Let ${\displaystyle L/K}$  be the quadratic extension ${\displaystyle \mathbb {Q} (i)/\mathbb {Q} }$ . The Galois group is cyclic of order 2, its generator ${\displaystyle \sigma }$  acting via conjugation:

${\displaystyle \sigma :c+di\mapsto c-di.}$ 

An element ${\displaystyle a=x+yi}$  in ${\displaystyle \mathbb {Q} (i)}$  has norm ${\displaystyle a\sigma (a)=x^{2}+y^{2}}$ . An element of norm one thus corresponds to a rational solution of the equation ${\displaystyle x^{2}+y^{2}=1}$  or in other words, a point with rational coordinates on the unit circle. Hilbert's Theorem 90 then states that every such element a of norm one can be written as

${\displaystyle a={\frac {c-di}{c+di}}={\frac {c^{2}-d^{2}}{c^{2}+d^{2}}}-{\frac {2cd}{c^{2}+d^{2}}}i,}$ 

where ${\displaystyle b=c+di}$  is as in the conclusion of the theorem, and c and d are both integers. This may be viewed as a rational parametrization of the rational points on the unit circle. Rational points ${\displaystyle (x,y)=(p/r,q/r)}$  on the unit circle ${\displaystyle x^{2}+y^{2}=1}$  correspond to Pythagorean triples, i.e. triples ${\displaystyle (p,q,r)}$  of integers satisfying ${\displaystyle p^{2}+q^{2}=r^{2}}$ .

The theorem can be stated in terms of group cohomology: if L^× is the multiplicative group of any (not necessarily finite) Galois extension L of a field K with corresponding Galois group G, then

${\displaystyle H^{1}(G,L^{\times })=\{1\}.}$ 

Specifically, group cohomology is the cohomology of the complex whose i-cochains are arbitrary functions from i-tuples of group elements to the multiplicative coefficient group, ${\displaystyle C^{i}(G,L^{\times })=\{\phi :G^{i}\to L^{\times }\}}$ , with differentials ${\displaystyle d^{i}:C^{i}\to C^{i+1}}$  defined in dimensions ${\displaystyle i=0,1}$  by:

${\displaystyle (d^{0}(b))(\sigma )=b/b^{\sigma },\quad {\text{ and }}\quad (d^{1}(\phi ))(\sigma ,\tau )\,=\,\phi (\sigma )\phi (\tau )^{\sigma }/\phi (\sigma \tau ),}$ 

where ${\displaystyle x^{g}}$  denotes the image of the ${\displaystyle G}$ -module element ${\displaystyle x}$  under the action of the group element ${\displaystyle g\in G}$ . Note that in the first of these we have identified a 0-cochain ${\displaystyle \gamma =\gamma _{b}:G^{0}=id_{G}\to L^{\times }}$ , with its unique image value ${\displaystyle b\in L^{\times }}$ . The triviality of the first cohomology group is then equivalent to the 1-cocycles ${\displaystyle Z^{1}}$  being equal to the 1-coboundaries ${\displaystyle B^{1}}$ , viz.:

${\displaystyle {\begin{array}{rcl}Z^{1}&=&\ker d^{1}&=&\{\phi \in C^{1}{\text{ satisfying }}\,\,\forall \sigma ,\tau \in G\,\colon \,\,\phi (\sigma \tau )=\phi (\sigma )\,\phi (\tau )^{\sigma }\}\\{\text{ is equal to }}\\B^{1}&=&{\text{im }}d^{0}&=&\{\phi \in C^{1}\ \,\colon \,\,\exists \,b\in L^{\times }{\text{ such that }}\phi (\sigma )=b/b^{\sigma }\ \ \forall \sigma \in G\}.\end{array}}}$ 

For cyclic ${\displaystyle G=\{1,\sigma ,\ldots ,\sigma ^{n-1}\}}$ , a 1-cocycle is determined by ${\displaystyle \phi (\sigma )=a\in L^{\times }}$ , with ${\displaystyle \phi (\sigma ^{i})=a\,\sigma (a)\cdots \sigma ^{i-1}(a)}$  and:

${\displaystyle 1=\phi (1)=\phi (\sigma ^{n})=a\,\sigma (a)\cdots \sigma ^{n-1}(a)=N(a).}$ 

On the other hand, a 1-coboundary is determined by ${\displaystyle \phi (\sigma )=b/b^{\sigma }}$ . Equating these gives the original version of the Theorem.

