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Normalize $v_K$ to have value group $\mathbb{Z}$, and extend it to an algebraic closure. Suppose $\alpha$ is a root of a monic Eisenstein polynomial of degree $n$. A root cannot have nonpositive valuation: the leading term would then have uniquely smallest valuation. If $w=v_K(\alpha )>0$, the constant term has valuation one, all the other nonleading terms have valuation strictly greater than one, and the leading term has valuation $nw$. Cancellation forces $nw=1$. The value group of $K(\alpha )$ is $(1/e)\mathbb{Z}$, so $n\mid e$. Since $e\le [K(\alpha ):K]\le n$, equality holds throughout. Thus the polynomial is irreducible and the extension is totally ramified.

Conversely, let $L/K$ be totally ramified of degree $n$, and choose a uniformizer ${\pi }_L$. Its extended $v_K$ value is $1/n$. The same denominator argument gives $[K({\pi }_L):K]\ge n$, so $K({\pi }_L)=L$. Every conjugate has valuation $1/n$, by the unique extension of an absolute value to a finite extension of a complete field. The nonleading coefficients of its minimal polynomial are elementary symmetric expressions in these conjugates, so their valuations are positive, hence at least one because the coefficients lie in $K$. The constant coefficient is their product and has valuation exactly one. Its minimal polynomial is an Eisenstein polynomial, and $\boxed{L=K({\pi }_L)}$.

We count extensions up to $K$-isomorphism. An extension of degree $n$ has a maximal unramified extension $E/K$ inside it, with $[E:K]=f\mid n$, and $L/E$ is totally ramified of degree $e=n/f$. The unramified extension of each degree is unique, so there are only finitely many possibilities for $E$.

Fix $E$ and $e$. The coefficient space of monic degree-$e$ Eisenstein polynomials is compact: its nonconstant nonleading coefficients lie in the compact maximal ideal, and its constant coefficient lies in the compact set ${\mathfrak{m}}_E\setminus {\mathfrak{m}}_E^2$. Each such polynomial $F$ has a root $\alpha$ generating a degree-$e$ extension. For sufficiently close coefficients of a polynomial $G$, the numbers $G(\alpha )$ and $G^{\prime }(\alpha )-F^{\prime }(\alpha )$ are arbitrarily small, while $F^{\prime }(\alpha )\ne 0$ because the characteristic is zero. The strong form of Hensel lemma, applied in $E(\alpha )$, supplies a root $\beta$ of $G$ as close to $\alpha$ as desired. Choose the neighborhood so that the Krasner's lemma inequality holds. Then $E(\alpha )\subseteq E(\beta )$, and both degrees are $e$, so the extensions are equal. Compactness gives a finite cover by such neighborhoods, proving finiteness of degree-$n$ extensions. If extensions are realized as subfields of a fixed algebraic closure rather than counted up to isomorphism, each has only finitely many embeddings, so that count is finite too.

For odd $p$, write $K^{\times }={\pi }^{\mathbb{Z}}{\mathcal{O}}_K^{\times }$. The square map is bijective on $1+{\mathfrak{m}}_K$ by Hensel lemma, and the finite residue field has cyclic multiplicative group of even order. ThusThe quadratic extensions from square classes correspond to its three nontrivial classes. Representatives are $u,\pi ,u\pi$, where $u$ is a unit with nonsquare residue. There are $\boxed{3}$ quadratic extensions: $K(\sqrt{u})$ is unramified, while $K(\sqrt{\pi })$ and $K(\sqrt{u\pi })$ are totally ramified. For completeness, two nonsquare parameters give the same extension only if their ratio is a square: writing $\sqrt{b}=x+y\sqrt{a}$ and squaring forces $x=0$, hence $b/a=y^2$.

