Let $f\colon X \rightarrow S$ be a semistable curve of genus $g \ge 0$. Being a semistable curve means that $f$ is a morphism of schemes such that

  1. $f$ is proper, flat, and of finite presentation;
  2. The geometric fibers of $f$ are reduced, connected, one-dimensional, of arithmetic genus $g$, and smooth away from finitely many points at which the singularities are required to be ordinary double points.

Does $f$, Zariski locally on $S$, come as a base change of a semistable curve $f' \colon X' \rightarrow S'$ with $S'$ regular?

A lemma of this sort seems to be used in the proof of 9.4/1 of "Neron models," i.e., in the proof of the representability of $\mathrm{Pic}^0_{X/S}$ by a scheme. The authors cite the paper of Deligne and Mumford for this reduction to a regular base. I can't see how the citation justifies the claim because Deligne and Mumford seem to deal with a narrower class of curves, namely, they impose an additional condition:

  1. If a geometric fiber of $f$ has an irreducible component isomorphic to $\mathbb{P}^1$, then that component meets the other components in $\ge 3$ points.

EDIT. The real question is: for a semistable curve $f$ as above, is $\mathrm{Pic}^0_{X/S}$ a scheme, as claimed in 9.4/1 of "Neron models"? The proof given there seems incomplete (see the comments below for some discussion). Any ideas are welcome.

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    $\begingroup$ For your real purposes (not explained in the above question, the aim really being construction of a certain canonical relatively ample line bundle) it is harmless to work etale-locally on the base. But any semistable curve becomes a stable "marked" curve etale-locally on the base, so one can use the $\mathbf{Z}$-smooth Deligne-Mumford stack of stable $n$-pointed genus-$g$ curves for suitable $n$ with $2g-2+n > 0$ (studied in a paper of Knudsen, building on the work of Deligne and Mumford). So this renders the likely erroneous reference to Deligne-Mumford in BLR moot. $\endgroup$
    – user74230
    Jan 26, 2015 at 3:43
  • $\begingroup$ Thank you! I will familiarize myself with the work of Knudsen to be able to fix the proof. Do you know, by the way, how do BLR get the locus $S_0 \subset S$ where $X\rightarrow S$ is smooth to be dense, and what inputs from Knudsen would I need to get this density claim to hold? $\endgroup$ Jan 26, 2015 at 4:35
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    $\begingroup$ Knudsen shows that $\overline{M}_{g,n}$ is an iterated curve fibration over the stack built by D-M when $g > 1$, over the D-R stack $\overline{M}_{1,1}$ for $g=1$, and over $\overline{M}_{0,3} = {\rm{Spec}}(\mathbf{Z})$ for $g=0$. From this one reduces to the universal curve $X$ over these "base stacks". The last is $\mathbf{P}^1$. For the others, we want $X$ over the formally smooth deformation ring at a geometric point to have smooth generic fiber. That in turn is clear from how non-smoothness in the universal deformation is encoded in parameters describing the deformation ring. $\endgroup$
    – user74230
    Jan 26, 2015 at 4:54
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    $\begingroup$ By the way, Knudsen's paper is not an easy read, or rather the technical details of his study of contraction and clutching morphisms require quite a bit of care. $\endgroup$
    – user74230
    Jan 26, 2015 at 4:58
  • $\begingroup$ @user74230: Thanks for your remarks. It seems that for the density needed by BLR, Thm. 2.7 in Knudsen's "The projectivity ... II" paper may suffice, since it describes the substack of singular curves as a normal crossings divisor in $M_{g, n}$. Regarding the readability and correctness of Knudsen, I'll take your word that it all works out in the end, at least until I finish reading BLR. $\endgroup$ Jan 26, 2015 at 5:35

1 Answer 1


The answer to the original problem (not the one in the edit) is: No.

Let $S$ be a nodal curve over the complex numbers with normalization $T$. Say $t, t' \in T$ are the inverse images of the node. Let $E$ be an elliptic curve. Take $E \times T$ and glue the fibres $E \times t$ and $E \times t'$ by translating with a nontorsion point to get $X \to S$.

This is the standard example of a smooth genus $1$ fibration $X \to S$ which is not locally projective.

Such a thing can never come from $X' \to S'$ with $S'$ regular. If it did then you would be able to Zariski shrink $S'$ and assume that $X' \to S'$ is smooth and proper. Then you would take a divisor on the generic fibre and take the closure $D'$ in $X'$. A small argument then shows $D'$ is relatively ample hence the pullback to $X$ is relatively ample over $S$. Contradiction.

  • $\begingroup$ Aha, Count Dracula! Your example is definitely a valid example with algebraic spaces, but why is it a scheme? $\endgroup$ Jun 16, 2015 at 18:39
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    $\begingroup$ By webusers.imj-prg.fr/~daniel.ferrand/Conducteur.pdf $\endgroup$ Jun 16, 2015 at 20:30
  • $\begingroup$ That works. Now I see that we can directly construct the local open affine covers. If $p$ is a closed point of $E$ such that $\underline{p}+\underline{\tau(p)} \sim 2\underline{\tau(\tau(p))}$ for your translation $\tau$, then the divisors $D=\underline{p} + 2\underline{\tau(p)}$ and $D'=\underline{\tau(p)}+2\underline{\tau(\tau(p))}$ are linearly equivalent. For an effective divsior $\mathcal{D}$ on $E\times T$ whose fibers over $t$, $t'$ are $D$, $D'$, the open complement of $\mathcal{D}$ maps to an open affine in $X$. $\endgroup$ Jun 16, 2015 at 20:58
  • $\begingroup$ My formula was wrong. Let $p$ and $q$ be points such that $\underline{p}+2\underline{q}$ is equivalent to $2\underline{\tau(p)}+\underline{\tau(q)}$, i.e. $\underline{q} +3\underline{o}= \underline{p}+3\underline{\tau(o)}$, where $o$ is the group identity. Then let $D$ be $2\underline{p}+\underline{q}$ and let $D'$ be $\underline{\tau(p)}+2\underline{\tau(q)}$. $\endgroup$ Jun 17, 2015 at 0:07

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