I saw the following commutative diagram in the book Number Theory 2: Introduction to Class Field Theory by Kato, Saito, Kurogawa.
And the authors says that the rightmost vertical map is the restriction of automorphisms of $(K_v)^{ab}$ to its subfield $K^{ab}$.
My question is
- How to view $K^{ab}$ as subfield of $(K_v)^{ab}$?
- Maybe it could be done by showing that $(K_v)^{ab}=(K^{ab})_v$ ? (by $(K^{ab})_v$ I mean the completion of $K^{ab}$ with respect to the unique extension of $v$). But I don't even don't know if it's true or not.
Thanks for any help in advance.
The construction is as follows: first, choose an embedding $i$ of the algebraic closure $\overline{K}$ of $K$ into the algebraic closure $\overline{K_v}$ of $K_v$.
Then, for any finite abelian extension $L/K$ (with $L \subset \overline{K}$), the extension $LK_v/K_v$ is finite abelian, so $i(L) \subset (K_v)^{ab}$. Now, note that $K^{ab}$ is the (filtered) reunion of the $i(L)$ where $K \subset L \subset \overline{K}$ is a finite abelian extension of $K$ – so that $i(K^{ab}) \subset (K_v)^{ab}$.