Let $K$ be a number field and consider the maximal abelian extension $K^{ab}$ of $K.$ For a finite prime $p,$ letting $K\_p$ be the completion of $K$ at $p,$ we have an extension $K\_p \subset K\_p K^{ab}$ where the latter denotes the compositum of the fields. 


Is $K\_p K^{ab} = K\_p^{ab}?$ That is, can I get the maximal abelian of extension of $K\_p$ from taking the compositum of $K^{ab}$ and $K\_p?$ 


This is clearly true for $\mathbb{Q}.$ I tried to see if I could use compatibility of local and global class field theory to show the statement. The thought was that if $N(C\_K^{ab}) \subset C\_K$ is the norm group in the idèle class group, then if we could show that the intersection $N(C\_{K^{ab}}) \cap K^\*\_{p}$ equals the image of the norm map from the non-zero elements $K\_p^{ab}$ to $K\_p^\ast,$ we would be done. But I couldn't prove the claimed equality.



Is the localization of the maximal abelian extension still a maximal abelian extension?
Your notations $N(C\_K^{\rm{ab}})$ and $N(C\_{K^{\rm{ab}}})$ are meaningless as written (and the first was probably a typo, meant to be the latter), and your comments about norms from $(K\_p)^{\rm{ab}}$ are also meaningless as written, since there is no sense of "norm" from an infinite-degree extension or useful notion of idele class group for $K^{\rm{ab}}$. Perhaps you meant to speak of some link between global and local norm groups from unspecified *finite* abelian extensions, but since there is confusion in the formulation it is unclear what calculations you have been trying to carry out.


By class field theory, it suffices that the continuous injection $K\_p^{\times} \rightarrow C\_K := \mathbf{A}\_K^{\times}/K^{\times}$ (whose induced effect on profinite completions is the canonical map 


$$G\_{K\_p}^{\rm{ab}} \rightarrow G\_K^{\rm{ab}}$$


whose kernel you want to be trivial) realizes every open subgroup of finite index in $K\_p^{\times}$ as arising from one in $C\_K$. The answer is affirmative. 


Since every finite abelian extension of a field is a compositum of cyclic extensions of finite degree, it suffices that every cyclic finite extension $F$ of $K\_p$ has the form $L\_{\mathfrak{p}}$ for an abelian finite extension $L$ of $K$ and a place $\mathfrak{p}$ of $L$ over $p$. One can even find a *cyclic* $L/K$ that does the job, with $[L:K]$ equal to $[F:K\_p]$ or $2[F:K\_p]$. Equality of global and local degrees may be impossible to achieve, seen already for $K = \mathbf{Q}$ and the unramified degree-8 extension $F$ of $\mathbf{Q}\_2$ (by quadratic reciprocity at 2). These matters are part of the Grunwald-Wang Theorem; see section 2 of Chapter X of the Artin-Tate book on class field theory.


