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For the integral closure $S$ of a Dedekind domain $R$ in a finite separable extension, the discriminant ideal is locally generated bywhere $({\alpha }_i)$ is a local $R$-basis of $S$. Intrinsically it is the image of $({\det }_{R}S{)}^{\otimes 2}$ under the determinant of the trace pairing.

A ramified prime makes the finite algebra ${\mathcal{O}}_K/p{\mathcal{O}}_K$ have a nonzero nilpotent element. Multiplication by its product with any other element is nilpotent and has trace zero, so the reduced trace pairing is degenerate. Therefore the field discriminant is divisible by $p$. This proves finiteness of the ramified rational primes without requiring explicit prime factorization in the field.

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.

The original PDF has cube roots throughout this question; the ninth roots in the converted TeX are transcription errors. Put $\alpha =\sqrt[3]{m_1{m}_2^2}$ and $\beta ={\alpha }^2/m_2$. The given condition excludes the trivial cube case. Both generators are integral, becauseMoreoverso $A=\mathbb{Z}+\mathbb{Z}\alpha +\mathbb{Z}\beta$ is an order in the cubic field. The field traces of $\alpha ,\beta ,{\alpha }^2,{\beta }^2$ are zero, while $\mathrm{Tr}(\alpha \beta )=3m_1{m}_2$. Thus the trace pairing matrix of this basis isWe must prove that this is the maximal order, not just compute the discriminant of a suborder. The discriminant-index formula for an integral lattice restricts any index prime to those dividing $3m_1{m}_2$. If $p\mid m_1$, the cubic for $\alpha$ is Eisenstein at $p$, while $m_2$ is a unit there and $A\otimes {\mathbb{Z}}_p={\mathbb{Z}}_p[\alpha ]$. If $p\mid m_2$, use instead the Eisenstein polynomial for $\beta$, and $\alpha ={\beta }^2/m_1$. In either case an Eisenstein generator is a uniformizer of the totally ramified local field and generates its full valuation ring. Indeed expansion in powers of that uniformizer using base residue representatives, and grouping powers modulo three, expresses every integral local element in its power basis.

These arguments also cover three when $3\mid m_1{m}_2$. If $3\nmid m_1{m}_2$, every unit modulo nine has cube congruent to $1$ or $-1$. Therefore the condition $m_1\not\equiv \pm m_2(\mathrm{mod}9)$ implies $m_1{m}_2^2\not\equiv \pm 1(\mathrm{mod}9)$: multiply a contrary congruence by the inverse of $m_2^2$ and use $m_2^3\equiv \pm 1$. Choose $r\in \{1,-1\}$ with $r^3\equiv m_1{m}_2^2(\mathrm{mod}3)$. The polynomialis Eisenstein at three, since its constant term is divisible by three but not nine. Its root $\alpha -r$ again generates the full local valuation ring. There is no index prime left. We have proved the integral basis of a cubefree pure cubic field and the requested discriminant:For $K=\mathbb{Q}(\sqrt[3]{6})$ write ${\alpha }^3=6$. Here ${\mathcal{O}}_K=\mathbb{Z}[\alpha ]$ and $d_K=-972$. The norm of $x+y\alpha +z{\alpha }^2$ is the determinant of its multiplication matrix, givingIn particularAn integral element of norm one has an integral inverse, by its monic characteristic polynomial. Here direct multiplication gives $u^{-1}=109+60\alpha +33{\alpha }^2$. Nontriviality also follows from linear independence of $1,\alpha ,{\alpha }^2$.

The signature is $(1,1)$, so the Minkowski bound for ideal classes isEvery class therefore has an integral representative of norm at most eight. Its prime factors have norms at most eight, and lie above $2,3,5,7$. Since the index of $\mathbb{Z}[\alpha ]$ is one, reduction of $X^3-6$ gives their factorizations by the Kummer-Dedekind theorem. At two and three the polynomial is $X^3$, so there is one prime of norm $2$ or $3$, respectively. They are principal, becauseAt five the factorization is $(X-1)(X^2+X+1)$ with the quadratic irreducible. The norm-five prime is $(\alpha -1)$ because $N(\alpha -1)=5$; the norm-twenty-five prime is outside the bound.

