(of course not, it's usually uncountable; I really mean is it the profinite completion of a finitely presented group).

By definition, $\pi_1^{\operatorname{et}}(\operatorname{Spec}(\mathbb Z)\setminus\{p_1,\ldots,p_n\})=\operatorname{Gal}(K/\mathbb Q)$ where $K$ is the maximal extension of $\mathbb Q$ unramified away from $\{p_1,\ldots,p_n\}$. In the standard analogy due to Mazur, this should be thought of as the fundamental group of a $3$-manifold (perhaps $\mathbb S^3$) minus some link $L=K_{p_1}\cup\cdots\cup K_{p_n}$. Now, we know from topology that for any link $L$ in any $3$-manifold $M$, $\pi_1(M\setminus L)$ is finitely presented, however the proof involves lots of "cutting" up of the manifold into simplices, and this type of argument seems unlikely to apply in the case of $\operatorname{Spec}(\mathbb Z)$.

My question is about $\pi_1^{\operatorname{et}}(\operatorname{Spec}(\mathbb Z)\setminus\{p_1,\ldots,p_n\})$: is it the profinite completion of a finitely presented group? What if we replace $\mathbb Q$ with a number field?

I guess I could really just forget about the primes, and ask the same question for $\pi_1^{\operatorname{et}}(\operatorname{Spec}\mathcal O_F)$ for any number field $F$ (it's just that this is trivial when $F=\mathbb Q$, though otherwise probably not).

Also, what types of consequences does/would this have, if true?

  • 1
    $\begingroup$ This is not an answer, but just a remark: CFT gives that $\pi_1(\text{Spec }O_F-S)^{ab}$ is top. finitely generated. $\endgroup$
    – shenghao
    May 26, 2011 at 0:20

1 Answer 1


The number theorists call this $G_S$, the Galois group of the maximal extension of a number field $k$ unramified outside a finite set of primes $S$. It is known that this group can be topologically generated by a finite number of conjugacy classes (see Neukirch, Schmidt, Wingberg, Cohomology of Number Fields, 10.9.11). In other words, there exists a finite subset $T$ of $G_S$ such that the only normal closed subgroup of $G_S$ containing $T$ is $G_S$ itself.

The absolute Galois group of Q is not finitely generated, but it is not known whether $G_S$ is finitely generated. According to the discussion ibid. p.532, it is not even known what to expect. The maximal pro-$p$-quotient of $G_S$ is finitely generated for every prime $p$. [This corrects my earlier answer.]

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    $\begingroup$ I thought that it was an unresolved conjecture of Shafarevich that $G_S$ was topologically finitely generated. Perhaps I'm confused. $\endgroup$ May 26, 2011 at 2:47
  • $\begingroup$ I think you are right, it is a Conjecture of Shafarevich whether $G_S$ is topologically finitely generated, but the result in Neukirch et al book says it is generated by the "conjugacy classes" of finitely many elements (of course the conjugacy classes are not finite, so this does not prove Shafarevich's conjecture) $\endgroup$
    – A. Pacetti
    May 26, 2011 at 14:29

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