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.

Thanks in advance

  • $\begingroup$ 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? $\endgroup$ – Filippo Alberto Edoardo May 21 '14 at 7:06
  • $\begingroup$ Yes, I am sure that this isomorphism is true since this is a trivial consequence of the article Factorial groups and Polya groups in Galoisian extensions of Q, in Commutative Ring Theory and Applications, Lecture Notes in Pure and Applied Mathematics, vol. 231, pp. 77-86, Marcel Dekker, New York, 2003. But I would like to get a direct proof of this isomorphism $\endgroup$ – joaopa May 21 '14 at 7:16

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 $\bigoplus_p \mathbb{Z}/e_p\mathbb{Z}$.


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