proof of Kummer theory
Proof.
Let $\zeta \in K$ be a primitive^{} ${n}^{\mathrm{th}}$ root of unity, and denote by ${\bm{\mu}}_{n}$ 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 $n\ne 0$ 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 ${\zeta}_{\sigma}\in {\bm{\mu}}_{n}$. Define a map
$$p:\mathrm{Gal}(L/K)\to {\bm{\mu}}_{n}:\sigma \mapsto {\zeta}_{\sigma}$$ |
We will show that $p$ is an injective^{} homomorphism^{}, which proves the result.
Since ${\bm{\mu}}_{n}\subset K$, 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 $K(\sqrt[n]{a})=K(u)=L$. ∎
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 |
---|---|
Canonical name | ProofOfKummerTheory |
Date of creation | 2013-03-22 18:42:07 |
Last modified on | 2013-03-22 18:42:07 |
Owner | rm50 (10146) |
Last modified by | rm50 (10146) |
Numerical id | 5 |
Author | rm50 (10146) |
Entry type | Proof |
Classification | msc 12F05 |