Let $K$ be a number field and let's denote with $O_K$ its ring of integers. Moreover we indicate with the letter $p$ the generic non-archimedean place of $K$ and with $\sigma$ the generic archimedean place.
A few days ago a speaker in a talk said that there is the following extension of topological groups:
$$(\ast)\quad 0\to\prod_{p}\mathcal O_p\to\mathbf A_K/K\to \left(\prod_\sigma K_\sigma\right)\big/ O_K\to 0$$
where
- $\mathbf A_K$ is the ring of adeles and $K\hookrightarrow \mathbf A_K$ diagonally.
- $\mathcal O_p$ is the completion of $O_K$ w.r.t. the place $p$.
- $K_\sigma$ is the completion of $K$ w.r.t. $\sigma$.
Where does the sequence $(\ast)$ come from? I don't understand theconstruction of the surjective map. The existence of such a sequence should be a consequence of the strong approximation theorem.
Thank you in advance.
$\mathbf A_K = \{a \in \prod_p K_p \prod_\sigma K_\sigma, a_p \in O_p$ for almost every $p\} $
There is $$ 0\to\prod_{p}\mathcal O_p\to\mathbf A_K/K\to \left(\prod_\sigma K_\sigma\right)/K\to 0$$ the arrows being the natural maps (sending $\prod_p a_p$ to $\prod_p a_p \prod_\sigma 0$ and $\prod_p a_p \prod_\sigma a_\sigma$ to $ \prod_\sigma a_\sigma$)
Let $o = \prod_p O_p \prod_\sigma K_\sigma$ your version is doing $$A_K/ K \cong o/(o \cap K) = o/O_K$$ (for every $a \in A_K$ there is $b \in K$ such that $a+b \in o$)
so that
$$ \begin{eqnarray}0\to\prod_{p}\mathcal O_p\to & \mathbf A_K/K & \\ & \cong \mathbf o/O_K& \to \left(\prod_\sigma K_\sigma\right)/O_K\to 0\end{eqnarray}$$