Algebraic number theory — number fields, Dedekind domains, and the class group

Intuition Beginner

The whole numbers have a property so basic it feels inevitable: every number bigger than one breaks apart into prime pieces in just one way. The number $60$ is $2\times 2\times 3\times 5$, and no other prime recipe makes $60$. This single fact — unique factorisation — is the spine of all of elementary number theory.

Algebraic number theory begins when you enlarge the number system. Adjoin a new number such as $\sqrt{-5}$, so the system holds every combination $a+b\sqrt{-5}$ with $a,b$ whole. In this bigger system the clean spine can snap. The number $6$ now splits in two genuinely different ways, and there is no way to declare one of them the real one.

The nineteenth-century repair is one of the turning points of mathematics. Stop factoring individual numbers and factor larger objects called ideals instead. At the level of ideals the lost uniqueness comes back perfectly. The leftover gap — measured by a finite group called the class group — counts exactly how badly the original factorisation failed.

Visual Beginner

The left panel records the breakdown at the level of numbers, where two prime recipes collide. The right panel records the same arithmetic at the level of ideals, where the collision dissolves into a single clean prime-ideal factorisation, and the class group is the small finite object recording that the two panels ever differed.

Worked example Beginner

Take the system $\mathbb{Z}[\sqrt{-5}]$, the elements $a+b\sqrt{-5}$ with $a,b$ whole. We show that $6$ splits in two genuinely different ways.

First recipe: $6=2\times 3$. Second recipe: $(1+\sqrt{-5})(1-\sqrt{-5})=1-(\sqrt{-5}{)}^2=1-(-5)=6$. So $6$ is also $(1+\sqrt{-5})(1-\sqrt{-5})$.

To see these are genuinely different, attach to each element $a+b\sqrt{-5}$ the size $a^2+5b^2$. This size multiplies: the size of a product is the product of the sizes. The sizes are $4$ for $2$, $9$ for $3$, and $1+5=6$ for each of $1+\sqrt{-5}$ and $1-\sqrt{-5}$.

If $2$ split further, its factors' sizes would multiply to $4$, forcing a factor of size $2$. But $a^2+5b^2=2$ has no whole-number solution: $b$ must be $0$, and then $a^2=2$, impossible. So $2$ is unbreakable; likewise $3$ and each of $1\pm \sqrt{-5}$ are unbreakable.

Yet $6$ is a product of the first pair and also of the second pair. Unique factorisation has failed, and the class group of this system has size two — the smallest possible record of a genuine failure.

Check your understanding Beginner

Formal definition Intermediate+

A number field is a finite field extension $K$ of $\mathbb{Q}$. Its degree is $n=[K:\mathbb{Q}]$. Because $\mathbb{Q}$ has characteristic zero, every finite extension is separable, so $K$ admits exactly $n$ embeddings ${\sigma }_1,\dots ,{\sigma }_n:K\hookrightarrow \mathbb{C}$. Of these, $r_1$ land inside $\mathbb{R}$ and the remaining $n-r_1$ occur in complex-conjugate pairs; writing $2r_2=n-r_1$ gives the signature $(r_1,r_2)$ ^{[Neukirch Ch. I §1]}.

An element $\alpha \in K$ is an algebraic integer if it is a root of some monic polynomial $f(x)\in \mathbb{Z}[x]$. The algebraic integers in $K$ are closed under addition, subtraction, and multiplication, so they form a subring ${\mathcal{O}}_K\subseteq K$, the ring of integers of $K$. Equivalently ${\mathcal{O}}_K$ is the integral closure of $\mathbb{Z}$ in $K$: the set of elements satisfying a monic polynomial equation with coefficients in $\mathbb{Z}$ ^{[Marcus Ch. 2]}. As an abelian group ${\mathcal{O}}_K$ is free of rank $n$; a $\mathbb{Z}$-basis ${\omega }_1,\dots ,{\omega }_n$ is an integral basis, and

$$d_K=\det ({\sigma }_i({\omega }_j){)}^2$$

is the (field) discriminant, a nonzero integer ^{[Lang Ch. I]}.

