Do all curves have Néron models - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T07:50:33Z http://mathoverflow.net/feeds/question/110359 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/110359/do-all-curves-have-neron-models Do all curves have Néron models Harry 2012-10-22T19:38:57Z 2012-10-24T22:55:50Z <p>Let $X$ be a smooth projective geometrically connected curve over a number field $K$. Assume that $g\geq 2$.</p> <p>Does there exist a Néron model $\mathcal X$ for $X$ over $O_K$?</p> <p>By a Néron model, I mean a smooth model (not necessarily proper) with the "Néron universal property": for any smooth $O_K$-scheme $\mathcal Y$,</p> <p>$$\mathrm{Hom}(\mathcal Y, \mathcal X) = X(\mathcal Y_K).$$</p> <p>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 $L/K$?</p> <p>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.)</p> <p>This question was asked on stackexchange four months ago:</p> <p><a href="http://math.stackexchange.com/questions/153369/do-neron-models-of-hyperbolic-curves-exist" rel="nofollow">http://math.stackexchange.com/questions/153369/do-neron-models-of-hyperbolic-curves-exist</a></p> http://mathoverflow.net/questions/110359/do-all-curves-have-neron-models/110592#110592 Answer by Qing Liu for Do all curves have Néron models Qing Liu 2012-10-24T22:47:20Z 2012-10-24T22:55:50Z <p>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 $\hat{\mathcal X}$. </p> <ul> <li><p>(1) If the Néron model exists, it is equal to the smooth locus $\mathcal X$ of the minimal regular model. </p></li> <li><p>(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 $X$. </p></li> <li><p>(3) (localization) If Néron models exist over DVRs, then they exist over any Dedekind domain. </p></li> <li><p>(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 $R$. </p></li> <li><p>(5) You were right to not include the case $g=0$. The projective line doesn't have Néron model. </p></li> <li><p>(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 $X$. </p></li> <li><p>(7) If $g=1$, then $\mathcal X$ is the Néron model of $X$. </p></li> </ul> <p>Proof. Sorry I can't give all details by lack of energy and because it would be pretty unreadable in MO. </p> <p>(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 $\mathcal N\simeq \mathcal X$. </p> <p>(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. </p> <p>(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 $O_K$. </p> <p>(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. </p> <p>(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. </p> <p>(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 $W_y/k(y)$. </p> <p>Another proof is to use $\Omega^1_{W/Y}$ and the fact that the image of a section is locally complete intersection. </p> <p>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 $\mathcal X'(O_K)$. </p> <p>(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. </p>