Discrete valuation rings and Dedekind domains

Written by GPT-6.1 Sol (OpenAI) in Codex, Ultra setting, October 2026. Self-checked by the writing AI. No complete independent source and dependency review is recorded. Original text: CC0.

The identity \(6=2\cdot3=(1+\sqrt{-5})(1-\sqrt{-5})\) defeats unique factorization of elements in \(\mathbf Z[\sqrt{-5}]\). Ideals give a finer description. The ideal generated by \(2\) is the square of a prime ideal, and the ideal generated by \(3\) is a product of two other prime ideals. Their exponents record divisibility even though the first prime ideal has no single generator.

We will construct those exponents by looking at one prime at a time. Each localization of a ring of integers at a nonzero prime is a discrete valuation ring. There, an ideal is generated by a power of one element. The main work is to recover a global ideal from these local powers, prove that only finitely many powers matter, and construct its inverse. The Chinese remainder theorem will then select elements with prescribed behavior at several primes.

We use Algebraic integers and rings of integers and Discriminants and integral bases. In particular, \(\mathcal O_K\) is integrally closed and is a free abelian group of rank \([K:\mathbf Q]\). We prove the local characterizations needed for ideal factorization, then examine how localization changes units and ideal classes. The principal references are J. S. Milne's freely available number-theory notes and the AI Integrated Stacks Project.

What one prime sees

All rings in this lesson are commutative with \(1\). A domain is nonzero. If \(A\) is a domain with fraction field \(F\), and \(\mathfrak p\) is a prime ideal, its localization is

It is a local ring: its unique maximal ideal is \(\mathfrak pA_{\mathfrak p}\). We use the convention that a Dedekind domain is a Noetherian, integrally closed domain of dimension exactly \(1\). Thus fields are excluded. Dimension \(1\) means that every nonzero prime is maximal and at least one such prime exists.

A discrete valuation ring, or DVR, is a local principal ideal domain that is not a field. Its maximal ideal has a generator \(\pi\), called a uniformizer. Every nonzero element of its fraction field has a unique expression

and \(x\in A\) exactly when \(m\geq0\). Uniqueness concerns the exponent and, once \(\pi\) is fixed, the unit. We write \(v(x)=m\), and \(v(0)=+\infty\). Then

The second inequality follows by taking the smaller power of \(\pi\) out of the sum. If the two valuations differ, the inequality is an equality: after that power is removed, one summand is a unit and the other belongs to the maximal ideal.

Every nonzero ideal of a DVR is \((\pi^m)\) for a unique \(m\geq0\). Indeed, take the smallest valuation of a nonzero element of the ideal. An element with that valuation generates it. The same argument shows that its fractional ideals are \(\pi^m A\) for \(m\in\mathbf Z\).

To pass from normality to valuations, we need a finite-module argument.

Nakayama's lemma. Let \(M\) be a finitely generated module over a commutative ring \(R\), and let \(J\) be contained in every maximal ideal. If \(JM=M\), then \(M=0\). More generally, if \(M=N+JM\), then \(M=N\).

Proof. Choose generators \(y_1,\ldots,y_t\) of \(M\). The equality \(JM=M\) gives a matrix \(C=(c_{ij})\) with entries in \(J\) and \(y_i=\sum_jc_{ij}y_j\). Multiplying \((1-C)y=0\) by its adjugate shows that \(\det(1-C)\) annihilates every generator. This determinant belongs to \(1+J\), whose elements are units: a maximal ideal containing \(1+j\) would also contain \(j\), and hence \(1\). Thus \(M=0\). Apply the same argument to the finitely generated quotient \(M/N\) for the second assertion. \(\square\)

DVR characterization. For a Noetherian local domain \(R\) of dimension \(1\), with maximal ideal \(\mathfrak m\) and residue field \(k\), the following are equivalent:

• \(R\) is a DVR.
• \(R\) is integrally closed.
• \(\mathfrak m\) is principal.
• \(\dim_k(\mathfrak m/\mathfrak m^2)=1\).
• Every nonzero ideal is a power of \(\mathfrak m\).

Proof. Suppose first that \(R\) is integrally closed. Choose \(0\ne a\in\mathfrak m\). Every prime containing \(a\) is nonzero and hence equals \(\mathfrak m\), so \(\sqrt{(a)}=\mathfrak m\). If \(x_1,\ldots,x_t\) generate \(\mathfrak m\), choose \(d_i\geq1\) with \(x_i^{d_i}\in(a)\). Every monomial of degree \(1+\sum_i(d_i-1)\) has such a factor. Hence \(\mathfrak m^N\subseteq(a)\) for some \(N\).

