Proof. ${\mu }_n\subset K$ gives $L∕K$ normal, and $\mathrm{char}K\nmid n$ gives that $L∕K$ is separable. So $L∕K$ is Galois.

Define the Kummer pairing

Well-defined: Suppose $\alpha ,\beta \in L$ with ${\alpha }^n={\beta }^n=x$. Then ${\left(\frac{\alpha }{\beta }\right)}^n=1$, so $\frac{\alpha }{\beta }\in {\mu }_n\subset K$. Then $\sigma \left(\frac{\alpha }{\beta }\right)=\frac{\alpha }{\beta }$ for all $\sigma \in \mathrm{Gal}(L∕K)$, hence $\frac{\sigma (\alpha )}{\alpha }=\frac{\sigma (\beta )}{\beta }$ for all $\sigma \in \mathrm{Gal}(L∕K)$.

Bilinear:

Non-degenerate: Let $\sigma \in \mathrm{Gal}(L∕K)$. If $⟨\sigma ,x⟩=1$ for all $x\in \mathrm{\Delta }$, then

adn hence $\sigma$ fixes $L$ pointwise, i.e. $\sigma =1$.

Let $x{(K^{\ast })}^n\in \mathrm{\Delta }$. If $⟨\sigma ,x⟩=1$ for all $\sigma \in \mathrm{Gal}(L∕K)$, then

So $\sqrt[n]{x}\in K$, so $x\in {(K^{\ast })}^n$, i.e. $x{(K^{\ast })}^n\in \mathrm{\Delta }$ is the identity element.

We get injective group homomorphisms

• (i) $\mathrm{Gal}(L∕K)\hookrightarrow \mathrm{Hom}(\mathrm{\Delta },{\mu }_n)$.
• (ii) $\mathrm{\Delta }\hookrightarrow \mathrm{Hom}(\mathrm{Gal}(L∕K),{\mu }_n)$.

(i) implies $\mathrm{Gal}(L∕K)$ is abelian and of exponent dividing $n$.

Fact: If $G$ is a finite abelian group of exponent dividing $n$ then $\mathrm{Hom}(G,{\mu }_n)\cong G$ (non canonically).

So

Therefore (i) and (ii) are isomorphisms. □

Proof.

• (i) Let $\mathrm{\Delta }\subset K^{\ast }∕{(K^{\ast })}^n$ be a finite subgroup. Let $L=K(\sqrt[n]{\mathrm{\Delta }})$ and ${\mathrm{\Delta }}^{\prime }=\frac{{(L^{\ast })}^n\cap K^{\ast }}{{(K^{\ast })}^n}$.

  We must show $\mathrm{\Delta }={\mathrm{\Delta }}^{\prime }$.

  Clearly $\mathrm{\Delta }\subset {\mathrm{\Delta }}^{\prime }$. So

  $$L=K(\sqrt[n]{\mathrm{\Delta }})\subset K(\sqrt[n]{{\mathrm{\Delta }}^{\prime }})\subset L.$$

  So $K(\sqrt[n]{\mathrm{\Delta }})=K(\sqrt[n]{{\mathrm{\Delta }}^{\prime }})$. So Lemma 11.1 gives $|\mathrm{\Delta }|=|{\mathrm{\Delta }}^{\prime }|$.

  Since $\mathrm{\Delta }\subset {\mathrm{\Delta }}^{\prime }$, it follows that $\mathrm{\Delta }={\mathrm{\Delta }}^{\prime }$.

• (ii) Let $L∕K$ be a finite abelian extension of exponent dividing $n$. Let $\mathrm{\Delta }=\frac{{(L^{\ast })}^n\cap K^{\ast }}{{(K^{\ast })}^n}$. Then $K(\sqrt[n]{\mathrm{\Delta }})\subset L$ and we aim to prove this inclusion is an equality.

  Let $G=\mathrm{Gal}(L∕K)$. Thu Kummer pairing gives an injection $\mathrm{\Delta }\hookrightarrow \mathrm{Hom}(G,{\mu }_n)$.

  Claim: This map is surjective.

  Granted the claim,

  $$\begin{array}{llll}\hfill [K(\sqrt[n]{\mathrm{\Delta }}):K]& \stackrel{\text{Lemma }\text{11.1}}{=}|\mathrm{\Delta }|& \hfill & \\ \hfill & \stackrel{\text{by claim}}{=}|G|& \hfill & \\ \hfill & =[L:K]& \hfill & \end{array}$$

  Since $K(\sqrt[n]{\mathrm{\Delta }})\subset L$ it follows that $L=K(\sqrt[n]{\mathrm{\Delta }})$.

  Proof of claim: Let $\chi :G\to {\mu }_n$ be a group homomorphism. Distinct automorphisms are linearly independent. So there exists $a\in L$ such that

  $$\underset{=y}{\underbrace{\sum _{\tau \in G}\chi {(\tau )}^{-1}\tau (a)}}\ne 0.$$

  Let $\sigma \in G$. Then

  $$\begin{array}{llll}\hfill \sigma (y)& =\sum _{\tau \in G}\chi {(\tau )}^{-1}\sigma \tau (a)& \hfill & \\ \hfill & =\sum _{\tau \in G}\chi {({\sigma }^{-1}\tau )}^{-1}\tau (a)& \hfill & \\ \hfill & =\chi (\sigma )\sum _{\tau \in G}\chi {(\tau )}^{-1}\tau (a)& \hfill & \\ \hfill & =\chi (\sigma )\cdot y& \hfill \text{(}\ast \text{)}\end{array}$$

  Therefore $\sigma (y^n)=y^n$ for all $\sigma \in G$. Hence $y^n\in K$.

  Let $x=y^n$. Then $x\in K^{\ast }\cap {(L^{\ast })}^n$. Then $x{(K^{\ast })}^n\in \mathrm{\Delta }$.

  Also, by ($\ast$), $\chi :\sigma \mapsto \frac{\sigma (y)}{y}=\frac{\sigma \sqrt[n]{x}}{\sqrt[n]{x}}$. So the map $\mathrm{\Delta }\hookrightarrow \mathrm{Hom}(G,{\mu }_n)$ sends $x\mapsto \chi$. This proves the claim. □

Proof. Theorem 11.2 gives $L=K(\sqrt[n]{\mathrm{\Delta }})$ for some $\mathrm{\Delta }\subset K^{\ast }∕{(K^{\ast })}^n$ a finite subgroup.

Let $𝔭$ be a prime of $K$.

for $P_1,\dots ,P_r$ some distinct primes in $O_L$.

If $x\in K^{\ast }$ represents an element of $\mathrm{\Delta }$ then

If $𝔭\notin S$ then all $e_i=1$, so $v_{𝔭}(x)\equiv 0(modn)$. Therefore $\mathrm{\Delta }\subset K(S,n)$ where

The proof is completed by the next lemma. □

Proof. The map

is a group homomorphism with kernel $K(\varnothing ,n)$.

Since $|S|<\infty$, it suffices to prove the lemma with $S=\varnothing$.

If $x\in K^{\ast }$ represents an element of $K(\varnothing ,n)$ then $(x)={𝔞}^n$ for some fractional ideal $𝔞$. There is a short exact sequence

$|{\mathrm{Cl}}_K|<\infty$ and $O_K^{\ast }$ being a finitely generated abelian group (Dirichlet’s unit theorem) gives us that $K(\varphi ,n)$ is finite. □