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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.

Let $B$ be the integral closure of a Dedekind domain $R$ in a finite separable extension $L/K$ of degree $n$. For a nonzero prime ideal $\mathfrak{p}$ of $R$, writeand let $f_{\mathfrak{q}/\mathfrak{p}}=[B/\mathfrak{q}:R/\mathfrak{p}]$. Then

An integral domain is a nonzero commutative ring without zero divisors. Equivalently, a product of two nonzero elements is always nonzero. The zero ring is excluded, and $\mathbb{Z}/6\mathbb{Z}$ fails because $2\cdot 3=0$. Every field is an integral domain, but the integers show that the converse fails: $2$ has no multiplicative inverse in $\mathbb{Z}$. See MIT's Basic Properties of Rings, definitions 15.9–15.10 and example 15.11.

The practical benefit is multiplicative cancellation. In the integers, $2x=6$ has at most one solution and indeed $x=3$ works. Yet $2x=1$ has no integer solution. Thus multiplicative cancellation guarantees uniqueness when a solution exists; it does not guarantee division by every nonzero element. Passing to the fraction field $\mathbb{Q}$ supplies the missing solution $1/2$.

The polynomial ring $\mathbb{Z}[X]$ is also an integral domain. For example, the product of $2X+1$ and $3X-4$ has leading coefficient $6$ and cannot vanish. Generally the leading coefficient of a product is the product of the two leading coefficients; an integral domain prevents that coefficient from vanishing. This connects integer arithmetic to polynomial arithmetic through one reusable principle.

Historically, this connection grew far beyond elementary examples. Keith Conrad identifies the ring of integers of a number field and a ring of polynomial functions on a smooth plane curve as early important examples of Dedekind domains. This shared structure connects number theory with algebraic geometry; see Conrad's Ideal Factorization, section 6.

A Dedekind domain is a Noetherian integrally closed domain in which every nonzero prime ideal is maximal. The equivalent local characterization is that its localizations at nonzero prime ideals are discrete valuation rings. A field is harmless as the dimension-zero case.

Write $R$ for the domain, $F$ for its fraction field and $I$ for the nonzero fractional ideal. Its proposed inverse is an $R$-submodule of $F$. Choose $0\ne d\in R$ with $dI\subseteq R$, so $d\in I^{-1}$ and the inverse is nonzero. Choose also $0\ne a\in I$. Every $x\in I^{-1}$ satisfies $xa\in R$, hence $I^{-1}\subseteq a^{-1}R$. Writing $a=r/s$ with $r,s\in R$ gives $r{I}^{-1}\subseteq sR\subseteq R$. Thus $I^{-1}$ is a fractional ideal; the Noetherian hypothesis also makes it finitely generated.

For a nonzero maximal prime ideal $\mathfrak{p}$, the discrete valuation ring $R_{\mathfrak{p}}$ makes $I_{\mathfrak{p}}$ principal, say $I_{\mathfrak{p}}=a_{\mathfrak{p}}{R}_{\mathfrak{p}}$. We need to justify that inversion commutes with this localization. One inclusion is immediate. Conversely, if $x{I}_{\mathfrak{p}}\subseteq R_{\mathfrak{p}}$, finite generation of $I$ supplies a common $s\notin \mathfrak{p}$ such that $sxI\subseteq R$. Therefore $sx\in I^{-1}$, provingIt follows that $(I{I}^{-1}{)}_{\mathfrak{p}}=R_{\mathfrak{p}}$ at every such prime ideal. Globally $I{I}^{-1}\subseteq R$. If this were a proper ideal, it would be contained in a maximal ideal and would remain proper after localizing there, a contradiction. In the field case every nonzero fractional ideal is the whole field. Hence in all casesThis proves that every nonzero fractional ideal of a Dedekind domain is an invertible fractional ideal, rather than presupposing its inverse exists.

For pairwise comaximal ideals $I_1,\dots ,I_r$ of $R$, reduction gives the Chinese remainder theorem for ideals:Its kernel is ${\bigcap }_i{I}_i$. For two comaximal ideals, write $1=u+v$ with $u\in I$, $v\in J$. If $x\in I\cap J$, then $x=xu+xv\in IJ$, showing $I\cap J=IJ$. Induction gives the product formula: a product of ideals individually comaximal with $J$ is still comaximal with $J$, by multiplying identities modulo $J$.