Choose the least \(N\geq1\) and an element \(b\in\mathfrak m^{N-1}\setminus(a)\). In the fraction field, \(z=b/a\) satisfies \(z\notin R\) and \(z\mathfrak m\subseteq R\). If \(z\mathfrak m\subseteq\mathfrak m\), multiplication by \(z\) preserves the finite module \(\mathfrak m\). Writing this multiplication on generators and taking the adjugate gives a monic polynomial \(\det(T-C)\in R[T]\) whose value at \(z\) annihilates \(\mathfrak m\). Since this ideal contains a nonzero element in a domain, the polynomial vanishes at \(z\). That makes \(z\) integral over \(R\), a contradiction. Thus the ideal \(z\mathfrak m\) is not contained in the unique maximal ideal; it equals \(R\). Consequently \(\mathfrak m=z^{-1}R\) is principal.

Now suppose \(\mathfrak m=(\pi)\). A nonzero \(x\in R\) cannot belong to every \(\pi^iR\). Otherwise the elements \(x_i=x/\pi^i\) belong to \(R\), and the ideals \((x_i)\) form an ascending chain. Equality at any step would imply \(1=c\pi\) by cancellation, contradicting \(\pi\in\mathfrak m\). Noetherianity excludes the resulting strict infinite chain. There is therefore a greatest \(r\) with \(x\in\pi^rR\), and \(x/\pi^r\) is outside \(\mathfrak m\), hence is a unit. This gives the valuation description above, first for elements and then for fractions. The least valuation in any nonzero ideal makes that ideal a power of \(\mathfrak m\). In particular \(R\) is a DVR.

A DVR is integrally closed. If \(v(z)<0\), then in a monic equation

the first term has strictly smaller valuation than every other nonzero term. The unequal-valuation rule forbids their sum from vanishing. Thus every integral \(z\) has nonnegative valuation and belongs to \(R\).

Nakayama's lemma gives \(\mathfrak m\ne\mathfrak m^2\). If \(\dim_k(\mathfrak m/\mathfrak m^2)=1\), lift a basis element to \(x\in\mathfrak m\). Then \(\mathfrak m=(x)+\mathfrak m^2\), and Nakayama gives \(\mathfrak m=(x)\). Conversely a principal nonzero maximal ideal has a one-dimensional cotangent space. Finally, if every nonzero ideal is a power of \(\mathfrak m\), choose \(x\in\mathfrak m\setminus\mathfrak m^2\). Writing \((x)=\mathfrak m^r\) forces \(r=1\). These implications prove all five equivalences. \(\square\)

Dedekind localization characterization. A Noetherian domain \(A\) that is not a field is Dedekind if and only if \(A_{\mathfrak p}\) is a DVR for every nonzero prime \(\mathfrak p\).

Proof. Normality localizes: if \(z\) satisfies a monic equation with coefficients in \(S^{-1}A\), choose \(s\in S\) clearing their denominators. Then \(sz\) satisfies a monic equation over \(A\), so normality gives \(sz\in A\) and \(z\in S^{-1}A\). For Dedekind \(A\), primes of \(A_{\mathfrak p}\) correspond to primes contained in \(\mathfrak p\); these are \(0,\mathfrak p\). Thus \(A_{\mathfrak p}\) is normal, Noetherian and one-dimensional, and the DVR characterization applies.

Conversely, the prime correspondence shows that no nonzero prime of \(A\) can contain a smaller nonzero prime: such a chain would survive in its DVR localization. Since \(A\) is not a field, it has a nonzero maximal ideal, giving dimension exactly one. An element of the fraction field integral over \(A\) belongs to every \(A_{\mathfrak m}\), since each is normal. The local membership test proved next then puts it in \(A\). This proves normality and the characterization. \(\square\)

For example, \(\mathbf Z_{(p)}\) has uniformizer \(p\) and valuation the exponent of \(p\) in a rational number. The ring \(k[t]_{(t)}\) has uniformizer \(t\). In \(k[[t]]\), a nonzero series is \(t^m\) times a series with nonzero constant coefficient; the latter has a formal inverse, found recursively. Taking the smallest order in an ideal shows that \(k[[t]]\) is a DVR as well.

Proposition 3.1. The ring of integers of every number field is a Dedekind domain.

Proof. Put \(A=\mathcal O_K\), where \(n=[K:\mathbf Q]\). An ideal of \(A\) is an additive subgroup of the free abelian group \(A\cong\mathbf Z^n\), so it is finitely generated over \(\mathbf Z\). Those generators also generate it as an \(A\)-ideal: their integer span is already the whole ideal, and their \(A\)-span is contained in it. Hence \(A\) is Noetherian. It is integrally closed by the integrality theorem.

Let \(\mathfrak p\ne0\) be prime, and choose \(0\ne\alpha\in\mathfrak p\). Its monic minimal polynomial has integer coefficients and nonzero constant coefficient \(c\). Evaluating that polynomial shows

The group \(A/cA\) is finite, so the domain \(A/\mathfrak p\), a quotient of it, is finite. Multiplication by a nonzero element of a finite domain is injective and therefore surjective; thus every nonzero element has an inverse. Consequently \(\mathfrak p\) is maximal.

