How can I prove this: Let $K$ be a real abelian number field, $K\_1$ be the Hilbert Class Field of $K$, and $J=K\_1\cap K(\zeta\_b)$. If a prime $p$ divided $[J:K]$ but did not divide $[K:\mathbb{Q}]$, then there would be an unramified extension of $\mathbb{Q}$ of degree $p$. ?



About real abelian number fields
Let $J^{(p)}\subseteq J$ be the subfield fixed by the $p$-Sylow subgroup of $\operatorname{Gal}(J/K)$ which is also the $p$-Sylow subgroup of the (abelian!) group $\operatorname{Gal}(J/\mathbb{Q})$ since, by assumption, $p\nmid [K:\mathbb{Q}]$. Then, by the structure theorem of abelian groups, we find a direct-product decomposition
$$\operatorname{Gal}(J/\mathbb{Q})=\operatorname{Gal}(J/J^{(p)})\times \operatorname{Gal}(J^{(p)}/\mathbb{Q}).$$
This corresponds to the existence of a finite abelian $p$-extension $\mathbb{Q}^{(p)}$ which is the fixed field of $\operatorname{Gal}(J^{(p)}/\mathbb{Q})$.


Let now $\ell$ be any rational prime: the extension $J/\mathbb{Q}$ being abelian, we can speak of its inertia subgroup $I\_\ell(J/\mathbb{Q})\subseteq\operatorname{Gal}(J/\mathbb{Q})$, of order $e\_\ell(J/\mathbb{Q})$. Since $J/K$ is contained in the Hilbert class field, $J/K$ is everywhere unramified: the same must hold for the subextension $J/J^{(p)}$. It follows that $I\_\ell(\operatorname{Gal}(J/\mathbb{Q}))$ intersects trivially the subgroup $\operatorname{Gal}(J/J^{(p)})$ and therefore $p\nmid e\_\ell(J/\mathbb{Q})$. In particular, by multiplicativity of ramification indexes in towers, the ramification index $e\_\ell(\mathbb{Q}^{(p)}/\mathbb{Q})$ cannot be divisible by $p$ since $e\_\ell(J/\mathbb{Q})=e\_\ell(J/\mathbb{Q}^{(p)})\cdot e\_\ell(\mathbb{Q}^{(p)}/\mathbb{Q})$, and is therefore $1$ because it is the order of a subgroup of the $p$-group $\operatorname{Gal}(\mathbb{Q}^{(p)}/\mathbb{Q})$.


This shows that $\mathbb{Q}^{(p)}/\mathbb{Q}$ is an abelian $p$-extension everywhere unramified and so $p$ divides the class number of $\mathbb{Z}$ (which is absurd, but this you know).


