Hilbert class field of $\mathbb Q(\sqrt{-39})$

Question

Is the Hilbert class field of $K=\mathbb Q\left(\sqrt{-39}\right)$ is equal to $K\left(\sqrt{-39},\frac{\sqrt{1+\sqrt{13}}}{2}\right)$

I don't know, is it? What have you tried? What are your ideas? It at least has $\Gal(K(\frac{1}{2}(\sqrt{\sqrt{13}+1})/K)=\mathbb{Z}_4\mathbb{Z}$ and $\text{Cl}(K)=\mathbb{Z}/4\mathbb{Z}$. Have you gotten anywhere on the splitting yet? — Alex Youcis, Oct 03 '13 at 07:47

Dietrich Burde · Accepted Answer · 2013-10-03T09:48:29.130

5

According to Lemmermayer's computation it is $\mathbb{Q}(\sqrt{-39},\sqrt{-2+2\sqrt{13}})$, see "Class field towers".

edited Oct 03 '13 at 09:48

answered Oct 03 '13 at 08:57

Dietrich Burde

@ Dietrich Burde Thank you very much for the answer and for suggesting a book too.. – Math123 Oct 03 '13 at 13:03

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