proof of Kummer theory
Proof.
(1) Let L=K(n√a); then L/K is Galois since K contains all nth roots of unity and thus is a splitting field for xn-a, which is separable
since n≠0 in K. Thus the elements of Gal(L/K) permute the roots of xn-a, which are
n√a,ζn√a,ζ2n√a,…,ζn-1n√a |
and thus for σ∈Gal(L/K), we have σ(n√a)=ζσn√a for some ζσ∈𝝁n. Define a map
p:Gal(L/K)→𝝁n:σ↦ζσ |
We will show that p is an injective homomorphism
, which proves the result.
Since 𝝁n⊂K, each nth root of unity is fixed by Gal(L/K). Then for σ,τ∈Gal(L/K),
ζστn√a=στ(n√a)=σ(ζτn√a)=ζτ(σ(n√a))=ζσζτn√a |
so that ζστ=ζσζτ and p is a homomorphism. The kernel of the map consists of all elements of Gal(L/K) which fix n√a, so that p is injective and we are done.
(2) Note that NL/K(ζ)=1 since ζ is a root of xn-1, so that by Hilbert’s Theorem 90,
ζ=σ(u)/u,for some u∈L |
But then σ(u)=ζu so that σ(un)=σ(u)n=ζnun=un and a=un∈K since it is fixed by a generator of Gal(L/K). Then clearly K(u) is a splitting field of xn-a, and the elements of 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]≥n=[L:K]. But K(u)⊂L, so K(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 |