a simple example of crossed homomorphism for the proof of Hilbert Theorem 90 For a field extension $K/F$ and a subgroup $G$ of $Aut(K)$, A crossed homomorphism is defined to be a function $f:G \rightarrow K^*$ satisfying $$f(\sigma\tau)= f(\sigma)\cdot \sigma(f(\tau)) $$ 
I want to illustrate this definition with a simple example. 
We use this definiton to prove Hilbert Theorem 90.
So, to obtain a simple example, I thought of a cyclic extension;
Consider the extension $\mathbb{Q}(\eta)/\mathbb{Q}, $  where $\eta^5=1$, primitive $5^{th}$ root of unity. 
Clearly, the extension will be cyclic; $G=Gal(\mathbb{Q}(\eta)/\mathbb{Q})=<\sigma>$, where $\sigma: \eta\rightarrow\eta^2$, and therefore $G\cong\mathbb{Z}_4$, and cyclic.
Now, an element $\alpha\in\mathbb{Q}(\eta)$ will be of the form $\alpha=a+b\eta+c\eta^2+d\eta^3+d\eta^4$, where $a,b,c,d,e\in \mathbb{Q}$.
The Norm of $\alpha$ relative to the extension will be $N(\alpha)=\sqrt{a^2+b^2+c^2+d^2+e^2}$, (I am not right here, I need to fix my definition of Norm, $N(\alpha)=\prod_{\sigma\in G}\sigma(\alpha)$)
Now I consider $u\in\mathbb{Q}(\eta)$  to be an element with $N(u)=1$. (So I need to have $N(u)=\prod_{\sigma\in G}\sigma(u)=u\sigma(u)\sigma^2(u)\sigma^3(u)=1$ ...what does it mean for $u$?)
I define $f:G\to K^*$ as $f(id)=1, f(\sigma)=u$, and $f(\sigma^i)=u\sigma(u)\cdots\sigma^{i-1}(u)$.
The problem is now that, I can not understand how this $f$ will be a crossed homomorphism.
I can not understand the step where the proof says; $$f(\sigma^i\sigma^j)=u\sigma(u)\cdots\sigma^{i+j-1}(u) \\ =(u\sigma(u)\cdots\sigma^{i-1}(u))\cdot \sigma^i(u\sigma(u)\cdots\sigma^{j-1}(u) $$
Can anyone help me to understand this step with the help of my example?
Thanks in advance,
 A: You are right up to $G \simeq \mathbb Z/4\mathbb Z$.
For the next line, note that the fifth cyclotomic polynomial is $\Phi_5 = x^4+x^3+x^2+x+1$, so that $\eta$ has degree $4$ (not $5$) over the rationals. This means that any element $a$ of $\mathbb Q(\eta)$ is a polynomial in $\eta$ of degree $3$ (not $4$).
Then, the norm is defined as the product of all (4) conjugates of $a$ : it is therefore a homogeneous polynomial of degree $4$ in the coefficients of $a$. Writing it in full would be a bit too long, so I give the particular case
$$ N(a_0 + a_1 \eta) = a_0^4 - a_1 a_0^3 + a_1^2 a_0^2 - a_1^3 a_0 + a_1^4.$$
(Note: there can never be a $\sqrt{}$ in a Galois norm, since $\sqrt{}$ is an analytic construction over the reals, while a Galois norm is a purely algebraic object).
Then, a $1$-cocycle (= crossed homomorphism) from the cyclic group $G = \mathbb Z/4/\mathbb Z$ to $K^{\times}$ is, as you suspected, determined by its value $c = f(\sigma)$ where $\sigma$ is a generator. By the cocycle relation, the next values are given by
$$ f(\sigma^2) = f(\sigma) \cdot \sigma (f(\sigma)) = c \sigma(c), \quad
 f(\sigma^3) = f(\sigma) \cdot \sigma (f(\sigma^2)) = c \sigma(c) \sigma^2(c), \quad
f(\sigma^4) = c \sigma(c) \sigma^2(c) \sigma^3(c), \dots$$
However,
 - since $G$ is cyclic, there is actually nothing more in this “$\dots$”;
 - since $\sigma^4 = 1$, we must have $f(\sigma^4) = 1$, which means $c\, \sigma(c) \sigma^2(c) \sigma^3(c) = 1$;
 - this last expression is exactly the (Galois) norm of $c$.