Put $J_i={\prod }_{j\ne i}{I}_j$. Since $I_i+J_i=R$, choose $e_i\in J_i$ with $e_i\equiv 1(\mathrm{mod}{I}_i)$. The element ${\sum }_i{a}_i{e}_i$ has any prescribed residues $a_i(\mathrm{mod}{I}_i)$, proving surjectivity. Its residue class modulo the product is unique because we already identified the kernel. In a Dedekind domain, powers of distinct nonzero prime ideals are pairwise comaximal, giving the prime-power form. The theorem requires comaximality; it is not a statement about arbitrary lists of ideals.

Use $R={\mathcal{O}}_k$, $S={\mathcal{O}}_L$, $T={\mathcal{O}}_K$, and denote their relative codifferents by $C_{L/k}$, $C_{K/L}$ and $C_{K/k}$. Transitivity of the field trace givesFor the forward direction, testing an element $t\in T$ against every $s\in S$ is legitimate because $st\in T$. For the reverse direction, take $s=1$ and use trace transitivity.

Localize at a nonzero base prime ideal. The intermediate ring is semilocal Dedekind domain, so its invertible fractional ideal $C_{L/k}$ is principal there, say $C_{L/k}=d^{-1}S$ locally. This principality follows by choosing, with the Chinese remainder theorem for ideals, a generator at each of its finitely many maximal ideals. The displayed criterion now becomesInvert these fractional ideals using Solution 1. Equality at every base localization gives the global different in a tower formulaThe second factor is extended to the upper ring of integers; that extension is implicit in the printed formula. At a local prime the exponents consequently satisfy $d(K/k)=d(K/L)+e(K/L)d(L/k)$.

First include the trivial absolute value if that convention is allowed. For a nontrivial Non-Archimedean absolute value on $\mathbb{Q}$, every integer has absolute value on a field at most one, and some nonzero integer has absolute value on a field less than one: otherwise every nonzero rational number, being a quotient of two integers, would have absolute value on a field one. Factoring that integer shows that some prime number $p$ has $|p|<1$.

There cannot be two such prime numbers. If $|p|<1$ and $|q|<1$ for distinct prime numbers, choose integers $a,b$ with $ap+bq=1$. Then the ultrametric inequality givesEvery other prime number therefore has absolute value on a field one. Factoring numerator and denominator of a rational number givesThus the nontrivial equivalence classes are precisely the p-adic absolute values, one for each prime number.

Now let $F$ be a number field, with ring of integers of a number field $A$. If the restriction of an absolute value on a field to $\mathbb{Q}$ is trivial, a monic polynomial over $\mathbb{Q}$ satisfied by $x\in F$ forces $|x|\le 1$: if $|x|>1$, its leading term would strictly dominate the other terms. Applying the same argument to $x^{-1}$ forces $|x|=1$ for $x\ne 0$. Thus a nontrivial absolute value on a field defined on $F$ has a nontrivial restriction, associated with some prime number $p$.

Every $a\in A$ has $|a|\le 1$, by applying the same leading-term argument to its monic integral equation. Consequentlyis a proper prime ideal: it is an ideal by the ultrametric inequality, and $|ab|<1$ with $|a|,|b|\le 1$ implies that at least one factor has absolute value on a field less than one. It is nonzero because it contains $p$.

The localization $A_{\mathfrak{p}}$ lies in the valuation ring of the absolute value on a field: a denominator outside $\mathfrak{p}$ has absolute value on a field one. Since $A$ is a Dedekind domain, $A_{\mathfrak{p}}$ is a discrete valuation ring. Choose its uniformizer $\pi$. Each $x\in F^{\times }$ has a unique expressionBoth $u$ and $u^{-1}$ lie in the valuation ring, so $|u|=1$. Also $\pi$ belongs to its maximal ideal, hence $|\pi |<1$. Therefore $|x|=|\pi {|}^{v_{\mathfrak{p}}(x)}$.

Conversely each nonzero prime ideal of $A$ gives the Non-Archimedean absolute value $|x|=c^{v_{\mathfrak{p}}(x)}$ for any $0<c<1$, using the discrete valuation of $A_{\mathfrak{p}}$. All choices of $c$ are equivalent, and different prime ideals are distinguished by the elements of $A$ having absolute value on a field less than one. We have proved Non-Archimedean absolute values on a number field:The trivial class is additional when trivial absolute values on a field are admitted.

