I saw a rather sleek proof of the following fact:

Let $K$ be a number field. Let $L/K$ be a maximal abelian unramified outside $S$ extension of exponent $m$ of $K$ where $S$ is a finite set of places of $K$. Then $L/K$ is finite.

The above result is used, for example, in the proof of the weak Mordell-Weil theorem.
Here's the proof. Let me know if it looks good.

By GCFT we have surjection $$ \frac{\mathbb I_K }{K^* K_{\infty +}^*\prod _{v\not \in S}\mathcal O_{K_v}\prod _{v\in S \backslash \infty } {K_v^*}^m}\rightarrow \operatorname{Gal}(L/K)$$
So it suffices to show that the LHS is finite. We have by strong approximation an injection $$\frac{\mathbb I_K }{K^* K_{\infty +}^*\prod _{v\not \in S}\mathcal O_{K_v}\prod _{v\in S \backslash \infty } {K_v^*}^m} \hookrightarrow \frac{\prod _{v\in S \backslash \infty}K_v^*}{\prod _{v\in S \backslash \infty}{K_v^*}^m} $$ The RHS is finite since each $K_v^*/{K_v^*}^m$ is $\mu_m(K_v)$ -finite by kummer theory.

Edit The above inclusion is clearly wrong as pointed out by @Aphelli. However, this may work $$\frac{\mathbb I_K}{K^* \prod _v \mathcal O_{K_v}}$$ is finite since it is isomorphic to the ideal class group. There is a finite index quotient of $K^*\prod _v \mathcal O_{K_v}$, namely by the subgroup $K^* K_{\infty +}^* \prod _{v\not \in S} \mathcal O_{K_v}^* \prod _{v\in S \backslash \infty } U_n(K_v)$ call this $H$. Since the quotient in the surjection is smaller than the quotient by $H$ of $\mathbb I_K$, it must be finite.

  • $\begingroup$ This is wrong, since with an empty $S$ you get $L=K$, so you’re not accounting for the class group of $K$. Also, I’d be wary of this proof being essentially circular. There are a lot of pieces in global CFT… $\endgroup$
    – Aphelli
    Feb 12 at 13:34
  • $\begingroup$ @Aphelli right! Since I am using the strong approximation, I need to assume that $S$ is non-empty. $\endgroup$ Feb 13 at 10:47
  • $\begingroup$ Apart from that, I don't see what I messed up. $\endgroup$ Feb 13 at 10:54
  • $\begingroup$ My point hasn’t changed. Somehow the class group of $\mathrm{Gal}(L/K)$ has not made an apparition – but you need to prove its finiteness to have global CFT! $\endgroup$
    – Aphelli
    Feb 13 at 11:06
  • 1
    $\begingroup$ My point is the following: suppose I take $m=2$, $K$ an imaginary quadratic field and $S$ a subset containing one odd place. Your argument shows that $|\mathrm{Gal}(L/K)|$ is at most $4$, which implies that the $2$-torsion in the class group of $K$ has rank at most $2$. It’s easy to check that this is false. $\endgroup$
    – Aphelli
    Feb 13 at 12:19


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