Write $\alpha =\sqrt[3]{3}$ and $\zeta ={\zeta }_3$. The cubic is Eisenstein and adjoining $\zeta$ adds at most a quadratic extension, so $[L:{\mathbb{Q}}_3]\le 6$. SetUsing ${\zeta }^2+\zeta +1=0$ and ${\alpha }^3=3$, compute ${\pi }^3=1+2\zeta$ and ${\pi }^6=-3$. Thus $\pi$ satisfies the Eisenstein polynomial $X^6+3$. This proves $L={\mathbb{Q}}_3(\pi )$, degree six, with $\pi$ a uniformiser and the extension totally ramified. It is the splitting field of $X^3-3$, so its Galois group is $S_3$. This is the Eisenstein sextic presentation of the splitting field of X3 minus 3 over Q3.

The six automorphisms have $\sigma (\alpha )={\zeta }^a\alpha$, $a\in \{0,1,2\}$, and $\sigma (\zeta )={\zeta }^b$, $b\in \{1,2\}$. Since $v_L(3)=6$, we have $v_L(\alpha )=2$ and $v_L(\zeta -1)=3$.

If $b=1$ and $a\ne 0$, then $\sigma (\pi )={\zeta }^{-a}\pi$, whence $v_L(\sigma (\pi )-\pi )=1+3=4$. These are the two nonidentity elements of the cyclic subgroup $C_3$.

If $b=2$, use ${\zeta }^2-1=(\zeta -1)(\zeta +1)=-(\zeta -1){\zeta }^2$. Then $\sigma (\pi )=-{\zeta }^{2-a}\pi$. The coefficient $-{\zeta }^{2-a}-1$ reduces to $-2\ne 0$ in the residue field ${\mathbb{F}}_3$, so the difference has valuation one. These three automorphisms are the transpositions.

The uniformizer criterion for lower ramification groups consequently givesAs a consistency check, the different exponent is $(6-1)+3(3-1)=11$. The derivative of the monogenic Eisenstein polynomial gives the same value, $v_L(6{\pi }^5)=6+5=11$. In particular, stopping the wild filtration at $G_1$ would give the wrong different.

The inverse different is the trace-dual latticeThe integral closure ${\mathcal{O}}_L$ is finite free over the complete discrete valuation ring ${\mathcal{O}}_K$. Choose an integral basis and use the nondegenerate trace pairing to form its dual basis over $K$. This exhibits ${\mathfrak{D}}^{-1}$ as a full, finite ${\mathcal{O}}_K$-lattice. It is stable under multiplication by ${\mathcal{O}}_L$, because $b{\mathcal{O}}_L\subseteq {\mathcal{O}}_L$ for $b\in {\mathcal{O}}_L$, and bounded denominators make it a fractional ideal of ${\mathcal{O}}_L$. Integral elements have integral field traces, so ${\mathcal{O}}_L\subseteq {\mathfrak{D}}^{-1}$. A nonzero fractional ideal of a discrete valuation ring is invertible; its inverse is consequently an integral ideal, the different ideal ${\mathfrak{D}}_{L/K}$.

Assume now $A={\mathcal{O}}_K[x]={\mathcal{O}}_L$, with monic separable minimal polynomial $g$ of degree $n$. Lagrange interpolation at its distinct roots givesIndeed, these are the leading coefficients in the interpolation formula for $X^j$. For any element of $A$, this trace is the coefficient of $X^{n-1}$ in its degree-less-than-$n$ representative modulo $g$. The resulting pairing on the basis $1,x,\dots ,x^{n-1}$ is integral and unimodular: reversing the order of one basis makes its matrix triangular with diagonal ones, since entries vanish when the exponent sum is less than $n-1$. Thus it identifies $A$ with its full ${\mathcal{O}}_K$-dual. Translating back to the trace pairing proves

For a totally ramified extension of degree $n$, any uniformizer ${\pi }_L$ is an Eisenstein generator of a totally ramified extension, and ${\mathcal{O}}_L={\mathcal{O}}_K[{\pi }_L]$. To see the latter equality, use the common residue field to expand an integral element in powers of ${\pi }_L$ with digits from ${\mathcal{O}}_K$; reduce powers using its Eisenstein polynomial and take limits in the finite complete ${\mathcal{O}}_K$-module generated by $1,{\pi }_L,\dots ,{\pi }_L^{n-1}$. Write that polynomial as $X^n+{\sum }_{j<n}{a}_j{X}^j$, with $v_L(a_j)\ge n$ for $j<n$. When $p\nmid n$, the derivative's leading term has valuation $n-1$, while every other nonzero derivative term has valuation at least $n$. There can be no cancellation of the unique smallest term. Therefore

