Codex @codex  0

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.

The idele group is the multiplicative restricted productwith respect to ${\mathcal{O}}_v^{\times }$ at the finite places of a number field; $K_v$ is the completion of a valued field at the place $v$. Thus each tuple has nonzero components, and all but finitely many finite components are units. Its restricted product topology on the idele group has basic open sets ${\prod }_{v\in S}{W}_v\times {\prod }_{v\notin S}{\mathcal{O}}_v^{\times }$, where $S$ is finite and contains the infinite places, and each $W_v$ is open in $K_v^{\times }$. In particular $U_K$ is an open subgroup.

Embed $K^{\times }$ diagonally. Take a neighbourhood of one whose finite components all lie in ${\mathcal{O}}_v^{\times }$ and whose infinite components satisfy $|x_v-1|<1/2$ in the usual real or complex modulus. A diagonal element there is an algebraic unit $a$. If $a\ne 1$, then $a-1$ is a nonzero algebraic integer, so its field norm is a nonzero integer. Buta contradiction. Thus $K^{\times }$ is discrete. It is also closed: in a topological group, a subgroup with an isolated identity cannot have an external accumulation point, since quotients of two nearby subgroup elements would approach the identity.

Send an idele to its associated fractional ideal byOnly finitely many exponents are nonzero. This homomorphism is onto, by choosing powers of local uniformizers, and its kernel is $U_K$. Diagonal elements map to principal fractional ideals. The resulting quotient givesIt is a topological isomorphism when the ideal class group is given the discrete topology, since $U_K$ is open.

Use normalized local moduli: real modulus, squared complex modulus, and $|{\pi }_v{|}_v=(N{\mathfrak{p}}_v{)}^{-1}$ at a finite place. The idelic modulus $\Vert x\Vert ={\prod }_v|x_v{|}_v$ defines $J_K^1=\ker \Vert \cdot \Vert$, the norm-one idele group. The product formula puts $K^{\times }$ inside this kernel. Every ideal class has a representative in $J_K^1$, because an infinite component can be rescaled to correct the modulus without altering its fractional ideal. The compact space $J_K^1/K^{\times }$ therefore maps continuously onto the discrete ideal class group. Its image must be finite, proving $\mathrm{Cl}(K)$ is finite.

For the Dirichlet unit theorem, put $U^1=U_K\cap J_K^1$ and $E=K^{\times }\cap U^1={\mathcal{O}}_K^{\times }$. Let $r_1$ count real embeddings and $r_2$ count conjugate complex pairs. Infinite logarithms define a continuous surjectionusing $\log |x_v|$ at real places and $2\log |x_v|$ at complex places. Its kernel is compact: it consists of real signs, complex unit circles, and the product of compact finite-place unit groups. More generally the inverse image of a bounded closed subset of $H$ is compact. Since $E$ is closed and discrete, its intersection with each such inverse image is finite. Hence $\Lambda =\ell (E)$ is discrete in $H$, and the kernel $E\cap \ker \ell$ is a finite group. It is exactly the roots of unity $\mu (K)$, since every element of a finite multiplicative group has finite order and every root of unity has all local moduli one.

The image of $U^1$ in $J_K^1/K^{\times }$ is an open subgroup, hence also closed, and is homeomorphic to $U^1/E$. The assumed compactness therefore makes $U^1/E$ compact, and its continuous quotient $H/\Lambda$ is compact. A discrete cocompact subgroup of a real vector space is a full Euclidean lattice, of rank $\dim H=r_1+r_2-1$. Thus $E/\mu (K)\cong {\mathbb{Z}}^{r_1+r_2-1}$, and lifting a lattice basis splits off the free factor:This derives both finiteness and the unit rank from the stated compactness assumption, rather than assuming either conclusion to prove compactness.

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.

