I want to calculate 1-cocycles, as they are described in Neukirch's Class field theory. In what follows the Galois group of $L/\mathbb{Q}$ is acting on $L^{\times}$.
The first comment in this question, answers my question (but it doesn't answer the "Edited version"): A very simple problem on 1-cocycles
But now I have another question, actually I did something but I don't know how much of it is correct, and my question is kind of proof-verification or solution verification.
Let $D < 0$ be a discriminant, and consider the imaginary quadratic field $\mathbb{Q}(\sqrt{D})$, and its Galois group $G:=Gal(\mathbb{Q}(\sqrt{D})/\mathbb{Q})=\{\sigma_0, \sigma_1 \}$, where $\sigma_0$ is the identity automorphism, and $\sigma_1$ is the complex conjugation ($\Big(\sigma_1(\sqrt{D})=-\sqrt{D}\Big)$. Now, I want to calculate 1-cocycles $\Big( f(\sigma\tau)=\sigma f(\tau)+f(\sigma) \Big)$ in this case. By the same way, as described in that question, we have:
$$f(\sigma_{0}\sigma_{0})=\sigma_{0}f(\sigma_{0})+f(\sigma_0) \Longrightarrow f(\sigma_{0})=\sigma_{0}f(\sigma_{0})+f(\sigma_0) \Longrightarrow \sigma_{0}f(\sigma_{0}) =0 \Longrightarrow f(\sigma_{0}) =0, $$
and as it is mentioned in the first comment there: $f(\sigma_0)=0$ (which is $1$), and nothing is new. Now I want to calculate $f(\sigma_1)$, and my question starts here (is my conclusion right? And how can I finish it?):
$$f(\sigma_{1}\sigma_{1})=\sigma_{1}f(\sigma_{1})+f(\sigma_1) \Longrightarrow f(\sigma_{0})=\sigma_{1}f(\sigma_{1})+f(\sigma_1) \Longrightarrow 0 =\sigma_{1}f(\sigma_{1})+f(\sigma_1), \ (\star\ calculation)$$
now suppose $f(\sigma_1)=a+b\sqrt{D}$, $$0 =\sigma_{1}f(\sigma_{1})+f(\sigma_1) \Longleftrightarrow 0=(a-b\sqrt{D})+(a+b\sqrt{D}) \Longleftrightarrow 0=2a \Longleftrightarrow a=0,$$
or equivalently $f(\sigma_1)=b\sqrt{D}$. And here I stopped and I am not able to continue. Are my calculations correct? I think it is not correct, but then how can I calculate $f(\sigma_1)$? I have too many doubts about what I am going to write here: $f(\sigma_1)=b\sqrt{D}$, and we have to switch into $e^{\theta\sqrt{D}}=\cos(\theta\sqrt{-D})+i\sin(\theta\sqrt{-D})$, and then what? For which values of $\theta$ we can say $\cos(\theta\sqrt{-D})+i\sin(\theta\sqrt{-D}) \in \mathbb{Q}^{\times}(\sqrt{D})$?
But it seems very unusual and weird to me, and I think I am doing big mistakes, but I don't have any idea how to calculate $f(\sigma_1)$. Are my "$\star$ calculations" true?
Edit: The correct form of calculations after the answer by Ted (thanks to the answer by Ted, now I feel much better):
$$f(\sigma_{0}\sigma_{0})=\sigma_{0}f(\sigma_{0}).f(\sigma_0) \Longrightarrow f(\sigma_{0})=f(\sigma_{0}).f(\sigma_0) \Longrightarrow f(\sigma_{0}) =1, $$ and also
$$f(\sigma_{1}\sigma_{1})=\sigma_{1}f(\sigma_{1}).f(\sigma_1) \Longrightarrow f(\sigma_{0})=\sigma_{1}f(\sigma_{1}).f(\sigma_1) \Longrightarrow 1 =\sigma_{1}f(\sigma_{1}).f(\sigma_1), \ (\star\star \ calculation)$$
now suppose $f(\sigma_1)=a+b\sqrt{D}$, $$1 =\sigma_{1}f(\sigma_{1}).f(\sigma_1) \Longleftrightarrow 1=(a-b\sqrt{D}).(a+b\sqrt{D}) \Longleftrightarrow 1=Norm(a+b\sqrt{D}),$$
So we can say that $f(\sigma_1)=a+b\sqrt{D}$, where $a^2-Db^2=1$. Also for any rational solution of $a^2-Db^2=1$, there is a 1-cocyle $f$, with $f(\sigma_1)=a+b\sqrt{D}$.
You need to clarify what module $G$ is acting on: is it the additive group, or the multiplicative group, of $\mathbb{Q}(\sqrt{D})$ ? If it is the multiplicative group, then your calculations are wrong because the operation in the cocycle identity should be multiplication instead of addition: $f(\sigma \tau) = \sigma f(\tau) \cdot f(\sigma)$.
In Hilbert's Theorem 90, $G$ is acting on the multiplicative group.