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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 $(6-1)+(3-1)=7$.

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.

Suppose first that $L=K(\alpha )$ and the degree-$n$ minimal polynomial of $\alpha$ is an Eisenstein polynomial. It is irreducible, $\alpha$ is a uniformizer of $L$, and its valuation shows that $e(L/K)\ge n=[L:K]$. Equality follows, so $L/K$ is a totally ramified extension.

Conversely, suppose $L/K$ is totally ramified of degree $n$ and choose a uniformizer $\alpha$ of $L$. The field $K(\alpha )$ already has ramification index at least $n$, so it equals $L$. Letbe the minimal polynomial. All conjugates of $\alpha$ have $L$-valuation one. Each $a_i$ with $i<n$ is an elementary symmetric polynomial in products of at least one conjugate and therefore has positive $L$-valuation. Since valuations of elements of $K$ are multiples of $n$, every $a_i$ lies in the maximal ideal of ${\mathcal{O}}_K$. Moreoverso $a_0$ is not divisible by the square of that ideal. Thus $f$ is Eisenstein, proving the Eisenstein generator of a totally ramified extension criterion.

Put $\alpha =\sqrt[p]{p}$. The polynomial $X^p-p$ is Eisenstein, so ${\mathbb{Q}}_p(\alpha )/{\mathbb{Q}}_p$ is totally ramified of degree $p$. The cyclotomic extension ${\mathbb{Q}}_p({\zeta }_p)/{\mathbb{Q}}_p$ is totally ramified of degree $p-1$. Their coprime degrees make their intersection trivial, sohas degree $p(p-1)$ and is totally ramified. It is the splitting field of $X^p-p$, hence Galois.

Normalize $v_K$ by $v_K(K^{\times })=\mathbb{Z}$. Thensois a uniformizer. Write an automorphism aswhere $a\in {\mathbb{F}}_p^{\times }$ and $b\in {\mathbb{F}}_p$. Sincethe uniformizer criterion for lower ramification groups gives valuation one for $\sigma (\varpi )-\varpi$ when $a\ne 1$, and valuation $p+1$ when $a=1$, $b\ne 0$. Thereforeand $G_i=1$ for $i\ge p+1$. These are the ramification groups of the splitting field of Xp minus p over the p-adic numbers.

Since $f(X)=X(\pi +X^{q-1})$, its iterates satisfyThe polynomial ${\Phi }_m$ is Eisenstein: it is monic, every nonleading coefficient is divisible by $\pi$, and its constant term is $\pi$. It is also separable, since $q-1$ is prime to the residue characteristic and the iterates have nonzero derivative.

If $\alpha \ne 0$ and $m$ is least with $f_m(\alpha )=0$, then ${\Phi }_m(\alpha )=0$. Eisenstein irreducibility makes ${\Phi }_m$ its minimal polynomial, so $K(\alpha )/K$ is totally ramified and separable. For $\alpha =0$ the extension is trivial and has the same properties. This is the Eisenstein layers of Lubin–Tate torsion argument.

The cyclotomic extension of a p-adic fieldhas degree $p-1$, is Galois with group $(\mathbb{Z}/p\mathbb{Z}{)}^{\times }$, and is totally ramified. Indeed, ${\zeta }_p-1$ is a root of the Eisenstein polynomial

Put $\alpha =\sqrt[p-1]{-p}$. Its polynomial $X^{p-1}+p$ is Eisenstein, so $L_2={\mathbb{Q}}_p(\alpha )$ is totally ramified of degree $p-1$. Every $(p-1)$st root of unity lies in ${\mathbb{Q}}_p$ by the Teichmuller lifts, so every root $\omega \alpha$ of this polynomial lies in $L_2$. Hence $L_2/{\mathbb{Q}}_p$ is also Galois.

For either $L_i/{\mathbb{Q}}_p$, the norm has every possible valuation because the residue-field degree is one. The norm units in a tamely totally ramified extension lie in the principal units $1+p{\mathbb{Z}}_p$: reduction of a unit norm is the $(p-1)$st power of its residue, hence is $1$. Part (b) says that this unit norm subgroup has index $p-1$, exactly the index of $1+p{\mathbb{Z}}_p$ in ${\mathbb{Z}}_p^{\times }$. ConsequentlyThe uniqueness clause in the existence theorem of local class field theory now gives

Suppose first that $L/K$ is a totally ramified extension of degree $n$, and let $\alpha$ be a uniformizer of $L$. If the valuation on $L$ is normalized by $v_L(\alpha )=1$, then $v_L(K^{\times })=n\mathbb{Z}$. The value group of $K(\alpha )$ already contains both $n\mathbb{Z}$ and $1$, so its ramification index over $K$ is at least $n$. Hence $[K(\alpha ):K]\ge n$, and therefore $L=K(\alpha )$.