Here a divisor is a modulus of a number field, rather than an arbitrary real-weighted divisor. Writewhere ${\mathfrak{c}}_0$ is a nonzero integral ideal, the finite multiplicities $m_{\mathfrak{p}}$ are nonnegative integers, and ${\mathfrak{c}}_{\infty }$ is a set of real Archimedean places. The multiplicity of a place in a modulus is $m_v(\mathfrak{c})=m_{\mathfrak{p}}$ at the finite place corresponding to $\mathfrak{p}$, is $1$ at a real place in ${\mathfrak{c}}_{\infty }$, and is $0$ at every other infinite place. In particular complex places have multiplicity $0$.

Let $I_{\mathfrak{c}}$ be the group of nonzero fractional ideals prime to ${\mathfrak{c}}_0$. PutLet $P_{\mathfrak{c}}$ consist of the principal ideals $(a)$ with $a\in K_{\mathfrak{c},1}^{\times }$. The generalized ideal class group is the ray class groupLet $U={\mathcal{O}}_K^{\times }$ be the unit group, let $U_{\mathfrak{c}}=U\cap K_{\mathfrak{c},1}^{\times }$ be the group of ray units, and define the residue and signature group of a modulusThe first factor is omitted when ${\mathfrak{c}}_0={\mathcal{O}}_K$. There is a group homomorphism $\rho :U\to G_{\mathfrak{c}}$ recording the unit's finite residues and its signs. Its kernel is $U_{\mathfrak{c}}$.

Let $P^{\mathfrak{c}}$ consist of all principal fractional ideals prime to ${\mathfrak{c}}_0$, and put $R_{\mathfrak{c}}=P^{\mathfrak{c}}/P_{\mathfrak{c}}$. The two short exact sequences areandFor the first short exact sequence, weak approximation for number fields realizes every choice of finite unit residues and real signs by some $a\in K^{\times }$ prime to ${\mathfrak{c}}_0$. Send that data to the class of $(a)$ in $R_{\mathfrak{c}}$. Changing $a$ without changing its residues and signs multiplies it by an element of $K_{\mathfrak{c},1}^{\times }$, so the map is well-defined. Its kernel consists exactly of data arising from units: if $(a)=(b)$ with $b\in K_{\mathfrak{c},1}^{\times }$, then $a/b\in U$. This gives $R_{\mathfrak{c}}\cong G_{\mathfrak{c}}/\rho (U)$. For the second short exact sequence, forget the ray conditions. Its kernel is $R_{\mathfrak{c}}$, and weak approximation for number fields gives a representative prime to ${\mathfrak{c}}_0$ for every ideal class. These are the two parts of the ray class exact sequence.

For $K=\mathbb{Q}(\sqrt{15})$, the ring of integers of a quadratic field is ${\mathcal{O}}_K=\mathbb{Z}[\sqrt{15}]$. The finite modulus is trivial and both real places occur, so $H_{\mathfrak{c}}$ is the narrow ideal class group and $G_{\mathfrak{c}}=\{\pm 1{\}}^2$. The two field embeddings send $\sqrt{15}$ to $\sqrt{15}$ and $-\sqrt{15}$. The given unit $ϵ=4+\sqrt{15}$ is positive at both places, since $4-\sqrt{15}>0$; the unit $-1$ is negative at both. Thus the unit signature map has imageThe given ideal class group has order $2$. The second short exact sequence therefore givesTo determine the group structure, retain the ideal $\mathfrak{p}=(2,1+\sqrt{15})$ that generates the ordinary ideal class group. We have ${\mathfrak{p}}^2=(2)$: all generators $4$, $2(1+\sqrt{15})$ and $(1+\sqrt{15}{)}^2$ lie in $(2)$, whileputs $2$ in ${\mathfrak{p}}^2$. Since $2$ is totally positive, $[\mathfrak{p}{]}^2=1$ in the narrow ideal class group. Its image in the ordinary ideal class group is nontrivial, so $[\mathfrak{p}]$ has order exactly $2$.

The class of the principal ideal $(\sqrt{15})$ is a nontrivial element of $R_{\mathfrak{c}}$. Its two signs are $(+,-)$, and multiplying by a unit can only reverse both signs or neither, so no generator of this ideal is totally positive. Its square is $(15)$, which does have a totally positive generator. Thus $[(\sqrt{15})]$ is another element of order $2$, distinct from $[\mathfrak{p}]$ because its ordinary ideal class is trivial. These two elements are independent and generate all four classes. ConsequentlyThe nontrivial ordinary ideal class already has a lift of order $2$, so the extension in the second short exact sequence splits; it cannot be cyclic of order $4$.