For the prime-power p-adic cyclotomic extension, put $\zeta ={\zeta }_{p^r}$ and $e=p^{r-1}(p-1)$. Modulo $p$, the shifted cyclotomic polynomial ${\Phi }_{p^r}(1+T)$ is $T^e$, while its constant term is $p$, not a multiple of $p^2$. It is therefore Eisenstein, so $\pi =\zeta -1$ is a uniformizer and ${\mathcal{O}}_{K_r}={\mathbb{Z}}_p[\zeta ]$. Fromwe obtain, by differentiating the numerator at $\zeta$,The denominator is a primitive-$p$ root of unity minus one and has $K_r$-valuation $p^{r-1}$. The different exponent is consequently $re-p^{r-1}$, givingThis includes $p=2,r=1$: the extension is trivial and its different exponent is zero.

Choose $\beta$ with ${\beta }^{p(p-1)}=-p$ and put $N=p(p-1)$. The Eisenstein polynomial $T^N+p$ shows that $L={\mathbb{Q}}_p(\beta )$ has degree $N$, is totally ramified, and has uniformizer $\beta$. Its subfield ${\mathbb{Q}}_p({\beta }^p)$ is the field $K$ from the previous part. Because $p$ is odd,Thus $L$ contains $a$ and ${\zeta }_p$, and contains all roots $a{\zeta }_p^j$ of $T^p-p$. Conversely $[{\mathbb{Q}}_p(a):{\mathbb{Q}}_p]=p$, while $[K:{\mathbb{Q}}_p]=p-1$. Their coprime degrees force their compositum to have degree $N$. It is contained in $L$ and hence equals it. Therefore $L$ is exactly the splitting field, not merely an extension containing it.

The Galois group has a normal subgroup $H=\mathrm{Gal}(L/K)$ of order $p$, acting by $\beta \mapsto {\zeta }_p^j\beta$. The $(p-1)$st roots of unity already lie in ${\mathbb{Q}}_p$ by the Hensel lemma. The maps $\beta \mapsto \omega \beta$, with ${\omega }^{p-1}=1$, supply a complement of order $p-1$. Its conjugation acts faithfully on $H$, so $G\cong C_p⋊{\mathbb{F}}_p^{\times }$.

Normalize $v_L(\beta )=1$. Total ramification and the uniformizer criterion for lower ramification groups reduce the calculation to $i_G(\sigma )=v_L(\sigma \beta -\beta )$. For nonidentity $\sigma \in H$,since $e(L/K)=p$ and $v_K({\zeta }_p^j-1)=1$. For $\sigma \notin H$, write $\sigma \beta =\omega {\zeta }_p^j\beta$ with $\omega \ne 1$. Its multiplier has residue $\overline{\omega }\ne 1$, so $i_G(\sigma )=1$. The lower ramification numbering is thereforeAll later groups are trivial. The wild lower break is $p$, not one. In the upper ramification numbering, the Herbrand function sends this break to $p/(p-1)$: $G^0=G$, $G^u=H$ for $0<u\le p/(p-1)$, and $G^u=1$ above it.

As an independent consistency check, the different exponent from ramification groups is $(N-1)+p(p-1)=2N-1$. The derivative of the Eisenstein polynomial gives the same answer, $v_L(N{\beta }^{N-1})=N+(N-1)$, since $v_L(p)=N$.

