proof of Kummer theory


Proof.

Let ζK 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 n0 in K. Thus the elements of Gal(L/K) permute the roots of xn-a, which are

an,ζan,ζ2an,,ζn-1an

and thus for σGal(L/K), we have σ(an)=ζσan for some ζσ𝝁n. Define a map

p:Gal(L/K)𝝁n:σζσ

We will show that p is an injectivePlanetmathPlanetmath homomorphismPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath, which proves the result.

Since 𝝁nK, each nth root of unity is fixed by Gal(L/K). Then for σ,τGal(L/K),

ζστan=στ(an)=σ(ζτan)=ζτ(σ(an))=ζσζτan

so that ζστ=ζσζτ 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 uL

But then σ(u)=ζu so that σ(un)=σ(u)n=ζnun=un and a=unK 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