A Dedekind domain is an integral domain $A$ satisfying three axioms: it is Noetherian; it is integrally closed in its field of fractions; and every nonzero prime ideal is maximal, equivalently $\dim A=1$ ^{[Noether 1921]}. For every number field $K$ the ring ${\mathcal{O}}_K$ is a Dedekind domain. The three axioms are exactly what is needed to rescue factorisation: they force the unique factorisation of ideals, meaning every nonzero ideal $\mathfrak{a}\subseteq {\mathcal{O}}_K$ admits a factorisation

$$\mathfrak{a}={\mathfrak{p}}_1^{e_1}{\mathfrak{p}}_2^{e_2}\cdots {\mathfrak{p}}_g^{e_g},e_i\ge 1,$$

into nonzero prime ideals ${\mathfrak{p}}_i$, with the primes and their exponents determined by $\mathfrak{a}$.

A fractional ideal of $K$ is a nonzero finitely generated ${\mathcal{O}}_K$-submodule of $K$. The fractional ideals form an abelian group ${\mathcal{I}}_K$ under ideal multiplication, with identity ${\mathcal{O}}_K$; the inverse of $\mathfrak{a}$ is ${\mathfrak{a}}^{-1}=\{x\in K:x\mathfrak{a}\subseteq {\mathcal{O}}_K\}$. The principal fractional ideals ${\mathcal{P}}_K=\{x{\mathcal{O}}_K:x\in K^{\times }\}$ form a subgroup. The quotient

$$\mathrm{Cl}(K)={\mathcal{I}}_K/{\mathcal{P}}_K$$

is the ideal class group, and $h_K=|\mathrm{Cl}(K)|$ is the class number. The ring ${\mathcal{O}}_K$ is a principal ideal domain — equivalently, a unique factorisation domain — precisely when $h_K=1$.

The norm of a nonzero ideal is the finite index $N(\mathfrak{a})=|{\mathcal{O}}_K/\mathfrak{a}|$; it is multiplicative, $N(\mathfrak{ab})=N(\mathfrak{a})N(\mathfrak{b})$. The unit group ${\mathcal{O}}_K^{\times }$ consists of the elements of ${\mathcal{O}}_K$ whose inverse also lies in ${\mathcal{O}}_K$, equivalently those of absolute norm $\pm 1$.

Counterexamples to common slips

• ${\mathcal{O}}_K$ need not equal $\mathbb{Z}[\alpha ]$ for a chosen generator $\alpha$. For $K=\mathbb{Q}(\sqrt{5})$ one has ${\mathcal{O}}_K=\mathbb{Z}\left[\frac{1+\sqrt{5}}{2}\right]$, strictly larger than $\mathbb{Z}[\sqrt{5}]$, because $\frac{1+\sqrt{5}}{2}$ satisfies the monic equation $x^2-x-1=0$ over $\mathbb{Z}$.
• Unique factorisation of ideals does not imply unique factorisation of elements. In $\mathbb{Z}[\sqrt{-5}]$ ideals factor uniquely, yet $6=2\cdot 3=(1+\sqrt{-5})(1-\sqrt{-5})$ gives two distinct element factorisations.
• A nonprincipal ideal is a perfectly definite set, not a missing number. The ideal $\mathfrak{p}=(2,1+\sqrt{-5})$ of $\mathbb{Z}[\sqrt{-5}]$ cannot be written as $x{\mathcal{O}}_K$ for any single $x$, but it exists as a subset and satisfies ${\mathfrak{p}}^2=(2)$.

Key theorem with proof Intermediate+

Theorem (unique factorisation of ideals in a Dedekind domain). Let $A$ be a Dedekind domain with field of fractions $K$. Every nonzero ideal $\mathfrak{a}\subseteq A$ factors as a finite product of nonzero prime ideals, and this factorisation is unique up to the order of the factors.

Proof. The argument proceeds in four stages: every nonzero ideal contains a product of nonzero primes; every nonzero prime is invertible; existence of factorisation follows by Noetherian induction; uniqueness follows by cancellation.

