Welcome to MathOverflow

Question

1

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?

edited May 5 2010 at 14:13

asked May 5 2010 at 13:55

BY
1,287●1●5●11

1

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

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

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.