An algebraic proof of Mumford's smoothness criterion for surfaces?

Question

(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?

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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].

Answer 1

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 residue 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.

Answer 2

Helene Esnault (and Eckart Viehweg) have a recent preprint on the arxiv: https://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.