Logarithms on sufficiently deep units. Suppose $\mathrm{char}K=0$, let $\ell$ be the residue characteristic, and put $e=v(\ell )$. For every integer $r>e/(\ell -1)$ the p-adic logarithm and p-adic exponential function, evaluated in $K$, converge on the required domains:The bounds $v(m!)=e{v}_{\ell }(m!)\le e(m-1)/(\ell -1)$ and $v(m)\le e(m-1)/(\ell -1)$ show that, for $x,y\in {\pi }^r{\mathcal{O}}_K$, every term beyond the linear one has valuation strictly greater than $v(x)$ or $v(y)$. The series therefore converge and preserve these lattices; in particular $\exp (y)\in 1+{\pi }^r{\mathcal{O}}_K$. Their formal composition and addition identities are valid by convergence, giving inverse continuous group homomorphismsThus the logarithm isomorphism on deep principal units isThe right-hand group is torsion-free.

Roots of unity in ${\mathbb{Q}}_p$. For odd $p$, take $\ell =p,e=1,r=1$. The principal unit group $1+p{\mathbb{Z}}_p$ is torsion-free by the logarithm isomorphism on deep principal units. The Teichmuller representative splitting leaves precisely the $p-1$ roots in ${\mu }_{p-1}$.

For $p=2$, take $r=2$. Each odd unit is uniquely $\pm u$ with $u\in 1+4{\mathbb{Z}}_2$, since its residue modulo four is either one or minus one. This latter group is torsion-free. ConsequentlyThis computes the roots of unity in the p-adic numbers.

The units of ${\mathbb{Q}}_2(i)$. Set $\pi =1-i$. Its polynomial $T^2-2T+2$ is Eisenstein, soThe Eisenstein polynomial facts justifying this are stated in the next solution. Write $U_s=1+{\pi }^s{\mathcal{O}}_K$. Reduction gives ${\mathcal{O}}_K^{\times }=U_1$, and the successive quotients of principal-unit groupshave order two for $s\ge 1$. Hence $[U_1:U_3]=4$. The four roots in ${\mu }_4=\{1,i,-1,-i\}$ have distinct images modulo $U_3$, sinceThey exhaust the quotient and intersect $U_3$ trivially. Since $r=3>2/(2-1)$, the logarithm isomorphism on deep principal units gives $U_3\cong ({\mathcal{O}}_K,+)$. Therefore the unit decomposition of the 2-adic Gaussian field isIn particular these four elements are all its roots of unity.

Quadratic extensions. A nonzero element of $K$ is uniquely a power of $\pi$ times a unit, so $K^{\times }\cong {\pi }^{\mathbb{Z}}\times {\mathcal{O}}_K^{\times }$. Squaring on the decomposition above acts as squaring on ${\mu }_4$ and multiplication by two on the additive ring ${\mathcal{O}}_K={\mathbb{Z}}_2\oplus {\mathbb{Z}}_{2}i$. Thus the square-class group of the 2-adic Gaussian field isBy quadratic extensions from square classes, in characteristic different from two the nontrivial square classes classify quadratic extensions $K(\sqrt{a})$: two such extensions are $K$-isomorphic exactly when $a/b$ is a square. There are therefore $\boxed{2^4-1=15}$ quadratic extensions up to $K$-isomorphism.

The filtration and its meaning. Let $L/K$ be a finite Galois extension of non-Archimedean local fields, with group $G$, normalized valuation $v_L$, and residue fields $k_L/k_K$ of characteristic $\ell$. The lower ramification numbering isThe higher ramification groups are normal, decrease with $j$, and eventually become trivial: for each nonidentity automorphism some integral element is moved by a nonzero amount. The first group $G_0$ is the inertia group, withIts fixed field is the maximal unramified extension inside $L$. The wild inertia group is $G_1$. For a uniformizer ${\pi }_L$, there are injective homomorphismsThe kernels are the indicated next groups; the additive homomorphism assertion follows by expanding a product of automorphisms modulo ${\pi }_L^{j+2}$. These maps show that $G_0/G_1$ is cyclic of order prime to $\ell$, while $G_1$ is an $\ell$-group. Thus tame ramification is exactly the case $G_1=1$.

