4

I'm trying to compute the Hilbert class field of the extension $\mathbb{Q}(\zeta_{5}, \sqrt{-43})$. I know that it has class number 7. I would like to show that it is contained in $\mathbb{Q}(\zeta_{215})$.

  • 2
    Is that true? I'm not fully conversant with this, but what would be wrong with the following argument. The prime $q=43$ is totally inert in $\Bbb{Q}(\zeta_5)$ and totally ramified ($e=42$) in $\Bbb{Q}(\zeta_{43})$. Therefore it has ramification index $e=2$ in $K=\Bbb{Q}(\zeta_5,\sqrt{-43})$. So there is a unique prime ideal $\mathfrak{q}$ of $K$ above $43$. Because $[\Bbb{Q}(\zeta_{215}):K]=\phi(215)/8=21=42/2$ it follows that $\mathfrak{q}$ is totally ramified in $\Bbb{Q}(\zeta_{215})/K$ as well as in all the intermediate fields. But it should be unramified in the Hilbert class field? – Jyrki Lahtonen Feb 07 '18 at 11:25
  • I think I understand your argument. How did you compute that $\mathfrak{q}$ is totally ramified in $\mathbb{Q}(\zeta_{215})$? –  Feb 08 '18 at 14:06
  • $q=43$ is totally ramified in $\Bbb{Q}(\zeta_{43})/\Bbb{Q}$ ($e=42$) and inert in $\Bbb{Q}(\zeta_{5})/\Bbb{Q}$ ($e=1,f=4$). So there is a unique prime ideal $\mathfrak{Q}$ above $43$ in the compositum $\Bbb{Q}(\zeta_{215})$, $(e=42, f=4)$. By multiplicativity of $e$ it follows that $$e(\mathfrak{Q}|\mathfrak{q})=e(\mathfrak{Q}|q)/e(\mathfrak{q}|q)=42/2=21=168/8=[\Bbb{Q}(\zeta_{215}):K].$$ – Jyrki Lahtonen Feb 08 '18 at 14:20

1 Answers1

0

The class field of $K = {\mathbb Q}(\zeta_5,\sqrt{-43})$ is not abelian, and in particular not contained in ${\mathbb Q}(\zeta_{43 \cdot 5})$ since its subfields have too much ramification.

The class number $7$ of $K$ is inherited from $k = {\mathbb Q}(\sqrt{-5 \cdot 43})$, which has class number $14$. The quadratic unramified extension of $k$ is $k(\sqrt{5}) \subset K$; the unramified extension of degree $7$ (with Galois group $D_7$ over the rationals by class field theory, where $D_7$ is the dihedral group of order $14$) is given by a root of $$ f(x) = x^7 - 3x^6 + 3x^5 - 5x^4 + 6x^3 - x^2 + 5x - 1 $$ by pari; it also says that the class field of $k$ has class number $1$.

The same root of $f$ generates the class field of $K$.