Hilbert class field of $\mathbb Q(\sqrt{-5})$ and $\mathbb Q(\sqrt{-10})$

Question

This might be a very silly doubt, as I am making an attempt to understand Hilbert Class field, I would like to know this. In Cohen,"Advanced topics in computational number theory" Hilbert class field for $K=\mathbb Q(\sqrt{-5})$ and $K=\mathbb Q(\sqrt{-10})$ is given as $X^2-X-1$, which I thought Hilbert class field for $K= \mathbb Q(\sqrt{-10})$ and $K=\mathbb Q(\sqrt{-5})$ are same. The Hilbert class field for $K=\mathbb Q(\sqrt{-5})=K(\sqrt{-1})=K(\sqrt{5})$. For $d=-10$, primes ramifying in $\mathbb Q(\sqrt{-10})$ are 2 and 5. also, $2$ is ramified in $\mathbb Q(\sqrt{-1})$ too! So I was wondering is the Hilbert class field of $\mathbb Q(\sqrt{-10})$ is same as $\mathbb Q(\sqrt{-5})$ or the Hilbert class field of $\mathbb Q(\sqrt{-10})=\mathbb Q(\sqrt{-2}, \sqrt{5})$. Any hint is welcomed

Answer 1

Class groups of $K_1=Q(\sqrt{-5})$ and $K_2=Q(\sqrt{-10})$ has order $2$. So, $[H(K_i):K_i]=2$, where $H(K)$ is the Hilbert class field of $K$.

Hilbert class field of $K_1=Q(\sqrt{-5})$ is $K_1(\sqrt{-1})=K_1(\sqrt{5})$.

Also, note that for $K_2=Q(\sqrt{-10})$, $Q(\sqrt{-2},\sqrt{5})$ is every where unramified. So again for $K_2=Q(\sqrt{-10})$, the Hilbert class field is $K_2(\sqrt{5})$.