Stage 1 — containment by a product of primes. Suppose some nonzero ideal contains no product of nonzero primes. Since $A$ is Noetherian, the collection of such ideals has a maximal member $\mathfrak{a}$. Then $\mathfrak{a}$ is not itself prime, so there exist $x,y\in A\setminus \mathfrak{a}$ with $xy\in \mathfrak{a}$. The ideals $\mathfrak{a}+(x)$ and $\mathfrak{a}+(y)$ each strictly contain $\mathfrak{a}$, so by maximality each contains a product of nonzero primes. Their product satisfies $(\mathfrak{a}+(x))(\mathfrak{a}+(y))\subseteq \mathfrak{a}+(xy)=\mathfrak{a}$, so $\mathfrak{a}$ contains a product of primes — a contradiction.

Stage 2 — invertibility of nonzero primes. Let $\mathfrak{p}$ be a nonzero prime and set ${\mathfrak{p}}^{-1}=\{x\in K:x\mathfrak{p}\subseteq A\}$, a fractional ideal containing $A$. Pick $0\ne a\in \mathfrak{p}$. By Stage 1 the principal ideal $(a)$ contains a product of nonzero primes; choose such a product ${\mathfrak{p}}_1\cdots {\mathfrak{p}}_r\subseteq (a)$ with $r$ minimal. Since $(a)\subseteq \mathfrak{p}$ and $\mathfrak{p}$ is prime, some ${\mathfrak{p}}_i\subseteq \mathfrak{p}$; both are maximal, so relabel ${\mathfrak{p}}_1=\mathfrak{p}$. Set $\mathfrak{c}=a^{-1}{\mathfrak{p}}_2\cdots {\mathfrak{p}}_r$, so $\mathfrak{pc}\subseteq A$. Minimality of $r$ gives ${\mathfrak{p}}_2\cdots {\mathfrak{p}}_r⊈(a)$, hence $\mathfrak{c}⊈A$, and any $x\in \mathfrak{c}\setminus A$ lies in ${\mathfrak{p}}^{-1}\setminus A$. So ${\mathfrak{p}}^{-1}⊋A$.

Now $\mathfrak{p}\subseteq \mathfrak{p}{\mathfrak{p}}^{-1}\subseteq A$, and $\mathfrak{p}{\mathfrak{p}}^{-1}$ is an ideal. If $\mathfrak{p}{\mathfrak{p}}^{-1}=\mathfrak{p}$, then every $x\in {\mathfrak{p}}^{-1}$ satisfies $x\mathfrak{p}\subseteq \mathfrak{p}$, so $x$ carries the finitely generated $A$-module $\mathfrak{p}$ into itself and is therefore integral over $A$; integral closedness forces $x\in A$, contradicting ${\mathfrak{p}}^{-1}⊋A$. Hence $\mathfrak{p}⊊\mathfrak{p}{\mathfrak{p}}^{-1}$, and maximality of $\mathfrak{p}$ gives $\mathfrak{p}{\mathfrak{p}}^{-1}=A$. The prime $\mathfrak{p}$ is invertible.

Stage 3 — existence. Assume, for contradiction, that some nonzero ideal admits no prime factorisation, and let $\mathfrak{a}$ be maximal among them. Then $\mathfrak{a}$ is not prime, so it lies strictly inside a maximal (hence prime) ideal $\mathfrak{p}$. Because $\mathfrak{p}$ is invertible, $\mathfrak{a}{\mathfrak{p}}^{-1}\subseteq \mathfrak{p}{\mathfrak{p}}^{-1}=A$ is an integral ideal, and $\mathfrak{a}\subseteq \mathfrak{a}{\mathfrak{p}}^{-1}$. The containment is strict: equality would force $\mathfrak{a}{\mathfrak{p}}^{-1}=\mathfrak{a}$, and every $x\in {\mathfrak{p}}^{-1}$ would carry the finitely generated module $\mathfrak{a}$ into itself, giving $x$ integral over $A$ and hence $x\in A$, so ${\mathfrak{p}}^{-1}=A$ — contradicting Stage 2. By maximality the strictly larger ideal $\mathfrak{a}{\mathfrak{p}}^{-1}$ factors, say $\mathfrak{a}{\mathfrak{p}}^{-1}={\mathfrak{p}}_1\cdots {\mathfrak{p}}_s$; multiplying by $\mathfrak{p}$ exhibits $\mathfrak{a}=\mathfrak{p}{\mathfrak{p}}_1\cdots {\mathfrak{p}}_s$, a contradiction.

