# 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. Independent full-lesson AI review is pending. Public domain (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 \([K:\mathbf Q]\). Two local characterizations from commutative algebra are stated precisely below and used as prerequisites. The factorization theorem, its consequences, and all the arithmetic examples are proved here. The principal references are [Milne ANT], [Stacks], and [Hecke 1923].

## 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

\[
A_{\mathfrak p}=\{a/s:a\in A,\ s\in A\setminus\mathfrak p\}\subset F.
\]

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

\[
x=u\pi^m,\qquad u\in A^\times,\quad m\in\mathbf Z,
\]

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

\[
v(xy)=v(x)+v(y),\qquad v(x+y)\geq\min(v(x),v(y)).
\]

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 \(m\in\mathbf Z\).

Here are the two local inputs that we import.

**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 \(\mathfrak m\).

The first four equivalences are [Stacks, Tag 00PD](https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/algebra.html#algebra-lemma-characterize-dvr). The assertion about ideals follows from the valuation description; conversely, the last condition forces \(\mathfrak m\) to be principal. For that converse, [Nakayama's lemma, Tag 00DV](https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/algebra.html#algebra-lemma-NAK) gives \(\mathfrak m\ne\mathfrak m^2\); choose \(x\in\mathfrak m\setminus\mathfrak m^2\). If \((x)=\mathfrak m^a\), then \(a=1\).

**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 \(\mathfrak p\). See [Stacks, Tag 034X](https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/algebra.html#algebra-lemma-characterize-Dedekind). That source allows fields among Dedekind domains and states its dimension condition as \(\dim A\leq1\); excluding fields gives the convention used here. We use its equivalence of normality and the local DVR condition, and give our own ideal factorization proof below.

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

\[
0\ne c\in\alpha A\subseteq\mathfrak p.
\]

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,

\[
T=\{s\in A:sx\in J\}
\]

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

\[
I\subseteq J\quad\Longleftrightarrow\quad
I_{\mathfrak m}\subseteq J_{\mathfrak m}\text{ for every maximal }\mathfrak m. \tag{1}
\]

In particular, equality can be checked locally. This argument is the concrete fraction-field form of [Stacks, Tag 00HN](https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/algebra.html#algebra-lemma-characterize-zero-local).

**Finite support test.** Suppose \(A\) is Dedekind and \(0\ne a\in A\). There are only finitely many primes containing \(a\). The quotient \(A/(a)\) is Noetherian and all its primes are maximal, so it has dimension \(0\). The standard zero-dimensional Noetherian theorem says that it is Artinian and has finitely many prime ideals; see [Stacks, Tag 00KH](https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/algebra.html#algebra-lemma-Noetherian-dimension-0), whose proof also establishes that finiteness. Prime ideals of the quotient correspond to primes of \(A\) containing \(a\).

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

\[
IJ=\left\{\sum_{\ell=1}^r x_\ell y_\ell:
x_\ell\in I,\ y_\ell\in J,\ r\geq0\right\},\qquad
I^{-1}=\{z\in F:zI\subseteq A\}.
\]

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,

\[
(I^{-1})_{\mathfrak p}=(I_{\mathfrak p})^{-1}. \tag{2}
\]

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

\[
(II^{-1})_{\mathfrak p}=A_{\mathfrak p}.
\]

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

\[
I=\prod_{\mathfrak p\ne0}\mathfrak p^{v_{\mathfrak p}(I)},
\qquad v_{\mathfrak p}(I)\in\mathbf Z,
\]

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

\[
I\subseteq J\quad\Longleftrightarrow\quad
v_{\mathfrak p}(I)\geq v_{\mathfrak p}(J)
\text{ for every }\mathfrak p. \tag{3}
\]

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

\[
I_{\mathfrak p}=\pi_{\mathfrak p}^{v_{\mathfrak p}(I)}A_{\mathfrak p}.
\]

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

\[
J=\prod_{\mathfrak p}\mathfrak p^{v_{\mathfrak p}(I)}
\]

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

\[
I=(dI)(dA)^{-1}
\]

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

\[
A=\{x\in F:v_{\mathfrak p}(x)\geq0\text{ for all }\mathfrak p\},
\quad
A^\times=\{x\in F^\times:v_{\mathfrak p}(x)=0\text{ for all }\mathfrak p\}.
\]

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

\[
\begin{aligned}
v_{\mathfrak p}(IJ)&=v_{\mathfrak p}(I)+v_{\mathfrak p}(J),\\
v_{\mathfrak p}(I+J)&=\min(v_{\mathfrak p}(I),v_{\mathfrak p}(J)),\\
v_{\mathfrak p}(I\cap J)&=\max(v_{\mathfrak p}(I),v_{\mathfrak p}(J)).
\end{aligned}
\]

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

\[
A/(J_1\cdots J_s)\ \cong\ \prod_{i=1}^s A/J_i.
\]

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

\[
A/\mathfrak p^m\ \cong\
A_{\mathfrak p}/\mathfrak p^mA_{\mathfrak p}. \tag{4}
\]

**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

\[
e_i\equiv1\pmod{J_i},\qquad e_i\equiv0\pmod{J_j}\ (j\ne i).
\]

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

\[
b-x_{\mathfrak p}\in\mathfrak pI\qquad(\mathfrak p\in S).
\]

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

\[
\mathcal P_A=\{xA:x\in F^\times\}.
\]

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:

\[
A\text{ is a UFD}\quad\Longleftrightarrow\quad
A\text{ is a PID}\quad\Longleftrightarrow\quad
\operatorname{Cl}(A)=1.
\]

**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. \(\square\)

## Two quadratic arithmetic tests

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

\[
\mathfrak p=(2,1+s),\qquad
\mathfrak q_+=(3,1+s),\qquad
\mathfrak q_-=(3,1-s).
\]

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

\[
\mathfrak p^2=(4,2+2s,-4+2s)=(2).
\]

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

\[
\mathfrak q_+\mathfrak q_-=(9,3+3s,3-3s,6)=(3),
\]

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

\[
(6)=\mathfrak p^2\mathfrak q_+\mathfrak q_-.
\]

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.

