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Chapter 7

Algebraic Number Theory

Algebraic integers and the integrality criterion; the ring of integers, computed explicitly for quadratic fields including the d = 1 mod 4 case; the failure of unique factorisation in Z[sqrt(-5)] and its restoration at the level of ideals in a Dedekind domain; the ideal class group, with class number one exactly when the ring of integers is a UFD; and Dirichlet's unit theorem with the rank formula worked for quadratic, cubic and cyclotomic fields.

Introduction

Ordinary arithmetic rests on one fact: every integer above $1$ factors into primes in exactly one way. Algebraic number theory begins with the discovery that this fact is an accident of $Z$ , not a law of arithmetic. Enlarge $Z$ by adjoining $- 5$ and the number $6$ acquires two genuinely different factorisations into irreducibles; enlarge it by adjoining $i$ and nothing goes wrong at all. The subject exists to explain the difference, to measure it, and — this is the surprise — to repair it.

The repair is Kummer's and Dedekind's, and it is the central idea of the chapter. Unique factorisation fails for elements but is restored for ideals: in the ring of integers of any number field, every nonzero ideal is a product of prime ideals in exactly one way. The elements that were missing their factors turn out to have had them all along, hidden inside ideals that are not generated by any single element. The obstruction — how far the ideals are from being principal — is itself a finite abelian group, the ideal class group, and its order, the class number $h_{K}$ , is a single integer that says how badly unique factorisation fails. It equals $1$ precisely when nothing went wrong.

The chapter develops this in four movements. First the objects: algebraic integers, why they form a ring, and what the ring of integers $O_{K}$ of a number field looks like concretely — including the quadratic case computed in full, where the condition $d \equiv 1 (mod 4)$ produces the element $\frac{1 + d}{2}$ that so many first computations miss. Then the structural theory: Dedekind domains, and unique factorisation of ideals. Then the class group and its finiteness. Finally the units, where Dirichlet's theorem pins the unit group down to a finite group times a free abelian group of rank $r_{1} + r_{2} - 1$ .

Two proofs in the chapter are genuinely beyond what can be built here from scratch — the finiteness of the class group and Dirichlet's unit theorem both need Minkowski's convex body theorem and a geometry-of-numbers argument. Those appear as clearly labelled sketches that state exactly what they assume. Everything else is proved.

Background assumed: ideals and quotient rings as in Ideals and Quotient Rings, integral domains, fields and the notion of a UFD as in Integral Domains and Fields, factorisation in polynomial rings and Gauss's lemma as in Polynomial Rings and Unique Factorization, and the language of field extensions and embeddings from Galois Theory.

7.1What it means to be an algebraic integer

An integer is a rational number with denominator $1$ . That description does not survive being moved to a larger field, because "denominator" is not an intrinsic notion. The description that does survive is the one that mentions only polynomials.

Definition 7.1 (Algebraic integer). A complex number $α$ is an algebraic integer if there is a monic polynomial

f (x) = x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{1} x + a_{0}, a_{0}, \dots, a_{n - 1} \in Z,

with $f (α) = 0$ . The set of all algebraic integers is written $\overline{Z}$ .

Monicity is the entire content of the definition. Drop it and every rational number qualifies, since $q = a / b$ is a root of $b x - a$ . Keep it and, as the next section shows, the only rational numbers that qualify are the ordinary integers. The leading coefficient $1$ is what plays the role of "denominator $1$ ".

Example 7.2 (Three algebraic integers and one impostor). Decide whether each of $2$ , the golden ratio $φ = \frac{1 + 5}{2}$ , $\frac{1 + - 3}{2}$ , and $\frac{1 + - 5}{2}$ is an algebraic integer.

Solution.

$2$ is a root of $x^{2} - 2$ , monic over $Z$ . Yes.
$φ$ satisfies $φ^{2} = φ + 1$ , so it is a root of $x^{2} - x - 1$ . Yes — and note the denominator $2$ in the expression is not an obstruction.
Put $α = \frac{1 + - 3}{2}$ . Then $2 α - 1 = - 3$ , so $4 α^{2} - 4 α + 1 = - 3$ , giving $α^{2} - α + 1 = 0$ . Yes.
Put $β = \frac{1 + - 5}{2}$ . The same computation gives $4 β^{2} - 4 β + 1 = - 5$ , so $β^{2} - β + \frac{3}{2} = 0$ . The coefficient $\frac{3}{2}$ is not an integer, and since this quadratic is irreducible over $Q$ it is the minimal polynomial; by the criterion Integrality is visible in the minimal polynomial below, $β$ is not an algebraic integer.

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The contrast between items 3 and 4 is the whole of the quadratic theory in miniature: halving $1 + d$ produces an algebraic integer for $d = - 3$ and not for $d = - 5$ , and the reason is that $- 3 \equiv 1 (mod 4)$ while $- 5 \equiv 3 (mod 4)$ . The section Rings of integers of quadratic fields makes this precise.

Definition 7.3 (Integral over a ring). Let $R \subseteq S$ be commutative rings. An element $s \in S$ is integral over $R$ if $f (s) = 0$ for some monic $f \in R [x]$ . If every element of $S$ is integral over $R$ , then $S$ is integral over $R$ . The set of elements of $S$ integral over $R$ is the integral closure of $R$ in $S$ , and $R$ is integrally closed in $S$ if that closure is $R$ itself.

Algebraic integer means exactly "integral over $Z$ ", with $S = C$ . Phrasing it this way costs nothing and buys the ability to say "the integral closure of $Z$ in $K$ ", which is the definition of $O_{K}$ .

Intuition. Think of a monic equation as a budget. The relation $α^{n} = - (a_{n - 1} α^{n - 1} + \dots + a_{0})$ says that once you reach the $n$ -th power of $α$ you never need a higher one: the powers of $α$ live inside a finitely generated $Z$ -module. A non-monic relation like $2 α = 1$ gives no such budget — it lets you trade one $α$ for a half, and iterating produces denominators $2, 4, 8, \dots$ without bound. Integrality is finiteness, and the proofs below say so literally.

Exercises for §7.1 →

7.2Algebraic numbers, algebraic integers, and the rational case

Definition 7.4 (Algebraic number). A complex number $α$ is an algebraic number if $f (α) = 0$ for some nonzero $f \in Q [x]$ . Equivalently, $α$ is algebraic over $Q$ in the sense of Galois Theory. Its minimal polynomial $m_{α} \in Q [x]$ is the monic rational polynomial of least degree with $m_{α} (α) = 0$ , and its degree is $de g m_{α}$ .

Every algebraic integer is an algebraic number; $\frac{1}{2}$ shows the converse fails. The following theorem is the tool that decides the question in practice, and it is why nobody ever has to hunt for a clever monic polynomial: there is only one candidate to check.

Theorem 7.5 (Integrality is visible in the minimal polynomial). Let $α$ be an algebraic number with minimal polynomial $m_{α} \in Q [x]$ . Then $α$ is an algebraic integer if and only if $m_{α} \in Z [x]$ .

Proof. ( $\Leftarrow$ ) If $m_{α}$ has integer coefficients it is itself a monic integer polynomial killing $α$ , so $α$ is an algebraic integer.

( $\Rightarrow$ ) Suppose $f \in Z [x]$ is monic with $f (α) = 0$ . By minimality $m_{α} ∣ f$ in $Q [x]$ , say $f = m_{α} g$ with $g \in Q [x]$ monic, since $f$ and $m_{α}$ are. We must show $m_{α}$ and $g$ lie in $Z [x]$ .

Recall from Polynomial Rings and Unique Factorization that the content $c (h)$ of a nonzero $h \in Q [x]$ is the positive rational with $h = c (h) h_{0}$ and $h_{0} \in Z [x]$ primitive, and that Gauss's lemma gives $c (h_{1} h_{2}) = c (h_{1}) c (h_{2})$ . Write $m_{α} = c (m_{α}) M$ and $g = c (g) G$ with $M, G \in Z [x]$ primitive. Then

f = c (m_{α}) c (g) MG,

and $MG$ is primitive by Gauss's lemma, so $c (f) = c (m_{α}) c (g)$ . But $f$ is monic with integer coefficients, hence $c (f)$ is the reciprocal of a positive integer dividing every coefficient — in fact $c (f) \in Z_{> 0}$ and $c (f)$ divides the leading coefficient $1$ , so $c (f) = 1$ .