A further generalization is to cohomology with non-abelian coefficients: that if H is either the general or special linear group over L, including ${\displaystyle \operatorname {GL} _{1}(L)=L^{\times }}$ , then

${\displaystyle H^{1}(G,H)=\{1\}.}$ 

Another generalization is to a scheme X:

${\displaystyle H_{\text{et}}^{1}(X,\mathbb {G} _{m})=H^{1}(X,{\mathcal {O}}_{X}^{\times })=\operatorname {Pic} (X),}$ 

where ${\displaystyle \operatorname {Pic} (X)}$  is the group of isomorphism classes of locally free sheaves of ${\displaystyle {\mathcal {O}}_{X}^{\times }}$ -modules of rank 1 for the Zariski topology, and ${\displaystyle \mathbb {G} _{m}}$  is the sheaf defined by the affine line without the origin considered as a group under multiplication. ^[1]

There is yet another generalization to Milnor K-theory which plays a role in Voevodsky's proof of the Milnor conjecture.

Let ${\displaystyle L/K}$  be cyclic of degree ${\displaystyle n,}$  and ${\displaystyle \sigma }$  generate ${\displaystyle \operatorname {Gal} (L/K)}$ . Pick any ${\displaystyle a\in L}$  of norm

${\displaystyle N(a):=a\sigma (a)\sigma ^{2}(a)\cdots \sigma ^{n-1}(a)=1.}$ 

By clearing denominators, solving ${\displaystyle a=x/\sigma ^{-1}(x)\in L}$  is the same as showing that ${\displaystyle a\sigma ^{-1}(\cdot ):L\to L}$  has ${\displaystyle 1}$  as an eigenvalue. We extend this to a map of ${\displaystyle L}$ -vector spaces via

${\displaystyle {\begin{cases}1_{L}\otimes a\sigma ^{-1}(\cdot ):L\otimes _{K}L\to L\otimes _{K}L\\\ell \otimes \ell '\mapsto \ell \otimes a\sigma ^{-1}(\ell ').\end{cases}}}$ 

The primitive element theorem gives ${\displaystyle L=K(\alpha )}$  for some ${\displaystyle \alpha }$ . Since ${\displaystyle \alpha }$  has minimal polynomial

${\displaystyle f(t)=(t-\alpha )(t-\sigma (\alpha ))\cdots \left(t-\sigma ^{n-1}(\alpha )\right)\in K[t],}$ 

we can identify

${\displaystyle L\otimes _{K}L{\stackrel {\sim }{\to }}L\otimes _{K}K[t]/f(t){\stackrel {\sim }{\to }}L[t]/f(t){\stackrel {\sim }{\to }}L^{n}}$ 

via

${\displaystyle \ell \otimes p(\alpha )\mapsto \ell \left(p(\alpha ),p(\sigma \alpha ),\ldots ,p(\sigma ^{n-1}\alpha )\right).}$ 

Here we wrote the second factor as a ${\displaystyle K}$ -polynomial in ${\displaystyle \alpha }$ .

Under this identification, our map becomes

${\displaystyle {\begin{cases}a\sigma ^{-1}(\cdot ):L^{n}\to L^{n}\\\ell \left(p(\alpha ),\ldots ,p(\sigma ^{n-1}\alpha ))\mapsto \ell (ap(\sigma ^{n-1}\alpha ),\sigma ap(\alpha ),\ldots ,\sigma ^{n-1}ap(\sigma ^{n-2}\alpha )\right).\end{cases}}}$ 

That is to say under this map

${\displaystyle (\ell _{1},\ldots ,\ell _{n})\mapsto (a\ell _{n},\sigma a\ell _{1},\ldots ,\sigma ^{n-1}a\ell _{n-1}).}$ 

${\displaystyle (1,\sigma a,\sigma a\sigma ^{2}a,\ldots ,\sigma a\cdots \sigma ^{n-1}a)}$  is an eigenvector with eigenvalue ${\displaystyle 1}$  iff ${\displaystyle a}$  has norm ${\displaystyle 1}$ .

Thus, we may choose ${\displaystyle x=1+\sigma a+\sigma a\sigma ^{2}a+\ldots +\sigma a\cdots \sigma ^{n-1}a}$ .

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