# Isomorphism related to the first cohomology group

I would like to prove the following:

Let $K_1$, (resp. $K_2$) be a finite Galoisian extension of $\mathbb Q$ of degree $[K_1:\mathbb Q]=n_1$ with ring of integers $\mathcal O_{K_1}$ and Galois group $G_1$ (resp. $\mathcal O_{K_2}$ with degree $n_2=[K_2: \mathbb Q]$ and Galois group $G_2$). One assumes that $GCD(n_1,n_2)=1$. Denote by $L$ the compositum of $K_1$ and $K_2$ and its ring of integers $\mathcal O_L$. Then, there exists an isomorphism.

$H^1(G_1,\mathcal O_{K_1}^{\times})/(\oplus_{p\text{ prime}}\mathbb Z/e_p\mathbb Z)\times H^1(G_2,\mathcal O_{K_2}^{\times})/(\oplus_{p\text{ prime}}\mathbb Z/e_p\mathbb Z)\simeq H^1(G_1\times G_2,\mathcal O_{L}^{\times})/(\bigoplus_{p\text{ prime}}\mathbb Z/e_p\mathbb Z)$

where $e_p$ denotes the ramification index of the prime $p$ in the corresponding extension.

Any help would be welcome.

• Are you sure of this isomorphism? It sounds a bit strange in this generality: one a one hand, $\mathcal{O}_1^\times\times\mathcal{O}_2^\times$ is not isomorphic to the units of $L$ (to convince yourself, take $K_i$ to be totally real and compare $\mathbb{Z}$-ranks), and the inertia groups may very well not be cyclic...Do you have any reference where this is at least stated? – Filippo Alberto Edoardo May 21 '14 at 7:06
Each of the quotients in your expression is 0, so yes, there is an isomorphism, but it's between trivial groups. Here is the proof: for a Galois extension $K/\mathbb{Q}$, write down the short exact sequence $$1\rightarrow \mathcal{O}_K^\times \rightarrow K^\times \rightarrow {\rm PId(K)}\rightarrow 1,$$ where ${\rm PId(K)}$ denotes the group of principal fractional ideals of $K$. Now hit this with Galois invariants, to get (using Hilbert 90) $$1\rightarrow \{\pm 1\} \rightarrow \mathbb{Q}^\times \rightarrow {\rm PId(K)}^G\rightarrow H^1(G,\mathcal{O}_K^\times)\rightarrow 0.$$ In other words, $H^1(G,\mathcal{O}_K^\times)\cong {\rm PId(K)}^G/{\rm PId(\mathbb{Q})}$. But ${\rm PId(K)}^G$ is generated by elements of the form $\prod_{\sigma\in G}\mathfrak{p}^\sigma$, as $\mathfrak{p}$ ranges over the prime ideals of $K$, while ${\rm PId(\mathbb{Q})}$ inside it is generated by elements of the form $(p) = \prod_{\sigma\in G}(\mathfrak{p}^\sigma)^{e_p}$. So the quotient is isomorphic to $\bigoplus_p \mathbb{Z}/e_p\mathbb{Z}$.