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The ideal class group of $\mathbb{Q}(\sqrt{-30})$ is $C_2\times C_2$. Its four reduced binary quadratic forms have coefficients $(1,0,30)$, $(2,0,15)$, $(3,0,10)$ and $(5,0,6)$. The displayed multiquadratic field has degree four over the base. The seven quadratic-subfield discriminants are $8,-3,5,-24,40,-15,-120$, whose product in absolute value is $120^4$. The conductor-discriminant formula and discriminant tower formula imply that the relative discriminant over $\mathbb{Q}(\sqrt{-30})$ is the unit ideal. This unramified abelian degree-four extension is therefore its Hilbert class field.

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 inducesThe 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 $b\ge 0$. 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 $\pm 1$, 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 factorizationThe 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 $H_k$, givingThis 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 $1\le a\le 6$, $|b|\le a$, and $c=(b^2+120)/(4a)\ge a$ gives exactlyThus $h_k=4$. 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 $H=\mathbb{Q}(\sqrt{2},\sqrt{-3},\sqrt{5})$. 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 discriminantsFor 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 givesThe 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: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.

Put $t=\sqrt{-30}$ and $K=\mathbb{Q}(t)$. The ring of integers is $\mathbb{Z}[t]$, and the fundamental discriminant is $D_K=-120$. For a reduced positive definite binary quadratic form $(a,b,c)$ of discriminant $-120$, reduction givesThe coefficient $b$ is even; checking $1\le a\le 6$ and $|b|\le a$, with the usual nonnegative sign choice at boundary equalities, leaves preciselyThe correspondence between proper quadratic forms and ideal classes therefore gives $h_K=4$. All four classes equal their inverses, since inversion replaces $b$ by $-b$. HenceThe three nontrivial classes may be represented by the ramified prime ideals ${\mathfrak{p}}_2=(2,t)$, ${\mathfrak{p}}_3=(3,t)$ and ${\mathfrak{p}}_5=(5,t)$. They square to principal ideals and satisfy ${\mathfrak{p}}_2{\mathfrak{p}}_3{\mathfrak{p}}_5=(t)$. None is principal: an integral generator of norm $2$, $3$ or $5$ would require $a^2+30b^2$ to equal that number, which is impossible. The relation then makes the three classes distinct, consistently with the four-form calculation.

ConsiderThe three rational square classes are independent, so $[H:\mathbb{Q}]=8$. Their radical product is $\sqrt{-30}$, so $K\subset H$ and $[H:K]=4$; the extension $H/K$ is abelian. To verify that it is unramified, rather than assuming that radical adjunctions are harmless at $2,3,5$, use the conductor-discriminant formula. The seven quadratic subfields have fundamental discriminantsThe formula givesThe discriminant tower identity now givesThus the relative discriminant is the unit ideal and no finite prime ramifies. The base has no real places, so there is no infinite ramification to check. This unramified abelian degree-four extension has degree $h_K$, and part (i) provesThis is the Hilbert class field of Q of square root minus thirty.

Next translate prime representation into an ideal class. For $p\nmid 30$, we claimIf a representation exists, $\alpha =2x+yt$ lies in ${\mathfrak{p}}_2$ and has $N_{K/\mathbb{Q}}\alpha =2p$. The integral ideal $(\alpha ){\mathfrak{p}}_2^{-1}$ has norm $p$, so it is a prime ideal $\mathfrak{P}$. Its class is $[{\mathfrak{p}}_2{]}^{-1}=[{\mathfrak{p}}_2]$. Conversely, if $[\mathfrak{P}]=[{\mathfrak{p}}_2]$, the product ${\mathfrak{p}}_2\mathfrak{P}$ is principal, say $(\alpha )$. Since this ideal lies in ${\mathfrak{p}}_2$, its generator has the form $\alpha =2x+yt$ with integers $x,y$. Taking norms gives $4x^2+30y^2=2p$, proving the reverse implication. This is prime representation via an ideal class, and it proves sufficiency as well as necessity.

Identify the required class in $\mathrm{Gal}(H/K)$ through the Artin map. We can use the explicit representation $17=2+15$: $2+t$ has norm $34$, so the prime above $17$ obtained from $(2+t){\mathfrak{p}}_2^{-1}$ has class $[{\mathfrak{p}}_2]$. Its Frobenius in the multiquadratic extension acts on the three radicals with signsTheir product is one, so this automorphism fixes $K$. Because $H/\mathbb{Q}$ is abelian, for any unramified rational prime $p$ that splits in $K$, the Frobenius at either prime of $K$ of norm $p$ is its rational Frobenius restricted to $H/K$. The Artin isomorphism therefore makes the required class condition equivalent toConversely, these signs already imply splitting in $K$, since their product is $(\frac{-30}{p})=1$. No extra splitting condition is missing.

The supplementary law for $2$, quadratic reciprocity for $-3$ and $5$, and then the Chinese remainder theorem turn these sign conditions intoTheir four simultaneous residue classes modulo $120$ are $17,23,47,113$. Finally, $2$ is represented by $(x,y)=(\pm 1,0)$, whereas $3$ and $5$ cannot be represented: $y\ne 0$ gives a value at least $15$, and $y=0$ gives twice a square. We have proved the complete criterion