proof of Kummer theory


Proof.

Let ζ∈K be a primitivePlanetmathPlanetmath nth root of unity, and denote by 𝝁n the subgroupMathworldPlanetmathPlanetmath of K⋆ generated by ζ.

(1) Let L=K⁢(an); then L/K is Galois since K contains all nth roots of unity and thus is a splitting fieldMathworldPlanetmath for xn-a, which is separablePlanetmathPlanetmath since n≠0 in K. Thus the elements of Gal⁡(L/K) permute the roots of xn-a, which are

an,ζ⁢an,ζ2⁢an,…,ζn-1⁢an

and thus for σ∈Gal⁡(L/K), we have σ⁢(an)=ζσ⁢an for some ζσ∈𝝁n. Define a map

p:Gal⁡(L/K)→𝝁n:σ↦ζσ

We will show that p is an injectivePlanetmathPlanetmath homomorphismPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath, 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),

ζσ⁢τ⁢an=σ⁢τ⁢(an)=σ⁢(ζτ⁢an)=ζτ⁢(σ⁢(an))=ζσ⁢ζτ⁢an

so that ζσ⁢τ=ζσ⁢ζτ and p is a homomorphism. The kernel of the map consists of all elements of Gal⁡(L/K) which fix an, 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)=ζ⁢u so that σ⁢(un)=σ⁢(u)n=ζn⁢un=un and a=un∈K since it is fixed by a generatorPlanetmathPlanetmathPlanetmath 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⁢(an)=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