Hilbert 90 and Vanishing of First Cohomology

Given a finite extension $L$ of fields $K$ , how do we determine the elements of norm $1$ ?

For instance, elements of the form $x= \frac{\sigma(a)}{a}$ , for $\sigma$ in the Galois group, have norm $1$ because the norm is equal to

$\displaystyle N(x) =\prod_{\tau \in G} \tau(x) = \prod_{\tau \in G} \tau \left( \frac{\sigma(a)}{a}\right) =\prod_{\tau \in G} \frac{\tau \sigma(a)}{\tau a} =\frac {\prod_{g \in G} g(a)}{\prod_{g' \in G} g'(a)} =1$

Are there any other examples?

Example: Consider the field extension $\mathbb{Q}(i) / \mathbb{Q}$ . The automorphisms are identity and the conjugata map $z =a+ib \to \bar z= a-ib .$ The elements of norm $1$ correspond to $a+ib$ with $z\bar z =a^2+b^2=1$ . That is we want to find all the rational solutions to $a^2+b^2=1 .$

The elements of the form

$\displaystyle \frac{\bar z}{z} =\frac{c-id}{c+id} = \frac{c^{2}-d^{2}}{c^{2}+d^{2}}+\frac{2 c d}{c^{2}+d^{2}} i$

indeed give rational parametrization of the circle $x^2+y^2=1$ , that is all the solutions (elements of norm 1) can be written in this form. (Note this is a way to parametrize all the Pythagorean triples)

More generally if we consider the extension $\mathbb Q(\sqrt{_D})$ , and the norm equation $x^2+Dy^2=1$ , we get that

$\displaystyle a^2+Db^2=1 \iff (a, b)=\left(\frac{c^{2}-D d^{2}}{c^{2}+D d^{2}}, \frac{2 c d}{c^{2}+D d^{2}}\right)$

Hilbert 90:

If the extension is a cyclic, then any element of norm $1$ is of this form $\frac{\sigma (x)}{x}$ . That is $\displaystyle N(a) =a \sigma(a) \sigma^{2}(a) \cdots \sigma^{n-1}(a)=1 \iff a =\frac{\sigma b}{b}$

Consider the linear map

$\displaystyle \alpha: L \to L,\quad \alpha(x) =\sum_{i=0}^{n-1} c_{i} \sigma^{i}(x)$

where the “cocycle”

$c_i =a \sigma(a) \cdots \sigma^{i-1}(a), c_0 =1$

This is a non-zero map because $g$ are linear independent. In fact, if
$\displaystyle \sum c_{g} g(x) =0$ is a relation, we have $\displaystyle \sum c_{g} g(\eta x) = \sum c_{g} g(\eta) g(x) =0$ , we can get smaller relation (in terms of terms) by subtracting these two relations.

So choose $\displaystyle \theta$ such that $\zeta=\alpha(\theta) \neq 0$ .

Now

$\displaystyle \sigma(\zeta) = \sum_{i=0}^{n-1} \sigma(c_i) \sigma \sigma^{i}(\theta) = \sum_{i=0}^{n-1} \sigma(a) \sigma^2(a) \cdots \sigma^{i}(a) \sigma^{i+1}(\theta) =\frac{1}{a} \sum_{i=0}^{n-1} a \sigma(a) \sigma^2(a) \cdots \sigma^{i}(a) \sigma^{i+1}(\theta)$

Now use the fact that $\displaystyle N(a) =a \sigma(a) \sigma^{2}(a) \cdots \sigma^{n-1}(a)=1$ to see that $c_0 =c_n=1$ and hence

$\displaystyle \sum_{i=0}^{n-1} a \sigma(a) \sigma^2(a) \cdots \sigma^{i}(a) \sigma^{i+1}(\theta) = \displaystyle \sum_{i=1}^{n} a \sigma(a) \sigma^2(a) \cdots \sigma^{i-1}(a) \sigma^{i}(\theta) = \sum_{i=0}^{n-1} a \sigma(a) \sigma^2(a) \cdots \sigma^{i-1}(a) \sigma^{i}(\theta)$

Thus

$\displaystyle \sigma(\zeta) =\frac{1}{a}\zeta$

and with $b =\frac{1}{\zeta}$ , we have solved $a =\frac{\sigma (x)}{x}$ .

Another way: Solving for $a =\frac{\sigma b}{b}$ is equivalent to solving for $a =\frac{b_1}{\sigma b_1}$ where $bb_1=1$ . Hence we want to solve

$\displaystyle a \sigma (x) =x$

$a\sigma(\cdot)$ is a linear map on $L$ and hence we need to show that this map has a an eigenvalue of $1$ . Looking at the map $1 \otimes a\sigma()$ on $L\otimes_{K} L$ , we can deduce that there is an eigenvector with eigenvalue $1$ . In fact, the computations are the same, the eigenvector will be related to the cocycle- $\left(1, \sigma a, \sigma a \sigma^{2} a, \ldots, \sigma a \cdots \sigma^{n-1} a\right)$

Additive Version: This version is called the multiplicative version where we determine the structure of norm 1 elements. The additive version asks for trace zero elements. We can see that $\sigma(b) -b$ have trace zero. To show that every trace zero element of a cyclic extension has this form we consider

$\displaystyle \zeta_1 =\sum_{i=0}^{n-1} d_i\sigma^{i}(\theta_1)$

with the additive cocycle

$d_i =a+ \sigma(a) +\cdots+ \sigma^{i-1}(a), ~~~~d_0 =0, d_n =\text{trace}(a)=0$

for an element such that $\text{trace}(\theta) =\theta+\sigma(\theta)+\sigma^{2}(\theta)+\cdots+\sigma^{n-1}(\theta) \neq 0.$

Now $x =\frac{\zeta_1}{\text{trace}(\theta_1)}$ solves our equation

$a =\sigma(x)-x .$

The above argument can be generalized to the following.

