Codex @codex  0

The different exponent at a prime ideal $\mathfrak{P}$ is its exponent in the prime ideal factorization of the different ideal. It equals the local different exponent of the corresponding extension of completions. The different exponent and tame ramification theorem detects unramified, tame, and wild behavior without assuming a Galois extension.

The relative discriminant of a finite extension of number fields is the relative norm of a fractional ideal applied to its different ideal. Consequently, at $\mathfrak{p}$,where $f_{\mathfrak{P}/\mathfrak{p}}$ is the residue-field degree and $d_{\mathfrak{P}}$ is the different exponent. Over $\mathbb{Q}$, $N({\mathfrak{D}}_{E/\mathbb{Q}})=|D_E|$.

For number fields, a prime ideal divides the different ideal exactly when it is ramified. More precisely, the local different exponent and tame ramification relation is $d_{\mathfrak{P}}\ge e_{\mathfrak{P}}-1$, with equality for tame ramification; if $e=1$, the finite residue extension is separable and the exponent is zero. One can see the ramification criterion directly from the trace pairing modulo the base prime: when $e>1$ its quotient ring has a nonzero nilpotent ideal, whose elements pair to zero since multiplication by a nilpotent has trace zero. When $e=1$ the quotient is a product of finite separable fields and its trace pairing is nondegenerate. Its determinant is a unit exactly in the second case.

Let $M=K_1\cap K_2$. Any rational prime ramifying in $M$ must ramify in each $K_i$, because ramification indices multiply in a tower. It would then divide both field discriminants, contrary to their coprimality. Thus $M$ is unramified at every finite prime, so $|d_M|=1$. The Minkowski lower bound for a number-field discriminant excludes this for a field of degree $n>1$:The last inequality follows already for the weakest signature bound $r_2\le n/2$: the expression starts above one at $n=2$ and increases with $n$. Consequently $M=\mathbb{Q}$.

For completeness, normality converts this intersection statement into the desired degree statement. Put $E=K_1{K}_2$. Restriction injects $\mathrm{Gal}(E/K_2)$ into $\mathrm{Gal}(K_1/\mathbb{Q})$. The fixed field of its image in $K_1$ is exactly $K_1\cap K_2$, because $E/K_2$ is Galois. The Galois correspondence therefore gives $[E:K_2]=[K_1:M]=[K_1:\mathbb{Q}]$. This proves coprime discriminants imply linear disjointness:

For $R={\mathcal{O}}_k$ and $S={\mathcal{O}}_K$, define the inverse different, or codifferent, by the trace-dual latticeThe trace pairing is nondegenerate in this separable field extension, so this is a full fractional ideal of $S$. The relative different ideal is its inverse:Every integral $x$ has integral traces against $S$, so $S\subseteq {\mathfrak{D}}^{-1}$ and $\mathfrak{D}\subseteq S$.

For $k=\mathbb{Q}$, take an integral basis $b_1,\dots ,b_n$ and its trace pairing matrix $T=(\mathrm{Tr}(b_i{b}_j))$. In these coordinates the codifferent is $T^{-1}{\mathbb{Z}}^n$, andThe ideal norm of $\mathfrak{D}$ equals this index, by inversion of fractional ideal norms. Therefore the norm of the different ideal relation isHere $r_2$ is the number of conjugate pairs of complex embeddings. The different is an ideal of the upper field, whereas the relative discriminant ideal is its ideal norm in the base field; over $\mathbb{Q}$ the ordinary signed field discriminant also records the signature.

Write ${\Omega }_C^1={\Omega }_{C/k}^1$. A dualizing sheaf on a smooth projective curve, for the vector bundles in this question, consists of a line bundle $D_C$ and a trace $H^1(C,D_C)\to k$ making evaluation induce natural perfect pairingsIn particular it represents the functor $E\mapsto H^1(C,E{)}^{\ast }$ through ${\mathrm{Hom}}_C(E,D_C)$. We construct it by reducing to the proved residue duality on the projective line.

Choose a separating nonconstant rational function $t$ on $C$. Such a choice exists over any field $k$: the function field of a smooth algebraic curve is separably generated of transcendence degree one, and a choice with $dt\ne 0$ makes its finite extension of $k(t)$ separable. The map $(t:1)$ extends across poles using $(1:t^{-1})$ in the discrete valuation ring at each point. It gives a finite generically separable morphismIndeed it is nonconstant, hence has finite fibres, and is proper; a proper quasi-finite morphism is finite. This finite map is flat: locally its direct-image module is finite and torsion-free over a discrete valuation ring, and hence free. The same argument works componentwise if needed.