At seven the roots are $3,5,6$, giving three primes ${\mathfrak{q}}_r=(7,\alpha -r)$ of norm seven. The elements $\alpha +1$ and ${\alpha }^2+\alpha -5$ have norm seven and vanish at roots six and three, respectively; hence they generate ${\mathfrak{q}}_6$ and ${\mathfrak{q}}_3$. Since ${\mathfrak{q}}_3{\mathfrak{q}}_5{\mathfrak{q}}_6=(7)$, the remaining prime is principal too. Every prime which can divide a Minkowski representative is now principal, soThis proves class number one for Q of cube root six with all primes within the bound accounted for.

There is an error in the printed request: the cubic is not irreducible at $n=2$, and three elements cannot be a field basis in that case. The integral-basis assertion is valid under the necessary hypothesis $[\mathbb{Q}(\alpha ):\mathbb{Q}]=3$.

Under that hypothesis, $\alpha$ is an algebraic integer and $\mathbb{Z}[\alpha ]\subseteq {\mathcal{O}}_k$ has finite index $m$. Change of basis in the trace pairing gives the discriminant-index identityThe field discriminant is an integer. If the left side is square-free, $m^2$ cannot divide it unless $m=1$. Thus $1,\alpha ,{\alpha }^2$ is an integral basis when the cubic is irreducible and its discriminant is square-free.

For any root, ${\alpha }^3=n\alpha +1$ givesSince $\alpha +\beta +\gamma =0$, $\alpha \beta +\beta \gamma +\gamma \alpha =-n$ and $\alpha \beta \gamma =1$,On the other hand, the product of $r{f}^{\prime }(r)$ is $-\Delta$: the product of the roots is one, and each of the three unordered root pairs contributes a negative squared difference. Therefore, without needing irreducibility for this polynomial calculation,The three-root product is the field norm only when the cubic is the minimal polynomial.

At $n=1$, the polynomial has no rational root (the only candidates are $\pm 1$), so it is irreducible, and its discriminant is $-23$. Hence $1,\alpha ,{\alpha }^2$ is an integral basis.

At $n=2$,If $\alpha =-1$, then $k=\mathbb{Q}$ with integral basis $\{1\}$. If $\alpha =(1\pm \sqrt{5})/2$, then $k=\mathbb{Q}(\sqrt{5})$ and $\{1,\alpha \}$ is an integral basis, since this quadratic order has square-free discriminant five; also ${\alpha }^2=\alpha +1$. In neither case is the three-element list a basis. Its discriminant five is instead that of the degree-three algebra $\mathbb{Q}[x]/(x^3-2x-1)$; a basis there does not repair the printed assertion about the field $k$.

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.

Let $L/K$ be a finite Galois extension of non-Archimedean local fields, with Galois group $G$, normalized valuation $v_L$, integer ring ${\mathcal{O}}_L$, and residue characteristic $p$. The lower ramification numbering isThese higher ramification groups measure successively finer agreement with the identity. The group $G_0$ is the inertia group, the kernel of the action on the residue field. Each $G_i$ is normal in $G$: conjugation permutes integral elements and preserves their valuations. The groups eventually become trivial, since each nonidentity automorphism moves some integral element and its displacement has finite valuation.

Let $K_0=L^{G_0}$, the maximal unramified subextension of a local field extension. Then $G/G_0$ is the Galois group of the residue extension, cyclic and generated by a Frobenius automorphism. For a uniformizer $\pi$ of $L$, the uniformizer criterion for lower ramification groups givesTo see why one uniformizer suffices, the extension $L/K_0$ is totally ramified and its integer ring is ${\mathcal{O}}_{K_0}[\pi ]$. Elements of $G_0$ fix its coefficients. Differences of powers of $\pi$ are divisible by $\sigma \pi -\pi$, so the valuation bound for $\pi$ implies it for every integral element. The converse follows by testing $\pi$ itself.

The tame inertia quotient is seen explicitly fromIt is a homomorphism because inertia acts trivially on the residue field, and its kernel is $G_1$. Thus $G_0/G_1$ is cyclic of order not divisible by $p$. For $i\ge 1$, the homomorphismhas kernel $G_{i+1}$. Indeed composing two automorphisms adds the leading displacements, since $\sigma \pi /\pi \equiv 1$ and inertia fixes residue coefficients. Hence each $G_i/G_{i+1}$ is an elementary abelian p-group. Eventual triviality makes $G_1$ a p-group; the prime-to-p quotient makes it the unique Sylow subgroup of $G_0$. This is the wild inertia group. Tame ramification means $G_1=1$.