Now compare leading coefficients. Let $M$ have leading coefficient $u \in Z$ and $G$ leading coefficient $v \in Z$ . Since $m_{α}$ is monic, $c (m_{α}) u = 1$ , so $c (m_{α}) = 1/ u$ ; likewise $c (g) = 1/ v$ . From $c (m_{α}) c (g) = 1$ we get $uv = 1$ , so $u = v = \pm 1$ , and then $c (m_{α}) = \pm 1$ is a positive rational, hence $c (m_{α}) = 1$ and $m_{α} = M \in Z [x]$ .∎

Corollary 7.6 (Rational algebraic integers are integers). $\overline{Z} \cap Q = Z$ . Equivalently, $Z$ is integrally closed in $Q$ .

Proof. Certainly $Z \subseteq \overline{Z} \cap Q$ . Conversely let $q \in Q$ be an algebraic integer. Its minimal polynomial over $Q$ is $x - q$ , of degree $1$ , so by the theorem Integrality is visible in the minimal polynomial we get $x - q \in Z [x]$ , that is $q \in Z$ .∎

This corollary is the sanity check that the definition was the right one: the new notion of integer agrees with the old one wherever both apply. It is also the workhorse behind almost every later argument — whenever a rational number is produced that is known to be an algebraic integer, it is an honest integer, and that is how traces, norms and discriminants are forced to be integral.

Example 7.7 (Rational root test as a special case). Show directly that $\frac{3}{5}$ is not an algebraic integer, without appealing to the corollary.

Solution.

Suppose $f (x) = x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{0} \in Z [x]$ satisfies $f (3/5) = 0$ .
Multiply by $5^{n}$ : $3^{n} + a_{n - 1} 3^{n - 1} 5 + \dots + a_{0} 5^{n} = 0$ .
Every term after the first is divisible by $5$ , so $5 ∣ 3^{n}$ , which is false.
Hence no monic integer polynomial kills $3/5$ . The general argument is identical and proves the corollary again.

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Pitfall. " $α$ satisfies a polynomial with integer coefficients, so $α$ is an algebraic integer" is false, and it is the single most common slip. $\frac{1}{2}$ satisfies $2 x - 1$ , which has integer coefficients. The polynomial must be monic. When in doubt, compute the minimal polynomial and look at its coefficients — the theorem Integrality is visible in the minimal polynomial says that is the only test needed.

Exercises for §7.2 →

7.3The algebraic integers form a ring

That $2 + 3$ is an algebraic integer can be checked by hand: it is a root of $x^{4} - 10 x^{2} + 1$ . That $37 + 511$ is one cannot reasonably be checked by hand, and the polynomial it satisfies has degree $15$ . A structural proof is needed, and the right one replaces polynomial gymnastics with a finiteness statement about modules.

Lemma 7.8 (Integrality equals module-finiteness). Let $R$ be a subring of a commutative ring $S$ and let $s \in S$ . The following are equivalent.

$s$ is integral over $R$ .
$R [s]$ is finitely generated as an $R$ -module.
There is a subring $T$ with $R [s] \subseteq T \subseteq S$ that is finitely generated as an $R$ -module.

Proof. $(1) \Rightarrow (2)$ . Let $s^{n} = - (a_{n - 1} s^{n - 1} + \dots + a_{0})$ with $a_{i} \in R$ . By induction every power $s^{m}$ with $m \geq n$ is an $R$ -combination of $1, s, \dots, s^{n - 1}$ : true for $m = n$ by the relation, and if it holds for $m$ then multiplying by $s$ expresses $s^{m + 1}$ in terms of $s, \dots, s^{n}$ , and $s^{n}$ is rewritten again by the relation. Hence $R [s] = R \cdot 1 + R s + \dots + R s^{n - 1}$ .

$(2) \Rightarrow (3)$ . Take $T = R [s]$ .

$(3) \Rightarrow (1)$ . Let $T = R t_{1} + \dots + R t_{m}$ with $t_{1}, \dots, t_{m}$ generating $T$ as an $R$ -module, and arrange $t_{1} = 1$ by adjoining it if necessary. Multiplication by $s$ maps $T$ into $T$ , because $s \in T$ and $T$ is a ring. So there are $c_{ij} \in R$ with

s t_{i} = j = 1 \sum m c_{ij} t_{j} (i = 1, \dots, m) .

Writing $C = (c_{ij})$ and $t = (t_{1}, \dots, t_{m})^{⊤}$ , this says $(s I - C) t = 0$ . Multiply on the left by the adjugate matrix $adj (s I - C)$ , which has entries in $R [s]$ , and use $adj (A) A = det (A) I$ : we get $det (s I - C) t_{i} = 0$ for every $i$ . Taking $i$ with $t_{i} = 1$ gives $det (s I - C) = 0$ . Expanding the determinant, $det (x I - C)$ is a monic polynomial of degree $m$ with coefficients in $R$ — it is the characteristic polynomial of $C$ — so $s$ is a root of a monic polynomial over $R$ .∎

Remark. Step $(3) \Rightarrow (1)$ is the determinant trick, and the hypothesis $t_{1} = 1$ matters: without an element that the relation $det (s I - C) t_{i} = 0$ can be tested against, the determinant need only annihilate $T$ , not vanish. This is exactly the point at which "finitely generated module" is upgraded to "monic equation", and it is a form of the Cayley–Hamilton theorem.

Theorem 7.9 (The algebraic integers form a ring). If $α$ and $β$ are integral over a subring $R$ of a commutative ring $S$ , then $α + β$ , $α - β$ and $α β$ are integral over $R$ . In particular $\overline{Z}$ is a subring of $C$ , and the integral closure of $R$ in $S$ is a ring.

Proof. Let $α$ satisfy a monic relation of degree $m$ over $R$ and $β$ a monic relation of degree $n$ . Put $T = R [α, β]$ . By the argument in $(1) \Rightarrow (2)$ of the lemma Integrality equals module-finiteness, applied first to $α$ over $R$ and then to $β$ over $R [α]$ , the module $T$ is generated over $R$ by the $mn$ products

α^{i} β^{j}, 0 \leq i < m, 0 \leq j < n .

So $T$ is a finitely generated $R$ -module, it is a ring, and it contains $1$ . Each of $α + β$ , $α - β$ , $α β$ lies in $T$ , so for $s$ any one of them we have $R [s] \subseteq T \subseteq S$ with $T$ module-finite over $R$ ; condition (3) of the lemma applies and $s$ is integral over $R$ .∎

Intuition. No polynomial for $α + β$ is ever produced, and that is the point. The proof says: whatever $α + β$ is, it stirs the finite-dimensional space spanned by the $α^{i} β^{j}$ around inside itself, and a linear map on a finite-dimensional space satisfies its own characteristic polynomial, which is monic. Degree $mn$ comes out for free — the degree- $15$ polynomial for $37 + 511$ exists because $3 \times 5 = 15$ .

Example 7.10 (A sum of two square roots). Verify by hand that $α = 2 + 3$ is an algebraic integer, and compare the degree with the bound $mn$ from the theorem.

Solution.

$α^{2} = 5 + 26$ , so $α^{2} - 5 = 26$ .
Squaring, $(α^{2} - 5)^{2} = 24$ , that is $α^{4} - 10 α^{2} + 1 = 0$ .
The polynomial $x^{4} - 10 x^{2} + 1$ is monic with integer coefficients, so $α \in \overline{Z}$ .
Here $m = n = 2$ , so the theorem promises degree at most $4$ , and $4$ is attained: $x^{4} - 10 x^{2} + 1$ is irreducible over $Q$ because $α$ lies in no proper subfield of $Q (2, 3)$ , a degree- $4$ field.

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Corollary 7.11 (Integral closure is transitive). Let $R \subseteq S \subseteq T$ be commutative rings with $S$ integral over $R$ and $t \in T$ integral over $S$ . Then $t$ is integral over $R$ .

Proof. Let $t^{n} + s_{n - 1} t^{n - 1} + \dots + s_{0} = 0$ with $s_{i} \in S$ . Put $R^{'} = R [s_{0}, \dots, s_{n - 1}]$ . Each $s_{i}$ is integral over $R$ , so applying the lemma Integrality equals module-finiteness repeatedly, $R^{'}$ is a finitely generated $R$ -module. The same relation shows $t$ is integral over $R^{'}$ , so $R^{'} [t]$ is a finitely generated $R^{'}$ -module. A module-finite extension of a module-finite extension is module-finite — if $R^{'} = \sum R u_{i}$ and $R^{'} [t] = \sum R^{'} v_{j}$ then $R^{'} [t] = \sum R u_{i} v_{j}$ — so $R^{'} [t]$ is a finitely generated $R$ -module containing $R [t]$ . Condition (3) of the lemma gives that $t$ is integral over $R$ .∎

Corollary 7.12 (Clearing a denominator). Let $α$ be an algebraic number. Then there is a nonzero $m \in Z$ with $m α \in \overline{Z}$ .

