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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.

Let $G$ be the Galois group of a polynomial of the irreducible quartic. Irreducibility makes $G$ a transitive subgroup of $S_4$. The three roots of the given cubic resolvent of a quartic correspond to the three partitions of the four quartic roots into two unordered pairs, and the kernel of the $S_4$-action on these partitions is the Klein four-group $V_4$. If the resolvent has Galois group $S_3$, the image of $G$ in this action is all of $S_3$. Among the transitive subgroups of the symmetric group on four points, only $S_4$ has this image. Hence

Now put $f_p(t)=t^4+pt+p$. The Eisenstein criterion at $p$ proves that $f_p$ is an irreducible polynomial, and its cubic resolvent isIts discriminant of a depressed cubic isFor odd $p$, the $p$-adic valuation of ${\Delta }_p$ is three because $p\nmid 256$; for $p=2$, ${\Delta }_2=2^4\cdot 101$. Thus ${\Delta }_p$ is never a square in $\mathbb{Q}$.

By the rational root theorem, a reducible $g_p$ has an integer root among $\pm 1,\pm p,\pm p^2$. Reduction modulo $p$ shows that the root is divisible by $p$; writing it as $pk$, with $k\in \{\pm 1,\pm p\}$, givesThe only possibilities are $k=-1,p=3$ and $k=1,p=5$, corresponding to roots $-3$ and $5$. For every other prime, $g_p$ is irreducible with nonsquare discriminant, so its Galois group is $S_3$ and the criterion above gives the quartic group $S_4$. For $p=3$ or $5$, the reducible resolvent makes the quartic group a proper subgroup. Therefore

Let $H$ denote a hyperplane-section divisor of a projective plane curve. The Adjunction formula for a smooth degree-$d$ projective plane curve statesSince $\mathrm{deg}(H{|}_C)=d$, this gives the canonical degree of a smooth plane curveCombining this with $\mathrm{deg}{K}_C=2g-2$ gives the genus of a smooth plane curve

Homogenization of the affine equation gives the projective completionThis is the plane model y plus x cubed plus xy cubed equals zero of the Klein quartic: after the coordinate relabelling $[A:B:C]=[Y:X:Z]$, its equation is $A^{3}B+B^{3}C+C^{3}A=0$.

Its first partial derivatives areIf one of $X,Y,Z$ vanishes at a common zero of these three derivatives, the displayed equations successively force all three coordinates to vanish, which is impossible in projective space. If $XYZ\ne 0$, multiplying the three derivative equations givesagain a contradiction. The Jacobian criterion therefore proves that $C$ is smooth.

The rational function $y=Y/Z$ defines a rational map of projective varieties $C⇢{\mathbb{P}}^1$. Because $C$ is a smooth projective curve, it extends uniquely to a morphismFor a generic finite value of $y$, its fibre is given by the cubicso the degree of a holomorphic map is $3$. The discriminant of a depressed cubic is

At $y=0$ the fibre consists of $(0,0)$. Since $F_y(0,0)=1$, the holomorphic implicit function theorem givesso $x$ is a local coordinate and $\pi$ has ramification index of a holomorphic map $3$ there. Each of the seven distinct roots of $4y^7+27$ gives one double, but not triple, root of the cubic in $x$, hence seven further ramification points of index $2$.

It remains to inspect the points at infinity. They areNear $Q$, set $u=X/Y$ and $v=Z/Y$. The equation becomes $u^{3}v+u+v^3=0$, so $u=-v^3+O(v^{10})$, while $y=1/v$ has a simple pole. Thus $e_Q=1$. Near $P$, set $s=Y/X$ and $t=Z/X$. Now $t+s^3+s{t}^3=0$, so $t=-s^3+O(s^{10})$ andThus $y$ has a double pole at $P$ and $e_P=2$. These calculations are the ramification of the x-coordinate on the Klein quartic after the coordinate relabelling above.

The total ramification contribution isThe Riemann-Hurwitz formula for the degree-three map to the projective line now givesand therefore

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 $A_3\cong C_3$.