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 $p$. ?

  • $\begingroup$ Write $J = KH$ for some $H \subseteq {\mathbb Q}(\zeta_b)$. If $p$ divides the degree of $H$, then there is a prime ideal with ramification index $p$ in $H$. If $H/K$ is unramified, this ramification must be killed by $K/{\mathbb Q}$, which is only possible if $p$ divides its degree. Look up Abhyankar's lemma and its proof for details. $\endgroup$ – Franz Lemmermeyer Oct 19 '18 at 15:32

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).

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  • $\begingroup$ ($Gal(J/\mathbb{Q})$ is abelian because $J$ is the compositium of two abelians extensions) $\endgroup$ – reuns Oct 16 '18 at 20:32

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