Let $K$ be a local field. Let us define $$ K^{ab} = \bigcup_{ \substack{ L \subseteq K^{sep} \\ L/K \ \text{ finite abelian } } } L . $$
1) I was just wondering are there any infinite abelian extensions $M/K$ contained inside $K^{ab}$? (other than $K^{ab}$ itself?)
2) For such $M$ does the corresponding Galois group $Gal(K^{ab}/M)$ finite index and open?
Thank you.
1) The typical example is the maximal unramified extension. For $K = \Bbb Q_p$, you have $K^{\rm unr} = \Bbb Q_p(\zeta_n \mid (n,p)=1)$. On the other hand, $\Bbb Q_p^{\rm ab} = \Bbb Q_p(\zeta_n \mid n \geq 1)$ has ramification because of the $p^k$-roots of unity. Therefore, $K^{\rm unr}$ is an infinite abelian extension of $K$, strictly contained in $K^{\rm ab}$.
2) If $K \subset M$ is an infinite abelian extension, then $$\mathrm{Gal}(K^{\mathrm{ab}} / M) \leq \mathrm{Gal}(K^{\mathrm{ab}} / K)$$ is normal with quotient $\mathrm{Gal}(M/K)$, which is infinite, so the index can't be finite. While it is a closed subgroup, it is not an open subgroup, for the open subgroups in a Galois group $\mathrm{Gal}(L/K)$ are exactly those subgroups of the form $\mathrm{Gal}(L/E)$ with $E$ being an intermediate field in $L/K$ of finite degree over $K$.