Finally, \(A\) is not a field: the inverse of \(2\) would be the rational algebraic integer \(1/2\), which is not an integer. Therefore \(A\) has a nonzero maximal ideal and has dimension exactly \(1\). \(\square\)

Recovering an ideal from its localizations

We need two elementary tools before multiplying prime powers. Let \(A\) be any domain and \(I,J\) be \(A\)-submodules of \(F\).

Local membership test. If \(x\in F\) belongs to \(J_{\mathfrak m}\) for every maximal ideal \(\mathfrak m\), then \(x\in J\). Indeed,

is an ideal. Membership in \(J_{\mathfrak m}\) says that \(T\) contains an element outside \(\mathfrak m\). If \(T\) were proper, it would lie in a maximal ideal, a contradiction. Thus \(1\in T\). Applying this to each \(x\in I\) proves

In particular, equality can be checked locally. This argument is the concrete fraction-field form of Stacks, Tag 00HN.

Finite support test. Suppose \(A\) is Dedekind and \(0\ne a\in A\). There are only finitely many primes containing \(a\).

Proof. If \(a\) is a unit, no prime contains it. Otherwise \(A/(a)\) is a nonzero Noetherian ring whose primes are all maximal.

In any Noetherian ring every radical ideal is a finite intersection of prime ideals. Indeed, if not, the ascending-chain condition provides a maximal counterexample \(J\). It is proper and not prime. Choose \(x,y\notin J\) with \(xy\in J\). Both \(\sqrt{J+(x)}\) and \(\sqrt{J+(y)}\) strictly contain \(J\), so are finite intersections of primes. Their intersection is \(J\): membership of \(z\) in both gives powers \(z^r\in J+(x)\), \(z^s\in J+(y)\), and hence \(z^{r+s}\in J\), since \((J+(x))(J+(y))\subseteq J\). As \(J\) is radical, \(z\in J\), contradicting the choice of \(J\).

Apply this fact to the nilradical of \(A/(a)\), writing it as \(\bigcap_{i=1}^tP_i\). Every prime contains the product \(P_1\cdots P_t\), so contains some \(P_i\). Maximality forces equality. The quotient therefore has finitely many primes, which correspond to the primes of \(A\) containing \(a\). \(\square\)

A fractional ideal is a nonzero finitely generated \(A\)-submodule \(I\) of \(F\). Equivalently, it is a nonzero submodule for which \(dI\subseteq A\) for some \(0\ne d\in A\). Finite generators have a common denominator. Conversely, \(dI\) is an ideal of a Noetherian ring, and multiplying by \(d\) identifies \(I\) with that finitely generated ideal. An ideal contained in \(A\) is called integral when we need the distinction.

For fractional ideals define

Products are fractional ideals: products of finite generators generate them, and denominators multiply. Associativity follows by expanding finite sums of triple products. The ring \(A\) is the identity.

We first prove that \(I^{-1}\) really is an inverse. It is nonzero, since \(dA\subseteq I^{-1}\) when \(dI\subseteq A\). Choosing \(0\ne c\in I\) gives \(I^{-1}\subseteq c^{-1}A\). Since \(A\) is Noetherian, this submodule is finitely generated; hence it is a fractional ideal.

Moreover,

One inclusion is immediate. For the other, let \(zI_{\mathfrak p}\subseteq A_{\mathfrak p}\) and choose generators \(x_1,\ldots,x_r\) of \(I\). For each \(j\) there is \(s_j\notin\mathfrak p\) such that \(s_jzx_j\in A\). Their product \(s\) satisfies \(szI\subseteq A\), so \(z=(sz)/s\) belongs to the left side. Products also commute with localization, directly from their finite-sum definition.

Every \(A_{\mathfrak p}\), for \(\mathfrak p\ne0\), is a DVR. Write \(I_{\mathfrak p}=\pi_{\mathfrak p}^{m}A_{\mathfrak p}\). Its inverse is \(\pi_{\mathfrak p}^{-m}A_{\mathfrak p}\). Thus (2) gives

All maximal ideals of \(A\) are nonzero. The membership test (1) now proves \(II^{-1}=A\). In particular, nonzero fractional ideals can be cancelled in an equality of products.

Theorem 3.2. In a Dedekind domain \(A\), each nonzero fractional ideal has a unique factorization

with only finitely many nonzero exponents. Every such ideal is invertible, and

The nonzero proper primary ideals are precisely \(\mathfrak p^m\), where \(\mathfrak p\ne0\) is prime and \(m\geq1\).

Proof. Invertibility has just been proved without assuming a global factorization.

First let \(I\subseteq A\). In each DVR define \(v_{\mathfrak p}(I)\geq0\) by

Choose \(0\ne a\in I\). If \(\mathfrak p\) does not contain \(a\), then \(a\) is a unit in \(A_{\mathfrak p}\), so \(I_{\mathfrak p}=A_{\mathfrak p}\). The finite support test therefore makes

a finite product. At a fixed prime \(\mathfrak q\), every other prime \(\mathfrak p\) becomes the unit ideal: distinct maximal ideals satisfy \(\mathfrak p\not\subseteq\mathfrak q\), so some element of \(\mathfrak p\) is inverted. The remaining factor becomes the required power of \(\mathfrak qA_{\mathfrak q}\). Thus \(J_{\mathfrak q}=I_{\mathfrak q}\) for all \(\mathfrak q\), and (1) gives \(J=I\).