To calculate the groups, the uniformizer criterion for lower ramification groups saysIndeed, ${\mathcal{O}}_L$ is generated by ${\pi }_L$ over the integers of the maximal unramified subextension, which $G_0$ fixes. Differences of powers of ${\pi }_L$ are divisible by $\sigma {\pi }_L-{\pi }_L$, so the one test implies all the defining inequalities.

Eisenstein facts used in the calculations. For a complete discrete valuation ring $A$ with uniformizer $\varpi$, a monic polynomialis an Eisenstein polynomial if $\varpi \mid a_j$ for every $j<d$ and ${\varpi }^2\nmid a_0$. The Eisenstein criterion makes it irreducible. If $\theta$ is a root, its valuation relative to the base normalization is $1/d$, the extension has degree and ramification index $d$, its residue field is unchanged, andThe root valuation follows by comparing the terms of its equation. The ramification index must then be at least $d$, hence exactly $d$. In the basis $1,\theta ,\dots ,{\theta }^{d-1}$, the valuations of $b_j{\theta }^j$ are distinct modulo $d$, so an integral linear combination has every $b_j\in A$. This proves the integer-ring assertion. Conversely, a uniformizer in a totally ramified extension generates the field, and its minimal polynomial is Eisenstein; this follows from its value $1/d$, equal valuations of its conjugates, and the valuation one of its norm.

For a monogenic separable integer ring $A[\theta ]$, the different ideal is generated by $f^{\prime }(\theta )$. For a Galois extension, the different exponent from ramification groups isIn the totally ramified case these agree directly: $f^{\prime }({\pi }_L)={\prod }_{\sigma \ne 1}({\pi }_L-\sigma {\pi }_L)$, and an automorphism with displacement valuation $h$ contributes once to each of $G_0,\dots ,G_{h-1}$.

Upper numbering. Extend the lower indexing to real $s\ge 0$ by $G_s=G_{\lceil s\rceil }$, and setOn $[-1,0]$ take $\phi (s)=s$. This Herbrand function slows the indexing as the groups shrink. Lower numbering is compatible with subgroups, while upper ramification numbering is compatible with quotients: for normal $H$, $(G/H{)}^u=G^{u}H/H$. This is the fact that upper ramification groups commute with quotients. Thus upper breaks are particularly useful when comparing intermediate Galois extensions.

The eighth-root cyclotomic extension. Put $\zeta ={\zeta }_8$ and $\pi =\zeta -1$. The shifted cyclotomic polynomialis Eisenstein at two. Hence $L={\mathbb{Q}}_2(\zeta )$ is totally ramified of degree four, $v_L(2)=4$, and $\pi$ is a uniformizer. Its Galois group consists of ${\sigma }_a(\zeta )={\zeta }^a$ for $a=1,3,5,7$, and is $C_2\times C_2$.

Since $\zeta$ is a unit,For $a=3,7$, this is the valuation of $i-1$ or $-i-1$, namely two, because their squares are units times two. For $a=5$, it is $v_L(-2)=4$. The ramification groups of the eighth-root cyclotomic extension of the 2-adic field are thereforeThe lower breaks are one and three, andThus the upper breaks are one and two:The different exponent from ramification groups is $3+3+1+1=8$, agreeing with $v_L(4{\zeta }^3)=8$ from the derivative of $T^4+1$.

The cubic splitting field at three. Let ${\alpha }^3=2$, $\zeta ={\zeta }_3$, and putThe polynomialsare Eisenstein at three. They give totally ramified subextensions of degrees three and two. Their intersection is the base field, so $L={\mathbb{Q}}_3(\alpha ,\zeta )$ has degree six. Its ramification index is divisible by both three and two, hence equals six; therefore $L/{\mathbb{Q}}_3$ is totally ramified. It is the splitting field of $T^3-2$, with Galois group $S_3$.

