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Keith Conrad's History of Class Field Theory notes has a comment that given any number field $K\neq \mathbb{Q}$, then $K$ has an abelian extension which is not contained in $K(\zeta_\infty)$, so the Kronecker-Weber theorem for $K$ is false. I haven't been able to find a proof of this. I assume that we have to look at the idelic formulation of global class field theory to find it. We have the following diagram of extensions (I can't figure out how to make vertical lines):

$K$ ------ $K(\zeta_\infty)$ ------ $K^{ab}$

$\mathbb{Q}$ ------ $\mathbb{Q}^{ab}$

and functoriality of idelic CFT, then gives the diagram:

$C_K$ ------> $\textrm{Gal}(K^{ab}/K)$

Norm ..... Restriction

$C_\mathbb{Q}$ ------> $\textrm{Gal}(\mathbb{Q}^{ab}/\mathbb{Q})$

Since any automorphism of a cyclotomic extension is uniquely determined by where it maps the different roots of unity, the restriction map would have to be injective if $K(\zeta_\infty)=K^{ab}$. If not, then the map clearly can't be injective, since if there's some $L/K$ that is abelian and not contained in a cyclotomic extension, then we can pick a non-trivial automorphism of $L$ that fixes the maximal cyclotomic subextension.

My question essentially has two parts:

  1. Is it enough to show that the norm map is not injective? What's confusing me is that the horizontal maps are not isomorphisms.

  2. How would one find the kernel of the norm map? I seem to get lost trying to do this computation.

EDIT: I just realized I've been stupid since I don't need to know the whole kernel just that it's noninjective... Pick any splitting prime $(p)=\mathfrak{p}_1\cdots \mathfrak{p}_n$, so

$$K_{\mathfrak{p}_i}=\mathbb{Q}_p.$$

Now just pick an element $(a_i)\in C_K$ where the element sitting in $K_{\mathfrak{p}_1}$ is $2$ and the element sitting in $\mathfrak{p}_2$ is $1/2$ and the rest are $1$. This should give norm $1$ if I understand the definition correctly?

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@Bora: I'm pretty sure I'm right. Take two elements $f,g$ of the Galois group of $K^{ab}/K$. Assuming $K^{ab}=K(\zeta_\infty)$, then $f\neq g$ if and only if $f(\zeta_n)\neq g(\zeta_n)$ for some $n$-th root of unity. Clearly the restrictions to $\mathbb{Q}^{ab}$ would be distinct. –  asdf Jan 15 '12 at 23:55
    
@Wanax: I'll have to think about that, since I can't totally follow your line of thought. I'm still interested in solving this through CFT, since question (1) is something I would like to know in general too. –  asdf Jan 16 '12 at 0:26
    
sorry, I was thinking of surjective... –  user5831 Jan 16 '12 at 0:38
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Kronecker-Weber is "false" even locally : the maximal abelian extension of a local field other than the $\mathbf{Q}_p$ themselves is not generated by roots of $1$. –  Chandan Singh Dalawat Jan 16 '12 at 2:25
    
@Wanax and @Dalawat: Thanks for the pointers!!! –  SGP Jan 16 '12 at 2:55

1 Answer 1

Edited following comments of Wanax:

This is just an amplification of Wanax's comments: All you need to do is to pick a prime $p$ of $Q$ which splits into (more than two) distinct primes $\mathfrak p_1$ and $\mathfrak p_2$ in $K$ and which is unramified in $K$. There are infinite number of such primes if $K$ is distinct from $Q$. Pick an element $\alpha$ of $K$ which is divisible by $\mathfrak p_1$ but not $\mathfrak p_2$. Then the quadratic extension $L= K(\sqrt{\alpha})$ is an abelian extension of $K$ and this $L$ gives the extension you want. Because if $L$ were obtained by adjoining a root of unity to $K$, both primes $\mathfrak p_1$ and $\mathfrak p_2$ would have to be ramified in $L$ but our choice ensures that only $\mathfrak p_1$ is ramified in $L$.

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@SGP: This is pretty much exactly where I'm stuck. I guess the issue is that the norm map induces a map on the profinite completion, but I don't know how noninjectivity of the original map translates to the completions. A reference on the relevant material on profinite groups (or an explanation) would be appreciated. –  asdf Jan 16 '12 at 2:21
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@Wanax: Could you please add your comments etc as an answer? All my attempt at an answer was to amplify your first comment. If you would please do so, I will delete my answer. –  SGP Jan 16 '12 at 17:05

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