Stage 4 — uniqueness. Suppose ${\mathfrak{p}}_1\cdots {\mathfrak{p}}_r={\mathfrak{q}}_1\cdots {\mathfrak{q}}_s$ with all factors prime. The prime ${\mathfrak{p}}_1$ contains the right-hand product, hence contains some ${\mathfrak{q}}_j$; both are maximal, so ${\mathfrak{p}}_1={\mathfrak{q}}_j$. Multiplying both sides by ${\mathfrak{p}}_1^{-1}$ (which exists by Stage 2) cancels the common factor, and induction on $r+s$ gives uniqueness up to reordering. $\square$

Bridge. Unique factorisation of ideals builds toward the spectrum of a ring 04.02.01, where prime ideals are the geometric points of an affine scheme, and appears again in the local-field theory of 21.02.07, where the factorisation of $p{\mathcal{O}}_K$ is read off one completion at a time. The foundational reason ideal factorisation repairs what element factorisation loses is that passing from elements to ideals linearises multiplication: the monoid of nonzero ideals is free on the prime ideals, while the monoid of nonzero elements need not be. This is exactly the mechanism by which a Dedekind scheme acquires a well-behaved divisor theory, and the central insight is that the class group measures the single obstruction — principality — between these two monoids. The bridge is that invertibility of prime ideals is dual to the integrally-closed condition: dropping either collapses unique factorisation, and putting these together, the repair generalises to the wider class of Krull domains, of which the Dedekind domains are the Noetherian one-dimensional case.

Exercises Intermediate+

Lean formalization Intermediate+

lean_status: none is recorded for this curriculum unit because the project-level bridge — narrating the failure of element factorisation in the quadratic ring $\mathbb{Z}[\sqrt{-5}]$ and its restoration by prime-ideal factorisation, and computing the class group $\mathrm{Cl}(\mathbb{Q}(\sqrt{-5}))\cong \mathbb{Z}/2$ end to end from the Dedekind axioms — is not assembled as a single formal development in Mathlib, although the individual pillars (IsDedekindDomain, RingOfIntegers, FractionalIdeal, ClassGroup, and the unique-factorisation result for IsDedekindDomain) are present. The gap is pedagogical assembly work, described in the unit metadata.

Advanced results Master

Theorem (Dirichlet unit theorem, statement). Let $K$ be a number field of signature $(r_1,r_2)$ and set $r=r_1+r_2-1$. Then the unit group decomposes as

$${\mathcal{O}}_K^{\times }\cong {\mu }_K\times {\mathbb{Z}}^r,$$

where ${\mu }_K$ is the finite cyclic group of roots of unity in $K$. The integer $r$ is the unit rank. The proof embeds the units into a real vector space by the logarithmic map $L(\epsilon )=(\log |{\sigma }_1(\epsilon )|,\dots ,\log |{\sigma }_{r_1}(\epsilon )|,2\log |{\tau }_1(\epsilon )|,\dots ,2\log |{\tau }_{r_2}(\epsilon )|)$, whose image lies in the trace-zero hyperplane of dimension $r$ and is a full lattice there; this is the analogue, for units, of the lattice geometry that governs the class group ^{[Neukirch Ch. I §7]}. The full development of the unit theorem and the regulator belongs to 21.02.07.

Theorem (finiteness of the class number, statement). For every number field $K$ the ideal class group $\mathrm{Cl}(K)$ is finite. Equivalently, there are only finitely many ideal classes. The standard proof converts the arithmetic question into a lattice-point count via Minkowski's convex-body theorem, which forces every ideal class to contain a representative of bounded norm; only finitely many ideals have norm below a given bound. The complete argument, with the Minkowski bound $M_K=(4/\pi {)}^{r_2}(n!/n^n)\sqrt{|d_K|}$, is given in 21.02.07. What matters here is the structural consequence: the failure of unique factorisation of elements, which a priori could be uncontrolled, is in fact measured by a single finite abelian group.

