MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X$ be a smooth projective geometrically connected curve over a number field $K$. Assume that $g\geq 2$.

Does there exist a Néron model $\mathcal X$ for $X$ over $O_K$?

By a Néron model, I mean a smooth model (not necessarily proper) with the "Néron universal property": for any smooth $O_K$-scheme $\mathcal Y$,

$$\mathrm{Hom}(\mathcal Y, \mathcal X) = X(\mathcal Y_K).$$

Is the smooth locus of the minimal regular model of $X$ over $O_K$ a Néron model? Does base change help? That is, does there exist a Néron model for $X_L$ after some suitable base change $L/K$?

Note that if $\mathcal{X}$ is the smooth locus of the minimal regular model of $X$ over $O_K$, we have the "Néron" property $\mathcal{X}(O_K) = X(K)$. (In fact, the image of a section in the minimal regular model lies in the smooth locus.)

This question was asked on stackexchange four months ago:

share|cite|improve this question
Why is it clear when $X$ has good reduction over ${\cal O}_K$ ? (I can see that for the "weak Néron property" - but why in general ?). – Damian Rössler Oct 23 '12 at 6:29
My apologies. In the course of editing the original question posted on stackexchange I reversed some of the statements. What I meant is that it's easy to see that the weak Neron property holds for $\mathcal{X}$ when $X$ has good reduction over $O_K$. This statement is only slightly harder to prove when $X$ doesn't have good reduction. I'll edit the question accordingly. – Harry Oct 23 '12 at 7:24

Here are some observations. I include the case $g=1$ (even if $X$ has no rational point). Denote by $\hat{\mathcal X}$ the (proper) minimal regular model of $X$ over the $O_K$ and let $\mathcal X$ be the smooth locus of $\hat{\mathcal X}$.

  • (1) If the Néron model exists, it is equal to the smooth locus $\mathcal X$ of the minimal regular model.

  • (2) If the fibers of $\mathcal X$ over $O_K$ have no rational irreducible component (e.g. if $X$ has good reduction), then $\mathcal X$ is the Néron model of $X$.

  • (3) (localization) If Néron models exist over DVRs, then they exist over any Dedekind domain.

  • (4) (base change) Let $R$ be a DVR. Let $R'/R$ be an extension of DVR such that an uniformizing element of $R$ is also an uniformizing element of $R'$ and such that the residue extension is separable (e.g. $R'$ can be the completion of a strict henselization of $R$). If the Néron model exists over $R'$, then the Néron model exists over $R$.

  • (5) You were right to not include the case $g=0$. The projective line doesn't have Néron model.

  • (6) Let $Y$ be a smooth scheme over a noetherian regular scheme $S$, let $Z$ be a regular scheme, flat and of finite type over $S$ and let $f: Y\to Z$ be a morphism. Then $f(Y)$ is contained in the smooth locus of $Z/S$. In particular, the canonical map ${\mathcal X}'(O_K)\to X(K)$ is bijective if $\mathcal X'$ is the smooth locus of (any) proper regular model of $X$.

  • (7) If $g=1$, then $\mathcal X$ is the Néron model of $X$.

Proof. Sorry I can't give all details by lack of energy and because it would be pretty unreadable in MO.

(1) Let $\mathcal N$ be the Néron model. Embedd it in a proper flat model, solve its singularity without touching to the regular locus (which contains $\mathcal N$). Then we get a proper regular model $\hat{\mathcal N}$ containing $\mathcal N$ as an open subset. The identity on $X$ extends to morphism $\hat{\mathcal N}\to \hat{\mathcal X}$. By (6), this morphism induces a morphism $\mathcal N\to \mathcal X$. Then $\mathcal X$ satisfies the universal Néron mapping property. By the uniqueness of Néron model, we get $\mathcal N\simeq \mathcal X$.

(2) Let $\mathcal Y -\to \mathcal X$ be a rational map defined over $K$ with $\mathcal Y$ smooth (regular is enough). The projection $p: \Gamma\to \mathcal Y$ is birational. Let $y\in Y$ and suppose $\Gamma_p$ is not finite. By a theorem of Abhyankar (the base scheme is excellent here, otherwise, localize and pass to the completion and use (4)), the components $E$ of $\Gamma_p$ are uniruled. But $E\to \mathcal X$ is a closed immersion, so $E$ is a rational curve in a close fiber of $\mathcal X$. Contradiction. Thus $p$ is quasi-finite biratonal and surjective. As $\mathcal Y$ is normal, $p$ is an isomorphism by Zariski's Main Theorem and the rational map we consider is actually defined everywhere. So $\mathcal X$ is the Néron model.

(3) The curve $X$ has good reduction away from finite many places. Using (2) for good reduction places and by gluing with Néron models over bad reduction places, we get a global Néron model over $O_K$.

(4) First the formation of the minimal regular model (and its smooth locus) is compatible with such base change. So if $\mathcal X\otimes R'$ satisfies the universal Néron mapping property over $R'$, then so does $\mathcal X$ over $R$ by faithfully flat descent for the definition domain of rational maps.

(5) Fix a model $\mathbb P^1_{O_K}$ of $\mathbb P^1_K$. They are plenty of endomorphisms of the generic fiber which don't extend to $\mathbb P^1_{O_K}$ (e.g. $[x,y]\mapsto [x, py]$). This shows that $\mathbb P^1_K$ has no proper smooth Néron model. The general case can be proved similarly with some extra works.

(6) Sketch: Consider $Y\times_S Z\to Y$. It is enough to show that its sections have images in the smooth locus (over $Y$), then use descent of smoothness (easy). The left hand side is regular because it is smooth over the regular scheme $Z$, and the right hand side si regular because it is smooth over the regular scheme $S$. So we can reduced to the case of flat morphism of finite type $W\to Y$ between two regular schemes. Let $y\in Y$ and let $w\in W$ be its image by a section $Y\to W$. Then $ O_{W,w}\to O_{Y,y}$ is a surjective map of regular local rings. Its kernel is generated by {$t_1, \dots, t_d$}, a part of a system of coordinates of $O_{W,w}$. So the maximal ideal $m_w$ of $O_{W,w}$ is generated by $t_1, \dots, t_d$ and $m_y$. Thus the maximal ideal of $O_{W_y, w}$ is generated by the images of $t_1, \dots, t_d$. The flatness of $W\to Y$ implies that $W_y$ has dimension $d$ at $w$. So $W_y$ is regular at $w$. It is in fact smooth because $w$ is a rational point of $W_y/k(y)$.

Another proof is to use $\Omega^1_{W/Y}$ and the fact that the image of a section is locally complete intersection.

Application: if $\hat{\mathcal X'}$ is a proper regular model over $O_K$, by the valuative criterion, $\hat{\mathcal X'}\to X(K)$ is bijective. But we just saw that the LHS is $\mathcal X'(O_K)$.

(7) We can work over a DVR $R$. If there exists a smooth $R$-scheme $\mathcal Y$ with non-empty special fiber and a morphism $\mathcal Y_K\to X$, then $\mathcal Y_K$ has a point in an étale extension of $R$. So $X$ has a point in an étale extension. By (4), we can thus suppose $X(K)\ne\emptyset$. So it is an elliptic curve, and Néron showed that $\mathcal X$ is the Néron model. If such $\mathcal Y$ doesn't exists, then $\mathcal X$ trivially satisfies the Néron mapping property.

share|cite|improve this answer
With Jilong Tong, we actually proved that the Néron model of any smooth projective geometrically connected curve of positive genus exists, and is equal to the smooth locus of the minimal proper regular model. – Qing Liu Jun 21 '13 at 7:05

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.