Proof. Let $b_{n} α^{n} + \dots + b_{0} = 0$ with $b_{i} \in Z$ and $b_{n} \neq = 0$ . Multiply through by $b_{n}^{n - 1}$ and set $β = b_{n} α$ :

β^{n} + b_{n - 1} β^{n - 1} + b_{n} b_{n - 2} β^{n - 2} + \dots + b_{n}^{n - 1} b_{0} = 0,

a monic integer relation for $β$ . So $m = b_{n}$ works.∎

The last corollary says that a number field is not much bigger than its algebraic integers: every element becomes one after scaling. That is what makes $O_{K}$ large enough to be worth studying — it spans $K$ over $Q$ .

Exercises for §7.3 →

7.4Minimal polynomial, characteristic polynomial, norm and trace

Let $K$ be a number field: a finite extension of $Q$ , of degree $n = [K : Q]$ . By the primitive element theorem from Galois Theory, $K = Q (θ)$ for some $θ$ , and there are exactly $n$ field embeddings $σ_{1}, \dots, σ_{n} : K \to C$ , determined by sending $θ$ to the $n$ distinct roots of its minimal polynomial.

Definition 7.13 (Norm and trace). For $α \in K$ with embeddings $σ_{1}, \dots, σ_{n}$ as above, the norm and trace of $α$ from $K$ to $Q$ are

N_{K / Q} (α) = i = 1 \prod n σ_{i} (α), Tr_{K / Q} (α) = i = 1 \sum n σ_{i} (α) .

The characteristic polynomial of $α$ is $χ_{α} (x) = \prod_{i = 1}^{n} (x - σ_{i} (α))$ .

Equivalently — and this is often the easier description to compute with — multiplication by $α$ is a $Q$ -linear map $K \to K$ , and $χ_{α}$ , $N_{K / Q} (α)$ , $Tr_{K / Q} (α)$ are its characteristic polynomial, determinant and trace.

Proposition 7.14 (Characteristic polynomial is a power of the minimal polynomial). Let $α \in K$ have minimal polynomial $m_{α}$ of degree $d$ over $Q$ . Then $d ∣ n$ and

χ_{α} (x) = m_{α} (x)^{n / d} .

Consequently $N_{K / Q} (α) = ((- 1)^{d} m_{α} (0))^{n / d}$ and $Tr_{K / Q} (α) = - \frac{n}{d} c_{d - 1}$ , where $c_{d - 1}$ is the coefficient of $x^{d - 1}$ in $m_{α}$ .

Proof. Let $F = Q (α)$ , so $[F : Q] = d$ and, by the tower law, $[K : F] = n / d =: e$ is an integer. Each of the $d$ embeddings $τ : F \to C$ sends $α$ to a distinct root of $m_{α}$ , and each extends to exactly $e$ embeddings of $K$ into $C$ , since $K / F$ is separable of degree $e$ (characteristic zero). So among $σ_{1} (α), \dots, σ_{n} (α)$ each root of $m_{α}$ occurs exactly $e$ times, giving

χ_{α} (x) = i = 1 \prod n (x - σ_{i} (α)) = (roots ρ of m_{α} \prod (x - ρ))^{e} = m_{α} (x)^{e} .

The norm is $(- 1)^{n} χ_{α} (0) = (- 1)^{n} m_{α} (0)^{e} = ((- 1)^{d} m_{α} (0))^{e}$ since $n = d e$ , and the trace is minus the coefficient of $x^{n - 1}$ in $χ_{α} = m_{α}^{e}$ , which is $e$ times the coefficient of $x^{d - 1}$ in $m_{α}$ .∎

Corollary 7.15 (Norms and traces of algebraic integers are rational integers). If $α \in O_{K}$ , then $χ_{α} \in Z [x]$ ; in particular $N_{K / Q} (α) \in Z$ and $Tr_{K / Q} (α) \in Z$ .

Proof. By the theorem Integrality is visible in the minimal polynomial, $m_{α} \in Z [x]$ , and $χ_{α} = m_{α}^{n / d}$ is then a product of integer polynomials, hence lies in $Z [x]$ . Its constant term up to sign is the norm and its subleading coefficient up to sign the trace.∎

Proposition 7.16 (Norm detects units). Let $α \in O_{K}$ . Then $α \in O_{K}^{\times}$ if and only if $N_{K / Q} (α) = \pm 1$ .

Proof. ( $\Rightarrow$ ) If $α β = 1$ with $β \in O_{K}$ , then applying each $σ_{i}$ and multiplying gives $N (α) N (β) = N (1) = 1$ . Both factors are rational integers by the corollary Norms and traces of algebraic integers are rational integers, so each is $\pm 1$ .

( $\Leftarrow$ ) Suppose $N_{K / Q} (α) = \pm 1$ . Write $χ_{α} (x) = x^{n} + c_{n - 1} x^{n - 1} + \dots + c_{0}$ with $c_{i} \in Z$ and $c_{0} = (- 1)^{n} N (α) = \pm 1$ . From $χ_{α} (α) = 0$ ,

α (α^{n - 1} + c_{n - 1} α^{n - 2} + \dots + c_{1}) = - c_{0} = \mp 1.

The bracket lies in $O_{K}$ , so $\mp$ that bracket is an inverse for $α$ in $O_{K}$ .∎

This is the single most useful computational fact in the chapter. It converts a question about invertibility — hard, because it is a question about the whole ring — into the evaluation of one integer.

Example 7.17 (Norm and trace in a cubic field). Let $K = Q (θ)$ where $θ^{3} - θ - 1 = 0$ . Compute $N_{K / Q} (θ)$ , $Tr_{K / Q} (θ)$ , and $N_{K / Q} (1 + θ)$ .

Solution.

$x^{3} - x - 1$ is irreducible over $Q$ (no rational root: $\pm 1$ fail), so it is $m_{θ}$ and $n = d = 3$ , $χ_{θ} = m_{θ}$ .
$N (θ) = (- 1)^{3} m_{θ} (0) = (- 1)^{3} (- 1) = 1$ . So $θ$ is a unit of $O_{K}$ by the proposition Norm detects units.
$Tr (θ) = - (coefficient of x^{2}) = 0$ .
For $1 + θ$ , use $N (α) = (- 1)^{n} m (0)$ with the general identity $N_{K / Q} (c - θ) = m_{θ} (c)$ for $c \in Q$ , which follows from $m_{θ} (x) = \prod (x - σ_{i} θ)$ . Here $1 + θ = - ((- 1) - θ)$ , so

N (1 + θ) = (- 1)^{3} N (- 1 - θ) = (- 1)^{3} m_{θ} (- 1) = - ((- 1)^{3} - (- 1) - 1) = - (- 1) = 1.

So $1 + θ$ is also a unit. Indeed $θ^{3} = θ + 1$ , so $(1 + θ)^{- 1} = θ^{- 3}$ , and $θ^{- 1} = θ^{2} - 1$ from $θ (θ^{2} - 1) = 1$ .

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Exercises for §7.4 →

7.5The ring of integers as a $Z$ -module

Definition 7.18 (Ring of integers). Let $K$ be a number field. Its ring of integers is

O_{K} = \overline{Z} \cap K = {α \in K : α is integral over Z},

the integral closure of $Z$ in $K$ .

It is a ring by the theorem The algebraic integers form a ring, and it is an integral domain because it sits inside the field $K$ . Two further facts make it tractable: it is large (its fraction field is all of $K$ ) and it is small (it is a free $Z$ -module of rank exactly $n$ ). Both are proved now.

Proposition 7.19 ( $O_{K}$ has fraction field $K$ ). Every $α \in K$ can be written $α = β / m$ with $β \in O_{K}$ and $m \in Z_{> 0}$ . Hence $K = Frac (O_{K})$ and $O_{K}$ contains a $Q$ -basis of $K$ .

Proof. Every $α \in K$ is algebraic over $Q$ , so by the corollary Clearing a denominator there is $m \in Z$ , $m \neq = 0$ , with $m α \in \overline{Z}$ ; and $m α \in K$ , so $m α \in O_{K}$ . Replacing $m$ by $- m$ if necessary makes it positive. Given any $Q$ -basis $α_{1}, \dots, α_{n}$ of $K$ , scaling each by a suitable positive integer gives a $Q$ -basis inside $O_{K}$ .∎

Definition 7.20 (Integral basis). An integral basis of $K$ is a tuple $ω_{1}, \dots, ω_{n} \in O_{K}$ such that every $α \in O_{K}$ is uniquely $α = a_{1} ω_{1} + \dots + a_{n} ω_{n}$ with $a_{i} \in Z$ ; equivalently $O_{K} = Z ω_{1} \oplus \dots \oplus Z ω_{n}$ as $Z$ -modules.

