Put
$$
H=\mathbb Q(\sqrt2,\sqrt{13}).
$$
It contains $K=\mathbb Q(\sqrt{26})$ and has degree two over $K$. The three quadratic subfields have discriminants $8$, $13$, and $104$, so the biquadratic discriminant formula gives
$$
d_H=8\cdot13\cdot104=104^2=d_K^2.
$$
The relative discriminant formula
$$
d_H=d_K^{[H:K]}N_{K/\mathbb Q}(\mathfrak d_{H/K})
$$
therefore gives $\mathfrak d_{H/K}=\mathcal O_K$: no finite prime ramifies. The extension is totally real, so no infinite prime ramifies either. Since $h_K=2$, the <Hilbert class field> has degree two over $K$. Consequently
$$
\boxed{H_K=\mathbb Q(\sqrt2,\sqrt{13}).}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 123 / 5 / d / Solution

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The Hilbert class field is the maximal unramified abelian extension of a number field. Its Galois group is canonically the ideal class group.


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