Let $L/K$ be a Galois extension with Galois group and $\mathfrak p$ be a prime of the ring of integers $\mathcal O\_K$.
I would like to prove that $H^1(G, I\_{\mathfrak p})=1$ where $I\_{\mathfrak p}$ is the free abelian group generated by the primes of $\mathcal O\_L$ above $\mathfrak p$.


Obviously, $I\_{\mathfrak p}$ is a $G$-module. I tried to adapt the proof of Hilbert 90, but I did not manage..
Thanks in advance. 



$H^1$ and fractional ideals group
This vanishing is very different from Hilbert's theorem 90. I will only sketch a proof as it is rather standard and it will be good for you to fill in the details.


Your module is a permutation module, which means that it has a Galois invariant basis. This is a special kind of induced representation, and the cohomology of induced representations can be computed using Shapiro's lemma. Let $H \subset G$ be the normal subgroup given by the kernel of the representation (note that this is exactly the decomposition group at $\mathfrak{p}$). The module $I\_\mathfrak{p}$ might not be irreducible (e.g. if $\mathfrak{p}$ splits), but we can reduce to this case as cohomology respects direct sums. Hence on applying Shapiro's lemma it suffices to show that
$$H^1(H,\mathbb{Z}) = 0,$$
where $H$ acts trivially on $\mathbb{Z}$. However we have
$$H^1(H,\mathbb{Z}) = \mathrm{Hom}(H,\mathbb{Z}) = 0,$$
as $H$ is finite. This completes the proof.


