Algebraic fundamental group of $X = Spec\mathbb{Z}\left[\frac{1}{2}\right]$

Question

I'm quite new to the notion of the fundamental group of schemes and I'm reading Lenstra's notes on Galois theory for schemes. One of the exercise asks to prove that the fundamental group $\pi(Spec\mathbb{Z}\left[\frac{1}{2}\right])$ is topologically generated by its 2-Sylow subgroups (as in the topologcal closure of the subgroup is the whole group). How do I prove this? I tried using a theorem saying that $\pi(Spec A\left[\frac{1}{n}\right]) = Gal(M/K)$ where $A$ is the ring of integers in a number field $K$ and $M$ is the maximal algebraic extension of $K$ that is unramified at all primes not dividing $n$.

In my scenario $A = \mathbb{Z}\,$, $K = \mathbb{Q}$ and $n = 2$, my guess is $M = \mathbb{Q}(i)$ since for every other algebraic extension there is at least a prime ideal $\mathfrak{p} \neq (2)$ that ramifies in that extension. In this way I would have $\pi(Spec\mathbb{Z}\left[\frac{1}{2}\right]) = \frac{\mathbb{Z}}{2\mathbb{Z}}$ and this is obviously generated by it's 2-Sylow. Is this right? It seems suspiciously easy...

Edit: my claim is obviously wrong, so my question now is, how can I compute the maximal field extension of $\mathbb{Q}$ which is unramified at every prime $\mathfrak{p} \neq (2)$

Answer 1

Let's first remark that if $G$ is a finite group, then $G$ is called a quasi-$p$-group if $G$ is generated by its $p$-Sylow subgroups. If $G$ is any finite subgroup, then the subgroup generated by all $p$-Sylow subgroups is quasi-$p$, let us denote it by $p(G)$. Clearly we have that $[G:p(G)]$ is always coprime to $p$. I claim that $p(G)$ is a characteristic subgroup of $G$, in particular it is normal. Indeed, let $H_1, \dots, H_n$ be the $p$-Sylow subgroups of $G$, then for any automorphism $\psi:G \to G$ and any $p$-Sylow subgroup $H_i$, we have that $\psi(H_i)$ is another $p$-Sylowsubgroup, so $\psi(H_i)=H_j$ for some $j$. Thus $p(G)=\psi(\langle H_1, \dots H_n\rangle)=\langle \psi(H_1), \dots, \psi(H_n)\rangle = \langle H_1, \dots, H_n\rangle$

We first prove the following:

Let $K/\Bbb Q$ be a a nontrivial finite Galois extension of odd order. Then there exists some odd prime $p$ that ramifies in $K$.

Proof By the Feit-Thompson theorem, $G=\mathrm{Gal}(K/\Bbb Q)$ is solvable. Thus there exists a proper normal subgroup $H \trianglelefteq G$ such that $G/H$ is abelian. Thus we can consider $K^H/\Bbb Q$ which is a nontrivial abelian Galois extension of odd order. Thus it suffices to prove the theorem in the abelian case, so we may assume that $G$ is itself abelian. So we get by Kronecker-Weber $K \subset \Bbb Q(\zeta_n)$, where $n$ is the conductor $\mathfrak{f}(K/\Bbb Q)$. It's well-known that a prime $p$ ramifies in $K$ if and only if $p$ divides $\mathfrak{f}(K/\Bbb Q)$. Thus it suffices to show that $n$ is not a power of $2$. But if $n=2^m$, then $[\Bbb Q(\zeta_n):\Bbb Q)]=\varphi(2^m)=2^{m-1}$ contradicting the fact that $K\subset \Bbb Q(\zeta_n)$ has odd degree over $\Bbb Q$.

Now we can proceed

Let $K/\Bbb Q$ be a finite Galois extension that is only ramified at $2$, then $G=\mathrm{Gal}(K/\Bbb Q)$ is a quasi-$2$-group.

Proof Consider the subgroup $2(G) \subset G$. By the remarks in the beginning, we get that $2(G)$ is a normal subgroup of odd index, thus $K^{2(G)}/\Bbb Q$ is a finite odd-degree Galois extension of $\Bbb Q$ that is only ramified at $2$. Thus by the previous result $K^{2(G)}=\Bbb Q$ and so $2(G)=G$

Now we're basically done. We just have to relate this to the infinite case. For this, we have the following result:

Let $G$ be a profinite group ans suppose each quotient $G/U$ is a quasi-$p$-group, where $U$ runs over all open normal subgroups of $G$, then the subgroup generated by all $p$-Sylow subgroups in $G$ is dense.

Proof We have that a closed subgroup $H \subset G$ is a $p$-Sylow subgroup of $G$ if and only if for all open normal subgroups $U$ of $G$ with projections $\pi_U:G\to G/U$, the image $\pi_U(H)$ is a $p$-Sylow subgroup of $G/U$. Furthermore, for any $p$-Sylow subgroup $P\subset G/U$, there is a $p$-Sylow subgroup $\widetilde{P} \subset G$ such that $\pi_U(\widetilde{P})=P$. Thus if we let $p(G)$ be the closure of the subgroup generated by all $p$-Sylow subgroups, we get that $\pi_U(p(G))=G/U$ for all $U$. This implies that $p(G)=G$, using that $p(G)$ is closed.