Easy computational example of first cohomology group: Is this how we do it? I'm an undergraduate student learning a little bit of group cohomology on my own.
I'd like to compute a few examples of the low-dimensional cohomology groups in some special cases to get some intuition on what I'm doing. Unfortunately, even though I get the definitions, I need some help in the computational side, specially computing $1$-cocycles.
I'm trying to compute $H^{1}$ for a few examples. Take a look at the first one I did:
Consider the $G$-module $M$, where $G=\operatorname{Gal}(\mathbb{Q}(i)\:/\:\mathbb{Q})$ and $M=[\mathbb{Q}(i)]^{\times}$.
I first looked for $1$-cocyles then, which are maps $f\colon G\to M$ such that $f(ab)=a\cdot f(b)+f(a)$ 
(Is it standard to write the operation on $M$ additively even when it is a multiplicative group, like in my example?)
So we need to figure out the possibilities for a function $f$ satisfying such functional equation. First, let $\varphi_{1}$ be the identity automorphism and we have $f(\varphi_{1})=f(\varphi_{1}\varphi_{1})=\varphi_{1}\:f(\varphi_{1})+f(\varphi_{1})$, where we conclude that $f(\varphi_{1})=1\in[\mathbb{Q}(i)]^{\times}$.
Similarly, I found that (denoting complex conjugation by $\varphi_{2}$) the  element $f(\varphi_{2})$ must have norm $1$, hence the units of $\mathbb{Z}[i]$ are the possibilities and hence there are $4$ possible $1$-cocycles. 
Is this correct? 
If so, how can one proceed in finding the $1$-cocyles for more complicated extensions? I feel that finding that the image of complex conjugation must have norm $1$ is luck and won't happen in general.
I also feel this is not the way people usually compute cohomology groups. Maybe using things I don't know much about such as derived functors and those lemmas from homological algebra, it gets easier.
Thanks for explanations.
 A: You’ve stumbled on the very simplest case of the original form of Hilbert’s Theorem 90. It says that if $K\supset k$ is a Galois extension with group $G$ that’s cyclic, say generated by $\sigma$, then every element of $K$ that’s of norm $1$ is of form $\frac{\sigma(\alpha)}{\alpha}$. The stronger, modern, version of Th. 90 says that if $K\supset k$ is Galois with group $G$, then $H^1(G,K^\times)$ is trivial. It’s a standard theorem in Galois Theory, not suitable for me to prove here, but you can find it in any text that’s advanced enough.
But I want to commend you for looking at things from the definition upwards, as you have done. I think that this is one of the best ways of understanding what’s really happening, and of appreciating the strength of the theorems that the general theory consists of.
Let’s take your specific case $K=\Bbb Q(i)$, and prove that if $\mathscr N^K_{\Bbb Q}(z)=1$, then $z$ is of the form $w/\overline w$ for a suitable Gaussian number $w$, in fact a Gaussian integer. Instead of mimicking the abstract proof, which has nothing to do with any special properties of the fields in question, I’ll give an elementary argument, using what we know about the arithmetic of the Gaussian integers.
If $\mathscr N^K_{\Bbb Q}(z)=1$, write $z=w/u$ with $w$ and $u$ Gaussian integers, so that $\mathscr N^K_{\Bbb Q}(w)=\mathscr N^K_{\Bbb Q}(u)$, an equation in $\Bbb Z$, so that the prime decomposition is the same on both sides. Since we’re assuming that $w\ne u$, though, it’s not the case that the prime decompositions of $w$ and $u$ in $\Bbb Z[i]$ are the same. Let’s look more closely. For a prime $\equiv3\pmod4$ like $7$ to appear in the equation of norms, it must do so to an even power, since the only Gaussian prime elements with norm $7$ or a power are $\pm7$ and $\pm7i$, related by units. Similarly, the only prime elements above $2$ are $\pm1\pm i$, all of these being related by units. The story is different for prime elements above natural primes $\equiv1\pmod4$ like $5$. These are $\pm2\pm i$ or $\pm1\pm2i$, some being related by units, but others like $2+i$ and $2-i$ not being so related. But every prime element dividing $5$ is a unit times one of those two. So for each $p\equiv1\pmod4$, we choose such a conjugate pair $a\pm bi$ of prime elements with $a^2+b^2=p$, and express $w$ and $u$ as products of these, and (say) positive primes $p\equiv3\pmod4$, as well as powers of $1+i$, with some unit factors. Now form $w/u$ and look at what we get: lots of cancellation, boiling down to:
$$
z=\frac wu=i^m\frac{A+Bi}{A-Bi}\,,
$$
but since $i=(1+i)/(1-i)$, we can drop off the unit factor and get your desired result.
There’s bonus here. Look at $\frac{2+i}{2-i}=\frac{3+4i}5$. Your number $z=\frac{a+bi}c$ of norm $1$ has $a^2+b^2=c^2$, a Pythagorean triple, and you’ve just seen that you can generate all such (ignoring negative signs on the $a$ and $b$) by taking any finite product of primes $p_i\equiv1\pmod4$, chose for each a pair of integers $a_i,b_i$ with $p_i=a_i^2+b_i^2$, multiply these Gaussian primes together with the previously given multiplicity, and get $A+Bi$, then divide by $A-Bi$ and get a Pythagorean triple. This is an arithmetic generation of the Pythagorean triples quite unlike the standard geometric generation of them.
