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Let $K=\mathbb{Q}(\sqrt{-31})$. Its ring of integers of a quadratic field is $\mathbb{Z}[(1+\sqrt{-31})/2]$, and its fundamental field discriminant is $-31$.

First compute the class number. The ideal-form correspondence for imaginary quadratic fields identifies the ideal classes with proper equivalence classes of primitive positive definite binary quadratic forms of discriminant $-31$. Each has a unique reduced positive definite binary quadratic form representative under the convention $|b|\le a\le c$, with $b\ge 0$ when $|b|=a$ or $a=c$. From $31=4ac-b^2\ge 3a^2$ we get $1\le a\le \sqrt{31/3}<4$. Checking these three possible leading coefficients gives exactlyFor $a=3$, neither odd residue choice $b=\pm 1,\pm 3$ makes $(b^2+31)/12$ integral. The listed forms are primitive and satisfy the reduction convention, soThis computes the ideal class group of Q of square root minus thirty-one.

Now take a root $\theta$ of $F(T)=T^3+T-1$, and let $H$ be its splitting field over $\mathbb{Q}$. There is no rational root: the only candidates are $\pm 1$, and neither vanishes. Thus the cubic is irreducible. Its polynomial discriminant isSince this is nonsquare, the Galois group of an irreducible cubic is $S_3$. Its quadratic subfield is $K$: the product of pairwise root differences squares to $-31$ and changes sign under odd permutations. Thus $H/K$ is cyclic of degree three, and $H=K(\theta )$. Indeed the cubic field and $K$ have coprime degrees, so their compositum already has the full splitting-field degree six. The squarefree polynomial discriminant also shows, by the discriminant-index formula for an integral lattice, that $\mathbb{Z}[\theta ]$ is the full integer ring of the cubic field.

It remains to prove that $H/K$ is unramified everywhere. At any rational prime $\ell \ne 31$, the reduction of $F$ is separable. Over a finite residue extension in which it splits, all residue roots are simple and lift by Hensel lemma to the corresponding unramified extension of ${\mathbb{Q}}_{\ell }$. Therefore the local splitting field is unramified over ${\mathbb{Q}}_{\ell }$, and its extension over a completion of $K$ is unramified as well.

At the only possible ramified prime, direct reduction givesThe simple root lifts to $r\in {\mathbb{Q}}_{31}$. Write $F(T)=(T-r)Q(T)$, with $Q$ monic quadratic over ${\mathbb{Q}}_{31}$. The discriminant product formula givesHere $Q(r)=F^{\prime }(r)$ is a unit, so the quadratic discriminant has square class $-31$. Hence the entire local splitting field isIt is quadratic over ${\mathbb{Q}}_{31}$, but its extension over the completion of $K$ is trivial. Thus the prime over $31$ splits completely in $H/K$; in particular it is not ramified there. This is the local splitting-field argument behind an unramified cyclic cubic extension from a prime-discriminant cubic. There are no real places of the imaginary quadratic field $K$, so no infinite-place ramification remains to check.

The Hilbert class field theorem identifies the maximal everywhere unramified abelian extension with an extension of degree $h_K$. Our cyclic unramified extension already has that degree, so it must be the whole Hilbert class field. ThereforeEquivalently, it is the splitting field of $T^3+T-1$, of degree six over $\mathbb{Q}$ and degree three over $K$. Replacing $\theta$ by ${\theta }^{-1}$ gives the alternative defining cubic $T^3-T^2-1$. This is the Hilbert class field of Q of square root minus thirty-one.

