(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?
Best Answer
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.]