Here are the prerequisites and the parts which cause me trouble (taken from Milne's Fields and Galois Theory):
In particular, I don't quite understand the exactness at $H^1(G,\mu_n)$ and what it has to do with Hilbert 90 (as mentioned in the text).
If I understood the underlying maps correctly, the map $F^\times \cap E^{\times n} \to H^1(G,\mu_n)$ maps an element $z \in F^\times \cap E^{\times n}$ to the crossed homomorphism $f_z: G \to \mu_n$, $\sigma \mapsto \frac{\sigma(c)}{c}$ for an arbitrary $c \in E^\times$ with $z = c^n$ (one can show that the maps do not depend of the choice of $c$).
We can see here that $f_z$ is a principal crossed homomorphism, so the image of $F^\times \cap E^{\times n} \to H^1(G,\mu_n)$ is trivial. However, this seems to contradict the exactness because the kernel of $H^1(G,\mu_n) \to 1$ is obviously $H^1(G,\mu_n)$ which is not trivial.
And then, I still don't see what all of this has to do with Hilbert 90. Hilbert 90 says that if the norm of an element $\alpha \in E$ is $1$, then there exists a $\beta \in E$ such that $\alpha = \beta/\sigma(\beta)$ (where $\sigma$ is a generator of the Galois group of $E/F$).
Could you please resolve any misunderstandings I have here? Thank you!
Best Answer
Hilbert 90 has two common formulations -- one is the form you are quoting (where it is only valid if $E/F$ is a cyclic extension, by the way), and the other says $H^1(\text{Gal}(E/F), E^\times) = 0$ (any $E/F$ Galois -- this can be translated to your cyclic example by using an explicit description of the cohomology of cyclic groups).
Now if we take your short exact sequence, we get a long exact sequence in cohomology. The next term after $H^1(G, \mu_n)$ is $H^1(G, E^\times)$, which vanishes by Hilbert 90.