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.

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    $\begingroup$ Note that even defining a (well-defined) compositum $K_pK^{\operatorname{ab}}$ is already a bit subtle. Indeed, we can only do this if both fields live in a common overfield. In this case, we can choose a prime $\mathfrak p$ of $K^{\operatorname{ab}}$ lying above $p$ to get an embedding $K^{\operatorname{ab}} \subseteq \overline{K_p}$, and by a standard transitivity argument the compositum does not depend on the prime $\mathfrak p$ lying above $p$. That is, all the residue fields at maximal ideals of $K_p \otimes_K K^{\operatorname{ab}}$ are isomorphic. $\endgroup$ – R. van Dobben de Bruyn Nov 24 '15 at 5:24
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    $\begingroup$ @R.vanDobbendeBruyn: If $k'/k$ is an arbitrary extension of fields (no number theory involved at all) then you know that $f:G_{k'}^{\rm{ab}} \rightarrow G_k^{\rm{ab}}$ is intrinsic to $k'/k$ (conjugation ambiguity wiped out by working with abelianizations) and correspondingly $k^{\rm{ab}} \rightarrow {k'}^{\rm{ab}}$ over $k \rightarrow k'$ is unique. That defines $k' k^{\rm{ab}}$ as an abelian extension of $k'$, corresponding to $\ker f$, without needing to link up algebraic closures. $\endgroup$ – nfdc23 Nov 24 '15 at 13:49
  • $\begingroup$ Nice. But at least there's something to be said. (E.g. try defining the compositum of $k(x)$ and $k(y) \cong k(x)$; should that be $k(x,y)$ or $k(x)$?) $\endgroup$ – R. van Dobben de Bruyn Nov 24 '15 at 14:51
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    $\begingroup$ @R.vanDobbendeBruyn: I agree something needs to be said when forming a compositum of a general pair of abstract fields over a ground field, but when one of the extensions is Galois we have no ambiguity about the resulting composite field (over the other extension). Also, I made a typo in my comment above (but I suppose the intent was clear): the image of 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
  • $\begingroup$ @Dedalus : in the title, do you mean "completion" instead of localization? $\endgroup$ – Watson Dec 3 '16 at 10:19

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.

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  • $\begingroup$ thank you for you answer. Regarding the notations $N(C_{K^{ab}})$ I mean that we take the intersection of the norm from all finite subextensions of $K \subset K^{ab}.$ The same goes for $C_{K^{ab}},$it should be the union of all finite extensions of $K \subset K^{ab}.$ But if I understand you correctly, my question should follow from Grunwald-Wang? $\endgroup$ – user44591 Nov 24 '15 at 16:26
  • $\begingroup$ Yes, the reference I gave takes care of this (as just a piece of the more refined story of the Grunwald-Wang Theorem which handles several places at once, etc.). Your notations involving $C_{K^{\rm{ab}}}$ are quite non-standard; it is best not to use those notations. $\endgroup$ – nfdc23 Nov 24 '15 at 18:29
  • $\begingroup$ I will avoid it, thank you for pointing it out. Does the theory simplify considerably if we just have one prime instead of several? $\endgroup$ – user44591 Nov 24 '15 at 18:54
  • $\begingroup$ The reason I'm asking if the single prime case is easier is because Neukirch has an exercise where one should prove it for one place, and his hint is to use that for a finite extension $L$ of $K$ $N_{L/K} C_L \cap K^*_p = N_{L_p|K_p} L_p^*.$ $\endgroup$ – user44591 Nov 24 '15 at 22:42
  • $\begingroup$ Correction to the preceeding comment: The extension $L$ should be finite abelian. $\endgroup$ – user44591 Nov 24 '15 at 22:50

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