For a fractional ideal choose \(d\ne0\) with \(dI\subseteq A\). Both \(dI\) and \(dA\) have the factorizations just proved, and

gives a finite product with integer exponents. Localizing any such product at \(\mathfrak p\) recovers its exponent there. This proves uniqueness and independence of \(d\). The local inclusion rule for powers of a uniformizer, together with (1), proves (3).

For the last assertion recall that a proper ideal \(Q\) is primary if \(xy\in Q\) and \(x\notin Q\) imply \(y^r\in Q\) for some \(r\geq1\). Its radical is then prime: if \(ab\in\sqrt Q\) and \(a\notin\sqrt Q\), apply primaryness to \(a^m b^m\in Q\) to obtain a power of \(b\) in \(Q\).

The radical of the factorization of a nonzero proper ideal is the intersection of the distinct prime factors. If there are at least two, that intersection is not prime. To see this, one prime contains the intersection; if the intersection itself were prime, primality applied to the product of the distinct factors would force one factor into the intersection. Maximality would then force all the factors to coincide. Thus a primary ideal can have only one distinct factor.

Conversely, suppose \(xy\in\mathfrak p^m\). If \(y\notin\mathfrak p\), then \(y\) is a unit in \(A_{\mathfrak p}\), so \(x\in\mathfrak p^mA_{\mathfrak p}\). At every other prime that ideal localizes to the unit ideal, and (1) gives \(x\in\mathfrak p^m\). Consequently \(x\notin\mathfrak p^m\) forces \(y\in\mathfrak p\), and then \(y^m\in\mathfrak p^m\). This proves primaryness. \(\square\)

For an element \(x\ne0\), write \(v_{\mathfrak p}(x)=v_{\mathfrak p}(xA)\). It agrees with the normalized valuation of the DVR \(A_{\mathfrak p}\). Thus

The first assertion is (1); the second applies it to both \(x\) and \(x^{-1}\). Factorization also yields

For the sum, localization commutes with taking sums and the formula holds in a DVR. For the intersection, let \(H\) be the product with the displayed maximum exponents. Rule (3) gives \(H\subseteq I\cap J\); any \(x\in I\cap J\) has all the required valuations, so (1) gives \(x\in H\). For integral ideals, \(I\) divides \(J\), meaning \(J=IC\) for an integral ideal \(C\), exactly when \(J\subseteq I\). Sum and intersection are therefore the greatest common divisor and least common multiple in the ideal divisibility order.

Choosing elements by congruences

Proposition 3.3 (Chinese remainder theorem). In any commutative ring, pairwise comaximal ideals \(J_1,\ldots,J_s\) satisfy

In a Dedekind domain this applies to positive powers of distinct nonzero primes. Moreover, for every maximal ideal \(\mathfrak p\) and \(m\geq1\),

Proof. If \(I+J=A\), choose \(u\in I,v\in J\) with \(u+v=1\). For \(x\in I\cap J\), the expression \(x=xu+xv\) belongs to \(IJ\); hence \(I\cap J=IJ\). Also \(av+bu\) has residues \(a\) modulo \(I\) and \(b\) modulo \(J\).

For several pairwise comaximal ideals, \(J_i+\prod_{j\ne i}J_j=A\). Indeed, choose elements of each \(J_j\) that are \(1\) modulo \(J_i\) and multiply them. Thus there are \(e_i\in A\) with

The sum \(\sum_i e_i a_i\) realizes any specified residues. The kernel is the intersection, which equals the product by the two-ideal identity and induction.

Distinct maximal ideals are comaximal. Their positive powers are also comaximal: a maximal ideal containing two such powers would contain both original primes, which is impossible.

To prove (4), observe that \(A/\mathfrak p^m\) has a unique maximal ideal \(\mathfrak p/\mathfrak p^m\), since \(\sqrt{\mathfrak p^m}=\mathfrak p\). Consequently the image of every \(s\notin\mathfrak p\) is already a unit in this quotient. If \(sa\in\mathfrak p^m\), invert that image to conclude \(a\in\mathfrak p^m\), proving injectivity. If \(a/s\in A_{\mathfrak p}\), choose \(t\in A\) representing the inverse of \(s\) modulo \(\mathfrak p^m\); then \(at\) maps to \(a/s\), proving surjectivity. \(\square\)

The same congruence elements \(e_i\) work in any \(A\)-module \(M\). Given \(x_i\in M\), the element \(\sum_i e_i x_i\) equals \(x_i\) modulo \(\mathfrak p_iM\), using the \(e_i\) for distinct maximal ideals \(\mathfrak p_i\).