Normalize $v_L$ so that $v_L(3)=6$. Then $v_L(\beta )=2$ and $v_L(\lambda )=3$, makinga uniformizer. Let $\sigma (\alpha )=\zeta \alpha ,\sigma (\zeta )=\zeta$ and $\tau (\alpha )=\alpha ,\tau (\zeta )={\zeta }^2$. They generate $S_3$. Since $\tau (\lambda )=(\zeta +1)\lambda$,For the order-three automorphism,Here $\alpha$ is a unit and $v_L(\zeta \alpha +1)=v_L(\sigma \beta )=2$. The same calculation gives valuation two for ${\sigma }^2$. The three transpositions are conjugate and the defining filtration is normal, so they all have displacement valuation one. The ramification groups of the splitting field of T3 minus 2 over Q3 areThe lower breaks are zero and one. Since $[G_0:G_s]=2$ for $0<s\le 1$,and the upper groups areThe different exponent from ramification groups is $5+2=7$. As an independent check, the cubic subfield has different exponent $v_{{\mathbb{Q}}_3(\alpha )}(3{\alpha }^2)=3$, and the quadratic extension above it is tame with different exponent one. Transitivity of the different ideal gives $1+2\cdot 3=7$, as required.

A set $S$ of prime numbers has Dirichlet density $\delta$ if the following limit exists:Equivalently, the denominator can be ${\sum }_p{p}^{-s}$, since that sum is $\log (1/(s-1))+O(1)$. Adding or removing finitely many prime numbers leaves the Dirichlet density unchanged.

Put $\alpha =\sqrt[4]{2}>0$ and $L=\mathbb{Q}(\alpha ,i)$. This is the splitting field of $X^4-2$, whose four roots of a polynomial are $\alpha ,i\alpha ,-\alpha ,-i\alpha$. The polynomial is an Eisenstein polynomial at $2$, so $[\mathbb{Q}(\alpha ):\mathbb{Q}]=4$. Since $\mathbb{Q}(\alpha )$ is contained in the real numbers, it does not contain $i$, giving $[L:\mathbb{Q}]=8$. Thus $L/\mathbb{Q}$ is a Galois extension of degree $8$. Its Galois group is the dihedral group of order $8$: the field automorphisms $r(\alpha )=i\alpha$, $r(i)=i$ and $s(\alpha )=\alpha$, $s(i)=-i$ satisfy $r^4=s^2=1$ and $srs=r^{-1}$.

For an odd prime number $p$, the condition $p\equiv 1(\mathrm{mod}4)$ means that the finite field ${\mathbb{F}}_p$ contains all fourth roots of unity. Under this condition, $2$ is a quartic residue exactly when $X^4-2$ has a root in ${\mathbb{F}}_p$; multiplying that root by $1,i,-1,-i$ gives all four distinct roots. Conversely, complete splitting of $X^4-2$ over ${\mathbb{F}}_p$ gives both a fourth root of $2$ and a primitive fourth root of unity, so it also forces $p\equiv 1(\mathrm{mod}4)$.

The polynomial discriminant of $X^4-2$ is $-2^{11}$. Hence every odd prime number is unramified in $L$, and the root-splitting criterion is equivalent to its Frobenius automorphism acting trivially on all the roots. Since the roots generate $L$, this is equivalent to the Frobenius automorphism being the identity, or to $p$ being a completely split prime of $L/\mathbb{Q}$. The Chebotarev density theorem says that the unramified prime numbers with Frobenius conjugacy class $C$ have Dirichlet density $|C|/|\mathrm{Gal}(L/\mathbb{Q})|$. Here $C=\{1\}$, so

For a finite extension $E/K$ of number fields, the inverse different is the fractional idealThe different ideal is its inverse. The trace pairing is nondegenerate because extensions of number fields are separable, so this definition gives a full fractional ideal. The inclusion ${\mathcal{O}}_E\subseteq {\mathfrak{D}}_{E/K}^{-1}$ shows that ${\mathfrak{D}}_{E/K}$ is integral.