Let $K=\mathbb{Q}(\alpha )$ with $\alpha$ an algebraic integer, let $m\in \mathbb{Z}[X]$ be its monic minimal polynomial of an algebraic element, and write $A=\mathbb{Z}[\alpha ]$. The Kummer-Dedekind theorem applies to a rational prime $p$ not dividing the finite index $[{\mathcal{O}}_K:A]$. Ifwith distinct monic irreducible polynomials $g_j$, choose monic integer lifts ${\tilde{g}}_j$. Then the distinct prime ideals above $p$ areIn particular, $p$ is unramified precisely when $\overline{m}$ has no repeated irreducible factor. The exclusion of primes dividing the index is essential.

Here is the quotient-ring proof of the Kummer-Dedekind theorem. The finite group ${\mathcal{O}}_K/A$ has order coprime to $p$. Localizing at the integers coprime to $p$ therefore gives $A\otimes {\mathbb{Z}}_{(p)}={\mathcal{O}}_K\otimes {\mathbb{Z}}_{(p)}$, and reduction yieldsThe second isomorphism follows from $A\simeq \mathbb{Z}[X]/(m)$. The Chinese remainder theorem givesEach factor is a local ring with unique maximal ideal generated by $g_j$ and residue field ${\mathbb{F}}_p[X]/(g_j)$. Pulling its maximal ideal back to ${\mathcal{O}}_K$ gives exactly ${\mathfrak{p}}_j$; these exhaust the primes above $p$. This proves the asserted residue degrees and their independence of the chosen lifts.

It remains to prove the exponents rather than simply identify the primes. Unique prime ideal factorization in the Dedekind domain ${\mathcal{O}}_K$ writes $(p)={\prod }_j{\mathfrak{p}}_j^{b_j}$. At ${\mathfrak{p}}_j$, its discrete valuation ring has $p=u{\pi }^{b_j}$, with $u$ a unit. Thus the maximal ideal of ${\mathcal{O}}_{K,{\mathfrak{p}}_j}/p{\mathcal{O}}_{K,{\mathfrak{p}}_j}$ has nilpotence index of an ideal $b_j$: its $b_j$th power is zero and its $(b_j-1)$st is nonzero. In ${\mathbb{F}}_p[X]/(g_j^{e_j})$ this index is $e_j$, since $g_j^{e_j}=0$ but $g_j^{e_j-1}\ne 0$. The displayed product isomorphism identifies these local factors, hence $b_j=e_j$. This completes the theorem's proof. The same local argument works over a base number field whenever ${\mathcal{O}}_K[\alpha ]$ equals the full relative ring of integers after localizing at the chosen base prime.

For the required cyclotomic field, the allowed ring of integers formula makes the index one. Its cyclotomic polynomial isFor $p\notin \{2,5\}$, the polynomial $X^{80}-1$ has derivative $80X^{79}$ coprime to it modulo $p$. Its divisor ${\overline{\Phi }}_{80}$ is therefore squarefree, so the Kummer-Dedekind theorem gives no ramification.

The cyclotomic reduction at a conductor prime is explicit at the remaining two primes:The quartic over ${\mathbb{F}}_2$ has no linear factor. Its only possible irreducible quadratic divisor would be $X^2+X+1$, whose roots have order three, whereas its own roots have order five. Hence it is irreducible. The two quartics over ${\mathbb{F}}_5$ are also irreducible: their roots have multiplicative order sixteen because their fourth powers are $2$ or $3$, both of order four; the finite-field Frobenius automorphism has orbit length four on an element of order sixteen, since $5^4\equiv 1(\mathrm{mod}16)$ and neither $5$ nor $5^2$ is $1$ modulo sixteen.

