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,


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 $c$.

  • $\begingroup$ Sorry, I have just fixed my understanding of Norm. So, could you please tell me in this case, why u, as an element with norm 1, affects the second line of the proof that way? $\endgroup$ – S.B. May 5 '15 at 9:50
  • $\begingroup$ It should be quite easy - once you balance your parentheses (a closing one is missing at the end) and figure out how $\sigma^i$ applies to the product $u \sigma(u) \dots \sigma^{j} (u)$. (Hint: $\sigma$ is a field automorphism). $\endgroup$ – Circonflexe May 5 '15 at 9:53
  • $\begingroup$ Thank you so much, I have got it now. $\endgroup$ – S.B. May 5 '15 at 9:56

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