most recent 30 from http://mathoverflow.net 2013-05-22T21:33:53Z http://mathoverflow.net/feeds/question/47620 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/47620/does-pi-1spec-mathbbz1-p-depend-on-p Makhalan Duff 2010-11-29T01:17:27Z 2010-11-29T14:10:18Z

<p>How difficult is it to know what $\pi_1(Spec(\mathbb{Z}[1/(p_1...p_r)]))$ is? Is it independent of the choice of $p_1,...,p_r$? When is it known, and what is known about it?</p>

http://mathoverflow.net/questions/47620/does-pi-1spec-mathbbz1-p-depend-on-p/47623#47623 Pete L. Clark 2010-11-29T01:27:23Z 2010-11-29T02:40:28Z

<p>The full etale fundamental groups in question are, I believe, complicated infinite profinite groups. (They are however "small" in the technical sense that they have only finitely many open normal subgroups of any given finite index, as follows from Hermite's finiteness theorem in algebraic number theory.)</p> <p>The abelianization of $\pi_1(\operatorname{Spec}(\mathbb{Z}[\frac{1}{p}])$ is the Galois group of the maximal abelian extension of $\mathbb{Q}$ which is ramified only at $p$ (and infinity). By Class Field Theory, this field is the direct limit of the ray class fields of conductor $p^n (\infty)$, i.e., the field generated by all $p$-power roots of unity. The Galois group is thus the inverse limit of the groups $(\mathbb{Z}/p^n \mathbb{Z})^{\times}$. When $p$ is odd, this is isomorphic to <code>$\mathbb{Z}_p \times \mu_{p-1}$</code> (where the second factor is cyclic of order $p-1$). So yes, this depends on $p$!</p>

http://mathoverflow.net/questions/47620/does-pi-1spec-mathbbz1-p-depend-on-p/47628#47628 Cam McLeman 2010-11-29T02:18:04Z 2010-11-29T14:10:18Z

<p>To add on to Pete's answer, let me comment that the differences are even more pronounced if we look at the maximal pro-$p$ quotient $\pi_1(\operatorname{Spec}(\mathbb{Z}\left[\frac{1}{p_1p_2\cdots p_r}\right])^{(p)}$ of this etale fundamental group. For example, if $r\geq 4$, then this group is infinite, in fact non-$p$-adic-analytic, if each $p_i\equiv 1\pmod{p}$ and is trivial if each $p_i\not\equiv1\pmod{p}$. The latter is basically for stupid reasons (only primes which are 1 mod p can ramify in a $p$-extension). But even ignoring stupid cases, there's a lot of fantastic arithmetic going on here. For example, $\pi_1(\operatorname{Spec}(\mathbb{Z}\left[\frac{1}{19\cdot 103}\right])^{(2)}$ is finite whereas $\pi_1(\operatorname{Spec}(\mathbb{Z}\left[\frac{1}{17\cdot 103}\right])^{(2)}$ is infinite, results which stem from simple quadratic residue calculations. Figuring out to generate these kinds of results more generally is an active and difficult area of research.</p>