Non-cyclotomic abelian extensions of a number field $K$ contained in $K_{\mathfrak p}$?

Let $K$ be a number field and $\mathfrak p$ a maximal ideal of its ring of integers $\mathcal O_K$. We let $K_{\mathfrak p}$ denote its completion at $\mathfrak p$ and $K^{ab}$ the maximal abelian extension of $K$. The maximal abelian extension of $K_{\mathfrak p}$ is given by $K_{\mathfrak p} \cdot K^{ab}$ (e.g. Neukirch p. 412). Now define the set $A = \bigcup_{\mathfrak p} (K_{\mathfrak p} \cap K^{ab}$), where the union is running over all maximal ideals of $\mathcal O _K$. Denote by $K(A)$ the field generated by $A$ over $K$, by construction it is contained in $K^{ab}$.

For example, if $K = \mathbb Q$, we see immediately $\mathbb Q(A) = \mathbb Q ^{ab}$, because, for every prime number p, $\mathbb Q _p$ contains the $(p-1)$-th roots of unity and (by Dirichlet's theorem about primes in arithmetic progressions) we know that these generate $\mathbb Q ^{ab}$. I guess, for every $K$ we have $K \cdot \mathbb Q ^{ab} \subset K(A)$. Is this last inclusion an equality for general $K$ or can $K(A)$ be even bigger?

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That "union" is a bit dodgy until you specify how to embed $K^{\rm{ab}}$ into an algebraic closure of $K_P$ for each $P$. For any algebraic extension $K'/K$ and any $x \in K'$ the minimal polynomial of $x$ over $K$ splits completely in $K_P$ for any prime $P$ totally split in $K(x)$ (of which there are zillions). So $K(A) = K^{\rm{ab}}$ always (bigger than $K \cdot \mathbf{Q}^{\rm{ab}}$ whenever $K \ne \mathbf{Q}$), but the intervention of abelian extensions is a red herring, could be any algebraic $K'/K$. – BCnrd May 5 2010 at 14:23
Thank you very much for your answer, this is what I was looking for! In fact I have something different in mind but I tried to formulate the most pessimistic formulation of my question, which indeed became somewhat dodgy. Sorry for that! – BY May 5 2010 at 16:35
The appeal to Dirichlet's theorem is more than necessary at the step where you refer to it: what you need is that for each m there are primes p = 1 mod m and that can be shown in an elementary way by playing around with cyclotomic polynomials. See Corollary 2.11 p. 13 of Washington's book on cycl. fields. – KConrad May 5 2010 at 17:52