proof of Kummer theory

Proof.

Let $\zeta \in K$ be a primitive $n^{\mathrm{th}}$ root of unity, and denote by ${𝝁}_n$ the subgroup of $K^{\star }$ generated by $\zeta$.

(1) Let $L=K(\sqrt[n]{a})$; then $L/K$ is Galois since $K$ contains all $n^{\mathrm{th}}$ roots of unity and thus is a splitting field for $x^n-a$, which is separable since $n\ne 0$ in $K$. Thus the elements of $\mathrm{Gal}(L/K)$ permute the roots of $x^n-a$, which are

$$\sqrt[n]{a},\zeta \sqrt[n]{a},{\zeta }^2\sqrt[n]{a},\mathrm{\dots },{\zeta }^{n-1}\sqrt[n]{a}$$

and thus for $\sigma \in \mathrm{Gal}(L/K)$, we have $\sigma (\sqrt[n]{a})={\zeta }_{\sigma }\sqrt[n]{a}$ for some ${\zeta }_{\sigma }\in {𝝁}_n$.Define a map

$$p:\mathrm{Gal}(L/K)\to {𝝁}_n:\sigma \mapsto {\zeta }_{\sigma }$$

We will show that $p$ is an injective homomorphism, which proves the result.

Since ${𝝁}_n\subset K$, each $n^{\mathrm{th}}$ root of unity is fixed by $\mathrm{Gal}(L/K)$. Then for $\sigma ,\tau \in \mathrm{Gal}(L/K)$,

$${\zeta }_{\sigma \tau }\sqrt[n]{a}=\sigma \tau (\sqrt[n]{a})=\sigma ({\zeta }_{\tau }\sqrt[n]{a})={\zeta }_{\tau }(\sigma (\sqrt[n]{a}))={\zeta }_{\sigma }{\zeta }_{\tau }\sqrt[n]{a}$$

so that ${\zeta }_{\sigma \tau }={\zeta }_{\sigma }{\zeta }_{\tau }$ and $p$ is a homomorphism. The kernel of the map consists of all elements of $\mathrm{Gal}(L/K)$ which fix $\sqrt[n]{a}$, so that $p$ is injective and we are done.

(2) Note that ${\mathrm{N}}_{L/K}(\zeta )=1$ since $\zeta$ is a root of $x^n-1$, so that by Hilbert’s Theorem 90,

$$\zeta =\sigma (u)/u,\text{for some }u\in L$$

But then $\sigma (u)=\zeta u$ so that $\sigma (u^n)=\sigma {(u)}^n={\zeta }^n{u}^n=u^n$ and $a=u^n\in K$ since it is fixed by a generator of $\mathrm{Gal}(L/K)$. Then clearly $K(u)$ is a splitting field of $x^n-a$, and the elements of $\mathrm{Gal}(L/K)$ send $u$ into distinct elements of $K(u)$. Thus $K(u)$ admits at least $n$ automorphisms over $K$, so that $[K(u):K]\ge n=[L:K]$. But $K(u)\subset L$, so $K(\sqrt[n]{a})=K(u)=L$.∎