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

\[
\mathfrak r=(2,1+t)
\]

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

\[
\mathfrak r^2=(4,2+2t,-2+2t)=(4,2+2t)=2\mathfrak r. \tag{5}
\]

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.

There is a different obstruction in \(\mathbf Z[x]\) and \(k[x,y]\). They are Noetherian by [Stacks, Tag 00FN](https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/algebra.html#algebra-lemma-Noetherian-permanence), UFDs by [Tag 0BC1](https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/algebra.html#algebra-lemma-polynomial-ring-UFD), and hence normal by [Tag 0AFV](https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/algebra.html#algebra-lemma-UFD-normal). They have dimension \(2\), with prime chains

\[
(0)\subsetneq(p)\subsetneq(p,x),\qquad
(0)\subsetneq(x)\subsetneq(x,y),
\]

respectively. For \(k[x,y]\), [Stacks, Tag 00OP](https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/algebra.html#algebra-lemma-dim-affine-space) says that every localization at a maximal ideal has dimension \(2\), so no prime chain is longer.

For \(\mathbf Z[x]\), we can check the upper bound directly. Primes with zero contraction to \(\mathbf Z\) correspond to primes of \(\mathbf Q[x]\), so a chain of them has at most one nonzero term. Primes contracting to \((p)\) correspond to primes of \(\mathbf F_p[x]\), giving just \((p)\) and the maximal ideals \((p,g)\), where \(\bar g\) is irreducible. A nonzero prime \(P\) with zero contraction cannot lie in \((p)\): choose \(0\ne f\in P\), divide out the integer content, and use primality and \(P\cap\mathbf Z=0\) to see that the primitive part also lies in \(P\); that primitive polynomial is not in \((p)\). Thus a chain cannot pass from a nonzero horizontal prime through \((p)\) to \((p,g)\). All chains have length at most \(2\). The displayed chain attains that length. Even normality and Noetherianity together therefore do not suffice; the dimension hypothesis is essential.

## What this lesson does not prove

The local characterizations are commutative-algebra prerequisites. [Stacks, Tag 00PD](https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/algebra.html#algebra-lemma-characterize-dvr) identifies the Noetherian normal local domains of dimension \(1\) with DVRs; [Tag 034X](https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/algebra.html#algebra-lemma-characterize-Dedekind) gives the corresponding global characterization. We also use [Nakayama's lemma, Tag 00DV](https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/algebra.html#algebra-lemma-NAK): if a finite \(R\)-module \(M\) satisfies \(IM=M\) for an ideal \(I\) contained in the Jacobson radical of \(R\), then \(M=0\). We do not reproduce those general proofs here.

We also use the zero-dimensional Noetherian theorem, including finiteness of its spectrum, as recorded in [Tag 00KH](https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/algebra.html#algebra-lemma-Noetherian-dimension-0). Basic localization includes extension and contraction of ideals and the correspondence between primes of \(S^{-1}A\) and primes of \(A\) disjoint from \(S\); see [Milne ANT, Propositions 1.11 and 1.12]. The local membership test and commutation of the fractional-ideal inverse with localization were 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 remaining basic algebra inputs are the subgroup theorem for finite free abelian groups, Hilbert's basis theorem, Gauss's UFD theorem for polynomial rings, normality of UFDs, and the dimension of polynomial rings over fields. The exact Stacks tags used for these polynomial facts are linked in the nonexamples above. The upper bound for \(\dim\mathbf Z[x]\) was proved there. 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

\[
\mathfrak p^2=(4,2+2s,(1+s)^2)=(4,2+2s,-4+2s).
\]

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

\[
\mathfrak p^{-1}=(1,(1+s)/2).
\]

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 \(2\),

\[
R/(2)\cong\mathbf F_2[T]/(T^2+1)
=\mathbf F_2[T]/((T+1)^2).
\]

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

\[
\mathfrak r^m=2^{m-1}\mathfrak r\subseteq2\mathfrak r\subsetneq(2).
\]

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

\[
(6)=\mathfrak p^2\mathfrak q_+\mathfrak q_-.
\]

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

\[
v_{\mathfrak l}((6))=
\begin{cases}
2,&\mathfrak l=\mathfrak p,\\
1,&\mathfrak l=\mathfrak q_+\text{ or }\mathfrak q_-,\\
0,&\text{otherwise}.
\end{cases}
\]

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

\[
z^n+c_1z^{n-1}+\cdots+c_n=0,\qquad c_i\in S^{-1}A,
\]

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

\[
x_i\in\mathfrak p_i^{m_i}\setminus\mathfrak p_i^{m_i+1}.
\]

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

\[
b\equiv x_i\pmod{\mathfrak p_i^{m_i+1}}
\qquad(1\leq i\leq s).
\]

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](https://www.jmilne.org/math/CourseNotes/ANT.pdf).
- **[Stacks]** *The Stacks Project*, “Commutative Algebra,” especially the tags linked in the text and [Tag 034W](https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/algebra.html#algebra-definition-dedekind-domain). The [official Stacks project](https://stacks.math.columbia.edu/) 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.
- **[Hecke 1923]** E. Hecke, [*Vorlesungen über die Theorie der algebraischen Zahlen*](https://archive.org/details/vorlesungenber00heckuoft), Leipzig, 1923, Chapter V, §§24–26, Satz 65–75.