Consequently the prime ideal factorization above $2$ has one prime with ramification index eight and residue degree four. Above $5$ it has two primes, each with ramification index four and residue degree four. In particular,

Let $L/K$ be an extension of number fields, let $\alpha \in {\mathcal{O}}_L$ be a primitive element of a field extension, and let $f\in {\mathcal{O}}_K[X]$ be its monic minimal polynomial of an algebraic element. Put $R={\mathcal{O}}_K[\alpha ]$ and $S={\mathcal{O}}_L$. The Kummer-Dedekind theorem applies at a nonzero prime ideal $\mathfrak{p}$ of ${\mathcal{O}}_K$ when $R_{\mathfrak{p}}=S_{\mathfrak{p}}$: equivalently, $\mathfrak{p}$ is outside the support of the finite-index quotient $S/R$. Write the factorization over the residue field $k={\mathcal{O}}_K/\mathfrak{p}$ aswith distinct monic irreducible polynomials ${\overline{g}}_j$, and choose monic lifts $g_j\in {\mathcal{O}}_K[X]$. ThenHere $f({\mathfrak{P}}_j/\mathfrak{p})$ denotes the residue-field degree, and the ramification index of a prime ideal is $m_j$.

To prove this, the local index condition givesThe final isomorphism is the Chinese remainder theorem. Each factor is a local ring with unique maximal ideal generated by ${\overline{g}}_j$, so its inverse image in $S$ is exactly ${\mathfrak{P}}_j$, and its residue field is $k[X]/({\overline{g}}_j)$. This proves the list of primes and the residue-field degrees. For the exponents, localize at ${\mathfrak{P}}_j$. The ring of integers of a number field is a Dedekind domain, so $S_{{\mathfrak{P}}_j}$ is a discrete valuation ring. If $\mathfrak{p}{S}_{{\mathfrak{P}}_j}={\mathfrak{P}}_j^{e_j}{S}_{{\mathfrak{P}}_j}$, the quotient has $e_j$ successive layers isomorphic to $S/{\mathfrak{P}}_j$, and hence $k$-dimension $e_j\mathrm{deg}{\overline{g}}_j$. The corresponding polynomial factor has $k$-dimension $m_j\mathrm{deg}{\overline{g}}_j$. Thus $e_j=m_j$, completing the proof rather than merely identifying the radicals.

For the calculation, set $K=\mathbb{Q}(\sqrt{5})$ and $t=(\sqrt{5}-1)/2$. The ring of integers of a quadratic field gives ${\mathcal{O}}_K=\mathbb{Z}[t]$, with $t^2+t-1=0$. The algebraic norm of $4+\sqrt{5}$ is $11$, so its principal ideal has ideal norm $11$. Its quotient ring has prime order and is therefore the finite field ${\mathbb{F}}_{11}$; consequently $(4+\sqrt{5})$ is a prime ideal. In that residue field, $\sqrt{5}=-4=7$ and $t=3$.

Let $L=\mathbb{Q}({\zeta }_5)$. Since $t={\zeta }_5+{\zeta }_5^{-1}$ lies in $\mathbb{Z}[{\zeta }_5]$, the permitted identity ${\mathcal{O}}_L=\mathbb{Z}[{\zeta }_5]$ gives ${\mathcal{O}}_L={\mathcal{O}}_K[{\zeta }_5]$. Thus the local index condition holds at every prime. The relative minimal polynomial of an algebraic element is $X^2-tX+1$, whose reduction isThe roots are distinct. Applying the Kummer-Dedekind theorem givesFor the last equality, evaluation at $5$ or $9$ in ${\mathbb{F}}_{11}$ gives ${\zeta }_5+{\zeta }_5^{-1}=3$ and therefore $4+\sqrt{5}=5+2t=0$. Both descriptions are the same kernels of these evaluation maps. Each prime has ideal norm $11$, residue-field degree one and ramification index of a prime ideal one.

A Dedekind domain is a Noetherian ring $D$ which is an integral domain, is an integrally closed domain, and has Krull dimension one. Explicitly, every ideal is finitely generated; every element of the fraction field that is an integral element over $D$ already belongs to $D$; and every nonzero prime ideal is a maximal ideal, with $D$ itself not a field.

With this convention, an equivalent local characterization is that $D$ is a Noetherian integral domain that is not a field and whose localization at a prime ideal $D_P$ is a discrete valuation ring for every nonzero prime ideal $P$. The integrally closed domain condition rules out singularities such as those represented by the subring $k[t^2,t^3]\subset k[t]$, where $t$ is integral but missing.

Some conventions also include fields by allowing Krull dimension at most one; the conclusions about ideals below remain valid under that convention.

