MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $V$?



share|cite|improve this question
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

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".

share|cite|improve this answer
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
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
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
@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

Your Answer


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.