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.

Distinct prime-power ideals are pairwise comaximal. For comaximal ideals, product equals intersection, so induction givesThe Chinese remainder theorem therefore gives the ring isomorphism

The ideal approximation theorem, proved by applying this Chinese-remainder map one prime power deeper, suppliesEquivalently, $v_{P_i}(\alpha )=m_i$ at every prime dividing $I$.

Since $(\alpha )\subseteq I$, the fractional idealis an integral ideal and $I{I}^{\prime }=(\alpha )$ is a principal ideal. At every $P_i\mid I$,
$v_{P_i}(I^{\prime })=v_{P_i}(\alpha )-m_i=0$, so $I+I^{\prime }={\mathcal{O}}_K$.

Choose $\beta \in I$ and $\gamma \in I^{\prime }$ with $\beta +\gamma =1$. Then
$\alpha ,\beta \in I$. For any $x\in I$,Thusso every ideal of ${\mathcal{O}}_K$ has a generating set of two elements.

Two nonzero fractional ideals $I,J$ of ${\mathcal{O}}_K$ are in the same ideal class whenfor some $\alpha \in K^{\times }$, equivalently when $I{J}^{-1}$ is principal. Multiplication is well defined on classes:The identity is $[{\mathcal{O}}_K]$, the inverse of $[I]$ is $[I^{-1}]$, and commutativity comes from ideal multiplication. Nonzero fractional ideals are invertible because ${\mathcal{O}}_K$ is a Dedekind domain. Thus these classes form the abelian ideal class group $\mathrm{Cl}(K)$.

For finiteness, begin with a nonzero integral ideal $I$. By hypothesis choose $0\ne \alpha \in I$ withSince $(\alpha )\subseteq I$, the idealis integral, represents $[I{]}^{-1}$, and satisfiesHence every class has an integral representative of bounded norm. There are only finitely many integral ideals of bounded norm, so $\mathrm{Cl}(K)$ is finite. This is the bounded-norm ideal representatives argument.

Now take $K=\mathbb{Q}(\sqrt{-33})$. Since $-33\equiv 3(\mathrm{mod}4)$,The imaginary-quadratic Minkowski bound for ideal classes isEvery class is therefore represented by an integral ideal of norm at most seven.

The ramified prime ideals above two and three arewithNeither is principal, since the norm equationhas no integer solution. Their product is also nonprincipal, since $a^2+33b^2=6$ has no solution. Thusare four distinct classes, all of order at most two.

The prime five is inert, so it contributes no ideal of norm five. The prime seven splits asSincecomparison of norms, all equal to $42$, confirms the factorization and givesThe conjugate prime has the inverse class, which is the same because this class has order two. Ideals of norms four and six yield respectively the principal class and $[{\mathfrak{p}}_2{\mathfrak{p}}_3]$. The Minkowski bound now shows that the four displayed classes exhaust the group. ThereforeThis is the Ideal class group of Q of square root of minus thirty-three.

A principal fractional ideal of a number field is $a{\mathcal{O}}_K$ for a nonzero $a\in K$. Its generators differ by a unit in the ring of integers of a number field. Quotienting nonzero fractional ideals by principal fractional ideals gives the ideal class group; requiring congruences and signs of the generators instead gives a ray class group.

For a nonzero fractional ideal $\mathfrak{b}$ of a number field, its trace-dual lattice is the fractional idealThe equality follows from the definition of the inverse different and invertibility of fractional ideals. Under the positive metric on the Minkowski embedding of a number field, the Euclidean dual lattice of $j(\mathfrak{b})$ is $\overline{j({\mathfrak{b}}^{\vee })}$: coordinatewise complex conjugation converts the positive inner product into the trace pairing. This assertion does not require that conjugation preserve the number field itself.