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I asked this question about a week ago here, but got no answer so I thought I'd aske here and see if I can get some help.

I was wondering if there are any conditions on a number field $K$ such that $$0 \longrightarrow K^{\times} \longrightarrow \mathbb{A}_{K}^{\times} \longrightarrow C_K \longrightarrow 0, $$ is a splits as a sequence of multiplicative abelian groups (here I have done the usual diagonal embedding of $K^{\times}$ into $\mathbb{A}_{K}^{\times}$)

I thought about looking at $H^{2}(C_K,K^{\times})$, but I'm not quite sure how to compute it.

Thank you.

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I don't have a good argument right now, but I am pretty sure, that the answer is never. But there is something called strong approximation. Probably this is sufficient for your application. – Marc Palm Feb 2 '13 at 16:57
Assume $K\ne\mathbf{Q}$ (splits for $\mathbf{Q}$). For any countably infinite direct product $G$ of finite cyclic groups, assume (H): there's no nontrivial homomorphism $G\rightarrow\mathbf{Z}$. If $r:\mathbf{A}^{\times}_K\rightarrow K^{\times}$ is a retraction and $r'$ its composite to $K^{\times}/\mu_K$ (torsion-free with no infinitely divisible elements $\ne 1$!), by (H) and "divisors" and the unit theorem $r'$ is trivial on $\prod_{v|\infty}K_v^{\times} \times \prod_{v\nmid\infty} O_{K,v}^{\times}$, so $r'(O_K^{\times})=1$. But $r(O_K^{\times})=O_K^{\times}$, so $K$ is imag. quadratic. – user30379 Feb 2 '13 at 17:02
@Marc Palm: It does split for $K = \mathbf{Q}$, since every rational idele admits a unique $\mathbf{Q}^{\times}$-multiple whose archimedean part is positive and finite part is everywhere an integral unit. – user30379 Feb 2 '13 at 17:03
I am confused. the finite part is not unique. strong approximation holds for any open subgroup. – Marc Palm Feb 2 '13 at 17:16
@Marc: Strong approximation has nothing to do with (discontinuous) splittings. Consider any $a \in \mathbf{A}^{\times}_{\mathbf{Q}}$ and the finitely many finite places $v$ such that $a_v$ is not an integral unit. Let $p_v$ be the residue characteristic at $v$ and $n_v = {\rm{ord}}_v(a_v)$, so $q(a) = \prod_v p_v^{n_v} \in \mathbf{Q}^{\times}_{>0}$ depends multiplicatively on $a$ and $a/q(a)$ is an integral unit at all finite places. Let $e(a)$ be the sign of $a_{\infty}/q(a)$, so $a \mapsto e(a)q(a)$ is a retraction of $\mathbf{Q}^{\times}\rightarrow\mathbf{A}^{\times}_{\mathbf{Q}}$. – user30379 Feb 2 '13 at 17:42

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