which is also a valuation ring (https://stacks.math.columbia.edu/tag/052K), and spreading out, we may assume that $v$ is the closed point of $\mathrm{Spec}(V)$. Then, by replacing $V$ by its strict Henselization, which is still a valuation ring https://stacks.math.columbia.edu/tag/0ASK, we assume that $V$ is strictly Henselian. In this case, by Hensel's lemma (EGA IV, 18.5.17), every connected component of $S_v$ meets some $V$-point $\mathrm{Spec}(V) \rightarrow S$. By EGA IV, 15.6.5, the fibral connected components of $S$ that meet this $V$-point comprise an open subset of $S$, so, by replacing $S$ by this open, we are reduced to the case when the fibers of $S$ are geometrically connected (we are using the well-known EGA IV, 4.5.13).

Let $s_0 \in S$ be the generic point of $S_v$. We claim that $\mathcal{O}_{S, s_0}$ is a valuation ring. For this, since $s_0$ maps to the generic point $a_0$ of the fiber $\mathbb{A}^n_v$ (generizations lift along flat maps!), the stability under étale maps of being a valuation ring (https://stacks.math.columbia.edu/tag/0ASJ) reduces us to considering $\mathcal{O}_{\mathbb{A}^n_V, a_0}$. The latter is a domain whose fraction field is the function field $K(\mathbb{A}^n_K)$ and, since the ideals of $V$ are totally ordered, one sees directly that every rational function in $K(\mathbb{A}^n_K)$, i.e., a quotient of polynomials with $K$-coefficients, can be expressed as a quotient of polynomials with $V$-integral coefficients and at least one coefficient in $V^\times$. Then either this quotient or its inverse lies in $\mathcal{O}_{\mathbb{A}^n_V, a_0}$, so that the latter is indeed a valuation ring (https://stacks.math.columbia.edu/tag/052K again), and hence so is $\mathcal{O}_{S, s_0}$.

Now we apply the valuative criterion of properness to obtain a $V$-map $\mathrm{Spec}(\mathcal{O}_{S, s_0}) \rightarrow A$ compatible with the map $S_K \rightarrow A$ that we want to extend. The former spreads out to an affine neighborhood of $\mathrm{Spec}(\mathcal{O}_{S, s_0})$, so we may find a quasicompact open $S_0 \subset S$ containing both $s_0$ and $S_K$ to which the initial map $S_K \rightarrow A$ extends.

Since $V$ is a filtered direct limit of normal domains $D$ that are of finite type over $\mathbb{Z}$, we may assume that $A$, $S$, $S_0$, $S_K \rightarrow A$, $S_0 \rightarrow A$ arise by base change from corresponding objects $A'$, $S'$, $S'_0$, $S'_{K'} \rightarrow A'$, $S'_0 \rightarrow A'$ (that are subject to analogous assumptions) defined over some such $D$, where $K' := \mathrm{Frac}(D)$. Since $s$ and $s_0$ have the same image in $\mathrm{Spec}(V)$ and we work locally at $s$, by replacing $S'$ by the preimage of the (open) image of $S_0'$ in $\mathrm{Spec}(D)$ and using the fact that $S'$ inherits geometric connectedness of fibers from $S$, we may assume that $S_0'$ is fiberwise dense in $S'$. Then $S_0'$ covers all the height $\le 1$ points of $S'$: indeed, it covers $S'_K$ and every other height $\le 1$ prime is the generic point of a fiber of $S'$ at a height $\le 1$ prime of $D$. Therefore, by Weil's result Thm. 4.4/1 in Bosch, Lutkebohmert, Raynaud "Neron models," the map $S_0' \rightarrow A$ extends to a map $S' \rightarrow A$. The base change of this extension back to $V$ is the desired extension $S \rightarrow A$ of $S_K \rightarrow A$. QED.

To end with some speculations, I am guessing that in the case of semiabelian reduction, Néron lft models should exist over arbitrary valuation rings--it would be interesting to see this worked out (or a counterexample given).

No, they needn't.

See: David Holmes: Neron models of jacobians over base schemes of dimension greater than 1., to appear in Journal fur die reine und angewandte Mathematik.

and

• I'm not sure where to find the answer in these two papers. Do these papers not assume the base scheme is locally noetherian (and thereby exclude certain valuation rings)? – Kriss Aug 5 '18 at 22:50
• For the non-existence remark, my paper even needs the base regular, so certainly locally noetherian. I guess Giulio has the same restriction. Note that even over normal base schemes, abelian schemes are not necessarily the Néron models of their generic fibres (e.g. E/k elliptic, base S = projective cone over E, then constant family $E\times S \to S$ is not a Néron model), so things can get pretty weird. – David Holmes Aug 7 '18 at 16:21