The filtration also measures the different exponent. For the totally ramified extension over $K_0$, the minimal polynomial of an algebraic element $g$ of $\pi$ has polynomial roots $\sigma \pi$, $\sigma \in G_0$. The derivative formula givesFor a fixed $\sigma$, its displacement valuation is the number of indices $i\ge 0$ for which it belongs to $G_i$. Interchanging the finite sums therefore gives different exponent from ramification groups:The unramified part contributes zero. More generally different in a tower follows by taking products of bases of a module dual for the trace pairing, and givesThus a tame extension has different exponent $e-1$, while nontrivial wild groups contribute additional positive terms.

The lower numbering restricts to subgroups: if $H=\mathrm{Gal}(L/E)$, its groups are $H_i=H\cap G_i$, directly from the definition using the same top-field valuation. Quotients need a different indexing. For real $s\ge 0$ put $G_s=G_{\lceil s\rceil }$ and define the Herbrand functionThis upper ramification numbering stretches or compresses intervals according to the inertia indices. The quotient theorem says that for a normal subgroup $H$, upper ramification groups commute with quotients:In particular the same upper index has a meaning compatible with passing to a smaller Galois extension, which lower indices generally lack.

For a tame totally ramified extension the picture is simple: $G_0=G$, all $G_i$ with $i\ge 1$ are trivial, and the different exponent is $|G|-1$. The extension ${\mathbb{Q}}_p({\zeta }_p)$ for odd $p$ illustrates this, giving exponent $p-2$.

A mixed-characteristic wild family is the lower ramification filtration of a prime-power cyclotomic extension. Take $L={\mathbb{Q}}_p({\zeta }_{p^r})$ with odd $p$ and $\pi ={\zeta }_{p^r}-1$. If ${\sigma }_a(\zeta )={\zeta }^a$ and $a\ne 1$ modulo $p^r$, thenThe last equality follows because ${\zeta }^{a-1}$ has order $p^{r-v_p(a-1)}$ and its uniformizer valuation scales by the ramification index in that cyclotomic tower. The lower breaks are $0,p-1,p^2-1,\dots ,p^{r-1}-1$. Integrating the successive indices $(p-1),p(p-1),\dots$ gives upper breaks $0,1,\dots ,r-1$.

In equal characteristic there is an equally explicit wild example. Start with $K={\mathbb{F}}_q((t))$ and an Artin–Schreier extensionThere is no polynomial root in $K$: a negative-valuation element has $v(x^p-x)=pv(x)$, while the right side has valuation $-m$. Adjoining a polynomial root splits the polynomial, whose polynomial roots are $y+c$ for $c\in {\mathbb{F}}_p$, so the extension is cyclic of degree $p$. The valuation equation forces total ramification, $v_L(t)=p$ and $v_L(y)=-m$. Choose $1\le b\le p-1$ with $-bm\equiv 1(\mathrm{mod}p)$ and set $a=(1+bm)/p$. Then $\pi =t^a{y}^b$ is a uniformizer. For a nonidentity automorphism $y\mapsto y+c$,the first binomial coefficient $b$ is nonzero modulo $p$, and all subsequent terms have strictly higher valuation. ThereforeThis is the ramification break of an Artin–Schreier pole. It shows concretely that wild breaks, unlike tame inertia, can occur arbitrarily far out in the filtration.

Put $a=\sqrt[3]{2}$ and $\zeta ={\zeta }_3$. The polynomial $X^3-2$ is irreducible over $\mathbb{Q}$ by the Eisenstein criterion at $2$, so $[\mathbb{Q}(a):\mathbb{Q}]=3$. This field is real and cannot contain the nonreal $\zeta$. Adjoining $\zeta$ gives a quadratic extension and hence $[F:\mathbb{Q}]=6$. Since $F$ is the splitting field of $X^3-2$, its Galois group acts faithfully on $a,\zeta a,{\zeta }^{2}a$ and has order six. Consequently

For a prime $\mathfrak{q}$ over a rational prime $p$, its absolute residue degree is the order of the cyclic quotient group $D_{\mathfrak{q}}/I_{\mathfrak{q}}$ from part (i). If that degree were six, then $D_{\mathfrak{q}}$, a subgroup of $S_3$, would have order at least six, so $D_{\mathfrak{q}}=S_3$ and $I_{\mathfrak{q}}=1$. This would make $S_3$ cyclic, a contradiction. The argument applies also at ramified primes. Thus

