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.

imageof the map $k^{\rm{ab}}$ in ${k'}^{\rm{ab}}$ is unique, not the map (though the map on Galois groups is canonical). $\endgroup$ – nfdc23 Nov 24 '15 at 18:26