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:

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

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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