Is there an example of a flat proper relative curve $X/S$ with geometrically connected fibres and with $\mathrm{dim} S > 1$ and $S$ regular and connected with $0$-dimensional locus of bad reduction $S_{\mathrm{bad}} = \{s \in S: X_s/s \text{ not smooth}\}$?
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1$\begingroup$ Yes. Start with the (everywhere smooth) family of lines in $\mathbb{P}^3$. Now consider the map $\mathbb{P}^3\to \mathbb{P}^3$ by $[x,y,z,w]\mapsto [x^3,y^3,z^3,w^3]$. Now consider the family of images under this map of the lines. Families such as these were studied by Matt DeLand in connection to extensions of Bend-and-Break (ala Chang-Ran). $\endgroup$– Jason StarrCommented Nov 6, 2015 at 12:53
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$\begingroup$ Thank you very much! Are there more examples? $\endgroup$– user19475Commented Nov 6, 2015 at 13:06
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$\begingroup$ For genus $0$ curves of degree $d$ in $\mathbb{P}^n$, I believe the sharp results are known by Chang-Ran and DeLand. In terms of your $S$, there are families with $\text{dim}(S) = n$. For higher genus curves, I believe this is open. I realize that you are asking a local question, but for the associated global question, the best known result is Diaz's theorem. $\endgroup$– Jason StarrCommented Nov 6, 2015 at 13:38
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$\begingroup$ Can you give me the precise reference for Chang-Ran, DeLand and Diaz? $\endgroup$– user19475Commented Nov 6, 2015 at 13:51
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$\begingroup$ Matt DeLand, Complete families of linearly non-degenerate rational curves, arxiv.org/abs/0710.5713v1. M. Chang and Z. Ran. Closed Families of Smooth Space Curves. Duke Mathematical Journal 52(1985), no. 3, 707-713. Steven Diaz. A bound on the dimensions of complete subvarieties of ${\cal M}_{g}$. Duke Math. J. 51 (1984), no. 2, 405–408. 14H10 (14H30) $\endgroup$– Jason StarrCommented Nov 6, 2015 at 14:08
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1 Answer
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The answer is no if $f:X\to S$ is locally projective and the genus $g$ of the general fiber is $\geq1$. (About these restrictions, see remarks at the end).
Assuming this, put $U:=S\smallsetminus S_\mathrm{bad}$. By assumption, $S$ is regular and $U$ contains all points of codimension $\leq1$ of $S$. I do not assume $\dim S_\mathrm{bad}=0$, but of course we may reduce to this case if we prefer.
The main point is:
$X_U$ extends uniquely to a proper and smooth $f':X'\to S$.
This follows from my paper: Un théorème de pureté pour les familles de courbes lisses, CRAS Paris vol 300 issue 14 (1985), 489-492.
Now let us prove that the identification $X_U\cong X'_U$ extends to an isomorphism $X\cong X'$. Fix an invertible sheaf $L$ on $X$, very ample relative to $S$ and such that $\mathscr{E}:=f_*L$ is locally free on $S$, and commutes with base change in the usual sense. Moreover, assume that the degree of $L$ in the fibers is large enough ($\geq2g+2$ ?) to automatically ensure these properties on all smooth curves of genus $g$. So, we have a closed immersion $i:X\hookrightarrow \mathbb{P}(\mathscr{E})$.
Claim: The pair $(X,L)$ (subject to the conditions of flatness, smoothness over $U$, and strong ampleness) is determined up to unique isomorphism by $(X_U,L_U)$.
Indeed, $(X_U,L_U)$ determines $\mathscr{E}_U$ and $i_U:X_U\hookrightarrow \mathbb{P}(\mathscr{E}_U)$. Then we recover $\mathscr{E}$ as $j_*\mathscr{E}_U$ (by the codimension assumption, $j$ being the inclusion of $U$), and then $X$ must be isomorphic to the schematic closure of $i_U(X_U)$ in $\mathbb{P}(\mathscr{E})$. Of course, $L$ is then the restriction of $\mathscr{O}_{\mathbb{P}(\mathscr{E})}(1)$ to $X$.
Now we can "do the same" with $X'$. More precisely, over $X'_U=X_U$ we have the sheaf $L_U$, and since $X'$ is a regular scheme this extends to an invertible sheaf $L'$ on $X'$. Since the degree in the fibers is locally constant, this $L'$ satisfies all our ampleness requirements. So we get a pair $(X',L')$ whose restriction to $U$ is isomorphic to $(X_U,L_U)$, and we conclude by the above claim.
Remark 1. One can probably get rid of the projective assumption: we may assume $S$ local henselian, in which case projectivity should be automatic.
Remark 2. My purity result might be true in genus $0$, but I never worked this out.
answered Nov 7, 2015 at 14:11
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$\begingroup$ Thank you very much! So you prove that the locus of bad reduction has codimension $\leq 1$? $\endgroup$– user19475Commented Nov 7, 2015 at 17:09
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2$\begingroup$ @TimoKeller: yes, that's right. $\endgroup$ Commented Nov 7, 2015 at 17:35
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$\begingroup$ @TimoKeller In the last part of Moret-Bailly's answer one can also argue as follows. Let $S$ be an integral noetherian regular scheme. Assume $g\geq 1$ and assume that $X_U\cong X_U^\prime$, where $X$ and $X^\prime$ are curves over $S$ in $\mathcal M_g(S)$ and $U\subset S$ is a dense open. The fact that this isomorphism extends to an isomorphism of $X$ and $X^\prime$ over $S$ follows from the separatedness of the stack $\mathcal M_g$ of smooth proper curves of genus $g$... $\endgroup$ Commented Nov 8, 2015 at 19:13
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$\begingroup$ ...More precicely, for all schemes $S$ and for all $X$ and $Y$ over $S$ (in $\mathcal M_g(S)$), the morphism $\mathrm{Isom}_S(X,Y)\to S$ is proper. (If $g\geq 2$, this morphism is finite. But this is not the case when $g=1$.) The given isomorphism $X_U \cong X^\prime_U$ induces a section of $Isom_S(X,Y)$ over $U$. Since $S$ is regular (noetherian) this "generic section" extends to a section over $S$. See Gabber-Liu-Lorenzini Prop 6.2 math.u-bordeaux1.fr/~qliu/articles/GLL2-Duke.pdf . $\endgroup$ Commented Nov 8, 2015 at 19:13
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1$\begingroup$ @AriyanJavanpeykar: But we don't know (yet) that $X\in\mathscr{M}_g(S)$. $\endgroup$ Commented Nov 8, 2015 at 21:46