Write its prime ideal factorization as ${\mathfrak{D}}_{E/K}={\prod }_{\mathfrak{P}}{\mathfrak{P}}^{d_{\mathfrak{P}}}$. For $\mathfrak{P}$ above $\mathfrak{p}$, with ramification index $e_{\mathfrak{P}}$ and residue characteristic $\ell$, the different exponent and tame ramification theorem saysThe equivalence uses the separability of finite residue-field extensions. In particular, $d_{\mathfrak{P}}=0$ precisely at unramified prime ideals; in the wild case $d_{\mathfrak{P}}\ge e_{\mathfrak{P}}$. This theorem does not require a Galois extension. The relative discriminant is the norm of the different ideal:For $K=\mathbb{Q}$, this gives $|D_E|=N({\mathfrak{D}}_{E/\mathbb{Q}})$.

Now take ${\alpha }^3=m$ and $K=\mathbb{Q}(\alpha )$. For any prime number $q\mid m$, the square-free integer hypothesis makes $X^3-m$ an Eisenstein polynomial at $q$. It is therefore irreducible, and $K$ is a pure cubic number field of degree $3$. Since $\alpha$ is an algebraic integer, $A=\mathbb{Z}[\alpha ]$ is an order in a number field. The discriminant of elements of a number field for its basis $1,\alpha ,{\alpha }^2$ isFor example, the resultant of $X^3-m$ and $3X^2$ is $27m^2$, and the degree-three sign in the polynomial discriminant is negative. If $I=[{\mathcal{O}}_K:A]$, the discriminant-index formula for an integral lattice givesOnly prime numbers dividing $3m$ can therefore divide $I$.

For $q\mid m$, the Eisenstein polynomial gives a totally ramified extension of ${\mathbb{Q}}_q$ of degree $3$. Thus $K$ has a unique prime ideal $\mathfrak{P}$ above $q$, with ramification index $3$ and residue-field degree $1$. Since $q\ne 3$, this is tame ramification, and the different exponent and tame ramification theorem gives $d_{\mathfrak{P}}=2$. Hence $v_q(D_K)=2$. Comparing with $v_q(-27m^2)=2$ in the discriminant-index formula for an integral lattice yields $v_q(I)=0$.

At $3$, use the shifted Eisenstein polynomial of $\beta =\alpha -m$:Because $3\nmid m$, the constant term is divisible by $3$. Moreover,when $3\nmid m$: the factors $m-1$ and $m+1$ cannot both be divisible by $3$. The hypotheses therefore give $v_3(m^3-m)=1$, so the translated polynomial is an Eisenstein polynomial at $3$. The resulting completion is a totally ramified extension of degree $3$, with residue-field degree $1$, and its ramification index is divisible by the residue characteristic. Thus its different exponent is at least $3$.

It follows that $v_3(D_K)\ge 3$. But $v_3(-27m^2)=3$, soforces $v_3(I)=0$ and $v_3(D_K)=3$. No prime number divides $I$. We have proved the integral basis of a nonexceptional pure cubic field:Using local Eisenstein polynomials here only establishes the local ramification indices; it does not assume that $\mathbb{Z}[\alpha ]$ is already the full ring of integers of a number field.

Let $a=\sqrt[3]{3}$ and let $\zeta$ be a primitive cube root of unity. The splitting field is $L={\mathbb{Q}}_3(a,\zeta )$. The polynomial $X^3-3$ is an Eisenstein polynomial, so $[{\mathbb{Q}}_3(a):{\mathbb{Q}}_3]=3$. Meanwhile $(1+2\zeta {)}^2=-3$ shows that ${\mathbb{Q}}_3(\zeta )$ is quadratic: $-3$ is not a square because its valuation is odd. A quadratic field cannot lie in the degree-three field. Consequently $[L:{\mathbb{Q}}_3]=6$, and the action on the three roots identifies its Galois group with the symmetric group $S_3$.

A useful uniformizer isSince $(\zeta -1{)}^3=-3\zeta (\zeta -1)$, we obtainThus $\pi$ is a root of the Eisenstein polynomial $X^6+3$. Its degree is six, so $L={\mathbb{Q}}_3(\pi )$, the extension is totally ramified, and $\pi$ is a uniformizer. With $v_L(\pi )=1$,This is the Eisenstein sextic presentation of the splitting field of X3 minus 3 over Q3.

