Galois Cohomology and Hilbert's Theorem 90

The setup and the two basic vanishing theorems

[[galois-cohomology|Galois cohomology]] is just group cohomology where G = Gal(L/K) is a Galois group and the module is something the field carries. Two foundational results: H^n(G, L) = 0 for n ≥ 1 (the additive group L is cohomologically trivial — this is the normal basis theorem in disguise), and the far more useful H^1(G, L*) = 0, which is [[hilbert-theorem-90|Hilbert's Theorem 90]] in its cohomological form. The additive vanishing makes the multiplicative result the interesting one.

Theorem 90 in the cyclic case, and why it is true

Hilbert's original statement is the cyclic case. Let L/K be a cyclic extension with Gal(L/K) = ⟨σ⟩ of order n. The cyclic recipe from guide 2 — applied to the multiplicative module L* — says a 1-cocycle is killed by the norm map and a coboundary is in the image of σ−1. Spelled out: if N_{L/K}(α) = 1 then α = β/σ(β) for some β in L*. Norm-one elements are exactly the multiplicative coboundaries.

Cyclic Theorem 90:  Gal(L/K) = <sigma>, order n,  N(x) = x . sigma(x) . sigma^2(x) ... sigma^{n-1}(x).
Claim: N(alpha) = 1  ==>  alpha = beta / sigma(beta).

Proof (the slick cocycle proof). Define the K-linear map on L:
   theta = id + alpha*sigma + (alpha*sigma(alpha))*sigma^2 + ... + (alpha*sigma(alpha)...sigma^{n-2}(alpha))*sigma^{n-1}.
By Dedekind's independence of characters the automorphisms 1, sigma, ..., sigma^{n-1}
are linearly independent over L, so theta is NOT the zero map: pick c in L with gamma = theta(c) != 0.
A direct computation using N(alpha)=1 gives
   alpha * sigma(gamma) = gamma,        i.e.   alpha = gamma / sigma(gamma).
Set beta = gamma. Done.   (gamma/sigma(gamma) is, in cocycle language, a 1-coboundary.)

WORKED:  L = Q(i), K = Q, Gal = {1, conj}, sigma = conjugation, n = 2.  N(a+bi) = a^2+b^2.
   alpha = (3+4i)/5 has N(alpha) = (9+16)/25 = 1.  Find beta = x+iy with alpha = beta/conj(beta):
   take beta = 1 + 2i :  beta/conj(beta) = (1+2i)/(1-2i) = (1+2i)^2/5 = (-3+4i)/5 ... = (-3+4i)/5.
   take beta = 2 + i  :  (2+i)/(2-i) = (2+i)^2/5 = (3+4i)/5 = alpha.   <-- beta = 2+i works.

That worked example is more than a curiosity: running over all β = x + iy, the formula β/σ(β) parametrizes every rational point on the unit circle — Theorem 90 over Q(i)/Q is the rational parametrization of Pythagorean triples. The abstract cohomological statement and a result you have known since school are the same theorem.

What Theorem 90 buys: Kummer theory

Feed Theorem 90 into the long exact sequence of the n-th power map. If K contains a primitive n-th root of unity, the sequence 1 → μ_n → L* → L* → 1 (last map x ↦ x^n) on cohomology gives, after H^1(G, L*) = 0 erases a term, the clean isomorphism K*/(K*)^n ≅ H^1(G, μ_n) ≅ Hom(G, μ_n). This is exactly [[kummer-extension|Kummer theory]]: abelian extensions of exponent n over K are governed by subgroups of K*/(K*)^n. Theorem 90 is the engine; Kummer theory is the vehicle.