# Maximal abelian unramified outside S extension of exponent m of K where S is a finite set of places of K is finite

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.

• 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… Feb 12 at 13:34
• @Aphelli right! Since I am using the strong approximation, I need to assume that $S$ is non-empty. Feb 13 at 10:47
• Apart from that, I don't see what I messed up. Feb 13 at 10:54
• 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! Feb 13 at 11:06
• 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. Feb 13 at 12:19