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

(Disclaimer: I'm a beginner in this area, so welcome corrections.)

Let $(X,x)$ be a germ of a complex surface (i.e. locally the zero set of some holomorphic functions) and assume that $x$ an isolated singular point. Mumford proved that if the local fundamental group of $X$ at $x$ is trivial, then in fact $x$ is smooth.

All the critters in the above paragraph have algebraic analogues, and the conversion was carried out (I believe) by Flenner: Let $A$ be a two-dimensional complete local normal domain containing an algebraically closed field of characteristic zero; if the \'etale fundamental group of [EDIT: the punctured spectrum of] $A$ is trivial, then $A$ is regular.

However, Flenner's proof is essentially by reduction to Mumford's theorem [as far as I, a non-German-speaker, can tell], rather than a new algebraic (or algebro-geometric) proof. So:

Does there exist a purely algebraic or algebro-geometric proof of Mumford's theorem?

Motivations include: (1) Mumford's proof is completely opaque to me; (2) No, I mean really really opaque; (3) I'm curious about extensions of the theorem to non-isolated singularities [which should probably be another question].

share|improve this question
I think you want to say the fundamental group of the punctured spectrum of $A$ –  Hailong Dao Feb 9 '10 at 0:42
Yep! Thanks, Hailong. –  Graham Leuschke Feb 9 '10 at 0:58
No problem, I hope you get a good answer! –  Hailong Dao Feb 9 '10 at 1:19

2 Answers 2

up vote 2 down vote accepted

I found what I think is the answer, in a paper by Cutkosky and Srinivasan called "Local fundamental groups of surface singularities in characteristic $p$". They prove, as Corollary 5: Suppose that $(A, m)$ is a complete normal local domain of dimension two, with algebraically closed resicue field $k$ of characteristic zero. (Slightly surprising, given the title of the paper.) Then $\pi_1(\operatorname{Spec} A -m)=0$ if and only if $A$ is smooth over $k$. They say that this gives "an arithmetic proof of the theorem of Mumford and Flenner."

The proof apparently uses Flenner's paper, but I don't think it uses Mumford's result. They get an expression for the local fundamental group in terms of a tree, and appeal to Flenner's Theorem 2.7 to know that the group is trivial iff $A$ is smooth. I haven't tried to read that section of Flenner's paper yet, but it seems to be independent of Mumford.

share|improve this answer

Helene Esnault (and Eckart Viehweg) have a recent preprint on the arxiv: http://arxiv.org/abs/1002.0024 for a characteristic $p$ version of Mumford's theorem. Perhaps your question is answered in the Flenner reference quoted there.

share|improve this answer
Thanks, but that article was actually one of the things that reminded me to ask this question. The Flenner paper in the references is the one linked in the question. –  Graham Leuschke Feb 9 '10 at 1:48

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.