Definition 7.21 (Discriminant of a tuple, discriminant of a field). For $α_{1}, \dots, α_{n} \in K$ with embeddings $σ_{1}, \dots, σ_{n}$ , set

disc (α_{1}, \dots, α_{n}) = det (σ_{i} (α_{j}))^{2} = det (Tr_{K / Q} (α_{i} α_{j})) .

The discriminant of $K$ is $d_{K} = disc (ω_{1}, \dots, ω_{n})$ for any integral basis.

The two expressions agree because if $M = (σ_{i} (α_{j}))$ then $M^{⊤} M$ has $(i, j)$ entry $\sum_{k} σ_{k} (α_{i}) σ_{k} (α_{j}) = Tr (α_{i} α_{j})$ , and $det (M^{⊤} M) = (det M)^{2}$ . The second form shows the discriminant is rational, and integral when the $α_{i}$ are.

Theorem 7.22 ( $O_{K}$ is a free $Z$ -module of rank $n$ ). Let $[K : Q] = n$ . Then $O_{K}$ is a free $Z$ -module of rank $n$ ; that is, $K$ has an integral basis of $n$ elements. Moreover $d_{K}$ is a nonzero rational integer, independent of the choice of integral basis.

Proof. By the proposition $O_{K}$ has fraction field $K$ choose a $Q$ -basis $α_{1}, \dots, α_{n}$ of $K$ lying in $O_{K}$ , and let $D = disc (α_{1}, \dots, α_{n})$ . Since the $σ_{i}$ are distinct field embeddings they are linearly independent as characters, so $det (σ_{i} (α_{j})) \neq = 0$ and $D \neq = 0$ ; and $D = det (Tr (α_{i} α_{j}))$ is a determinant of rational integers by the corollary Norms and traces of algebraic integers are rational integers, so $D \in Z ∖ {0}$ .

We claim

L := Z α_{1} + \dots + Z α_{n} \subseteq O_{K} \subseteq \frac{1}{D} L .

Only the right inclusion needs work. Let $α \in O_{K}$ and write $α = \sum_{j} c_{j} α_{j}$ with $c_{j} \in Q$ . Taking traces against each $α_{i}$ ,

Tr (α α_{i}) = j \sum c_{j} Tr (α_{i} α_{j}),

a linear system $T c = b$ where $T = (Tr (α_{i} α_{j}))$ has integer entries and determinant $D$ , and $b$ has integer entries because each $α α_{i}$ lies in $O_{K}$ . By Cramer's rule $c_{j} = det (T_{j}) / D$ with $det (T_{j}) \in Z$ , so $D c_{j} \in Z$ and $α \in \frac{1}{D} L$ .

Now $\frac{1}{D} L$ is a free $Z$ -module of rank $n$ , and a submodule of a free $Z$ -module of rank $n$ is free of rank at most $n$ (the structure theorem for modules over a PID). So $O_{K}$ is free of rank $\leq n$ ; and it contains $L$ , of rank $n$ , so its rank is exactly $n$ .

Finally, two integral bases differ by a matrix $U \in G L_{n} (Z)$ , which has $det U = \pm 1$ , and discriminants transform by $det (U)^{2} = 1$ . So $d_{K}$ is well defined.∎

Remark. The inclusions $L \subseteq O_{K} \subseteq \frac{1}{D} L$ do more than prove freeness: they bound the search. The index $[O_{K} : L]$ is finite and, comparing discriminants, $disc (L) = [O_{K} : L]^{2} d_{K}$ . So the index squared divides the discriminant of any convenient $Z$ -basis — the fact that drives every hand computation of an integral basis.

Proposition 7.23 (Index and discriminant). Let $L \subseteq O_{K}$ be a free $Z$ -submodule of rank $n$ with $Z$ -basis $α_{1}, \dots, α_{n}$ . Then $[O_{K} : L]$ is finite and

disc (α_{1}, \dots, α_{n}) = [O_{K} : L]^{2} d_{K} .

Proof. Write $α_{i} = \sum_{j} m_{ij} ω_{j}$ for an integral basis $ω_{j}$ , with $M = (m_{ij}) \in M_{n} (Z)$ . Since both modules have rank $n$ and $L \subseteq O_{K}$ , $det M \neq = 0$ and by the Smith normal form $[O_{K} : L] = ∣ det M ∣$ , finite. The embedding matrices satisfy $(σ_{i} (α_{j})) = (σ_{i} (ω_{k})) M^{⊤}$ , so taking determinants and squaring,

disc (α_{1}, \dots, α_{n}) = (det M)^{2} d_{K} = [O_{K} : L]^{2} d_{K} .

∎

Exercises for §7.5 →

7.6Rings of integers of quadratic fields: power basis or not

If $K = Q (θ)$ with $θ \in O_{K}$ , the obvious candidate for $O_{K}$ is $Z [θ] = Z + Z θ + \dots + Z θ^{n - 1}$ , a power basis. Sometimes this is correct; often it is not, and the quadratic case shows exactly why in a computation worth doing once, slowly.

Theorem 7.24 (Ring of integers of a quadratic field). Let $d \neq = 0, 1$ be squarefree and $K = Q (d)$ . Then

O_{K} = {Z [\frac{1 + d}{2}] = Z \oplus Z \frac{1 + d}{2}, Z [d] = Z \oplus Z d, d \equiv 1 (mod 4), d \equiv 2, 3 (mod 4),

and correspondingly

d_{K} = {d, 4 d, d \equiv 1 (mod 4), d \equiv 2, 3 (mod 4) .

Proof. Every $α \in K$ is $α = r + s d$ with $r, s \in Q$ . If $s = 0$ then $α \in Q$ and, by the corollary Rational algebraic integers are integers, $α \in O_{K} ⟺ α \in Z$ . So assume $s \neq = 0$ ; then $α \in / Q$ , the conjugate $\overset{α}{ˉ} = r - s d$ is the other root, and the minimal polynomial is

m_{α} (x) = x^{2} - 2 r x + (r^{2} - d s^{2}) .

By the theorem Integrality is visible in the minimal polynomial, $α \in O_{K}$ if and only if

t := 2 r \in Z and m := r^{2} - d s^{2} \in Z .

Step 1: the denominator of $s$ divides $2$ . From $t = 2 r \in Z$ and $m \in Z$ we get $d s^{2} = r^{2} - m = \frac{t ^{2}}{4} - m$ , so $4 d s^{2} = t^{2} - 4 m \in Z$ . Write $s = u / v$ in lowest terms with $v > 0$ . Then $v^{2} ∣ 4 d u^{2}$ , and $g cd (u, v) = 1$ forces $v^{2} ∣ 4 d$ . Since $d$ is squarefree, the only squares dividing $4 d$ are $1$ and $4$ , so $v \in {1, 2}$ . Hence $u := 2 s \in Z$ as well, and we may write

α = \frac{t + u d}{2}, t, u \in Z .

Step 2: the congruence condition. With this notation $4 m = t^{2} - d u^{2}$ , so the two integrality conditions are together equivalent to

t^{2} \equiv d u^{2} (mod 4) .

Step 3: case $d \equiv 2, 3 (mod 4)$ . Squares mod $4$ are $0$ and $1$ . Suppose $t$ or $u$ is odd. If $u$ is odd then $u^{2} \equiv 1$ and $d u^{2} \equiv d \equiv 2$ or $3 (mod 4)$ , while $t^{2} \equiv 0$ or $1$ — impossible. So $u$ is even, $d u^{2} \equiv 0 (mod 4)$ , hence $t^{2} \equiv 0 (mod 4)$ and $t$ is even too. Thus $t = 2 a$ , $u = 2 b$ and $α = a + b d \in Z [d]$ . Conversely every such $α$ is a root of $x^{2} - 2 a x + (a^{2} - d b^{2}) \in Z [x]$ . So $O_{K} = Z [d]$ .

Step 4: case $d \equiv 1 (mod 4)$ . Now $d u^{2} \equiv u^{2} (mod 4)$ , so the condition $t^{2} \equiv d u^{2} (mod 4)$ becomes $t^{2} \equiv u^{2} (mod 4)$ , which holds precisely when $t$ and $u$ have the same parity. Both even reproduces $Z [d]$ ; both odd gives the new elements, of which the simplest is $ω = \frac{1 + d}{2}$ (root of $x^{2} - x + \frac{1 - d}{4}$ , which has integer coefficients exactly because $4 ∣ d - 1$ ). And if $t, u$ are both odd then

\frac{t + u d}{2} = \frac{t - u}{2} + u \frac{1 + d}{2}

with $\frac{t - u}{2} \in Z$ , so $α \in Z \oplus Z ω$ . Both-even elements $a + b d$ equal $(a - b) + 2 b ω$ , also in $Z \oplus Z ω$ . So $O_{K} = Z \oplus Z ω = Z [ω]$ .

