# proof of Kummer theory

###### Proof.

Let $\zeta\in K$ be a primitive $n^{\mathrm{th}}$ root of unity, and denote by $\boldsymbol{\mu}_{n}$ 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 $n\neq 0$ in $K$. Thus the elements of $\operatorname{Gal}(L/K)$ permute the roots of $x^{n}-a$, which are

 $\sqrt[n]{a},\ \zeta\sqrt[n]{a},\ \zeta^{2}\sqrt[n]{a},\ \ldots,\ \zeta^{n-1}% \sqrt[n]{a}$

and thus for $\sigma\in\operatorname{Gal}(L/K)$, we have $\sigma(\sqrt[n]{a})=\zeta_{\sigma}\sqrt[n]{a}$ for some $\zeta_{\sigma}\in\boldsymbol{\mu}_{n}$. Define a map

 $p:\operatorname{Gal}(L/K)\to\boldsymbol{\mu}_{n}:\sigma\mapsto\zeta_{\sigma}$

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

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

 $\zeta_{\sigma\tau}\sqrt[n]{a}=\sigma\tau(\sqrt[n]{a})=\sigma(\zeta_{\tau}\sqrt% [n]{a})=\zeta_{\tau}(\sigma(\sqrt[n]{a}))=\zeta_{\sigma}\zeta_{\tau}\sqrt[n]{a}$

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

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

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

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 $\operatorname{Gal}(L/K)$. Then clearly $K(u)$ is a splitting field of $x^{n}-a$, and the elements of $\operatorname{Gal}(L/K)$ send $u$ into distinct elements of $K(u)$. Thus $K(u)$ admits at least $n$ automorphisms over $K$, so that $[K(u):K]\geq n=[L:K]$. But $K(u)\subset L$, so $K(\sqrt[n]{a})=K(u)=L$. ∎

## References

• 1 Dummit, D., Foote, R.M., Abstract Algebra, Third Edition, Wiley, 2004.
• 2 Kaplansky, I., Fields and Rings, University of Chicago Press, 1969.
Title proof of Kummer theory ProofOfKummerTheory 2013-03-22 18:42:07 2013-03-22 18:42:07 rm50 (10146) rm50 (10146) 5 rm50 (10146) Proof msc 12F05