Let $K$ be a local field. Let us define $$ K^{ab} = \bigcup_{ \substack{ L \subseteq K^{sep} \\ L/K \ \text{ finite abelian } } } L. $$
Suppose I have a finite extension $F/K$ where $F$ is contained inside $K^{ab}$. Does it then follow that $F$ is an abelian extension of $K$? (I think this should be true for an easy reason, but I'm not seeing it at the moment...)
Thank you!
Yes. Let $K$ be any field and let $M$ be any abelian extension (algebraic, but not necessarily finite). Let $F/K$ be any subextension of $M/K$. Then $F/K$ is abelian (as well as $M/F$).
Indeed, $H = \mathrm{Gal}(M/F) \leq G =\mathrm{Gal}(M/K)$ is a (closed) normal subgroup, since $G$ is abelian. In particular, $F/K$ is a Galois extension and moreover, the quotient $G/H$ is isomorphic (as topological group) to $\mathrm{Gal}(F/K)$ (see theorem 2.3.6), and this is an abelian group. Thus $F/K$ is abelian.