For the second request printed with the same subpart label, work over $E=\mathbb{Q}(\zeta )$. The Galois group of $F/E$ is cyclic of order three, with generator $\sigma (a)=\zeta a$ and $\sigma (\zeta )=\zeta$. The ideal norm of $(\zeta +3)$ is seven, sinceIn its residue field we have $\zeta \equiv -3\equiv 4(\mathrm{mod}{\mathfrak{p}}_1)$. The polynomial $X^3-2$ has no root in ${\mathbb{F}}_7$, because its cube residues are $0,1,6$. It is therefore irreducible there. Its discriminant $-108$ is a unit at ${\mathfrak{p}}_1$. To check the local index condition, the trace pairing matrix of $1,a,a^2$ has this unit determinant. For any integral element $b$ of $F$, its traces against these three basis elements are integral; solving with this invertible matrix shows that its basis coefficients are integral locally. Thus the relative ring of integers equals ${\mathcal{O}}_E[a]$ at ${\mathfrak{p}}_1$. The relative Kummer-Dedekind theorem gives one unramified prime above ${\mathfrak{p}}_1$, of residue degree three.

Use arithmetic Frobenius, characterized by $b\mapsto b^7$ on this residue field. As $a^3=2$,The three possible images $a,\zeta a,{\zeta }^{2}a$ remain distinct modulo $\mathfrak{q}$, since $a$ is nonzero and $1,4,2$ are distinct modulo seven. This congruence therefore identifies the actual automorphism uniquely:The geometric Frobenius convention would give its inverse. Since $F/E$ is abelian, the arithmetic element is independent of the choice of prime above ${\mathfrak{p}}_1$.

Set $K=\mathbb{Q}(\sqrt{pq})$ and $E=\mathbb{Q}(\sqrt{p},\sqrt{q})$. The distinct primes give independent square classes, so $[E:\mathbb{Q}]=4$ and $[E:K]=2$. In particular $K(\sqrt{p})=K(\sqrt{q})=E$: these are the same quadratic extension of $K$. We prove it is unramified by an explicit field discriminant calculation.

Since $p,q\equiv 1(\mathrm{mod}4)$, put $a=(1+\sqrt{p})/2$, $b=(1+\sqrt{q})/2$. Both are algebraic integers. The ring $R=\mathbb{Z}[a,b]\subseteq {\mathcal{O}}_E$ has basis $1,a,b,ab$. Up to the ordering of the basis, its trace pairing matrix is the tensor product of the quadratic trace matrices, because the four embeddings of $E$ choose the two radical signs independently. The two quadratic matrices areThusOn the other hand, $pq$ is squarefree and congruent to one modulo four, so $d_K=pq$. Let $j=[{\mathcal{O}}_E:R]$. The discriminant-index formula for an integral lattice and the relative discriminant tower formula giveThe ideal norm on the right is a positive integer, while $j$ is a positive integer. Hence $N({\mathfrak{d}}_{E/K})=1/j^2$ forces $j=1$ and ${\mathfrak{d}}_{E/K}={\mathcal{O}}_K$. Its prime divisors are exactly the ramified finite primes, so $E/K$ is unramified at every finite prime, including primes above $2,p,q$. This also proves $d_E=p^2{q}^2$, agreeing with the discriminant of a biquadratic field formula.

Every embedding of $E$ is real, so the real places of $K$ split as well. The extension is quadratic, hence abelian. The ordinary Hilbert class field characterization therefore impliesFinally, the Artin reciprocity map identifies $\mathrm{Cl}(K)$ with $\mathrm{Gal}(H_K/K)$. Restriction onto $\mathrm{Gal}(E/K)\simeq C_2$ is surjective, so the ideal class group has a quotient of order two. Therefore $\boxed{|\mathrm{Cl}(K)|\text{ is even}}$.

A place of a number field is an equivalence class of nontrivial absolute values on a field. Its finite places correspond to nonzero prime ideals of the ring of integers of a number field; its infinite places come from real embeddings and conjugate pairs of complex embeddings.

For a embedding of a number field into a p-adic algebraic closure $\iota :K\hookrightarrow {\overline{\mathbb{Q}}}_p$, pull back the p-adic absolute value. This gives a place above $p$. Every element of $\mathrm{Gal}({\overline{\mathbb{Q}}}_p/{\mathbb{Q}}_p)$ preserves that absolute value, by uniqueness of its extension to each finite local extension, so equivalent embeddings give the same place.