Step 5: discriminants. For the basis $1, d$ the embedding matrix is $(11 d - d)$ , of determinant $- 2 d$ , so $disc (1, d) = 4 d$ . For $1, ω$ the matrix is $(11 \frac{1 + d}{2} \frac{1 - d}{2})$ , of determinant $- d$ , so $disc (1, ω) = d$ . Substituting the integral bases found above gives the stated $d_{K}$ .∎

Pitfall. Writing $O_{K} = Z [d]$ for every $d$ is the most frequent error in the subject, and it is not a harmless one: it changes the discriminant by a factor of $4$ , changes which primes ramify, and produces a ring that is not integrally closed and therefore not a Dedekind domain. For $d = - 3$ it misses the sixth roots of unity; for $d = 5$ it misses the golden ratio, which is a unit. Always check $d mod 4$ first.

Figure 7.25. For

d \equiv 1 (mod 4)

the true ring of integers is strictly larger than

Z [d]

. With

d = - 3

, the lattice

Z [- 3]

(grey) is rectangular, while

O_{K} = Z [ω]

adds the orange points at the centres of the rectangles, producing a hexagonal lattice of index

2

Example 7.26 (Is $\frac{3 + 17}{2}$ an algebraic integer?). Decide, and identify $O_{K}$ for $K = Q (17)$ .

Solution.

$17$ is squarefree and $17 \equiv 1 (mod 4)$ , so by the theorem Ring of integers of a quadratic field, $O_{K} = Z [ω]$ with $ω = \frac{1 + 17}{2}$ , and $d_{K} = 17$ .
For $α = \frac{3 + 17}{2}$ we have $t = 3$ , $u = 1$ , both odd, so the parity condition of Step 4 holds and $α \in O_{K}$ .
Check directly: $m_{α} (x) = x^{2} - 3 x + \frac{9 - 17}{4} = x^{2} - 3 x - 2 \in Z [x]$ . ✓
In terms of the basis, $α = \frac{3 - 1}{2} + 1 \cdot ω = 1 + ω$ .

□

Example 7.27 (A field where the power basis is wrong for a subtler reason). Let $K = Q (320)$ and $θ = 320$ . Show $Z [θ] \neq = O_{K}$ by exhibiting an algebraic integer outside it.

Solution.

Put $α = \frac{θ ^{2}}{2}$ . Then $α^{3} = \frac{θ ^{6}}{8} = \frac{400}{8} = 50$ , so $α$ is a root of $x^{3} - 50$ , monic over $Z$ .
Hence $α \in O_{K}$ .
But $α = \frac{1}{2} θ^{2}$ has a coefficient $\frac{1}{2} \in / Z$ in the basis $1, θ, θ^{2}$ of $K$ , so $α \in / Z [θ]$ .
Therefore $[O_{K} : Z [θ]]$ is divisible by $2$ , and by the proposition Index and discriminant the discriminant of $Z [θ]$ exceeds $∣ d_{K} ∣$ by a factor of at least $4$ .

□

Remark. There exist number fields with no power basis at all — no single $θ$ with $O_{K} = Z [θ]$ ; such fields are called non-monogenic, the classical example being $Q (θ)$ with $θ^{3} - θ^{2} - 2 θ - 8 = 0$ , where the index $[O_{K} : Z [θ^{'}]]$ is even for every generator $θ^{'}$ . So "find the power basis" is not a strategy that always terminates; "find an integral basis" is.

Exercises for §7.6 →

7.7Discriminant and ramification

The discriminant is not merely a bookkeeping invariant. It is the exact record of which primes behave badly in $O_{K}$ .

Definition 7.28 (Ramification). Let $p$ be a rational prime. Writing $p O_{K} = p_{1}^{e_{1}} \dots p_{g}^{e_{g}}$ for the factorisation into distinct prime ideals guaranteed by the theorem Unique factorisation of ideals below, the integer $e_{i}$ is the ramification index of $p_{i}$ over $p$ , and $f_{i} = [O_{K} / p_{i} : F_{p}]$ is the residue degree. The prime $p$ is ramified in $K$ if some $e_{i} > 1$ , and unramified otherwise. If $g = 1$ and $f_{1} = n$ , $p$ is inert; if $g = n$ (so all $e_{i} = f_{i} = 1$ ), $p$ splits completely.

Theorem 7.29 (Degree relation). With the notation above, $\sum_{i = 1}^{g} e_{i} f_{i} = n = [K : Q]$ .

Proof. Both sides count the $F_{p}$ -dimension of $O_{K} / p O_{K}$ . On the one hand $O_{K} ≅ Z^{n}$ as a $Z$ -module by the theorem $O_{K}$ is a free $Z$ -module of rank $n$ , so $O_{K} / p O_{K} ≅ (Z / p)^{n}$ has dimension $n$ . On the other hand the Chinese remainder theorem applied to the pairwise comaximal ideals $p_{i}^{e_{i}}$ gives

O_{K} / p O_{K} ≅ i = 1 \prod g O_{K} / p_{i}^{e_{i}},

and the filtration $p_{i}^{k} / p_{i}^{k + 1}$ for $0 \leq k < e_{i}$ has successive quotients that are one-dimensional vector spaces over $O_{K} / p_{i}$ — one-dimensional because in the localisation the ideal $p_{i}$ is principal, as established in the theorem Localisations are discrete valuation rings. Hence $dim_{F_{p}} O_{K} / p_{i}^{e_{i}} = e_{i} f_{i}$ , and summing gives the claim.∎

Theorem 7.30 (Dedekind's discriminant criterion). A rational prime $p$ ramifies in $K$ if and only if $p ∣ d_{K}$ . In particular only finitely many primes ramify.

Proof. We prove the direction that is within reach and label the rest. Suppose $p$ is unramified, so $O_{K} / p O_{K} ≅ \prod_{i} O_{K} / p_{i}$ is a product of finite fields, hence a reduced ring (no nonzero nilpotents) and a separable $F_{p}$ -algebra. The discriminant of a finite-dimensional commutative algebra over a field is nonzero exactly when the trace form is nondegenerate, which for a product of separable field extensions it is. Reducing an integral basis modulo $p$ gives an $F_{p}$ -basis of $O_{K} / p O_{K}$ whose trace-form determinant is $d_{K} mod p$ , so $d_{K} \neq \equiv 0 (mod p)$ .

Conversely if $p$ ramifies, some $O_{K} / p_{i}^{e_{i}}$ with $e_{i} \geq 2$ has a nonzero nilpotent $ν$ . Multiplication by $ν$ on $O_{K} / p O_{K}$ is then nilpotent, so $Tr (ν β) = 0$ for all $β$ (a nilpotent endomorphism has zero trace, and $ν β$ is nilpotent since the ring is commutative). The trace form is therefore degenerate and $d_{K} \equiv 0 (mod p)$ .

Finally $d_{K} \neq = 0$ , so it has finitely many prime divisors.∎

Example 7.31 (Ramification in quadratic fields). Which primes ramify in $Q (- 5)$ , in $Q (5)$ and in $Q (i)$ ?

Solution.

$- 5 \equiv 3 (mod 4)$ , so $d_{K} = 4 (- 5) = - 20 = - 2^{2} \cdot 5$ . Ramified primes: $2$ and $5$ .
$5 \equiv 1 (mod 4)$ , so $d_{K} = 5$ . Ramified: $5$ only. Note $2$ is not ramified here even though $2$ divides $disc (1, 5) = 20$ — that number is the discriminant of the non-maximal order $Z [5]$ , not of $O_{K}$ .
$- 1 \equiv 3 (mod 4)$ , so $d_{K} = - 4$ and only $2$ ramifies: indeed $(2) = (1 + i)^{2}$ in $Z [i]$ .

□

Pitfall. Item 2 above is the trap the proposition Index and discriminant exists to catch. Computing $disc$ of a convenient basis and reading off ramified primes is only valid once you know that basis is integral. An index of $2$ inserts a spurious factor $4$ and a spurious ramified prime $2$ .

Exercises for §7.7 →

7.8Computing an integral basis

The proposition Index and discriminant turns the search for $O_{K}$ into a finite one. Start from $L = Z [θ]$ for a generator $θ \in O_{K}$ , compute $D = disc (1, θ, \dots, θ^{n - 1})$ , and note $D = [O_{K} : L]^{2} d_{K}$ . So the index is an integer whose square divides $D$ : only the primes $p$ with $p^{2} ∣ D$ can occur, and for each one it suffices to test whether $L$ can be enlarged $p$ -adically.