The only possible rational roots of the monic cubic are $1,-1$, and neither is a root. Thus $f$ is irreducible. The discriminant of a depressed cubic givesIt is not a rational square. By the permitted cubic discriminant criterion, the splitting field has Galois group $S_3$, degree six, and contains its quadratic subfield generated by $\sqrt{\Delta }$. Consequently

For a rational prime $r\ne 59$, the reduction of $f$ is separable because its discriminant is a unit at $r$. Choose a finite extension of ${\mathbb{F}}_r$ in which that reduction splits. Part 1(a) lifts it to an unramified extension of ${\mathbb{Q}}_r$, and the Hensel lemma lifts all the simple roots into that extension. The local splitting field is therefore unramified over ${\mathbb{Q}}_r$. In particular all primes of $L/K$ above such $r$ are unramified.

At $59$, the supplied simple residue root $-28$ lifts to a root in ${\mathbb{Q}}_{59}$. The remaining factor is quadratic, so the local splitting field has degree at most two over ${\mathbb{Q}}_{59}$. It contains $\sqrt{-59}$, which is not a square in ${\mathbb{Q}}_{59}$ because its valuation is odd. It is therefore exactly ${\mathbb{Q}}_{59}(\sqrt{-59})$. This is also the completion of $K$ at its unique prime above $59$: the polynomial $T^2+59$ is Eisenstein there. Thusfor a prime $\mathfrak{P}$ above $\mathfrak{p}$. All the other primes above it give the same local conclusion by Galois conjugacy. The relative extension is therefore unramified at $59$ as well; indeed that prime splits completely in $L/K$.

The field $K$ has only complex places, so there is no infinite-place ramification. Hence $\boxed{L/K\text{ is unramified everywhere}}$.

We use the Hilbert class field theorem: for a number field $E$, its maximal everywhere unramified abelian extension $H_E$ is finite, contains every such extension, and satisfies $\mathrm{Gal}(H_E/E)\cong {\mathrm{Cl}}_E$, so $[H_E:E]=h_E$. The cyclic degree-three extension just constructed is contained in $H_K$. ThereforeThis is the unramified cubic extension of Q of square root minus fifty-nine; only divisibility of the class number, rather than its exact value, is needed here.

Use the Hilbert class field theorem stated in part (a), and let $H_L$ be the Hilbert class field of $L$. Put $F=M\cap H_L$ inside a common algebraic closure.

Any intermediate extension of $M/L$ is totally ramified at the given prime. To verify this even when $M/L$ is not Galois, there is a unique prime of $F$ above $\mathfrak{p}$, its residue field embeds into the residue field at the unique prime of $M$ and therefore equals $k_{\mathfrak{p}}$, and the fundamental identity for prime decomposition gives ramification index $[F:L]$. On the other hand $F/L$ is a subextension of $H_L/L$, so it is unramified there. Its ramification index is consequently both $[F:L]$ and $1$, forcingBecause $H_L/L$ is Galois, this intersection identity givesThus $M{H}_L/M$ is abelian. We use the standard local fact that unramified extensions remain unramified under finite base extension: after base change each factor is obtained by lifting an extension of the residue field. Since $H_L/L$ is unramified at every finite prime, $M{H}_L/M$ is also unramified at every finite prime. There are no real places of $M$, since every embedding of $M$ restricts to an embedding of the totally imaginary field $L$. Hence there is no infinite-place obstruction.

The maximality of $H_M$ now implies $M{H}_L\subseteq H_M$. Taking degrees over $M$ proves the class number divisibility under total ramification:The argument uses the degree of an unramified class-field extension, and does not assume that extension of ideals induces an injection of ideal class groups.

The Hilbert class field $\tilde{K}$ is the maximal everywhere unramified abelian extension of the number field $K$, with complete splitting at real places if infinite places are included. Global class field theory gives the canonical Artin reciprocity mapso $[\tilde{K}:K]$ is the class number of $K$.

The Hilbert class field $H_K$ is the maximal unramified abelian extension of $K$, including the condition that real places do not become complex when infinite ramification is included. The Artin reciprocity law induces a canonical isomorphismConsequently $[H_K:K]=h_K$.