Corollary 3.4. Every ideal of a Dedekind domain is generated by at most two elements. More precisely, if \(I\ne0\) and \(0\ne a\in I\), then some \(b\in I\) satisfies \(I=(a,b)\). A Dedekind domain with finitely many nonzero prime ideals is a PID.

Proof. Let \(S\) be the finite set of primes containing \(a\). For each \(\mathfrak p\in S\), choose \(x_{\mathfrak p}\in I\) whose image generates \(I_{\mathfrak p}\). Such a choice exists: a generator of \(I_{\mathfrak p}\) is a fraction \(x/s\) with \(x\in I\) and \(s\notin\mathfrak p\), and \(x\) is also a generator. The module congruence construction gives \(b\in I\) with

In the DVR this means that \(b\) differs from a generator of \(I_{\mathfrak p}\) by an element of valuation at least \(v_{\mathfrak p}(I)+1\); thus \(b\) itself generates \(I_{\mathfrak p}\). At a prime outside \(S\), \(a\) is a unit and \(I_{\mathfrak p}=A_{\mathfrak p}\). It follows that \((a,b)_{\mathfrak p}=I_{\mathfrak p}\) everywhere. The local membership test gives \(I=(a,b)\). The zero ideal is generated by \(0\).

If there are only finitely many nonzero primes, apply the same construction to all of them. The resulting \(b\in I\) generates \(I_{\mathfrak p}\) at every maximal ideal, so \(I=bA\). Every ideal is principal. \(\square\)

The distinction between the two conclusions matters. An ordinary ring of integers has infinitely many prime ideals, so the second construction imposes conditions at an infinite set and cannot be carried out by the finite Chinese remainder theorem.

Measuring the failure of principal generation

The fractional ideals form an abelian group \(\mathcal I_A\), freely generated by the nonzero primes by Theorem 3.2. The principal fractional ideals form its subgroup

The ideal class group is \(\operatorname{Cl}(A)=\mathcal I_A/\mathcal P_A\). Its identity class contains all principal fractional ideals. Two ideals \(I,J\) have the same class precisely when \(I=xJ\) for some \(x\in F^\times\). The map \(F^\times\to\mathcal I_A\), \(x\mapsto xA\), has kernel \(A^\times\). No finiteness of the class group is asserted for a general Dedekind domain; for number fields it will follow from a geometric bound.

Proposition 3.5. For a Dedekind domain \(A\), the following are equivalent:

Proof. A trivial class group says that every nonzero fractional ideal is principal, hence every integral ideal is principal. Conversely, if integral ideals are principal, multiply any fractional ideal by a common denominator to prove it principal too.

Suppose \(A\) is a UFD. In a nonzero prime \(\mathfrak p\), choose \(a\ne0\). A factorization of \(a\) into irreducibles has a factor \(\pi\in\mathfrak p\). In a UFD irreducibles are prime, so \((\pi)\) is a nonzero prime ideal. Both \((\pi)\) and \(\mathfrak p\) are maximal, and their inclusion forces equality. All prime ideals are therefore principal; Theorem 3.2 makes every nonzero integral ideal principal.

Finally, in a PID a nonunit factors into irreducibles. Otherwise successive proper factorizations produce a strictly ascending chain of principal ideals, contradicting Noetherianity. An irreducible \(\pi\) generates a maximal ideal: a containing principal ideal \((d)\) means \(d\mid\pi\), so it is either \((\pi)\) or \(A\). Thus irreducibles are prime. Cancelling a prime factor from two factorizations and repeating proves uniqueness. \(\square\)

Localization, units and ideal classes

Inverting elements removes exactly the prime ideals that contain them. Ideal factorization lets us describe both the lost classes and the new units.

Let \(S\subseteq A\setminus\{0\}\) contain \(1\) and be multiplicatively closed, put \(B=S^{-1}A\), and let

Every ideal \(J\) of \(B\) is the extension of its contraction: if \(a/s\in J\), then \(a/1=s(a/s)\in J\). A prime \(P\) of \(B\) contracts to a prime \(\mathfrak p\) disjoint from \(S\), and \(P=\mathfrak pB\). Conversely, for such a prime,

is a domain, so \(\mathfrak pB\) is prime. Its contraction is \(\mathfrak p\): \(a/1\in\mathfrak pB\) implies \(sa\in\mathfrak p\) for some \(s\in S\), and primality gives \(a\in\mathfrak p\).

These correspondences also show that \(B\) is Noetherian and either Dedekind or a field. Normality follows from the denominator-clearing proof above. Its nonzero primes are maximal; it is a field precisely when every nonzero prime of \(A\) meets \(S\). For a surviving prime,

because localizing \(B\) again inverts exactly the elements outside \(\mathfrak p\). Thus its ideal valuations are the original ones.