Method 7.32 (Enlarging an order to the maximal one). Given $θ \in O_{K}$ with $K = Q (θ)$ :

Compute $D = disc (1, θ, \dots, θ^{n - 1})$ , the discriminant of $m_{θ}$ .
List the primes $p$ with $p^{2} ∣ D$ . Only these can divide $[O_{K} : Z [θ]]$ .
For each such $p$ , test the candidate elements $\frac{1}{p} (c_{0} + c_{1} θ + \dots + c_{n - 1} θ^{n - 1})$ with $c_{i} \in {0, \dots, p - 1}$ not all zero, and keep any whose characteristic polynomial lies in $Z [x]$ .
Adjoin all such elements, obtaining a larger module $L^{'}$ with $disc (L^{'}) = disc (L) / p^{2 k}$ where $p^{k}$ is the index gained. Repeat until no prime with $p^{2} ∣ disc$ admits an enlargement.
The resulting module is $O_{K}$ and its discriminant is $d_{K}$ .

Step 3 is finite — at most $p^{n}$ candidates — which is why the procedure terminates, and Dedekind's criterion refines it further by testing the factorisation of $m_{θ}$ modulo $p$ rather than enumerating.

Example 7.33 (An integral basis by the recipe). Let $K = Q (13)$ and start from $θ = 13$ . Run the recipe.

Solution.

$m_{θ} = x^{2} - 13$ , and $disc (1, θ) = 4 \cdot 13 = 52$ .
Primes with $p^{2} ∣ 52 = 2^{2} \cdot 13$ : only $p = 2$ .
Candidates $\frac{c _{0} + c _{1} θ}{2}$ with $c_{i} \in {0, 1}$ not both zero. For $\frac{1 + θ}{2}$ the characteristic polynomial is $x^{2} - x + \frac{1 - 13}{4} = x^{2} - x - 3 \in Z [x]$ . ✓ (The other two candidates, $\frac{1}{2}$ and $\frac{θ}{2}$ , give $x^{2} - x + \frac{1}{4}$ and $x^{2} - \frac{13}{4}$ , both rejected.)
So $O_{K} = Z \oplus Z \frac{1 + 13}{2}$ , index $2$ over $Z [13]$ , and $d_{K} = 52/ 2^{2} = 13$ .
This agrees with the theorem Ring of integers of a quadratic field, since $13 \equiv 1 (mod 4)$ .

□

Example 7.34 (A field where the power basis survives). Let $K = Q (32)$ , $θ = 32$ . Show $O_{K} = Z [θ]$ .

Solution.

For $f = x^{3} - a$ the discriminant is $- 27 a^{2}$ , so here $D = - 27 \cdot 4 = - 108 = - 2^{2} 3^{3}$ .
Primes $p$ with $p^{2} ∣ D$ : $p = 2$ and $p = 3$ .
The index $m = [O_{K} : Z [θ]]$ satisfies $m^{2} ∣ 108$ , so $m ∣ 6$ .
Testing $p = 2$ : any candidate $\frac{c _{0} + c _{1} θ + c _{2} θ ^{2}}{2}$ has norm $\frac{c _{0}^{3} + 2 c _{1}^{3} + 4 c _{2}^{3} - 6 c _{0} c _{1} c _{2}}{8}$ , and for $c_{i} \in {0, 1}$ the numerators are $1, 2, 4, 3, 5, 6, 1$ — none divisible by $8$ , so no enlargement at $2$ .
Testing $p = 3$ : the same norm formula over $c_{i} \in {0, 1, 2}$ never produces a numerator divisible by $27$ (the norm is $\equiv c_{0}^{3} + 2 c_{1}^{3} + 4 c_{2}^{3} (mod 3)$ , i.e. $\equiv c_{0} + 2 c_{1} + c_{2} (mod 3)$ , and vanishing mod $3$ is necessary but the full cube condition fails in every case).
So no enlargement is possible, $O_{K} = Z [32]$ and $d_{K} = - 108$ .

□

Exercises for §7.8 →

7.9Dedekind domains

Rings of integers are almost never UFDs, but they are always Dedekind domains, and that is enough. The definition packages exactly the three properties the factorisation theory needs.

Definition 7.35 (Dedekind domain). An integral domain $R$ that is not a field is a Dedekind domain if

$R$ is Noetherian: every ideal is finitely generated;
$R$ is integrally closed in its fraction field $Frac (R)$ ;
every nonzero prime ideal of $R$ is maximal (equivalently, $R$ has Krull dimension $1$ ).

Each condition rules out a specific pathology. Without (1) an ideal could fail to have a factorisation at all, for lack of a finiteness to induct on. Without (2) the ring has "missing" elements — $Z [- 3]$ is Noetherian and one-dimensional but omits $\frac{1 + - 3}{2}$ , and in it $(2)$ is not a product of primes. Without (3) one is in a ring like $Z [x]$ or $C [x, y]$ , where the prime $(x)$ sits strictly inside the prime $(x, y)$ and ideals have codimension- $2$ behaviour that no one-dimensional factorisation can see.

Proposition 7.36 (Every PID is a Dedekind domain). If $R$ is a principal ideal domain and not a field, then $R$ is a Dedekind domain.

Proof. (1) Every ideal is generated by one element, so certainly finitely generated. (2) Let $α = a / b \in Frac (R)$ be integral over $R$ , with $a, b \in R$ coprime — possible since a PID is a UFD, as shown in Polynomial Rings and Unique Factorization. From $α^{n} + c_{n - 1} α^{n - 1} + \dots + c_{0} = 0$ , multiplying by $b^{n}$ gives $a^{n} = - b (c_{n - 1} a^{n - 1} + \dots + c_{0} b^{n - 1})$ , so every prime factor of $b$ divides $a^{n}$ and hence $a$ ; coprimality forces $b$ to be a unit and $α \in R$ . (3) Let $(p)$ be a nonzero prime ideal and $(p) \subseteq (c) ⊊ R$ . Then $p = c d$ for some $d$ , and $p$ prime with $c$ a non-unit forces $d$ to be a unit, so $(c) = (p)$ . Hence $(p)$ is maximal.∎

Corollary 7.37 ( $Z$ and $F [x]$ are Dedekind). $Z$ and $F [x]$ for a field $F$ are Dedekind domains.

Proof. Both are Euclidean, hence principal ideal domains by Polynomial Rings and Unique Factorization, and neither is a field. Apply the proposition Every PID is a Dedekind domain.∎

Example 7.38 (A ring that fails exactly one axiom). Show that $R = Z [- 3]$ is Noetherian and one-dimensional but not a Dedekind domain.

Solution.

$R$ is a finitely generated $Z$ -algebra, hence Noetherian by the Hilbert basis theorem; and it is a free $Z$ -module of rank $2$ , so every quotient by a nonzero ideal is finite, which forces nonzero primes to be maximal (a finite integral domain is a field).
Its fraction field is $K = Q (- 3)$ , and $ω = \frac{1 + - 3}{2} \in K$ satisfies $ω^{2} - ω + 1 = 0$ , so $ω$ is integral over $R$ .
But $ω \in / R$ , since its coordinates in the basis $1, - 3$ are not integers.
So $R$ is not integrally closed, and axiom (2) fails. Its integral closure is $O_{K} = Z [ω]$ , which is Dedekind.

□

Remark. This example is the reason axiom (2) is stated in terms of the fraction field rather than as a convenience. $Z [- 3]$ is a perfectly reasonable ring in which $4 = 2 \cdot 2 = (1 + - 3) (1 - - 3)$ has two factorisations that even ideals cannot reconcile; passing to the integral closure repairs it, because in $Z [ω]$ one has $1 + - 3 = 2 ω$ and the two factorisations differ by a unit.

Exercises for §7.9 →

7.10Unique factorisation of ideals

This is the theorem the subject is built on. Elements need not factor uniquely; ideals always do.

Theorem 7.39 (Unique factorisation of ideals). Let $R$ be a Dedekind domain. Every nonzero proper ideal $a \subseteq R$ can be written

a = p_{1}^{a_{1}} p_{2}^{a_{2}} \dots p_{g}^{a_{g}}

with $p_{1}, \dots, p_{g}$ distinct nonzero prime ideals and $a_{i} \geq 1$ , and this expression is unique up to the order of the factors.

Proof. Labelled sketch. The full argument is standard but long; what follows is its skeleton, with the two lemmas it rests on stated explicitly as assumptions.

Assumed (A1). Every nonzero ideal contains a product of nonzero primes. If not, Noetherianity gives a maximal counterexample $a$ . It is not prime, so there are $x, y \in / a$ with $x y \in a$ ; then $a + (x)$ and $a + (y)$ strictly contain $a$ , hence each contains a product of primes, and their product lies in $a$ — a contradiction. (This step uses only axiom (1).)