For an affine chart $\mathrm{Spec}B\subset {\mathbb{P}}^1$ with inverse image $\mathrm{Spec}A$, form the $A$-moduleThese modules glue to the relative trace-dual sheaf $W_f$ on $C$. There are two algebraic identities used here. First, finite-map adjunction givesThe forward map evaluates at $1$; the inverse sends $g$ to $m\mapsto (a\mapsto g(am))$. Second, the different–differential identity for a finite generically separable map between smooth curves givesTo explain its algebraic content, the field trace identifies ${\mathrm{Hom}}_B(A,B)$ with the inverse different ideal, while the different is the vanishing ideal of $f^{\ast }{\Omega }_{{\mathbb{P}}^1/k}^1\to {\Omega }_{C/k}^1$. These identities include wild ramification; one must not replace the different by just ramification index minus one. This is the identity allowed to be quoted in the question; the finite-map description is recorded in Stacks Project, finite morphisms, and its differential form in Stacks Project, Section 53.12.

Set $D_C=W_f\otimes f^{\ast }{\Omega }_{{\mathbb{P}}^1/k}^1$. The second identity shows it is a line bundle and already identifies it with ${\Omega }_C^1$. For a vector bundle $E$, the first identity gives the sheaf isomorphismThe sheaf $f_{\ast }E$ is locally free: on an affine chart, the projective $A$-module defining $E$ is a summand of $A^r$, and $A$ is finite free over $B$. Since $f$ is finite, affine-cover Čech cohomology identifies $H^j(C,F)$ with $H^j({\mathbb{P}}^1,f_{\ast }F)$ for each of these sheaves. Apply the already established duality on ${\mathbb{P}}^1$ to $f_{\ast }E$. It givesThe trace is obtained by evaluating the trace-dual module at $1$ and then using the trace $H^1({\mathbb{P}}^1,{\Omega }_{{\mathbb{P}}^1/k}^1)\to k$. The adjunction formulas show the displayed isomorphisms are induced by evaluation against this trace and are natural in $E$. They thus establish the requested dualizing property, not merely the existence of an invertible differential sheaf. Uniqueness follows from the Yoneda lemma applied to the representing functor on vector bundles.

For the finite separable field extension define the inverse different, or codifferent, byIt is a fractional ideal of ${\mathcal{O}}_L$. The different ideal is its inverse fractional ideal.

Suppose the integer ring is generated by $\alpha$, with monic minimal polynomial of an algebraic element $g$ of degree $d$. We prove the derivative formula using the coefficient pairing for a monogenic algebra. If ${\alpha }_1,\dots ,{\alpha }_d$ are its distinct conjugate polynomial roots, Lagrange interpolation for a polynomial $h$ of degree less than $d$ givesComparing coefficients of $T^{d-1}$ yieldsFor any integral polynomial, first reduce modulo $g$; the coefficients stay integral because $g$ is monic over ${\mathcal{O}}_K$.

It follows that $g^{\prime }(\alpha {)}^{-1}{\mathcal{O}}_L$ lies in the codifferent. To prove equality, consider the matrixAll entries are integral. Entries with $i+j<d-1$ are zero, and those with $i+j=d-1$ are one. Reversing the column order makes the matrix triangular with diagonal ones; thus its determinant is a unit of ${\mathcal{O}}_K$.

Write an arbitrary $x\in L$ as $x=g^{\prime }(\alpha {)}^{-1}{\sum }_j{b}_j{\alpha }^j$ with $b_j\in K$. Membership in the codifferent says $A(b_j{)}_j\in {\mathcal{O}}_K^d$. Since $A$ is invertible over ${\mathcal{O}}_K$, this forces every $b_j$ integral. ThereforeSeparability ensures that $g^{\prime }(\alpha )\ne 0$; the trace pairing calculation also proves that no extra factor has been overlooked.

The polynomial is an Eisenstein polynomial at $5$, so it is irreducible and $[K:\mathbb{Q}]=3$. Its polynomial discriminant isOne must remove the index at $2$ before using this as the field discriminant. Put $\beta =({\alpha }^2+\alpha )/2$. Direct reduction using ${\alpha }^3=5\alpha -20$ givesThus $B=\mathbb{Z}+\mathbb{Z}\alpha +\mathbb{Z}\beta$ is a ring finite over $\mathbb{Z}$, so its elements are integral. Its basis has discriminant of elements of a number field $-2575=-5^2\cdot 103$. The discriminant-index formula for an integral lattice says $\mathrm{disc}(B)=[{\mathcal{O}}_K:B{]}^2{\Delta }_K$, so only $5$ could divide the remaining index.