Localization exact sequence. Write \(D_T=\bigoplus_{\mathfrak p\in T}\mathbf Z[\mathfrak p]\), the group of finitely supported formal sums of primes in \(T\). If \(B\) is a field, set \(\operatorname{Cl}(B)=0\). There is an exact sequence

Here \(\operatorname{div}_T(x)=\sum_{\mathfrak p\in T}v_{\mathfrak p}(x)[\mathfrak p]\), \(c\) takes a formal prime sum to the class of the corresponding fractional ideal, and \(E([I])=[IB]\).

Proof. The fractional-ideal group of \(A\) is free on its nonzero primes. Extension to \(B\) deletes factors in \(T\): such a prime contains an element of \(S\), so extends to \(B\); all other valuations are preserved by \(B_{\mathfrak pB}=A_{\mathfrak p}\). Every fractional ideal of \(B\) is therefore the extension of one of \(A\), and the kernel on ideal groups is exactly \(D_T\). This includes the field case, when the target ideal group is trivial.

Principal ideals extend to principal ideals, so \(E\) is well-defined and surjective. If \(IB=xB\), then \(x^{-1}I\) has zero valuations at all surviving primes; its class is represented by a divisor supported in \(T\). Conversely every such divisor extends to the unit ideal. Hence \(\ker E=\operatorname{im}c\).

An element \(x\in F^\times\) is a unit of \(B\) exactly when every surviving valuation of \(x\) is zero. For Dedekind \(B\), apply the local membership test to \(x\) and \(x^{-1}\); for field \(B=F\), the assertion is automatic. The divisor map is consequently defined and has finite support. Its kernel consists of elements whose valuations at every prime of \(A\) are zero, exactly \(A^\times\). Finally, a divisor in \(D_T\) maps to zero under \(c\) exactly when its ideal equals \(xA\) for some \(x\in F^\times\). Its surviving valuations vanish, so \(x\in B^\times\), and its divisor is the given one. This proves exactness at all terms. \(\square\)

In particular,

The map to \(D_T\) also explains why inverting one prime element can create new units without making every ideal principal.

Two quadratic arithmetic tests

Put \(s=\sqrt{-5}\). The quadratic integer theorem gives \(A=\mathcal O_{\mathbf Q(s)}=\mathbf Z[s]\). Define

Sending \(s\) to \(1\) over \(\mathbf F_2\) gives \(A/\mathfrak p\cong\mathbf F_2\), and sending it to \(-1\), respectively \(1\), over \(\mathbf F_3\) gives \(A/\mathfrak q_\pm\cong\mathbf F_3\). Thus these are prime ideals, and the two over \(3\) are distinct. Products of their generators give

All three displayed generators are divisible by \(2\); conversely their second minus third is \(6\), and \(6-4=2\). Similarly,

because every generator is divisible by \(3\) and \(9-6=3\). Consequently

The ideal \(\mathfrak p\) is not principal. If \(\mathfrak p=(\alpha)\), then \((\alpha^2)=(2)\), so \(\alpha^2=2u\) for a unit \(u\in A\). Writing \(\alpha=a+bs\), the norm is \(a^2+5b^2\). Units have norm \(1\), so \(u=\pm1\). Taking norms gives \(N(\alpha)^2=4\), hence \(a^2+5b^2=2\), which has no integer solutions. Thus \([\mathfrak p]\ne1\) but \([\mathfrak p]^2=1\). This proves the existence of a nontrivial class, without yet determining the whole group.

Inverting the ramified prime

For \(B=A[1/2]\), the only nonzero prime meeting \(\{2^j:j\geq0\}\) is \(\mathfrak p\), since \((2)=\mathfrak p^2\). The class map \(\mathbf Z[\mathfrak p]\to\operatorname{Cl}(A)\) has kernel \(2\mathbf Z[\mathfrak p]\), because \([\mathfrak p]\) has order exactly two. The element \(2\) has divisor \(2[\mathfrak p]\), and \(A^\times=\{1,-1\}\), as the norm computation shows. Exactness of (6) therefore gives

Indeed, dividing any \(B\)-unit by the appropriate power of \(2\) leaves zero valuation at every prime, hence an \(A\)-unit. We have killed the displayed nonprincipal class. This argument alone does not determine the other classes or imply that \(B\) is a PID; the class-number calculation comes later.

A nonnormal quadratic order

Now let \(t=\sqrt{-3}\) and \(R=\mathbf Z[t]\). This is a proper order: the integral element \((1+t)/2\) is missing, so \(R\) is not normal and hence not Dedekind. The ideal

is prime since its quotient is \(\mathbf F_2\). Direct multiplication gives

The last equality follows because the second and third generators differ by \(4\). If \(\mathfrak r\) had a fractional ideal inverse, cancelling it in (5) would give \(\mathfrak r=(2)\), contrary to \(1+t\notin2R\). The failure of normality has produced a concrete failure of invertibility.

Polynomial rings and the dimension obstruction

We supply the algebra needed for the following two nonexamples.