Assumed (A2) — invertibility. For every nonzero prime $p$ , setting $p^{- 1} = {x \in K : x p \subseteq R}$ where $K = Frac (R)$ , one has $p p^{- 1} = R$ . Proving $p^{- 1} ⊋ R$ uses (A1) and maximality of $p$ ; proving that $p p^{- 1}$ , which is an ideal containing $p$ , cannot equal $p$ is exactly where integral closedness enters: if $p p^{- 1} = p$ then every $x \in p^{- 1}$ stabilises the finitely generated module $p$ , so by the determinant trick of the lemma Integrality equals module-finiteness $x$ is integral over $R$ , hence $x \in R$ , contradicting $p^{- 1} ⊋ R$ . Since $p p^{- 1}$ is an ideal strictly between $p$ and $R$ and $p$ is maximal, it is $R$ .

Existence, granting (A1) and (A2). Suppose some nonzero proper ideal is not a product of primes and choose, by Noetherianity, a maximal such $a$ . It is contained in a maximal ideal $p$ . Then $a \subseteq a p^{- 1} \subseteq p p^{- 1} = R$ , so $a p^{- 1}$ is an ideal, and it strictly contains $a$ by the same determinant-trick argument as above. By maximality of the counterexample, $a p^{- 1} = p_{1} \dots p_{r}$ is a product of primes (or is $R$ ), and multiplying by $p$ gives $a = p p_{1} \dots p_{r}$ — a contradiction.

Uniqueness, granting (A2). Suppose $p_{1} \dots p_{r} = q_{1} \dots q_{s}$ . Then $p_{1} \supseteq q_{1} \dots q_{s}$ , and since $p_{1}$ is prime it contains some $q_{j}$ ; as $q_{j}$ is maximal, $p_{1} = q_{j}$ . Multiply both sides by $p_{1}^{- 1}$ , which cancels that factor exactly because of (A2), and induct on $r$ .∎

Intuition. Read the theorem as a change of vocabulary. Elements are too rigid: $2$ and $1 + - 5$ have no common factor in $Z [- 5]$ , yet they are not coprime in any reasonable sense, since they share the ideal $p = (2, 1 + - 5)$ . The ideal $p$ is the common factor, and it is not generated by any single element — which is exactly why element-level factorisation could not see it. Passing to ideals adds the missing atoms.

Corollary 7.40 (To contain is to divide). For nonzero ideals $a, b$ of a Dedekind domain $R$ : $a \supseteq b$ if and only if $a ∣ b$ , meaning $b = ac$ for some ideal $c$ .

Proof. ( $\Leftarrow$ ) If $b = ac$ then $b \subseteq a$ since $c \subseteq R$ .

( $\Rightarrow$ ) Suppose $b \subseteq a$ . Using invertibility (A2) extended multiplicatively, set $c = a^{- 1} b$ , a product of a fractional ideal with an ideal, which satisfies $c \subseteq a^{- 1} a = R$ , so $c$ is an honest ideal of $R$ , and $ac = a a^{- 1} b = b$ .∎

Corollary 7.41 (Exponents are valuations). Writing $v_{p} (a)$ for the exponent of $p$ in the factorisation of $a$ , we have for all nonzero ideals

v_{p} (ab) = v_{p} (a) + v_{p} (b), v_{p} (a + b) = min (v_{p} (a), v_{p} (b)) .

Thus $a + b$ is the gcd and $a \cap b$ the lcm of $a$ and $b$ .

Proof. Additivity is immediate from uniqueness applied to the product of the two factorisations. For the sum: $a + b$ is the smallest ideal containing both, and by the corollary To contain is to divide an ideal contains both $a$ and $b$ exactly when it divides both, i.e. when its exponent at each $p$ is at most $min (v_{p} (a), v_{p} (b))$ . The smallest such ideal — smallest in the containment order means largest exponents — is the one attaining the minimum at every $p$ . The statement for $\cap$ is the same argument with $max$ .∎

Figure 7.42. Element factorisation of the ideal

(6)

Z [- 5]

branches two incompatible ways, but both routes refine to the same four prime ideals. The ideal level is where the two stories meet.

Exercises for §7.10 →

7.11Localisation: every Dedekind domain is locally a DVR

Unique factorisation of ideals says the global arithmetic is controlled by one prime at a time. Localisation makes that literal.

Definition 7.43 (Discrete valuation ring). A discrete valuation ring is a principal ideal domain with exactly one nonzero maximal ideal. Equivalently, a domain $R$ carrying a surjective map $v : Frac (R)^{\times} \to Z$ with $v (x y) = v (x) + v (y)$ and $v (x + y) \geq min (v (x), v (y))$ , such that $R = {x : v (x) \geq 0} \cup {0}$ . A generator $π$ of the maximal ideal is a uniformiser, and every nonzero element is uniquely $u π^{k}$ with $u$ a unit and $k \geq 0$ .

Theorem 7.44 (Localisations are discrete valuation rings). Let $R$ be a Dedekind domain and $p$ a nonzero prime ideal. Then the localisation $R_{p} = {a / s : a \in R, s \in / p}$ is a discrete valuation ring with maximal ideal $p R_{p}$ .

Proof. $R_{p}$ is a domain with the same fraction field as $R$ . Its ideals are exactly the $a R_{p}$ for ideals $a$ of $R$ , and $q R_{p} = R_{p}$ whenever $q \neq \subseteq p$ — because then $q$ contains an element outside $p$ , which becomes a unit. Since distinct nonzero primes of $R$ are maximal, hence incomparable, the only prime surviving is $p$ , so $R_{p}$ is local with maximal ideal $m = p R_{p}$ .

Localisation preserves Noetherian and preserves integral closedness (a monic relation over $R_{p}$ can be cleared of denominators), and one-dimensionality is inherited, so $R_{p}$ is again Dedekind. By the theorem Unique factorisation of ideals applied in $R_{p}$ , where $m$ is the only nonzero prime, every nonzero ideal is $m^{k}$ for a unique $k \geq 0$ .

It remains to see $m$ is principal. Pick $π \in m ∖ m^{2}$ , possible since $m \neq = m^{2}$ by uniqueness of factorisation. Then $(π) = m^{k}$ for some $k$ , and $(π) \subseteq m$ gives $k \geq 1$ while $π \in / m^{2}$ gives $k < 2$ . So $(π) = m$ , every ideal is $(π^{k})$ , and $R_{p}$ is a PID with one maximal ideal. Setting $v (x) = k$ when $x = u π^{k}$ gives the valuation.∎

Corollary 7.45 (Local-global principle for ideals). Nonzero ideals $a, b$ of a Dedekind domain are equal if and only if $a R_{p} = b R_{p}$ for every nonzero prime $p$ .

Proof. One direction is trivial. Conversely, $a R_{p} = m_{p}^{v_{p} (a)}$ by the theorem Localisations are discrete valuation rings together with unique factorisation, so equality of all localisations gives $v_{p} (a) = v_{p} (b)$ for every $p$ , and uniqueness of factorisation gives $a = b$ .∎

Intuition. A Dedekind domain is a family of DVRs glued along a common fraction field, one for each prime, with the compatibility that a given element has valuation $0$ at all but finitely many of them. "Unique factorisation of ideals" is then just the statement that an ideal is determined by, and can be prescribed arbitrarily as, a finitely supported list of non-negative exponents.

Exercises for §7.11 →

7.12Fractional ideals and the ideal group

Invertibility is easier to state once denominators are allowed.

Definition 7.46 (Fractional ideal). Let $R$ be a Dedekind domain with fraction field $K$ . A fractional ideal is a nonzero finitely generated $R$ -submodule $a \subseteq K$ ; equivalently, $a = \frac{1}{d} b$ for some nonzero ideal $b \subseteq R$ and $d \in R ∖ {0}$ . Products are defined by $ab = {\sum_{i} a_{i} b_{i}}$ , and $a^{- 1} = {x \in K : x a \subseteq R}$ . A fractional ideal of the form $α R$ with $α \in K^{\times}$ is principal.

Theorem 7.47 (The fractional ideals form a free abelian group). Let $R$ be a Dedekind domain. The set $I_{K}$ of fractional ideals is an abelian group under multiplication, with identity $R$ and inverse $a^{- 1}$ , and it is free abelian on the set of nonzero prime ideals: every fractional ideal is uniquely

a = p \prod p^{v_{p} (a)}, v_{p} (a) \in Z, almost all 0.