For an unramified prime $\mathfrak{p}$, the ideal-theoretic decomposition law givesUnder the Hilbert class field isomorphism, this Artin symbol is the image of $[\mathfrak{p}]\in \mathrm{Cl}(K)$. It is trivial exactly when $[\mathfrak{p}]=1$, which means exactly that $\mathfrak{p}$ is a principal ideal.

Let $K=\mathbb{Q}(\sqrt{-6})$, whose discriminant is $-24$, and putThis is a biquadratic extension of $\mathbb{Q}$ whose three quadratic subfields have discriminants $-24$, $8$, and $-3$. The discriminant of a biquadratic field is thereforeThe relative discriminant formula givesso $H/K$ is unramified at every finite prime. Both fields are totally imaginary, so no infinite place ramifies. Since $[H:K]=2=h_K$, the extension has the full degree of the Hilbert class field, and hence

By Question 1, a prime ideal of $K$ is principal exactly when it splits completely in the Hilbert class field $H_K$. Question 3(4) therefore gives

PutIt 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 givesThe relative discriminant formulatherefore 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

Let $I$ be an integral ideal of an imaginary quadratic number field and $(e_1,e_2)$ an integral basis of $I$. Its normalized norm is a positive definite binary quadratic form. An unramified rational prime $p$ is represented by this form exactly when there is an integral prime ideal $P$ of norm $p$ with $[P]=[I{]}^{-1}$. Indeed, a representation gives an element $\alpha \in I$ with $N\alpha =pN(I)$, so $P=(\alpha )I^{-1}$ is integral of norm $p$. Conversely, $IP=(\alpha )$ with $\alpha \in I$ produces a representation by expressing $\alpha$ in the basis. Artin reciprocity then detects this ideal class by Frobenius in the Hilbert class field.

At least one finite prime ramifies in a Zp-extension; otherwise the entire tower would lie in a finite Hilbert class field. Ramification can occur only over $p$: local class field theory makes inertia away from $p$ an image of a unit group with finite maximal pro-$p$ quotient, whereas ${\mathbb{Z}}_p$ has no nontrivial finite subgroup. Nonzero inertia is open in ${\mathbb{Z}}_p$, so it becomes total after passing to a sufficiently high finite layer.

A number field is totally real when every embedding into the complex numbers has image in the real numbers. The maximal real subfield of a cyclotomic field is a basic example. Ordinary Hilbert class fields split at real places and therefore retain this property.

For distinct primes $p,q\equiv 1(\mathrm{mod}4)$, put $E=\mathbb{Q}(\sqrt{p},\sqrt{q})$ and $K=\mathbb{Q}(\sqrt{pq})$. The algebraic integers $a=(1+\sqrt{p})/2$ and $b=(1+\sqrt{q})/2$ generate an order whose basis has discriminant of elements of a number field $p^2{q}^2$: the trace pairing matrix is, up to basis order, the tensor product of the quadratic trace matrices of determinants $p$ and $q$. If its index in ${\mathcal{O}}_E$ is $j$, the discriminant-index formula for an integral lattice and the relative discriminant tower formula giveBoth $j$ and this ideal norm are positive integers, forcing $j=1$ and ${\mathfrak{d}}_{E/K}={\mathcal{O}}_K$. Thus $E/K$ is unramified at all finite primes. It is also totally real, so no real place ramifies. This quadratic abelian extension is contained in the ordinary Hilbert class field of $K$, and Artin reciprocity gives a quotient of its ideal class group of order two. The extensions $K(\sqrt{p})$ and $K(\sqrt{q})$ are equal to $E$, not two different extensions of $K$.

The splitting field has Galois group $S_3$ and quadratic subfield $\mathbb{Q}(\sqrt{-59})$. The extension over that subfield is abelian of degree three and unramified everywhere. The only possible rational ramified prime is $59$; at that prime a simple residue root lifts, so the local splitting field is quadratic and coincides with the quadratic subfield's completion. It follows from the Hilbert class field degree that $3\mid h_{\mathbb{Q}(\sqrt{-59})}$.