Hilbert's basis argument. If \(R\) is Noetherian and \(I\subseteq R[T]\) is an ideal, let \(L_d\) be the ideal of leading coefficients of its degree-\(d\) polynomials, including zero. Addition with leading-term cancellation and multiplication by ring elements show it is an ideal. Multiplication by \(T\) gives \(L_d\subseteq L_{d+1}\). The chain stabilizes at some \(N\). Choose finitely many polynomials giving generators of each \(L_d\), \(0\leq d\leq N\). Given \(f\in I\), subtract their \(T\)-multiples to cancel its leading coefficient: use \(L_N\) if its degree is at least \(N\), and \(L_d\) otherwise. Induction on degree expresses \(f\) in the ideal generated by that finite list. Thus \(R[T]\) is Noetherian.

Polynomial UFDs and normality. For a UFD \(R\), a polynomial is primitive if no irreducible divides all its coefficients. Irreducibles are prime by unique factorization, so reduction modulo any such prime is into a domain. Consequently a product of primitive polynomials is primitive. Factor a polynomial over \(F=\operatorname{Frac}(R)\), where \(F[T]\) is a Euclidean domain. Clear denominators in every factor and divide out its coefficient gcd to obtain primitive factors over \(R\). The remaining scalar is a unit for a primitive original polynomial: for each prime of \(R\), the least coefficient valuation of a primitive polynomial is zero, and the product property makes the scalar's valuation zero. A fraction with every prime valuation zero is a unit, by its coprime numerator and denominator. The same argument makes two primitive factors associated over \(F\) associated by a unit over \(R\). Existence and uniqueness over \(F[T]\), together with the factorization of the coefficient gcd in \(R\), therefore prove existence and uniqueness over \(R[T]\).

A UFD is normal. Write a fraction \(a/b\) with coprime numerator and denominator and suppose it satisfies a monic degree-\(m\) equation over \(R\). Multiplication by \(b^m\) shows that any prime dividing \(b\) divides \(a^m\), hence \(a\), a contradiction. Thus \(b\) is a unit. A field and \(\mathbf Z\) are Noetherian UFDs; polynomial division makes \(k[x]\) a PID. The preceding arguments prove that both \(\mathbf Z[x]\) and \(k[x,y]\) are Noetherian and normal.

There is a different obstruction in \(\mathbf Z[x]\) and \(k[x,y]\). They are Noetherian by Stacks, Tag 00FN, UFDs by Tag 0BC1, and hence normal by Tag 0AFV. They have dimension \(2\), with prime chains

respectively. Both upper bounds can be proved by the same argument. If \(R\) is a nonfield PID, primes of \(R[T]\) with zero contraction to \(R\) correspond to primes of \(F[T]\), so their chains have at most one nonzero term. Primes contracting to a nonzero prime \((\pi)\) correspond to primes of \((R/(\pi))[T]\); they are \((\pi)\) and the maximal ideals \((\pi,g)\) with irreducible \(\overline g\). A nonzero zero-contraction prime \(P\) cannot lie in \((\pi)\). Choose a nonzero polynomial in \(P\), divide out its coefficient gcd and use primality, since that gcd is outside \(P\). The resulting primitive polynomial still belongs to \(P\), but not to \((\pi)\). A chain therefore cannot contain a nonzero zero-contraction prime, then \((\pi)\), then \((\pi,g)\). All chains have length at most two. Apply this with \(R=\mathbf Z\) and \(R=k[x]\). The displayed chains attain the bounds, so both dimensions are exactly two. Even normality and Noetherianity together do not suffice; the dimension hypothesis is essential.

What this lesson does not prove

The finite-module argument, local DVR characterization, Dedekind localization characterization, finite-support test and localization exact sequence are proved in this lesson. The AI Integrated Stacks Project supplies broader commutative-algebra context: its treatments of DVRs, Dedekind domains and Nakayama's lemma include additional equivalent formulations.

The finite-support proof uses only the ascending-chain condition and primality; it does not need the full equivalence between Artinian rings and zero-dimensional Noetherian rings. Extension and contraction of ideals, the prime correspondence, the local membership test and commutation of the fractional-ideal inverse with localization were also proved explicitly.

From the preceding two lessons we import the integral-basis theorem, normality of \(\mathcal O_K\), rational integrality and the complete quadratic integer theorem. The preceding integral-basis lesson proves the subgroup theorem for finite free abelian groups. Hilbert's basis theorem, the polynomial UFD theorem, normality of UFDs, and both polynomial dimension bounds used here were proved in the nonexamples above. The linked Stacks tags provide broader context for these supplied proofs. The arithmetic results do not rely on a proof of ideal factorization in another course.

The global axiomatic approach to ideal factorization, its converse, and the more general primary-ideal theory are separate topics. Here we have proved the localization route and classified only the nonzero primary ideals in Dedekind domains. Ideal norms, modules over Dedekind domains, and finiteness of number-field class groups come next.

Exercises and complete solutions

Exercise 1 (easy). In \(A=\mathbf Z[\sqrt{-5}]\), verify \((2,1+\sqrt{-5})^2=(2)\), and determine the inverse of \(\mathfrak p=(2,1+\sqrt{-5})\).

