How to determine a Hilbert class field?

Question

I tried to solve the exercise VIII.XX in Number Fields by Marcus. It asks to find the Hilbert class field of $Q(\sqrt m)$ for $m=-6,-10,-21,-30$. And the emphasis of this question is on the first two.

Attempt to solve the first case.
I knew that the class number of $K=Q(\sqrt{-6})$ is $2$, so its Hilbert class field extension $H/K$ is of degree $2$. Since all groups of order $4$ are abelian, we know that $H$ is also abelian over $Q$. Further, by the computation of the conductor, we know that $H\subset Q(\zeta_{24})$, where $\zeta_{24}$ is a $24$-th primitive root of unity. After we compute explicitly the galois group of $Q(\zeta_{24})/Q$ is isomorphic with h$C_2\times C_2\times C_2$. Noting that $\sqrt{-6}=(\zeta_8+\zeta_8^7)(1+2\zeta_3)$, we deduce that the galois group of $Q(\zeta_{24})/K$ is isomorphic with $G=C_2\times C_2$. So there are $3$ subgroups of $G$ of index $2$. And I cannot decide which one of them is the subgroup fixing $H$. Of course, if we write out all fixed fields of the three subgroups, then, by examining the decomposition of ideals, we can determine, or conclude, that $H=Q(\sqrt2,\sqrt{-3})$. But I think there must be a way of deciding this without this trial and error.
In fact, in the second case, $K=Q(\sqrt{-10})$, there are $7$ possibilities of $H$! After some calculations moreover, I found that $H=Q(\sqrt{-10},\sqrt5)$. But my method is again trial and error. And I want to find a way to determine $H$ without such brute force.

Question
Given a specific field $K/Q$, what is the standard, or the most efficient, way in determining its Hilber class field $H$?

I looked into Algebraic Number Fields by Janusz, but the example I found there is the case $Q(\sqrt{-5})$, in which the galois group of $Q(\zeta_{20})/K$ is cyclic of order $4$, thus having a unique subgroup of index $2$, and so cannot help me here.

Any help is greatly appreciated. Thanks in advance.

Answer 1

I don't know if this is helpful, but for $E = \mathbb{Q}(\sqrt{-6})$ case....

You know that $H/E$ is quadratic, and thus $H = E(\sqrt{D})$ for some $D \in E$. $H/E$ is an unramified extension, however all number fields are ramified over $\mathbb{Q}$, so clearly $D$ must factor into ramified primes of $E$.

I looked over $2$ first. I don't remember why, but I convinced myself at one time that for a quadratic extension to be unramified, you need to find a $D$ whose prime factorization is square. (hopefully this is fresh in your mind so you'll be able to work out why!) So I observe

$$(2, \sqrt{-6})^2 = (2)$$

However, $2$ is clearly (?) not a square. Thus $E(\sqrt{2})$ is an unramified extension.

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I also vaguely recall a general theorem that the Hilbert class field of $\mathbb{Q}(\sqrt{m})$ consists of adjoining square roots of the prime factors of $m$, with appropriate signs.