The unique factorization of ideals in a Dedekind domain states that every nonzero proper ideal $I$ of a Dedekind domain $D$ has a factorizationwhere the $P_j$ are distinct nonzero prime ideals and the $n_j$ are positive integers. The prime ideals and their exponents are unique up to reordering. The unit ideal $D$ is the empty product; the zero ideal is excluded.

Write $v_P(I)$ for the prime-ideal valuation in a Dedekind domain, namely the exponent of $P$ in this factorization, with zero assigned when $P$ is absent. The associated containment rule isLarger exponents give smaller ideals, just as higher divisibility gives smaller principal ideals in $\mathbb{Z}$. The theorem concerns unique factorization of ideals; it does not assert unique factorization of elements or that these ideals are principal.

Let $D$ be a Dedekind domain. The zero ideal is generated by $0$, and $D$ is generated by $1$, so consider a nonzero proper ideal $I$. Choose $0\ne a\in I$. We use these consequences of the unique factorization of ideals in a Dedekind domain: every nonzero ideal has finitely many nonzero prime-ideal valuations in a Dedekind domain; containment reverses the valuations as in part (b); andThe last formula follows from the containment rule: $J+K$ is the smallest ideal containing both $J$ and $K$, and the product with the indicated minimum exponents is exactly that ideal.

Let $T$ be the finite set of prime ideals dividing $(a)$. For each $P\in T$, set $r_P=v_P(I)$. Since $(a)\subseteq I$, we have $r_P\le v_P((a))$. The containment rule also gives $I⊈P^{r_P+1}$, so chooseDistinct nonzero prime ideals of $D$ are distinct maximal ideals and hence comaximal ideals; their positive powers are also comaximal. Indeed, if $u+v=1$ with $u\in P$, $v\in Q$, for positive integers $m,n$ expanding $(u+v{)}^{m+n-1}$ shows $1\in P^m+Q^n$, since each term contains either at least $m$ factors of $u$ or at least $n$ factors of $v$. We use the Chinese remainder theorem in its ideal form: for pairwise comaximal ideals $J_1,\dots ,J_s$, the map $D\to {\prod }_{j}D/J_j$ is surjective. Apply it to the powers $P^{r_P+1}$ to choose $e_P\in D$ withThen $b={\sum }_{P\in T}{e}_P{b}_P$ belongs to $I$ and satisfies $b\equiv b_P(\mathrm{mod}{P}^{r_P+1})$ for every $P\in T$. Thus $b\notin P^{r_P+1}$, whereas $b\in I\subseteq P^{r_P}$, proving $v_P((b))=r_P$.

For $P\in T$, therefore,For $P\notin T$, $v_P((a))=0$, and $(a)\subseteq I$ forces $v_P(I)=0$ as well. Hence $(a,b)$ and $I$ have the same valuation at every nonzero prime ideal. Uniqueness of their factorizations givesThis proves two generators for an ideal in a Dedekind domain; in fact the first nonzero generator $a\in I$ can be chosen in advance.

TakeWe justify that this is a Dedekind domain. It is an integral domain inside the number field $K=\mathbb{Q}(\sqrt{-5})$. As an additive group it is free of rank two over $\mathbb{Z}$. The subgroup theorem for finitely generated abelian groups says that every subgroup of such a group is finitely generated. Thus every ideal has finitely many additive generators, and those same elements generate it as a $D$-ideal. This proves that $D$ is Noetherian.

We use the standard formula for the ring of integers of a quadratic field: for squarefree $d$, it is $\mathbb{Z}[(1+\sqrt{d})/2]$ if $d\equiv 1(\mathrm{mod}4)$, and $\mathbb{Z}[\sqrt{d}]$ otherwise. Since $-5\equiv 3(\mathrm{mod}4)$, $D$ is the ring of integers of a number field $K$, meaning all elements of $K$ integral over $\mathbb{Z}$. We also use transitivity of integral extensions. If an element of $K$ is integral over $D$, it is integral over $\mathbb{Z}$ because $D$ is integral over $\mathbb{Z}$, and therefore belongs to $D$. Thus $D$ is an integrally closed domain.