Solution. Write \(s=\sqrt{-5}\). Multiplying the two generators yields

Every generator lies in \((2)\). The difference of the last two is \(6\), and subtracting \(4\) gives \(2\); thus the reverse inclusion holds. Therefore \(\mathfrak p((1/2)\mathfrak p)=A\), and the inverse is

The half-integral element belongs to the fractional inverse, not to \(A\).

Exercise 2 (medium). Prove that \((2)\) is not a product of prime ideals in \(R=\mathbf Z[\sqrt{-3}]\).

Solution. Put \(t=\sqrt{-3}\). Modulo \(2\),

This ring has exactly one prime ideal, generated by the image of \(T+1\). Indeed every prime contains the nilpotent \(T+1\), and the quotient by it is a field. Therefore the only prime of \(R\) containing \((2)\) is \(\mathfrak r=(2,1+t)\).

If \((2)=P_1\cdots P_m\), then each \(P_i\) contains the product and hence contains \((2)\). Thus all \(P_i=\mathfrak r\). The empty product is \(R\), which is not \((2)\). For \(m=1\) the claimed equality fails because \(1+t\notin(2)\). For \(m\geq2\), (5) gives

The last inclusion is strict: \(2\in2\mathfrak r\) would imply \(1\in\mathfrak r\) after cancelling the nonzero element \(2\) in the domain. No value of \(m\) works.

Exercise 3 (medium). Compute \(v_{\mathfrak l}((6))\) for every nonzero prime \(\mathfrak l\) of \(\mathbf Z[\sqrt{-5}]\), and find the valuations of the sum and intersection of \((2)\) and \((3)\).

Solution. The checked products in the arithmetic example give

The three primes are distinct: their residue characteristics differ for \(\mathfrak p\) and the other two, and \(s\) has distinct images \(1,-1\) for \(\mathfrak q_-,\mathfrak q_+\). Uniqueness of factorization gives

The minimum exponents for the sum are all zero; the maximum exponents for the intersection are those of \((6)\). Thus \((2)+(3)=A\) and \((2)\cap(3)=(6)\). The first also follows from \(3-2=1\).

Exercise 4 (medium). Let \(S\subset A\setminus\{0\}\) be multiplicatively closed, where \(A\) is Dedekind. Prove that \(S^{-1}A\) is either Dedekind or a field. Determine which case occurs.

Solution. An ideal \(J\) of \(S^{-1}A\) is the extension of its contraction \(I\subseteq A\): if \(a/s\in J\), then \(a/1=s(a/s)\in J\), and the reverse inclusion is immediate. Generators of \(I\) therefore generate \(J\), so the localization is Noetherian.

It is normal as well. If \(z\in F\) satisfies a monic equation

choose \(s\in S\) clearing all the coefficient denominators. Then \(sz\) satisfies a monic equation with coefficients \(s^ic_i\in A\). Normality of \(A\) gives \(sz\in A\), so \(z\in S^{-1}A\).

The prime correspondence identifies the nonzero primes of the localization with nonzero primes of \(A\) disjoint from \(S\). Each is maximal, so the dimension is at most \(1\). If at least one exists, the dimension is exactly \(1\), giving a Dedekind domain. If none exists, the domain has no nonzero maximal ideal and is a field. Thus the field case occurs exactly when \(S\) meets every nonzero prime of \(A\). For example, \(A=\mathbf Z\) and \(S=\mathbf Z\setminus\{0\}\) give \(\mathbf Q\).

Exercise 5 (hard). Suppose \(A\) is Dedekind and its nonzero primes are \(\mathfrak p_1,\ldots,\mathfrak p_s\). Use ring congruences alone to construct a generator of any nonzero ideal \(I\).

Solution. Let \(m_i=v_{\mathfrak p_i}(I)\). Choose

Such an element exists because localizing the two powers gives different ideals in a DVR; if the powers were equal globally, their localizations would be equal. For \(m_i=0\) one may take \(x_i=1\). Proposition 3.3 supplies \(b\in A\) with

Then \(b\) has valuation exactly \(m_i\) at each \(\mathfrak p_i\). In particular \(b\ne0\). Since these are all the nonzero primes, Theorem 3.2 gives \((b)=I\). This construction also exhibits every fractional ideal as principal: multiply by a common denominator first and divide the resulting generator by it.

References

• [Milne ANT] J. S. Milne, Algebraic Number Theory, version 3.08, 2020, Chapter III, especially Propositions 3.2–3.6, Theorem 3.7, Corollaries 3.14–3.16, Theorem 3.20 and Remark 3.21(c). Author's lecture notes.
• [Stacks] The Stacks Project, “Commutative Algebra,” especially the tags linked in the text and Tag 034W. The official Stacks project is the upstream work. The AI Integrated Stacks Project is an edition with AI-proposed corrections and AI-written additions, not reviewed by the Stacks project's maintainers; the tag links lead to its English reader.