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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.

LetOver ${\mathbb{Q}}_2$, the Newton polygon has three length-one segments of slopes $-2,-1,0$. The resulting roots can also be obtained directly from Hensel lemma. There is one unit root because $F(1)$ is even and $F^{\prime }(1)$ is odd. For the two remaining roots, putwhose parenthesized polynomial has a simple root $Y\equiv 1(\mathrm{mod}2)$, andwhose parenthesized polynomial has a simple root $Z\equiv 1(\mathrm{mod}2)$. Thus $F$ splits completely over ${\mathbb{Q}}_2$, with roots of valuations $0,1,2$. There are three primes $q$ above $2$, and for each one

Over ${\mathbb{Q}}_5$, $F$ is an Eisenstein polynomial. Hence there is one prime $q$ above $5$, and if $\alpha$ is the chosen root thenwith $e=3$ and $f=1$.

The polynomial is irreducible over $\mathbb{Q}$ by the Eisenstein criterion at $5$. Its polynomial discriminant iswhich is not a square number, so the Galois group of an irreducible cubic isAt $2$, all three roots already lie in ${\mathbb{Q}}_2$, so the local splitting field is trivial andAt $5$, the cubic is totally and tamely ramified. The square class of its discriminant is represented by $-13807\equiv 3(\mathrm{mod}5)$, a nonsquare unit, so the quadratic resolvent field of a cubic is the unramified quadratic extension of ${\mathbb{Q}}_5$. The local splitting field therefore has degree six, withThese calculations are summarized by local factorization of X3 plus 25X2 minus 50X plus 40.

Let $\alpha$ be a root ofThe polynomial is Eisenstein at $3$, so $E={\mathbb{Q}}_3(\alpha )$ is totally ramified of degree three and ${\mathcal{O}}_E={\mathbb{Z}}_3[\alpha ]$. Its discriminant isIts odd valuation makes $\Delta$ nonsquare in ${\mathbb{Q}}_3$, so the Galois group of an irreducible cubic shows that the splitting field $L$ has Galois group $S_3$. The quadratic extension obtained by adjoining $\sqrt{\Delta }$ is ramified, so $L/{\mathbb{Q}}_3$ is totally ramified. Thereforebecause the wild inertia group is the unique Sylow $3$-subgroup of $S_3$.

It remains to find the wild break. Sinceand ${\alpha }^2+1$ is a unit, the different ideal of $E/{\mathbb{Q}}_3$ has exponent $v_E(f^{\prime }(\alpha ))=3$. The extension $L/E$ is a tamely ramified quadratic extension and has different exponent one. The different in a tower therefore gives different exponentOn the other hand, the different exponent from ramification groups iswhere $b$ is the last index for which $G_b=A_3$. Thus $b=1$, and

If a monic polynomial $f$ has roots ${\alpha }_1,\dots ,{\alpha }_n$ in a splitting field, its polynomial discriminant isIt is a symmetric polynomial in the roots, so it belongs to the base field; it is nonzero exactly when $f$ is a separable polynomial.

Putthe Vandermonde determinant. A root permutation $\sigma$ sends $D$ to $\mathrm{sgn}(\sigma )D$. Hence the Galois group of a polynomial is contained in the alternating group $A_n$ exactly when every Galois automorphism fixes $D$, which by the fixed-field property is equivalent to $D\in K$. This implies that $\mathrm{disc}(f)=D^2$ is a square in $K$. Conversely, if $\mathrm{disc}(f)=d^2$ for $d\in K$, then $D^2=d^2$, so $D=\pm d\in K$. The Galois group therefore fixes $D$ and consists of even permutations. This proves the discriminant criterion for an alternating Galois group.

Now letThe rational root theorem shows that none of $\pm 1,\pm 2$ is a root, so this cubic is irreducible over $\mathbb{Q}$. The discriminant of a depressed cubic isThis is not a square in $\mathbb{Q}$, so the Galois group of an irreducible cubic is $S_3$ over $\mathbb{Q}$.

Over $K=\mathbb{Q}(\sqrt{-19})$, the discriminant is the squareThe cubic remains irreducible: a root in the degree-two extension $K/\mathbb{Q}$ would generate over $\mathbb{Q}$ both a degree-three field, by irreducibility, and a subfield of a degree-two field, contradicting the tower law for field extensions. Its Galois group over $K$ is consequently $A_3\cong C_3$.

The rational-root test shows that $f=X^3-X-1$ is irreducible. Its discriminant iswhich is not a square in $\mathbb{Q}$. The Galois group of an irreducible cubic therefore gives

Quadratic subfields correspond to order-three subgroups. The unique such subgroup is $A_3$, and its fixed field is generated by the alternating product of root differences, whose square is $\Delta$. Thus the unique quadratic subfield is

Let the roots be $\alpha ,\beta ,\gamma$ and choose
$\delta =(\alpha -\beta )(\alpha -\gamma )(\beta -\gamma )$, so ${\delta }^2=-23$. Put $\theta =\alpha +\delta$. A nonidentity even permutation moves $\alpha$ and fixes $\delta$, so it cannot fix $\theta$. An odd permutation negates $\delta$. If it fixes $\alpha$, it plainly does not fix $\theta$. If it sends $\alpha$ to $\beta$ and fixed $\theta$, then $\beta -\alpha =2\delta$. Cancelling the factor $\beta -\alpha$ in the expression for $\delta$ would giveBut $f^{\prime }(\gamma )=3{\gamma }^2-1$, so this would make ${\gamma }^2$ equal to $1/6$ or $1/2$, contradicting the irreducibility of its cubic minimal polynomial. The other transpositions are identical. Hence $\theta$ has trivial stabilizer, so