How does one know that a number field $K$ has a maximal abelian extension (unique up to isomorphism) $K^{\text{ab}}$?
I've read proofs involving Zorn's lemma that it has an algebraic closure (And that algebraic closures are unique up to isomorphism.) $\bar{K}$ All these proofs involved ideals of the polynomial ring in variables $x_f$, $f$ an irreducible monic polynomial in $K[x]$, but I don't see any obvious way of "restricting" this proof to abelian extensions.
I tried proving that such an extension exists using Zorn's lemma: Let $\Sigma$ be the set of all abelian subgroups of $\text{Gal}(\bar{K}/K)$ partially ordered by inclusion. Any chain of subgroups $(G_\alpha)$ has an upper bound, namely, $\bigcup_\alpha G_\alpha$ (which is a [sub]group as each $G_\alpha$ is contained in another), so by Zorn's lemma $\Sigma$ has a maximal element. But I don't have that this element is unique. (and I don't think I proved that $\bigcup_\alpha G_\alpha$ is abelian, either).
Additionally, how does $\text{Gal}(K^\text{ab}/K)$ relate to $\text{Gal}(\bar{K}/K)$ ? My incomplete attempt at a Zorn's lemme proof doesn't tell me what the maximal abelian galois group should be, and I don't know many ways of finding abelian subgroups.