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It is known that a morphism of schemes $f\colon X \to S$ is smooth at a point $x \in X$ if and only if there is an open neighborhood $U$ of $x$ and an étale map $g \colon U \to \mathbb A^n_S$ such that $g \circ p=f_{|U}$, where $p \colon \mathbb A^n_S \to S$ is the natural projection.

I'm looking for a similar characterization for semistable curves $f \colon X \to S$. I'm interested in the case $S=Spec(k)$, with $k$ a field, and in the case $S=Spec(V)$, with $V$ a discrete valuation ring, where now $X$ is generically smooth.

In particular my question is: in the second case it is true that we can find $\lbrace Spec(R_i)\rbrace _{i \in I}$, an affine open covering of $X$, such that for each $i$, there is an étale map $V[x,y]/(xy-\pi) \to R_i$, where $\pi$ is a uniformizer of $V$?

Thanks.

Ricky

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The answer is essentially yes: should say $R_i$ has an etale neighborhood in common with $V[x,y]/(xy - a_i)$ for nonzero $a_i$ in $V$ (if by "generically smooth" you mean generic fiber is smooth). The case when $a_i$ can be taken of order at most 1 is when $X$ is regular. The real theorem you want is "structure theorem for ordinary double points", which is rigorously developed in the Freitag-Kiehl book as an application of Artin approximation (e.g., ensures that several ways to define "semistable curve" over a field $k$ are equivalent, including that non-smooth pts are always $k$-etale). –  BCnrd Jul 30 '10 at 2:29
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If you take the irreducible nodal cubic $y^2=x^2+x^3 \subset \mathbb{C}^2$, which is stable, no Zariski open set of it can be isomorphic to the reducible curve $xy=0$. For this reason, it seems to me that the answer to your question should be "no".

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I get your point, it could be that I have to work étale locally. Nevertheless I'm not sure that your example gives an answer: one should prove that there are no Zariski open subset of the nodal curve that are étale covering of $xy=0$. –  Ricky Jul 29 '10 at 14:23
    
It seems to me unlikely, at least over $\mathbb{C}$. In this case in fact xy=0 is a contractible topological space, so $\pi_1=\pi_1^{alg}=0$, so every étale cover is trivial. –  Francesco Polizzi Jul 29 '10 at 14:36
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But any étale morphism is open, so its image contains the generic points of the components passing through $x=y=0$. As the top curve is irreducible, this is impossible. To answer your original question, yes you have to work étale locally. If $V$ is excellent, using Artin's approximation theorem, for any singuar point $p$ of $X$, there exist two étale morphisms $U\to X$ and $U\to Spec(V[x,y]/(xy-\pi))$ containning $p$ and $(x=y=\pi=0)$ in their images. –  Qing Liu Jul 29 '10 at 15:42
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If one sets up the formulation of the result with a general base, or really local base ring (not just dvr), then limit arguments allow us to reduce to the case of a base essentially of finite type over $\mathbf{Z}$, so Artin approximation can be applied. That is, with the proper formulation one has a structure theorem over any base at all but the real content is the excellent case; all explained in the Freitag-Kiehl book. Once the general case is done, can then specialize to a dvr base, having now avoided any excellence hypotheses on it. –  BCnrd Jul 30 '10 at 2:59
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@Q: Yes (assuming you meant flat map to be of finite presentation, and base to be affine). Much of EGA IV$_3$, section 11 is devoted to the thorny issue of descent of flatness through the limit game, the key being a result of Raynaud. I am traveling and don't have EGA on my laptop (and numdam download is too slow here), but should be easy to find required results there; try 11.2. Appendix C of the Thomason-T. article in Grothendieck Festschrift proves the awe-inspiring fact that every qcqs scheme is an invlim of finite type $\mathbf{Z}$-schemes with affine transition maps. –  BCnrd Jul 31 '10 at 20:09
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