Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $R$ be a DVR, and $k$ residue field of $R$. We suppose that $X_{0}$ is a stable curve over Spec$k$.

Dose there exist a stable model $X$ over $R$ such that the special fiber isomorphic to $X_{0}$ ?

If we assume $R=C[[t]]$, where C is complex number field, how to find a deformation which make the generic fiber is smooth?

share|improve this question
    
If by a stable model you mean a flat family of stable curves, then the answer is "yes" as stable curves have unobstructed deformations (and all formal deformations are algebraic). Probably you can also make the generic fibre smooth by choosing the deformations carefully... –  anon Jan 24 '13 at 10:57
2  
@anon: Unfortunately the OP did not specify that his DVR is complete (or even just Henselian). Otherwise your comment is perfectly correct. –  Jason Starr Jan 24 '13 at 11:25
    
Ah, yes, I missed the potential non-completeness, thanks! Is the claim clear in the henselian case without excellence assumption? –  anon Jan 24 '13 at 12:11
    
@anon: I think the claim is okay in the Henselian case even without excellence (but I need to think it through). –  Jason Starr Jan 24 '13 at 13:24
    
@kiseki: Choose a pluri-canonical projective embedding $j:X_0\hookrightarrow \mathbf{P}^n_k$, and consider the Hilbert scheme $H$ of $\mathbf{P}^n$ over $R$. By the proof of smoothness of the moduli stack of stable curves (of arithmetic genus $\ge 2$), $H$ is smooth near the point $x_0 \in H(k)$ corresponding to $(X_0,j)$ and stability holds for the universal curve over an open neighborhood of $x_0$. An open $U$ in $H$ around $x_0$ is etale over an affine space, so by approximating an $\widehat{R}$-point of that with an $R$-point we win when R is henselian. –  user30180 Jan 24 '13 at 14:32

1 Answer 1

up vote 5 down vote accepted

The comments above fully answer the OP's question. I am simply collecting some of these into an answer.

First, the answer to the original question is "no" if one does not impose some additional hypothesis on $R$ such as being complete or Henselian. As with many similar such questions, one negative answer comes from the Harris-Mumford(-Eisenbud) theorem that $\overline{M}_g$ is non-uniruled for $g\geq 23$. If $X_0$ is a stable, genus $g$ curve that is reducible with a single node $p$, say $(X',p') \cup (X'',p'')$ where $p$ is identified with the point $p'$ in the first irreducible component $X'$ and with $p''$ in the second irreducible component $X''$, and if $(X',p')$ and $(X'',p'')$ are sufficiently general pointed curves, then there is no deformation to a smooth curve over the (non-complete, non-Henselian) DVR $R=\mathbb{C}[t]_{\langle t \rangle}$. If there were, this would give a rational curve in $\overline{M}_g$ that intersects a general point of a boundary divisor. This would imply that a general point of the "interior" is also contained in a rational curve, contradicting the Harris-Mumford(-Eisenbud) theorem.

On the other hand, if $R$ is complete, or just Henselian, then there does exist a deformation. It is clear from the comments that the OP is looking for a very explicit formulation of this result. Here is one such formulation. Every proper curve is projective, and for stable curves, there is even an explicit tensor power of the dualizing sheaf that is very ample. Thus, assume that $X_0$ is given as a closed curve in some projective space $\mathbb{P}^n$. Up to re-embedding by a $2$-uple Veronese embedding (only necessary in positive characteristic), a sufficiently general pencil of hyperplane sections is "Lefschetz". More precisely, for a sufficiently general codimension $2$ linear subspace $L \subset \mathbb{P}^n$ that is disjoint from $X_0$, for the associated linear projection $$\pi_L:(\mathbb{P}^n\setminus L) \to \mathbb{P}^1,$$ the restriction of $\pi_L$ to $X_0$, $$\pi:X_0\to \mathbb{P}^1,$$ has sheaf of relative differentials $\Omega_\pi$ that is the pushforward to $X_0$ of an invertible sheaf from an effective Cartier divisor $D$ of $X_0$ with (a) no two distinct points of $D$ are contained in a common fiber of $\pi$, (b) the length of $D$ at every double point of $X_0$ equals $2$, and (c) for every smooth point of $X_0$ contained in $D$, the length of $D$ equals $1$.

The branch divisor of $\pi$ is, by definition, $\pi_*D$: an effective Cartier divisor in $\mathbb{P}^1$ that has length $2$ at the image of every double point of $X_0$ and has length $1$ at the image of every other point of $D$. By the analysis in Stable Maps and Branch Divisors of B. Fantechi and R. Pandharipande (the map $\pi$ is a "stable map"), for every formal deformation of the divisor $\pi_* D$ in $\mathbb{P}^1$, there exists a unique formal deformation of the stable map $(X_0,\pi:X_0\to \mathbb{P}^1)$ to $\mathbb{P}^1$ such that the associated branch divisor of the deformation equals the deformation of the branch divisor. In particular, choosing a deformation of $\pi_*D$ to a reduced divisor in $\mathbb{P}^1$ gives a formal deformation of $X_0$ to a smooth, stable curve.

share|improve this answer

Your Answer

 
discard

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.