Theorem (ramification and the discriminant). For a rational prime $p$, the ideal $p{\mathcal{O}}_K$ factors as $p{\mathcal{O}}_K={\mathfrak{p}}_1^{e_1}\cdots {\mathfrak{p}}_g^{e_g}$ with ${\sum }_i{e}_i{f}_i=n$, where $f_i=[{\mathcal{O}}_K/{\mathfrak{p}}_i:{\mathbb{F}}_p]$ is the residue degree. A prime $p$ ramifies (some $e_i\ge 2$) if and only if $p$ divides the discriminant $d_K$ ^{[Neukirch Ch. I §8]}. Thus only finitely many primes ramify, namely the prime divisors of $d_K$. In $\mathbb{Z}[\sqrt{-5}]$, where $d_K=-20$, the ramified primes are $2$ and $5$: $2$ ramifies as $(2)={\mathfrak{p}}^2$ (Exercise 5) and $5$ ramifies as $(5)=(\sqrt{-5}{)}^2$.

Theorem (Hermite-Minkowski finiteness). For fixed degree $n$ and bound $B$, only finitely many number fields $K$ satisfy $[K:\mathbb{Q}]=n$ and $|d_K|\le B$; and $|d_K|>1$ for every $K\ne \mathbb{Q}$. The discriminant is therefore a coarse but faithful invariant separating number fields, and $\mathbb{Q}$ is the unique number field with no ramified prime. The geometry-of-numbers engine behind this statement is developed in 21.02.07.

Synthesis. Algebraic number theory is governed by the replacement of element factorisation by ideal factorisation, and the foundational reason the replacement works is exactly the three Dedekind axioms — Noetherian, integrally closed, dimension one — which jointly force every nonzero prime to be invertible. The central insight is that principality is the unique obstruction between the monoid of nonzero elements and the free monoid of nonzero ideals: this is exactly what the class group $\mathrm{Cl}(K)$ measures, and the same obstruction appears again in the Picard group of invertible sheaves on the Dedekind scheme $\mathrm{Spec}{\mathcal{O}}_K$ 04.02.01. The bridge is that invertibility and integral closedness are dual conditions, and putting these together, the whole theory generalises to Krull domains, of which ${\mathcal{O}}_K$ is the Noetherian one-dimensional instance; the class group builds toward the analytic class-number formula of 21.03.03, where $h_K$ is read off the residue of the Dedekind zeta function ${\zeta }_K$ at $s=1$.

Full proof set Master

Proposition 1 (the two element factorisations of $6$ reconcile at the ideal level). In $R=\mathbb{Z}[\sqrt{-5}]={\mathcal{O}}_{\mathbb{Q}(\sqrt{-5})}$ the principal ideals generated by the irreducible factors of $6$ satisfy $(2)={\mathfrak{p}}^2$, $(3)=\mathfrak{q}\bar{\mathfrak{q}}$, $(1+\sqrt{-5})=\mathfrak{pq}$, $(1-\sqrt{-5})=\mathfrak{p}\bar{\mathfrak{q}}$, with $\mathfrak{p}=(2,1+\sqrt{-5})$, $\mathfrak{q}=(3,1+\sqrt{-5})$, $\bar{\mathfrak{q}}=(3,1-\sqrt{-5})$. Consequently

$$(6)=(2)(3)={\mathfrak{p}}^2\mathfrak{q}\bar{\mathfrak{q}}=(\mathfrak{pq})(\mathfrak{p}\bar{\mathfrak{q}})=(1+\sqrt{-5})(1-\sqrt{-5}),$$

so the two distinct element factorisations of $6$ refine to a single prime-ideal factorisation.