We use these standard different ideal facts: the norm of the different is the absolute field discriminant; its localization is the local different; an unramified prime has different exponent zero, and a tamely ramified prime of ramification index $e$ has different exponent $e-1$. The Eisenstein polynomial at $5$ gives total ramification of degree three, which is tame because $5\nmid 3$. Its contribution to $v_5({\Delta }_K)$ is therefore two. This excludes a further index factor $5$, proving ${\mathcal{O}}_K=B$ and $\boxed{{\Delta }_K=-2575}$.

Since the index $[{\mathcal{O}}_K:\mathbb{Z}[\alpha ]]=2$ is prime to $5$ and $103$, Dedekind factorization theorem applies at both primes. Modulo $5$, $f=X^3$, giving a single prime $\mathfrak{p}=(5,\alpha )$ with residue degree one and ramification index three. Modulo $103$,Thus $\mathfrak{q}=(103,\alpha -6)$ has residue degree one and ramification index two; the other prime over $103$ is unramified. Both ramified primes are tame. No other prime ramifies because none divides ${\Delta }_K$. Consequently

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.

For this non-Archimedean local field, the Schwartz-Bruhat space $\mathcal{S}(F)$ consists of locally constant, compactly supported complex-valued functions. Choose a nontrivial continuous additive character $\psi :F\to {\mathbb{C}}^{\times }$ of modulus one and an additive Haar measure $dx$. We use the plus-sign convention for the Fourier transform over a local field:Compact support makes the integral absolutely convergent. The additive character and the scale of the Haar measure are part of the definition; without them there is no canonical numerical transform.

Let $q=|{\mathcal{O}}_F/{\pi }_F{\mathcal{O}}_F|$, and write $V=\mathrm{vol}({\mathcal{O}}_F)$. Define the integer $c$ by the annihilator of the valuation ring ${\mathcal{O}}_F^{\perp }={\pi }_F^c{\mathcal{O}}_F$. Equivalently, $\psi$ is trivial on ${\pi }_F^c{\mathcal{O}}_F$ and not on ${\pi }_F^{c-1}{\mathcal{O}}_F$. Such a conductor exists: continuity puts the image of some additive ball inside an arc containing no nontrivial circle subgroup, so the character is trivial on that ball; nontriviality bounds the possible ball exponents below. Then $H_n={\pi }_F^n{\mathcal{O}}_F$ has volume $V{q}^{-n}$ and character annihilator $H_n^{\perp }={\pi }_F^{c-n}{\mathcal{O}}_F$.

Translation givesThe integral is the volume if $y\in H_n^{\perp }$. Otherwise translate it by $t_0\in H_n$ with $\psi (t_{0}y)\ne 1$; Haar measure invariance multiplies the same integral by a nonidentity scalar, so it must vanish. ThusEvery Schwartz-Bruhat function is a finite linear combination of such coset indicators: compactness of its support supplies a common sufficiently small translation subgroup on which it is constant, and finitely many of its cosets cover the support. The transform of each indicator has compact support and is locally constant, since $\psi$ has open kernel. Consequently $\hat{f}\in \mathcal{S}(F)$ for every $f\in \mathcal{S}(F)$.

Apply the transform again to an indicator. The same character-orthogonality calculation, together with $(H_n^{\perp }{)}^{\perp }=H_n$, givesChoose the self-dual Haar measure, characterized here by $V=q^{c/2}$. By linearity the desired Fourier inversion isIn particular, one may rescale any nontrivial additive character to make $c=0$, and then take $V=1$. For a concrete construction, start with ${\psi }_0(x)={\psi }_{{\mathbb{Q}}_p}({\mathrm{Tr}}_{F/{\mathbb{Q}}_p}x)$, where the standard rational additive character has kernel ${\mathbb{Z}}_p$. Its annihilator is the inverse different ${\mathfrak{D}}_{F/{\mathbb{Q}}_p}^{-1}$. If that ideal is ${\pi }_F^{-d}{\mathcal{O}}_F$, the rescaled character $\psi (x)={\psi }_0({\pi }_F^{-d}x)$ has ${\mathcal{O}}_F^{\perp }={\mathcal{O}}_F$. This supplies the stated normalization and also explains how the different ideal enters local Fourier analysis.

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.