Every nonzero prime ideal $P$ is maximal. Indeed, choose $0\ne \alpha =a+b\sqrt{-5}\in P$. Multiplying by its algebraic conjugate puts the nonzero integer $N(\alpha )=a^2+5b^2$ in $P$. Hence the integral domain $D/P$ is a quotient of the finite ring $D/N(\alpha )D$, and is finite. Every finite integral domain is a field, since multiplication by a nonzero element is injective and therefore surjective. The element $2$ has no inverse in $D$, since $1/2\notin \mathbb{Z}[\sqrt{-5}]$, so $D$ is not a field. Consequently $(0)$ is not maximal, and a maximal ideal above it is nonzero. Together with the preceding result, this gives Krull dimension exactly one.

Finally, consider the idealReduction modulo $2$ with $\sqrt{-5}\mapsto 1$ identifies $D/I$ with ${\mathbb{F}}_2$, so the additive index $[D:I]$ is $2$. If $I$ were the principal ideal $(a+b\sqrt{-5})$, multiplication by its nonzero generator would have integer matrixin the basis $1,\sqrt{-5}$. The index of its image is the absolute determinant, soThere are no integer solutions: $b\ne 0$ makes the right side at least $5$, while $b=0$ would require $a^2=2$. Thus $I$ is a nonprincipal ideal in the integers adjoined a square root of minus five, and $D$ is a Dedekind domain that is not a principal ideal domain.

We use the following standard facts about cyclotomic fields. For an odd prime $\ell$, the unique quadratic subfield of $\mathbb{Q}({\zeta }_{\ell })$ isAlso $\mathbb{Q}({\zeta }_{{\ell }^r})$ has degree $\phi ({\ell }^r)={\ell }^{r-1}(\ell -1)$, and $\ell$ is totally ramified, with a unique prime and residue field ${\mathbb{F}}_{\ell }$. These are the quadratic subfield of a prime cyclotomic field and total ramification in a prime-power cyclotomic field results.

TakeSince $59\equiv 3(\mathrm{mod}4)$, every $E_r$ contains $K=\mathbb{Q}(\sqrt{-59})$, through its subfield $E_1$. The unique prime above $59$ is totally ramified in $E_r/K$: the absolute ramification index is $58\cdot 59^{r-1}$, the index in $K/\mathbb{Q}$ is $2$, and the relative index is thereforeThe field $K$ has no real places. By part (b), $h_K\mid h_{E_r}$, and part (a) showed $3\mid h_K$. ThusThe ring of integers of a number field is a Dedekind domain, and is a principal ideal domain exactly when its ideal class group is trivial. Hence none of these rings of integers is principal. Finally the degrees $58\cdot 59^{r-1}$ strictly increase, so the fields are pairwise distinct. The explicit infinite family of nonprincipal cyclotomic fields of conductor a prime power is

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.

The ring $\mathbb{Z}[T]$ is Noetherian and integrally closed, but it is not a Dedekind domain because it has Krull dimension two. Concretely, the nonzero prime ideal $(2)$ is properly contained in the prime ideal $(2,T)$ and is therefore not maximal.

Because $0\notin S$, the localization $S^{-1}A$ remains an integral domain. Part (a) makes it Noetherian. Integral closedness is preserved by localization: if $x$ in the common fraction field is integral over $S^{-1}A$, clearing the finitely many denominators in a monic equation shows that $sx$ is integral over $A$ for some $s\in S$, whence $sx\in A$ and $x\in S^{-1}A$.

The prime ideal correspondence for localization shows that every chain of primes in $S^{-1}A$ comes from a chain in $A$, soIf its dimension is one, it is a Noetherian integrally closed domain of dimension one and hence a Dedekind domain. If its dimension is zero, its zero ideal is maximal, so the domain is a field. This proves the Localization of a Dedekind domain alternative.

A discrete valuation on a field $K$ is a surjective group homomorphismsatisfyingwhenever $x+y\ne 0$. Its discrete valuation ring is

One standard characterization defines a Dedekind domain as a Noetherian integrally closed domain of Krull dimension one. Equivalently, all its localizations at nonzero prime ideals are discrete valuation rings.

For a nonzero ideal $I$ in a Dedekind domain, $v_P(I)$ is the exponent of the nonzero prime ideal $P$ in the unique factorization of ideals in a Dedekind domain. Then $I\subseteq J$ if and only if $v_P(I)\ge v_P(J)$ for every $P$, and $v_P(I+J)=\min (v_P(I),v_P(J))$.