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

I was describing Manish Kumar's work a few weeks ago to a fellow graduate student, and she stumped me with a big-picture question I couldn't answer.

Manish Kumar proved that the commutator subgroup of $\pi_1(\mathbb{A}^1_K)$, where $K$ is a characteristic $p$, algebraically closed field, is pro-finite free. He proved this, in fact, for any smooth affine curve over $K$.

(He proved this for algebraically closed fields of char $p$ which are uncountable in his thesis: http://www.math.msu.edu/~mkumar/Publication/thesis.pdf; and without the cardinality restriction in: http://arxiv.org/PS_cache/arxiv/pdf/0903/0903.4472v2.pdf)

As I explained to my colleague, this is a geometric analogue of Shafarevich's conjecture, that $Gal(\mathbb{Q}^{ab})$ is pro-finite free. Indeed $Gal(\mathbb{Q}^{ab})=\pi_1^c(Spec(\mathbb{Q}))$ ($c$ stands for the commutator subgroup). But why is $\mathbb{A}^1_K$, for $K$ an algebraically closed characteristic $p$ field (or indeed, any smooth affine curve over $K$), an analogue of $Spec(\mathbb{Q})$? Usually $\mathbb{A}^1_K$ (for $K$ an algebraically closed field) is an analogue of $Spec(\mathbb{Z})$. I came up with some partial explanations, but no full heuristic. Can you think of one?

share|improve this question

3 Answers 3

One can see that the commutator subgroup of the topological fundamental group of a complex curve is free for elementary reasons, but this is a pretty weak analogy. In fact I don't have a good heuristic of why it should be true, other than the fact that is. I was Manish's adviser, and I was pretty surprised by the result when he proved it.

share|improve this answer

I guess I'm not sure I agree they're analogous. First of all, extensions of Q can be ramified anywhere, while covers of A^1_K are unramified away from the infinite place. Q is much more like Spec K(T), the generic point of A^1_K -- but even here, to get a good analogy, you perhaps want K to be finite instead of algebraically closed.

share|improve this answer

Alright, I think I should write my 2 cents here:

Obviously $Spec(\mathbb{Q})$ and $\mathbb{A}_K$ are not directly analogous, but they do appear to be in relation to this problem. It seems that they are related through the intermediary $Spec(F)$ where $F$ is a function field over an algebraically closed field. Shafarevich's conjecture holds for $Spec(F)$ (this is an earlier result).

If we take any smooth affine curve, $C$, over an algebraically closed field of positive characteristic; and use the result that $\pi_1^c(C)$ is pro-finite free. Every abelian covering of $C$ will give an abelian extension of $\kappa(C)$ ($C$'s function field). However, there's no reason to think we get all abelian extensions of $\kappa(C)$ that way. However $\kappa(C)$ is also $\kappa(D)$ for different smooth affine curves, so may use their abelian unramified covers.

To make some order of this, start with an abelian extension of $\kappa(C)$, $L$. We may take $C$'s normalization in $L$. This may be branched at some points in $C$, but we may discard those. So any abelian extension of $\kappa(C)$ comes from an abelian unramified cover of some possibly different smooth affine curve whose function field is $\kappa(C)$.

It seems, however, extraordinary to expect that since $\pi_1^c(Spec(\kappa(C)))$ is pro-finite free, $\pi_1^c(C)$ should be; for any affine curve $C$. Is there some secret motivation for thinking this that I'm missing?

share|improve this answer
1  
I would not necessarily say that one is free "since" the other is free. –  JSE Apr 9 '10 at 1:24

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.