3
$\begingroup$

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 $p$. ?

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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).

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.