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.

Normalize the discrete valuation by $v_L({\pi }_L)=1$, and put $v_L(0)=+\infty$. For $\sigma \ne 1$, define its ramification numberand set $i_G(1)=+\infty$. The minimum exists: all finite values are nonnegative integers, and a nonidentity automorphism moves some integral element. The lower ramification numbering isThus $G_s=G_{\lceil s\rceil }$, $G_0$ is the inertia group, and $G_1$ is the wild inertia group. This states the convention at integer endpoints as well as between them. Automorphisms preserve $v_L$, so these are normal subgroups of $G$.

Define the Herbrand function and its inverse byThe integrand is positive and piecewise constant. Since $G_t=1$ for sufficiently large $t$, the function is continuous, strictly increasing and unbounded, and its inverse is defined on all nonnegative reals. The upper ramification numbering isIn particular, the denominator in the Herbrand integral involves $G_0$, even when $L/K$ is not totally ramified. These conventions are also given in Milne's Class Field Theory, Chapter I, §4.

Now use the monogenic hypothesis ${\mathcal{O}}_L={\mathcal{O}}_K[\alpha ]$. Since $L/K$ is Galois, its minimal polynomial has the distinct conjugates $\sigma (\alpha )$ as roots, soTaking valuations proves the first equality.

For every polynomial $P\in {\mathcal{O}}_K[T]$, the difference $P(\sigma (\alpha ))-P(\alpha )$ is divisible in ${\mathcal{O}}_L$ by $\sigma (\alpha )-\alpha$. Every element of ${\mathcal{O}}_L$ is such a polynomial value, and taking the element $\alpha$ attains the resulting lower bound. HenceFor an integer $i_G(\sigma )=m\ge 0$, the element $\sigma$ lies in exactly the groups $G_0,\dots ,G_{m-1}$; if $m=0$, it lies in none. Summing this count over the nonidentity elements and interchanging the two finite sums givesThis proves the ramification-group sum for a monogenic integer ring. Equivalently it is the different exponent from ramification groups, since the different ideal is generated by $f^{\prime }(\alpha )$ under the same monogenic hypothesis.

Write $n=[L:K]$. The field trace gives a nondegenerate $K$-bilinear trace pairingThe inverse different, or codifferent, is the trace-dual latticeIt contains $S$, because the trace of an element integral over $R$ belongs to the integrally closed ring $R$. It is stable under multiplication by $S$: if $s,t\in S$ and $x\in {\mathfrak{D}}_{S/R}^{-1}$, then $\mathrm{Tr}(sxt)=\mathrm{Tr}(x(st))\in R$. Nondegeneracy of the trace pairing and finite generation of $S$ show that this trace dual is a finitely generated $R$-module spanning $L$. Therefore it is a fractional ideal of $S$.

Its inverseis the different ideal. Since $S\subseteq {\mathfrak{D}}_{S/R}^{-1}$, every such $x$ lies in $S$; hence ${\mathfrak{D}}_{S/R}$ is an integral ideal of $S$.

The discriminant ideal $\mathrm{disc}(S/R)$ is locally generated bywhere ${\alpha }_1,\dots ,{\alpha }_n$ is a local $R$-basis of $S$. Equivalently, it is the image of the determinant of the trace pairingThis formulation makes the definition independent of a basis, since changing a basis multiplies its discriminant by the square of the determinant of the change-of-basis matrix.

The asserted identity of ideals can be checked after localization at every nonzero prime ideal of $R$. We may therefore assume that $R$ is a discrete valuation ring and choose a basis ${\alpha }_1,\dots ,{\alpha }_n$ of $S$. Let ${\beta }_1,\dots ,{\beta }_n$ be its trace-dual basis, so $\mathrm{Tr}({\alpha }_i{\beta }_j)={\delta }_{ij}$; this is a basis of ${\mathfrak{D}}_{S/R}^{-1}$. If $T=(\mathrm{Tr}({\alpha }_i{\alpha }_j))$, thenThus the determinant measuring the inclusion $S\subseteq {\mathfrak{D}}_{S/R}^{-1}$ is $\det T$. The determinant description of the norm of a fractional ideal consequently givesLocalization then proves the equality over the original Dedekind domain.

Finally, the determinant-of-pairing map identifies the invertible $R$-module $({\det }_{R}S{)}^{\otimes 2}$ with the discriminant ideal. Hence in the ideal class groupup to the harmless inverse caused by the convention used to identify invertible modules with fractional ideals. In either convention the class is a square.