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 $$p$$. ?

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