= Solution
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 $|b|\leq a\leq c$, with $b\geq0$ when $|b|=a$ or $a=c$. From $31=4ac-b^2\geq3a^2$ we get $1\leq a\leq\sqrt{31/3}<4$. Checking these three possible leading coefficients gives exactly
$$
[1,1,8],\qquad[2,1,4],\qquad[2,-1,4].
$$
For $a=3$, neither odd residue choice $b=\pm1,\pm3$ makes $(b^2+31)/12$ integral. The listed forms are primitive and satisfy the reduction convention, so
$$
\boxed{h_K=3,\qquad\operatorname{Cl}(K)\cong C_3.}
$$
This 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 $\pm1$, and neither vanishes. Thus the cubic is irreducible. Its <polynomial discriminant> is
$$
\operatorname{disc}(F)=-4-27=-31.
$$
Since 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 $\ell\ne31$, 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 gives
$$
F(T)\equiv(T-17)^2(T-28)\pmod{31},\qquad F'(28)\equiv28\not\equiv0\pmod{31}.
$$
The 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 gives
$$
\operatorname{disc}(F)=\operatorname{disc}(Q)\,Q(r)^2.
$$
Here $Q(r)=F'(r)$ is a unit, so the <quadratic discriminant> has square class $-31$. Hence the entire local splitting field is
$$
\mathbb Q_{31}(\sqrt{\operatorname{disc}(Q)})=\mathbb Q_{31}(\sqrt{-31})=K_{31}.
$$
It 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>. Therefore
$$
\boxed{H_K=\mathbb Q(\sqrt{-31},\theta),\qquad\theta^3+\theta-1=0.}
$$
Equivalently, 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>.
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