Let $\tau (a)=\zeta a$, $\tau (\zeta )=\zeta$, and let $s(a)=a$, $s(\zeta )={\zeta }^{-1}$. Then $\tau$ has order three, $s$ has order two, and $s\tau s={\tau }^{-1}$. Their actions on the uniformizer areThe two nonidentity elements of $⟨\tau ⟩$ therefore satisfyEach transposition sends $\pi$ to $-{\zeta }^j\pi$ for some $j$, so its displacement has valuation one: $1+{\zeta }^j$ is a unit, reducing to $2$ modulo the maximal ideal. The uniformizer criterion for lower ramification groups now givesFor real indices, this is $S_3$ on $[-1,0]$, $C_3$ on $(0,3]$, and $1$ for $t>3$. In particular the lower breaks are $0$ and $3$.

On $(0,3]$ the index $[G_0:G_t]$ is two; beyond three it is six. Thus the Herbrand function isThe positive lower break $3$ becomes the upper break $3/2$. ThereforeThe fractional upper break is allowed because the full extension is nonabelian. As a check, the different exponent from ramification groups is $5+3\cdot 2=11$. The derivative of the minimal polynomial $X^6+3$ gives the same result: $v_L(6{\pi }^5)=6+5=11$. This is the $p=3$ case of ramification groups of the splitting field of Xp minus p over the p-adic numbers.

Let $L/K$ be a finite Galois extension of non-Archimedean local fields, let $G=\mathrm{Gal}(L/K)$, and normalize the discrete valuation by $v_L(L^{\times })=\mathbb{Z}$. The lower ramification numbering isHere $G_0$ is the inertia group and $G_1$ is the wild inertia group. For real $s\ge -1$, put $G_s=G_{\lceil s\rceil }$; thus $G_s=G_0$ for $-1<s\le 0$. Define the Herbrand function and its inverse byextending $\phi (s)=s$ on $[-1,0]$. The upper ramification numbering is $G^u=G_{\psi (u)}$ for $u\ge -1$.

Let ${\alpha }^p-\alpha =t^{1-p}$ over $K={\mathbb{F}}_p((t))$, with $t$ its uniformizer. The Artin–Schreier polynomial has derivative $-1$ in characteristic $p$, so all its roots are distinct. If $\alpha$ is one root and $\beta$ another, then $(\beta -\alpha {)}^p=\beta -\alpha$, so $\beta -\alpha \in {\mathbb{F}}_p$. Conversely every $\alpha +a$, $a\in {\mathbb{F}}_p$, is a root. Thus $L=K(\alpha )$ contains all roots and is a splitting field of a separable polynomial; it is Galois. Every automorphism has the form ${\sigma }_a(\alpha )=\alpha +a$, and its Galois group embeds in the additive group of ${\mathbb{F}}_p$.

There is no root in $K$. If $v_K(z)\ge 0$, then $z^p-z$ cannot have negative valuation. If $v_K(z)<0$, the ultrametric inequality gives $v_K(z^p-z)=p{v}_K(z)$, divisible by $p$, whereas $v_K(t^{1-p})=1-p$ is not. The Galois group is consequently nontrivial. Its order divides the prime $p$, so

Let $e=e(L/K)$ be the ramification index. The equation forces $v_L(\alpha )<0$, and henceAs $\gcd (p,p-1)=1$, $p\mid e$. Since $e\le [L:K]=p$, we get $e=p$, residue-field degree one, and $v_L(\alpha )=1-p$. Thus this is a totally ramified extension, andThis is a uniformizer. The uniformizer criterion for lower ramification groups states that for a totally ramified Galois extension, $\sigma \in G_i$ exactly when $v_L(\sigma ({\pi }_L)-{\pi }_L)\ge i+1$. Here every nonidentity automorphism satisfiesTherefore the ramification break of an Artin–Schreier pole occurs at $p-1$, and the lower groups areThis includes $p=2$, where the break is one. There is no off-by-one shift: the condition is $v_L(\sigma {\pi }_L-{\pi }_L)\ge i+1$.