Surprising Analogue of Q - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T15:00:24Z http://mathoverflow.net/feeds/question/20107 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/20107/surprising-analogue-of-q Surprising Analogue of Q H. Hasson 2010-04-01T20:39:00Z 2010-04-08T17:07:35Z <p>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.</p> <p>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 $K$.</p> <p>(He proved this for algebraically closed fields of char $p$ which are uncountable in his thesis: <a href="http://www.math.msu.edu/~mkumar/Publication/thesis.pdf" rel="nofollow">http://www.math.msu.edu/~mkumar/Publication/thesis.pdf</a>; and without the cardinality restriction in: <a href="http://arxiv.org/PS_cache/arxiv/pdf/0903/0903.4472v2.pdf" rel="nofollow">http://arxiv.org/PS_cache/arxiv/pdf/0903/0903.4472v2.pdf</a>)</p> <p>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, <em>any</em> 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?</p> http://mathoverflow.net/questions/20107/surprising-analogue-of-q/20686#20686 Answer by JSE for Surprising Analogue of Q JSE 2010-04-08T03:14:54Z 2010-04-08T03:14:54Z <p>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.</p> http://mathoverflow.net/questions/20107/surprising-analogue-of-q/20737#20737 Answer by H. Hasson for Surprising Analogue of Q H. Hasson 2010-04-08T14:53:30Z 2010-04-08T16:39:57Z <p>Alright, I think I should write my 2 cents here:</p> <p>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).</p> <p>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 <em>their</em> abelian unramified covers.</p> <p>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 <em>possibly different</em> smooth affine curve whose function field is $\kappa(C)$.</p> <p>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?</p> http://mathoverflow.net/questions/20107/surprising-analogue-of-q/20761#20761 Answer by Donu Arapura for Surprising Analogue of Q Donu Arapura 2010-04-08T17:07:35Z 2010-04-08T17:07:35Z <p>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.</p>