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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.

The polynomial is Eisenstein at $3$, so a root $\alpha$ generates a totally ramified cubic field $F={\mathbb{Q}}_3(\alpha )$ and is a uniformizer of $F$. Its cubic discriminant isIts $3$-adic valuation is $3$, so it is not a square. The cubic discriminant criterion therefore givesThe unique quadratic subfield is generated by $\sqrt{\Delta }$. Since the permitted square root $\sqrt{-5}$ is in ${\mathbb{Q}}_3$, this subfield isIt is totally ramified of degree two, by the Eisenstein polynomial $T^2-3$. The ramification index of $M/{\mathbb{Q}}_3$ is divisible by both the index $3$ of $F$ and the index $2$ of $E$. Since the whole degree is $6$, it follows thatIn particular $G_0=G$.

We use the following standard facts about totally ramified extensions: every uniformizer generates the valuation ring over the base valuation ring, and for a totally ramified Galois extensionThus we can use the uniformizer criterion for lower ramification groups. Put $s=\sqrt{3}$. Since $v_M(\alpha )=2$ and $v_M(s)=3$,is a uniformizer of $M$.

Let ${\alpha }_1,{\alpha }_2,{\alpha }_3$ be the three roots. The Galois group acts transitively on their unordered pairs and preserves the valuation, so all differences between distinct roots have the same valuation $d$. The discriminant identity giveshence $d=3$. Thus, for any $\sigma$, either $\sigma (\alpha )=\alpha$ or $\sigma (\alpha )-\alpha$ has valuation $3$.

A nontrivial three-cycle fixes $s$, since it fixes the quadratic subfield, and sends $\alpha$ to a different root. Also $v_M(\sigma (\alpha )+\alpha )=2$: write the sum as $2\alpha +(\sigma (\alpha )-\alpha )$, whose two terms have unequal valuations $2,3$. ThereforeA transposition changes the sign of $s$. ConsequentlyThe numerator has valuation $4$: it is $2{\alpha }^2$ plus terms of strictly higher valuation if the root changes, and exactly $2{\alpha }^2$ if the root is fixed. Thus every transposition has $v_M(\sigma (\pi )-\pi )=1$. We conclude that the ramification groups of the splitting field of T3 minus 3T plus 3 over Q3 are

As an independent check, the cubic field has ${\mathcal{O}}_F={\mathbb{Z}}_3[\alpha ]$ and different exponent $v_F(g^{\prime }(\alpha ))=v_F(3({\alpha }^2-1))=3$. The extension $M/F$ is tamely totally ramified of degree two, so its different exponent is $1$. The different in a tower gives $d(M/{\mathbb{Q}}_3)=1+2\cdot 3=7$, agreeing with the ramification sum $(6-1)+(3-1)=7$.