Proof. Reduce $x^2+5$ modulo $2$ and $3$. Modulo $2$, $x^2+5\equiv (x+1{)}^2$, so $2$ ramifies and $(2)={\mathfrak{p}}^2$ with $\mathfrak{p}=(2,1+\sqrt{-5})$; the quotient $R/\mathfrak{p}$ has two classes ($\sqrt{-5}$ is identified with $1$ and reduced modulo $2$), so $N(\mathfrak{p})=2$ and $\mathfrak{p}$ is maximal. Modulo $3$, $x^2+5\equiv (x-1)(x+1)$, so $3$ splits and $(3)=\mathfrak{q}\bar{\mathfrak{q}}$ with $\mathfrak{q}=(3,1+\sqrt{-5})$, $\bar{\mathfrak{q}}=(3,1-\sqrt{-5})$; each quotient has three classes, so $N(\mathfrak{q})=N(\bar{\mathfrak{q}})=3$ and both are maximal, hence prime. The element $1+\sqrt{-5}$ has absolute norm $1+5=6$, so $N((1+\sqrt{-5}))=6=2\cdot 3$. Because $1+\sqrt{-5}$ lies in both $\mathfrak{p}$ and $\mathfrak{q}$, the principal ideal $(1+\sqrt{-5})$ is divisible by $\mathfrak{pq}$, and equality of norms forces $(1+\sqrt{-5})=\mathfrak{pq}$. Symmetrically $(1-\sqrt{-5})=\mathfrak{p}\bar{\mathfrak{q}}$. Multiplying yields the displayed identity. $\square$

Proposition 2 (the ring of integers is integrally closed). For any number field $K$, the ring ${\mathcal{O}}_K$ is integrally closed in $K$: an element of $K$ that is integral over ${\mathcal{O}}_K$ already lies in ${\mathcal{O}}_K$.

Proof. Let $\alpha \in K$ be integral over ${\mathcal{O}}_K$, so ${\alpha }^m+a_{m-1}{\alpha }^{m-1}+\cdots +a_0=0$ for some $a_i\in {\mathcal{O}}_K$. The subring $B=\mathbb{Z}[a_0,\dots ,a_{m-1}]\subseteq {\mathcal{O}}_K$ is a finitely generated $\mathbb{Z}$-module: each $a_i$ is an algebraic integer, so $\mathbb{Z}[a_i]$ is finite over $\mathbb{Z}$, and adjoining finitely many such elements preserves finiteness. The relation above shows that $\alpha$ is integral over $B$, so $B[\alpha ]$ is a finitely generated $B$-module, hence a finitely generated $\mathbb{Z}$-module. Integrality is transitive, so $\alpha$ is integral over $\mathbb{Z}$, which places $\alpha$ among the algebraic integers. Hence $\alpha \in {\mathcal{O}}_K$ by definition. $\square$

Proposition 3 (in a number field, the units are exactly the elements of absolute norm $\pm 1$). For $\alpha \in {\mathcal{O}}_K$, one has $\alpha \in {\mathcal{O}}_K^{\times }$ if and only if $|N_{K/\mathbb{Q}}(\alpha )|=1$.

Proof. If $\alpha \beta =1$ with $\beta \in {\mathcal{O}}_K$, multiplicativity of the absolute norm gives $N_{K/\mathbb{Q}}(\alpha )N_{K/\mathbb{Q}}(\beta )=N_{K/\mathbb{Q}}(1)=1$; both factors are rational integers, so $|N_{K/\mathbb{Q}}(\alpha )|=1$. Conversely, suppose $|N_{K/\mathbb{Q}}(\alpha )|=1$, and let $f(x)=x^d+c_{d-1}{x}^{d-1}+\cdots +c_0\in \mathbb{Z}[x]$ be the monic minimal polynomial of $\alpha$ over $\mathbb{Q}$. Its constant term is $c_0=(-1{)}^d{N}_{K/\mathbb{Q}}(\alpha )=\pm 1$. Substituting $\alpha$ and dividing by $\alpha$ (which is nonzero, since its norm is nonzero) yields

$$\alpha ({\alpha }^{d-1}+c_{d-1}{\alpha }^{d-2}+\cdots +c_1)=-c_0=\mp 1,$$

so ${\alpha }^{-1}=\mp ({\alpha }^{d-1}+c_{d-1}{\alpha }^{d-2}+\cdots +c_1)$, which lies in $\mathbb{Z}[\alpha ]\subseteq {\mathcal{O}}_K$. Therefore $\alpha$ possesses an inverse in ${\mathcal{O}}_K$ and is a unit. $\square$

