proof of Hilbert Theorem 90

Remember that two cocyclesMathworldPlanetmathPlanetmath a,a:GL* are called cohomologous, denoted by aa, if there exists bL*, such that a(τ)=ba(τ)τ(b-1) for all τG. Then

H1(G,L*)={a:GL*|a is a cocycle}/.

Now let a:GL* be a cocycle. Then consider the map


Since elements of the Galois group are linearly independentMathworldPlanetmath, α is not 0. So we can choose cL, such that b=α(c)0. Then for σG we have

σ(b) =τGσ(a(τ)τ(c))

since a is a cocycle, i.e. a(στ)=a(σ)σ(a(τ)). Then we get

σ(b) =a(σ)-1τGa(στ)(στ)(c)

Thus we have that a(σ)=bσ(b)-1 is a 1-coboundary.

Now we prove the corollary. Denote the norm by N. Now if x=yσ(y), we have


Now let N(x)=1, n=|G|. Since G is assumed cyclic, let σ be a generatorPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath of G. G is isomorphicPlanetmathPlanetmathPlanetmath to /n. We define the map x~:/nL* by


where [i] denotes the class of i in /n. Since N(x)=1, x~ is well defined. We have

x~([i+k]) =0ji+k-1σj(x)

Therefore x~ is a cocycle. Because of Hilberts Theorem 90, there exists yL*, such that x=x~([1])=yσ(y)-1.

Title proof of Hilbert Theorem 90
Canonical name ProofOfHilbertTheorem90
Date of creation 2013-03-22 15:19:27
Last modified on 2013-03-22 15:19:27
Owner mathcam (2727)
Last modified by mathcam (2727)
Numerical id 8
Author mathcam (2727)
Entry type Proof
Classification msc 11R32
Classification msc 11S25
Classification msc 11R34