Conversely, a place $v$ above $p$ gives a completion of a number field at a prime ideal $K_v$, a finite extension of ${\mathbb{Q}}_p$. Embed it into ${\overline{\mathbb{Q}}}_p$ and restrict to $K$. If two embeddings give the same place, they extend to two ${\mathbb{Q}}_p$-embeddings of this same completion. They are conjugate under the absolute Galois group: their finite separable images lie in a finite normal closure, and an isomorphism of such images extends to an automorphism of the algebraic closure. The completion really is the closure of the embedded $K$, since rational coefficients are dense in the ${\mathbb{Q}}_p$-span of a primitive element. Thus we obtain the embedding orbit description of finite places:

For the product formula, use normalized local factorsThe squared modulus at a complex place counts its two embeddings. Equivalently one can use ordinary complex modulus and put exponent two in the product. These normalized local factors for the product formula must be specified; arbitrary representatives of place classes would not satisfy the unweighted printed formula.

The fractional principal ideal $(x)$ has only finitely many nonzero prime exponents. Its ideal norm is $|N_{K/\mathbb{Q}}(x)|$: for integral $x$, multiplication by $x$ on an integral basis has this determinant in absolute value, equal to the index of $(x)$ in ${\mathcal{O}}_K$; a quotient gives the fractional case. Hence the finite-place product is $|N_{K/\mathbb{Q}}(x){|}^{-1}$. The infinite-place product is $|N_{K/\mathbb{Q}}(x)|$, by the embedding expression for the field norm. Therefore all but finitely many local factors are one, and

Finally, let ${\Delta }_K\ne 0$ be the integer field discriminant, defined by the determinant of the trace pairing on an integral basis. If a rational prime $p$ ramifies, the algebra ${\mathcal{O}}_K/p{\mathcal{O}}_K$ has a nonzero nilpotent element: use prime ideal factorization and the Chinese remainder theorem to see a nonzero nilpotent in a factor ${\mathcal{O}}_K/{\mathfrak{p}}^e$ with $e>1$. If $y$ is nilpotent, multiplication by $yz$ is nilpotent for every $z$ in this commutative algebra, and therefore has trace zero. Thus $y$ is in the radical of its trace pairing, making the reduced discriminant zero. We have proved the discriminant obstruction to ramification:Only finitely many rational primes divide this nonzero integer, so only finitely many primes ramify.

Put $K=\mathrm{Frac}(A)$ and let $B$ be the integral closure of $A$ in $L$. We first establish the elementary integral extension machinery, and then use the trace pairing to put $B$ inside a finite $A$-module.

An integral element $x$ satisfies a monic polynomial over the base ring. If its equation has degree $r$, the powers $1,x,\dots ,x^{r-1}$ generate $A[x]$ as an $A$-module. More generally, adjoining finitely many integral elements gives a finitely generated module: reduce the exponent of each generator using its monic equation. Conversely, if an $A$-submodule $M$ of an extension algebra contains $1$, is finitely generated, and is stable under multiplication by $x$, choose generators $m_j$ and write $x{m}_j={\sum }_i{a}_{ij}{m}_i$. Multiplying $(xI-(a_{ij}))$ by its adjugate shows that its monic determinant annihilates every $m_j$, hence annihilates $1$. This finite-module criterion for integrality is the determinant trick.

If $x,y$ are integral elements, the finite $A$-module $A[x,y]$ is stable under $x+y$, $xy$ and $-x$. The criterion proves that these are integral elements. Thus the integral closure really is a subring. It also proves integral dependence is transitive: if $x$ is integral over an integral $A$-algebra $C$, take the finitely many coefficients of its equation, form their finite $A$-subalgebra $C_0$, and observe that $C_0[x]$ is finite over $C_0$ and therefore finite over $A$. The criterion applies to $x$. In particular, an algebra generated by finitely many integral elements is a finite integral extension, even though an arbitrary integral extension need not be finite.

Choose a $K$-basis $v_1,\dots ,v_d$ of $L$. Each basis element is algebraic over $K$. Ifchoose a nonzero $a_i\in A$ clearing all coefficient denominators. Then $e_i=a_i{v}_i$ satisfies a monic equation with coefficients $a_i^j{c}_{ij}\in A$. Hence all $e_i$ belong to $B$ and still form a $K$-basis. This is integral field basis by denominator clearing.

We next prove the trace of an integral element over a normal domain property. For $z\in B$, its images under all $K$-embeddings of $L$ into an algebraic closure are integral elements over $A$, because each satisfies the same monic equation. Their sum is integral by the subring property just proved. Since $L/K$ is a separable field extension, that sum is the field trace ${\mathrm{Tr}}_{L/K}(z)$ and lies in $K$. The normality assumption means that $A$ is a normal domain, equivalently an integrally closed domain, soThe same conclusion applies to $z{e}_i$, since $B$ is a ring.