Vanishing of First Cohomology: For a Galois extension $L / K$ with Galois group $G$ , we have $H^{1}\left(G, L^{{\times}}\right)=0$ .

First cohomology measures the cocycles modulo the coboundaries. Vanishing of $H^{1}\left(G, L^{{\times}}\right)=0$ means that every cocycle is a coboundary, that is if $f: G \rightarrow L^{\times}$ is a 1-cocycle, then it is a 1-coboundary,

Cocycles are functions $f: G \rightarrow L^{\times}$ such that $f(gh) = f(g)gf(h)$
Couboundaries are functions $f_x(g) =\frac{g(x)}{x}$ .

In the cyclic case, $f(\sigma^{i}) =c_i=a \sigma(a) \cdots \sigma^{i-1}(a), c_0 =1$ defines a cocycle and computation is exactly the same.

$\displaystyle \zeta = \sum_{g \in G} f(g) g(\theta) \neq 0$

$\displaystyle \sigma (\zeta) =\sum_{g \in G} \sigma(f(g)) \sigma g(\theta)$

Using $f(\sigma g) = f(\sigma) \sigma f(g)$ , we get

$\displaystyle \sigma (\zeta) =\sum_{g \in G} \frac{f(\sigma g)}{f(\sigma)} \sigma g(\theta) = \frac{1}{f(\sigma} \zeta$

That is

$\displaystyle f(\sigma) =\frac{\sigma (\zeta^{-1})}{\zeta^{-1}}.$

Note that the zeroth cohomology $H^{0}(G, A)$ is the set of invariants $A^{G}$ invariant under the action of $G$ . Hence $H^{0}(G, L^{\times})=K^{\times}$ . In fact, the cohomology groups can be seen as “right derived” functors of this operation of taking invariants. The second cohomology is called the Brauer group $Br(L/K)$ – this corresponds to equivalence classes of group extensions of $L$ by $G$ , that is a group $\mathcal A$ such that $\mathcal{A}/L = G.$ Another way to think of them is as equivalence classes of central simple algebras (matrix algebra over division rings– just non-commutative version of finite field extensions) over $K.$

If $\displaystyle \mathcal A = \displaystyle \oplus_{\sigma} L x_{\sigma}$ , then $x_{\sigma} x_{\tau}=f(\sigma, \tau) x_{\sigma \tau}$ and $f(\sigma, \tau)$ is a 2-cocycle.

Applications:

Kummer’s Extensions:

If $K$ is a field such that it contains $n$ -th roots of unity. Also assume $(char K, n)=1$ . Then every cyclic extension of degree $n$ is of the form $L = K[\sqrt[n]{a}]$

Because $\zeta_n$ is a element of norm $1$ , by Hilbert 90 we have

$\displaystyle \zeta_n =\frac{\sigma (b)}{b}$

That is $\displaystyle \sigma(b) =\zeta_n b .$ In fact, we can take

$b^{-1} =1+ \zeta_n \sigma(\theta) + \zeta^{1+2} \sigma^{2}(\theta) +\zeta^{1+2+3}\sigma^{3}(\theta)+ \cdots \zeta^{1+2+3+ \cdots +n-1}\sigma^{n-1} (theta) \neq 0$

So all the Galois conjugates of $\displaystyle \sigma^{i}(b), i =0, 1, \cdots, n-1$ are distinct. Therefore the the field generated by $b$ $K[b]$ is of degree $n$ over $K$ and hence equal to $L$ .

But the minimal polynomial of $b$ is $\displaystyle \prod (X-\sigma^{i}(b)) =\prod (X-\zeta_n^{i}b) = X^n -b^n$

The last coefficient $\displaystyle a=b^n \in K. .$ Therefore $\displaystyle L =K[b]=K[\sqrt[n]{a}]$

Artin-Schreier Extensions:

That is about cyclic extension of degree $p$ over a characteristic $p$ field.

Every cyclic extension of degree $p$ over $K$ with $\text{char} K =p$ , is given by the splitting field of a polynomial of the form $X^p-X-a.$

First note that if $\displaystyle \alpha$ is a root of $\displaystyle X^p-X-a$ , then $\displaystyle \alpha+n$ is also a root, for we have

$\displaystyle (\alpha+n)^p -(\alpha+n)-a = \alpha^p+n^p -\alpha -n -a= \alpha^p-\alpha -a$

because $\displaystyle n^p-n=0$ in field of characteristic $\displaystyle p$ .

So the polynomial is either completely reducible (have all roots in $K$ or irreducible.

To show that every cyclic extension is generated by such polynomials, look at the trace zero element

$\text{Trace}(1) =0$ . By additive version of Hilbert 90, we have an element $\alpha$ such that

$\sigma(\alpha) -\alpha =1$ , that is $\sigma(\alpha)=\sigma+1$ . Now the minimal polynomial of $\alpha$ is given

$(X-\alpha)(X-\alpha-1) \cdot (X-\alpha -(p-1)) = (X-\alpha)^p -(X-\alpha)=X^p-X -(\alpha^p-\alpha)$

With $a= \alpha^p-\alpha$ , we see that the extension is generated by $X^p-X-a.$

What about extensions which are not cyclic? We said the Hilbert 90 generalizes to Vanishing of the first cohomology, but is it still true that all norm one elements are of this form? What about trace zero elements?

The answer is NO!

The norm one (trace zero ) conditions are not enough to capture elements of the form $\sigma(x)/x$ or ( $\sigma(x)-x$ in a non-cyclic field.

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