# Maximal abelian extension with exponent $m$ unramified outside a finite set of places is finite

I'm studying the weak Mordell-Weil theorem in Silverman's "Arithmetic of elliptic curves" and I can't understand the proof of proposition 1.6 of chapter 8.

Here $K$ is a number field, $S$ a finite set of places containing the archimedean ones and $m>2$ an integer. I want to prove that the maximal abelian extension of exponent $m$ unramified outside $S$ is finite.

We can enlarge $K$ so we add the $m^{th}$ roots of unit to use kummer. We can also enlarge $S$ to make the ring o $S$-integers $R_S=\lbrace a\in K : v(a) \geq 0 \text{ for all} v\notin S\rbrace$ a PID.

The first question is, how can I do it? It says to add to $S$ the valuation corresponding to the prime dividing the integral representant of the ideal classes. Why does it make $R_S$ a PID?

Then we apply kummer and we deduce that the extension we are looking for is generated by the $m^{th}$ root of elements $a\in K^*/(K^*) ^m$ such that $\text{ord}_v(a) =0$ mod $m$ for every $v\notin S$.

We have to prove that this set called $T_S$ is finite by showing a surjective map from $R_S^*/(R_S^*) ^m$ which is finite thanks to Dirichlet theorem.

So we take a representant $a\in T_S$. $aR_S$ is an ideal (fractional?) which is the $m$ power of an ideal in $R_S$ given that the prime in $R_S$ correspond to the valuation not in $S$. Obviously the set of element $x\in R_S$ such that $v(x) >0$ is a prime but how can I show that every prime is in this form?

The last question could make the previous easier to answer. Given a set as $S$ and a $v_0\notin S$ is there an element $x\in K$ such that $v_0(x)=1$ and $v(x) =0$ for every other $v \notin S$?

• $R_S$ is the completion at a prime of $\mathcal{O}_{(S)}$ ? Commented Sep 8, 2017 at 13:42
• I don't understand your question completely, I hope the edit makes it clearer. About principalization I'll look into it Commented Sep 8, 2017 at 14:05

I keep your notations, adding $\mu_m$ := group of $m$-th roots of $1$, $G_S$ = Galois group of the maximal extension $K_S$ of $K$ which is unramified outside $S$ (the reference to $K$ is implicit), $X_S$ := its maximal abelian quotient of exponent $m$. To ease the notations, for any abelian group $A$, additive or multiplicative, write $A/m$ for $A/mA$ or $A/A^m$. You want to show that $X_S/m$ is finite.

In a first step, you enlarge $S$ so that the ring $R_S$ of $S$-integers becomes principal. How so ? The ideal class group $Cl_K (S)$ of $R_S$, usually called the $S$-class group of $K$, is the quotient of the usual class group $Cl(K)$ modulo the classes of the ideals with support in $S$ (Neukirch-Schmidt-Wingberg, chap.8). Since $Cl(K)$ is finite, you can finitely enlarge $S$ so that it contains the support of the ideals whose classes generate $Cl(K)$. This kills $Cl_K (S)$, i.e. $R_S$ is a PID.

In a second step, you add $\mu_m$ to $K$ in order to use Kummer theory. Let $T_S$ be the Kummer radical of $X_S=Hom(X_S,\mu_m)= Hom(G_S,\mu_m)$. If $S$ contains also the prime divisors of $m$ (which you forgot to mention), $T_S$ consists of the classes $[a] \in K^*/K^*{^m}$ s.t. $v(a) \equiv 0$ mod $m$ for all $v\notin S$. It is clear that $R_S^{*}/m$ injects naturally into $T_S$. Consider then the following composite map $\phi_S$ : $[a]\in T_S \to I_a= \prod_{v\notin S} P^{v(a)/m} \to$ class of $I_a \in Cl_K (S)$. In our situation, the triviality of $Cl_K (S)$ implies that $Ker \phi_S = T_S$. But by definition, $[a]\in Ker \phi_S$ iff $I_a = uR_S, u\in R_S^{*}$, which means that $Ker \phi_S=R_S^{*}/m$, so finally $R_S^{*}/m =T_S$, and Dirichlet's theorem implies the finiteness of $T_S$ as desired.

NB : A little bit of Galois cohomology allows to give a sleeker proof of a more precise result. Assume only that $S$ contains the archimedean primes, and let $U(K_S)$ denote the group of $S$-units of $K_S$, obtained by taking the inductive limit of the groups $R_S (L)$ when $L$ runs through the finite extensions of $K$ contained in $K_S$. Since $S$ contains the prime divisors of $m$, one knows (op. cit.) that $U(K_S)$ is $m$-divisible, i.e. raising to the $m$-th power gives an exact sequence of $G_S$-modules $1\to\mu_m \to U(K_S) \to U(K_S)\to 1$. Taking $G_S$-cohomology, one gets the so called Kummer exact sequence $1 \to R_S^{*}/m \to H^1(G_S, \mu_m) \to H^1(G_S,U(K_S))[m] \to 1$, where $(.)[m]$ denotes the kernel of multiplication by $m$. But $H^1(G_S, \mu_m)=Hom(G_S,\mu_m)$ because $G_S$ acts trivially on $\mu_m$, and it is known (op. cit.) that $H^1(G_S,U(K_S))\cong Cl_K (S)$. Since $Cl_K (S)$ is finite, we don't even need to enlarge $S$.

Let me give an example showing why the first claim is true. The number field $K = {\mathbb Q}(\sqrt{-5}\,)$ has class number $2$, and the ideal class is generated by the class of the ideal $I = (2,1+\sqrt{-5}\,)$. Now consider the ring ${\mathcal O}_2 = {\mathbb Z}[\frac12, \sqrt{-5}\,]$. In this (Dedekind) ring, the ideal $I = (1) = {\mathcal O}_2$ is the unit ideal since it contains the unit $2$; in particular, it is principal. If $J$ is any other ideal in the same class as $I$, then $I = \alpha J$ for some $\alpha \in K^\times$. In ${\mathcal O}_2$, this relation implies that $J = (\alpha)$. Thus every ideal in the class of $I$ in $K$ becomes principal in ${\mathcal O}_2$, which means that this ring has class number $1$.

• Thanks for your answer. I've not exactly clear how this answer relates to my question. Are you telling me that $R_S$ is in fact a localization of $O_K$ so I need to study only the images of the ideal of $O_K$ in $R_S$? How can i be sure that by enlarging S I've actually put the right inverse element in $R_S$ as it happens in your example? Is it because the answer to my last question is affirmative? Commented Sep 9, 2017 at 13:30
• You don't need an explicit description. If $P$ is a prime ideal in some ideal class of order $e$, then $P^e = (\pi)$, and you get the localization by adjoining the reciprocal of $\pi$. If you adjoin $1/p$, you localize at all primes above $p$; in your case, this doesn't hurt.
– user23365
Commented Sep 9, 2017 at 13:42