The trace pairing of a finite separable field extension is nondegenerate. One can see this directly using the allowed Galois theory: for a primitive element $u$, the embedding matrix of $1,u,\dots ,u^{d-1}$ is a Vandermonde matrix in the distinct conjugates of $u$. Its determinant is nonzero, and the trace Gram matrix is its transpose times itself. Nondegeneracy is unchanged by a change of basis.

Let $M={\sum }_{i}A{e}_i$. Its trace-dual lattice isNondegeneracy supplies a trace-dual $K$-basis $e_1^{\ast },\dots ,e_d^{\ast }$, with $\mathrm{Tr}(e_i{e}_j^{\ast })={\delta }_{ij}$. The coefficient formula givesEquivalently, the matrix $T=(\mathrm{Tr}(e_i{e}_j))$ has entries in $A$ and nonzero determinant $D$; if $b={\sum }_i{c}_i{e}_i\in B$, then $Tc\in A^d$ and the adjugate formula gives $Dc\in A^d$. Thus $B\subseteq D^{-1}M$ as well. Both descriptions exhibit a finite free ambient $A$-module.

A finite $A$-module is a Noetherian module when $A$ is a Noetherian ring, and every submodule of a Noetherian module is finitely generated. Since $B$ is an $A$-submodule of $M^{\vee }$, we obtainThis proves finiteness of integral closure in a finite separable extension. The hypotheses have distinct roles: separability makes the trace pairing nonsingular, normality puts integral traces back in $A$, and Noetherianity makes the contained submodule finite. Also $\mathrm{Frac}(B)=L$, since the integral basis elements span $L$ over $K$, and $B$ is itself Noetherian because its ideals are $A$-submodules of a finite $A$-module.

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.

A normalization is important here. I use the covolume-one theta function of a fractional ideal, for which the requested limit is $1$, and also give the formula for the unnormalized Gaussian theta sum.

Let $n=[K:\mathbb{Q}]$, let $d_v=[K_v:\mathbb{R}]\in \{1,2\}$, and choose one field embedding ${\sigma }_v$ for each Archimedean place. The Minkowski embedding of a number field identifieswith a real inner product space havingIts Euclidean Lebesgue measure is $d\mu ={\prod }_{v\text{ real}}d{x}_v{\prod }_{v\text{ complex}}2d\mathrm{Re}{x}_{v}d\mathrm{Im}{x}_v$. With this convention, the covolume of a fractional ideal lattice $j(\mathfrak{b})$ iswhere $D_K$ is the field discriminant and $N(\mathfrak{b})$ is the positive absolute norm of a fractional ideal. Thus the Euclidean lattice ${\Lambda }_{\mathfrak{b}}=C_{\mathfrak{b}}^{-1/n}j(\mathfrak{b})$ has covolume $1$. DefineThis Gaussian theta sum converges absolutely whenever every $y_v>0$.

The trace dual of a fractional ideal isHere ${\mathfrak{D}}_{K/\mathbb{Q}}^{-1}$ is the inverse different. Since $N({\mathfrak{D}}_{K/\mathbb{Q}})=|D_K|$, we have $C_{{\mathfrak{b}}^{\vee }}=C_{\mathfrak{b}}^{-1}$. There is a subtle distinction between the trace pairing and the positive inner product: the dual lattice of $j(\mathfrak{b})$ is $\overline{j({\mathfrak{b}}^{\vee })}$, where the bar conjugates the complex coordinates and fixes the real ones. Indeed,Consequently ${\Lambda }_{\mathfrak{b}}^{\ast }=\overline{{\Lambda }_{{\mathfrak{b}}^{\vee }}}$. Coordinatewise complex conjugation preserves the weighted squared lengths in the Gaussian theta sum.

Here are the precise analytic formulas used in the proof. For a Schwartz function on $K_{\infty }$, take the Fourier transform to beFor a full Euclidean lattice $\Lambda$ of covolume $C$, the Poisson summation formula for a Euclidean lattice isFor $f_y(x)=\exp (-\pi {\sum }_v{d}_v{y}_v|x_v{|}^2)$, the Gaussian Fourier transform, applied in orthonormal real coordinates, givesMore generally, $\widehat{\exp (-\pi x^{T}Ax)}(z)=(\det A{)}^{-1/2}\exp (-\pi z^T{A}^{-1}z)$ for a real positive-definite matrix $A$ that is symmetric. Each complex coordinate contributes two real coordinates, which explains its exponent $d_v=2$.