= Solution
For a <number field> $k$, its ordinary <Hilbert class field> $H_k$ is the maximal finite abelian extension unramified at every finite prime and with every real place splitting completely. The <Artin reciprocity map> induces
$$
\operatorname{Cl}(k)\cong\operatorname{Gal}(H_k/k),\qquad[H_k:k]=h_k.
$$
The image of an unramified <prime ideal> is its <Frobenius automorphism>. Thus a prime splits completely in $H_k$ precisely when its ideal class is trivial. This field packages ideal-class obstructions into splitting behavior. Ideals become principal in $H_k$ by the principal ideal theorem. Specifying the real-place convention matters for real fields: omitting it leads instead to the narrow class-field problem. Both illustrative fields below are imaginary quadratic, so there are no real places to discuss.
First consider $k=\mathbb Q(\sqrt{-23})$, whose <quadratic discriminant> is $-23$. The correspondence between <ideal classes> and reduced primitive positive definite <binary quadratic forms> lets us compute its class number without guessing a class field. A reduced form $(a,b,c)$ obeys $|b|\le a\le c$, $b^2-4ac=-23$, and $a\le\sqrt{23/3}<3$; a boundary case uses $b\ge0$. For $a=1$ the only class is $(1,1,6)$; for $a=2$ the two possibilities are $(2,1,3)$ and $(2,-1,3)$. Hence $h_k=3$ and its <ideal class group> is cyclic of order three.
Take $g(X)=X^3-X-1$. Its only possible rational roots are $\pm1$, neither a root, so it is irreducible. Its <polynomial discriminant> is $4-27=-23$, which is nonsquare. Thus its <splitting field> $H$ has <Galois group> $S_3$ and the quadratic fixed field of $A_3$ is $k$, generated by the square root of its discriminant. Therefore $H/k$ is cyclic of degree three and $H=k(\theta)$ for a root $\theta$ of $g$.
At a prime other than $23$, the separable residue polynomial splits over a finite <residue-field extension>; <Hensel's lemma> puts all its roots in an unramified local extension. At $23$ there is the exact factorization
$$
g(X)\equiv(X-10)^2(X-3)\pmod{23}.
$$
The simple factor lifts by <Hensel's lemma>. The remaining quadratic factor has discriminant valuation one, because the cubic discriminant is $-23$ and the resultant with the simple factor is a unit. Its local splitting field is ramified quadratic, with <inertia group> of order two. Its intersection with $A_3$ is trivial, so the relative extension $H/k$ is unramified at $23$ as well. There are no real places. An unramified abelian extension of degree $h_k=3$ must be all of $H_k$, giving
$$
\boxed{H_{\mathbb Q(\sqrt{-23})}=\mathbb Q(\sqrt{-23},\theta),\qquad\theta^3-\theta-1=0.}
$$
This is the <Hilbert class field of Q of square root minus twenty-three>.
The other alternative, $k=\mathbb Q(\sqrt{-30})$, has <quadratic discriminant> $-120$. Reduced forms have $a\le\sqrt{40}<7$ and even $b$. Testing $1\le a\le6$, $|b|\le a$, and $c=(b^2+120)/(4a)\ge a$ gives exactly
$$
(1,0,30),\quad(2,0,15),\quad(3,0,10),\quad(5,0,6).
$$
Thus $h_k=4$. Each class is its own inverse, because replacing $b$ by $-b$ leaves these representatives unchanged; consequently the <ideal class group> is $C_2\times C_2$.
Set $H=\mathbb Q(\sqrt2,\sqrt{-3},\sqrt5)$. The square classes $2,-3,5$ are independent, so $[H:\mathbb Q]=8$. Their product gives $\sqrt{-30}$, and $H/k$ is abelian of degree four. Its seven quadratic subfields have discriminants
$$
8,\quad-3,\quad5,\quad-24,\quad40,\quad-15,\quad-120.
$$
For a multiquadratic abelian extension, the <conductor-discriminant formula> takes the product of these quadratic character conductors, namely the absolute quadratic discriminants. Hence $|d_H|=120^4$. The <relative discriminant> tower formula gives
$$
120^4=|d_k|^4N_{k/\mathbb Q}(\mathfrak d_{H/k})=120^4N_{k/\mathbb Q}(\mathfrak d_{H/k}).
$$
The integral relative <discriminant ideal> therefore has norm one and is the unit ideal. Thus $H/k$ is everywhere unramified, and its degree equals the class number. This proves the other explicit answer:
$$
\boxed{H_{\mathbb Q(\sqrt{-30})}=\mathbb Q(\sqrt2,\sqrt{-3},\sqrt5).}
$$
It agrees with the <Hilbert class field of Q of square root minus thirty>. Both alternatives have been computed independently from their class numbers and ramification.
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