Connections Master

• Rings and modules 01.03.01. The entire edifice of this unit runs on the ideal-and-module machinery of 01.03.01: ideals are the submodules of ${\mathcal{O}}_K$ over itself, the quotient ${\mathcal{O}}_K/\mathfrak{a}$ that defines the ideal norm is a ring quotient, and the Noetherian and integrally-closed conditions in the Dedekind axiomatisation are imported directly from the ideal theory established there. The first isomorphism theorem for rings underlies the norm multiplicativity used throughout the exercises.

• Fields and Galois theory 01.04.01. A number field is by definition a finite extension of $\mathbb{Q}$, and the embeddings $K\hookrightarrow \mathbb{C}$, the signature $(r_1,r_2)$, and the norm and trace are the field-theoretic invariants of 01.04.01 specialised to characteristic zero. Ramification of primes is the ideal-theoretic shadow of how the polynomial $x^2+5$ factors modulo $p$, read through the Galois theory of the splitting field.

• Commutative algebra 01.05.01. The Dedekind domain is the commutative-algebra abstraction of the ring of integers: the three axioms Noetherian, integrally closed, dimension one are the structural conditions isolated by Noether and developed in 01.05.01, and unique factorisation of ideals is the concrete reward. The localisation ${\mathcal{O}}_K\to ({\mathcal{O}}_K{)}_{\mathfrak{p}}$ at a prime ideal, a discrete valuation ring, is the local ring on which the local-global analysis of ramification is built.

• Schemes 04.02.01. The prime ideals of ${\mathcal{O}}_K$ are the closed points of the affine scheme $\mathrm{Spec}{\mathcal{O}}_K$, a one-dimensional Noetherian integral scheme that is the geometric avatar of a Dedekind domain. Unique factorisation of ideals is the statement that every Weil divisor on this scheme is locally principal, and the class group $\mathrm{Cl}(K)$ coincides with the Picard group of invertible sheaves — the bridge from arithmetic to geometry.

• Analytic number theory: the class-number formula 21.03.03 and the Minkowski engine 21.02.07. The class group and class number constructed here are the input to the analytic class-number formula of 21.03.03, which reads $h_K$ off the residue of the Dedekind zeta function at $s=1$, and the finiteness of $h_K$ together with the Dirichlet unit theorem are proved in full via the geometry of numbers in 21.02.07. This unit supplies the structural definitions those units consume.

Historical & philosophical context Master

The subject was forced into existence by the failure of unique factorisation in algebraic number systems. Kummer, pursuing Fermat's Last Theorem in the cyclotomic fields $\mathbb{Q}({\zeta }_p)$, found that element factorisation collapses there; his 1847 paper Zur Theorie der complexen Zahlen ^{[Kummer 1847]} introduced "ideal numbers" — fictitious divisors that restore a form of unique factorisation where element factorisation fails. Dedekind's Supplement X to Dirichlet's Vorlesungen über Zahlentheorie (1871) ^{[Dedekind 1871]} gave the lasting reformulation: replace ideal numbers by honest sets, the ideals, and prove that in the ring of integers of any number field the nonzero ideals factor uniquely into prime ideals. The three properties of ${\mathcal{O}}_K$ that make the proof work — Noetherian, integrally closed, every nonzero prime maximal — were abstracted by Noether in Idealtheorie in Ringbereichen (1921) ^{[Noether 1921]} into the general notion now called a Dedekind domain. Hilbert's Zahlbericht (1897) ^{[Hilbert 1897]} codified the resulting theory of number fields and set the agenda for class field theory.

The conceptual move from elements to ideals is the founding gesture of modern commutative algebra: it reframes an obstruction, non-unique factorisation, as a structure, the class group. Dedekind's insistence that the objects of study are ideals rather than ideal numbers made the theory independent of any particular presentation of the field, and Noether's subsequent axiomatisation carried the ideal-theoretic viewpoint into commutative algebra and, eventually, into the scheme theory of 04.02.01.

Bibliography Master

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}

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}

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