Letbe the minimal polynomial of $\alpha$. Every conjugate of $\alpha$ has positive valuation, so each $a_i$ lies in the maximal ideal of ${\mathcal{O}}_K$. Moreoverwhich means $v_K(a_0)=1$. Thus $f$ is an Eisenstein polynomial.

Conversely, if $\alpha$ is a root of an Eisenstein polynomial of degree $n$, the Eisenstein criterion makes that polynomial irreducible and its Newton polygon gives $v_L(\alpha )=1/n$ when $v_K$ is normalized. Consequently $e(K(\alpha )/K)\ge n$; equality with the field degree forces $e=n$ and residue-field degree one. Thus $K(\alpha )/K$ is totally ramified.

A pure cubic number field is a degree-three number field generated by a root of $X^3-m$ for a rational number $m$ that is not a rational cube. For a positive square-free integer $m>1$, this polynomial is an Eisenstein polynomial at any prime dividing $m$. The ring of integers of a number field depends on congruences at $3$, as expressed by the integral basis of a nonexceptional pure cubic field.

For odd $p$, the splitting field $K={\mathbb{Q}}_p({\zeta }_p,\sqrt[p]{p})$ is totally ramified of degree $p(p-1)$. Withthe lower ramification groups are

To verify the filtration, write ${\alpha }^p=p$ and $\lambda ={\zeta }_p-1$. The Eisenstein polynomials for $\alpha$ and $\lambda$ give coprime degrees $p$ and $p-1$, so their compositum has degree $p(p-1)$. Its element $\varpi =\lambda /\alpha$ has $p$-adic valuation $1/[p(p-1)]$, forcing total ramification. For an automorphism ${\sigma }_{a,b}$ with $\alpha \mapsto {\zeta }_p^a\alpha$ and ${\zeta }_p\mapsto {\zeta }_p^b$,Its residue is $b$, so the normalized displacement valuation is one if $b\ne 1$. If $b=1$ and $a\ne 0$, that valuation is $1+v_K({\zeta }_p^{-a}-1)=p+1$. The uniformizer criterion for lower ramification groups now gives every group displayed above.

For an odd prime $p$, this Eisenstein polynomial generates a totally ramified cubic root field. Its discriminant is $-p^2(4p+27)$. For $p>3$ the unit part is a square exactly when $p\equiv 1(\mathrm{mod}3)$. The splitting field then has cyclic Galois group of order three; otherwise it has symmetric group $S_3$, with an unramified extension of degree two and inertia group $C_3$. Both cases are tamely ramified, so all positive ramification groups are trivial.

At $p=3$, adjoining $\beta =\sqrt{-39}$ gives a ramified quadratic subfield, and the splitting field has total ramification index six. A root $\alpha$ has normalized valuation two, while $\beta$ has valuation three, so $\pi =\beta /\alpha$ is a uniformizer. The derivative $3{\alpha }^2+3$ has valuation six, and all distinct root differences have valuation three. A three-cycle fixes $\beta$, giving $v_L(\sigma \pi -\pi )=2$. A transposition fixing $\alpha$ negates $\beta$, giving valuation one; conjugation handles the other transpositions. Hence $G_0=S_3$, $G_1=C_3$, and $G_i=1$ for $i\ge 2$. The different exponent is seven.

A polynomial may fail the Eisenstein polynomial test while a translate $f(T+a)$ passes it. Translation does not change the generated extension: if $f(\alpha )=0$, then $f(T+a)$ vanishes at $\alpha -a$. For $f(T)=T^3-m$, the translate by $m$ is Eisenstein at $3$ whenever $3\nmid m$ and $m\not\equiv \pm 1(\mathrm{mod}9)$.

The Eisenstein polynomial of $1-i$ is $T^2-2T+2$, so ${\mathbb{Q}}_2(i)$ has residue field ${\mathbb{F}}_2$ and ramification index two. The four roots $1,i,-1,-i$ have distinct cosets modulo $U_3$, which has index four by the successive quotients of principal-unit groups. The logarithm isomorphism on deep principal units gives $U_3\cong ({\mathbb{Z}}_2[i],+)$. Hence these four roots split the full unit group, and are all its torsion.

If $L/K$ is an unramified extension and $M/K$ is a totally ramified extension of non-Archimedean local fields inside a fixed algebraic closure, then $LM/L$ is totally ramified of degree $[M:K]$. A uniformizer of $M$ has an Eisenstein polynomial over $K$. Since a uniformizer of $K$ remains a uniformizer in $L$, that polynomial remains Eisenstein over $L$ and generates $LM$.