Proof. Multiplication of fractional ideals is associative and commutative with identity $R$ , directly from the definition. Writing $a = \frac{1}{d} b$ with $b \subseteq R$ an ideal, the theorem Unique factorisation of ideals factors both $b$ and $(d)$ into primes, so $a = \prod p^{v_{p} (b) - v_{p} ((d))}$ is a product of integer powers of primes; existence follows. Invertibility of each $p$ is (A2) of that theorem, so every such product is invertible, and $I_{K}$ is a group.

Uniqueness: suppose $\prod p^{n_{p}} = \prod p^{m_{p}}$ . Multiplying by $\prod p^{- m i n (n_{p}, m_{p})}$ puts the identity in the form $c = d$ with $c, d$ honest ideals of $R$ having no common prime factor, and uniqueness in $R$ forces both to be $R$ , hence $n_{p} = m_{p}$ for all $p$ . So the prime ideals form a free basis.∎

Definition 7.48 (Ideal norm). For a nonzero ideal $a \subseteq O_{K}$ , its norm is $N (a) = ∣ O_{K} / a ∣$ , a positive integer. The definition extends to fractional ideals by multiplicativity.

Proposition 7.49 (The ideal norm is multiplicative and extends the element norm). For nonzero ideals $a, b \subseteq O_{K}$ we have $N (ab) = N (a) N (b)$ , and for $α \in O_{K} ∖ {0}$ ,

N ((α)) = N_{K / Q} (α) .

Proof. For multiplicativity it suffices, by unique factorisation and induction, to treat $b = p$ prime. The exact sequence of finite abelian groups

0 \to a / ap \to O_{K} / ap \to O_{K} / a \to 0

gives $N (ap) = N (a) \cdot ∣ a / ap ∣$ . Now $a / ap$ is a vector space over the field $O_{K} / p$ , and it is one-dimensional: by unique factorisation there is no ideal strictly between $ap$ and $a$ , so the space has no proper nonzero subspace. Hence $∣ a / ap ∣ = N (p)$ .

For the second claim, let $ω_{1}, \dots, ω_{n}$ be an integral basis and let $M$ be the matrix of multiplication by $α$ in that basis, so $N_{K / Q} (α) = det M$ . The ideal $(α) = α O_{K}$ is the image of $O_{K} ≅ Z^{n}$ under $M$ , and the index of the image of an injective integer matrix equals $∣ det M ∣$ by the Smith normal form. So $N ((α)) = ∣ det M ∣ = ∣ N_{K / Q} (α) ∣$ .∎

Example 7.50 (Norm of a prime above 2 in $Z [- 5]$ ). Let $p_{2} = (2, 1 + - 5)$ . Compute $N (p_{2})$ and verify $p_{2}^{2} = (2)$ .

Solution.

Modulo $p_{2}$ we have $2 \equiv 0$ and $- 5 \equiv - 1 \equiv 1$ , so every element $a + b - 5$ is congruent to $a + b mod 2$ . The quotient is therefore $F_{2}$ and $N (p_{2}) = 2$ .
Expand: $p_{2}^{2} = (4, 2 (1 + - 5), (1 + - 5)^{2})$ and $(1 + - 5)^{2} = - 4 + 2 - 5$ .
Every listed generator is divisible by $2$ , so $p_{2}^{2} \subseteq (2)$ . Conversely $2 = (- 4 + 2 - 5) - (4) + 2 (1 + - 5) - 2 - 5 \cdot 1$ ; more cleanly, $2 (1 + - 5) - (- 4 + 2 - 5) = 2 + 2 - 5 + 4 - 2 - 5 = 6$ and $6 - 4 = 2$ , so $2 \in p_{2}^{2}$ .
Hence $p_{2}^{2} = (2)$ , consistent with norms: $N (p_{2})^{2} = 4 = ∣ N_{K / Q} (2) ∣$ .

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Exercises for §7.12 →

7.13Rings of integers are Dedekind domains

Everything above applies to $O_{K}$ once the three axioms are checked, and each check is short given the work already done.

Theorem 7.51 ( $O_{K}$ is a Dedekind domain). For every number field $K$ , the ring $O_{K}$ is a Dedekind domain.

Proof. Noetherian. By the theorem $O_{K}$ is a free $Z$ -module of rank $n$ , $O_{K} ≅ Z^{n}$ as a $Z$ -module. Any ideal $a$ is in particular a $Z$ -submodule, hence free of rank $\leq n$ over $Z$ , so generated by at most $n$ elements even as a $Z$ -module, and a fortiori finitely generated as an ideal.

Integrally closed. $Frac (O_{K}) = K$ by the proposition $O_{K}$ has fraction field $K$ . If $α \in K$ is integral over $O_{K}$ , then since $O_{K}$ is integral over $Z$ , the corollary Integral closure is transitive makes $α$ integral over $Z$ ; and $α \in K$ , so $α \in O_{K}$ by definition.

Dimension one. Let $p \neq = 0$ be prime. Pick $0 \neq = α \in p$ ; its norm $m = N_{K / Q} (α)$ is a nonzero rational integer lying in $p$ , since $m = α \cdot \prod_{i \geq 2} σ_{i} (α)$ and that second factor is $m / α \in K \cap \overline{Z} = O_{K}$ . Hence $m O_{K} \subseteq p$ , so $O_{K} / p$ is a quotient of $O_{K} / m O_{K} ≅ (Z / m)^{n}$ , a finite ring. So $O_{K} / p$ is a finite integral domain, and a finite integral domain is a field: for $0 \neq = x$ the map $y \mapsto x y$ is injective, hence surjective by finiteness, so $x$ has an inverse. Therefore $p$ is maximal.

Finally $O_{K}$ is not a field, since $Z \subseteq O_{K}$ and $2$ has no inverse in $O_{K}$ : $N (2) = 2^{n} \neq = \pm 1$ , so the proposition Norm detects units applies.∎

Corollary 7.52 (Residue fields are finite). For every nonzero prime $p$ of $O_{K}$ , the field $O_{K} / p$ is finite, of order $p^{f}$ where $p$ is the unique rational prime in $p$ and $f = [O_{K} / p : F_{p}]$ .

Proof. Finiteness was shown in the proof above. The intersection $p \cap Z$ is a nonzero prime ideal of $Z$ — nonzero because $m \in p$ , prime as the contraction of a prime — hence $(p)$ for a unique rational prime $p$ . Then $O_{K} / p$ is a finite field containing $Z / p = F_{p}$ , so it has $p^{f}$ elements with $f$ its degree over $F_{p}$ .∎

Remark. The corollary makes $N (p) = p^{f}$ and, combined with the theorem Degree relation, gives the identity $N (p O_{K}) = p^{n} = \prod_{i} N (p_{i})^{e_{i}} = p^{\sum e_{i} f_{i}}$ . Every numerical check on a factorisation of a rational prime can be run through this identity, and it catches most errors immediately.

Exercises for §7.13 →

7.14The ideal class group

Unique factorisation of ideals is unconditional; whether it says anything about elements depends on whether ideals are principal. The class group measures the gap.

Definition 7.53 (Ideal class group and class number). Let $K$ be a number field, $I_{K}$ the group of fractional ideals of $O_{K}$ and $P_{K} = {α O_{K} : α \in K^{\times}}$ the subgroup of principal fractional ideals. The ideal class group is the quotient

Cl (K) = I_{K} / P_{K},

and the class number is $h_{K} = ∣ Cl (K) ∣$ . Two fractional ideals $a, b$ lie in the same class, written $a \sim b$ , precisely when $b = α a$ for some $α \in K^{\times}$ .

Proposition 7.54 (The class group is well defined). $P_{K}$ is a subgroup of $I_{K}$ , so $Cl (K)$ is an abelian group; and $\sim$ as described is an equivalence relation whose classes are the cosets of $P_{K}$ . Moreover every class contains an honest (integral) ideal of $O_{K}$ .

Proof. $P_{K}$ is the image of the group homomorphism $K^{\times} \to I_{K}$ , $α \mapsto α O_{K}$ — a homomorphism because $(α O_{K}) (β O_{K}) = α β O_{K}$ — and the image of a homomorphism is a subgroup. Since $I_{K}$ is abelian by the theorem The fractional ideals form a free abelian group, the quotient exists and $a P_{K} = b P_{K}$ iff $a b^{- 1} \in P_{K}$ iff $b = α a$ for some $α$ , which is the stated relation.

For the last claim, let $a$ be a fractional ideal and write $a = \frac{1}{d} b$ with $b \subseteq O_{K}$ and $d \in O_{K} ∖ {0}$ , as in the definition Fractional ideal. Then $b = d a \sim a$ is integral.∎

Theorem 7.55 (The class number detects unique factorisation). For a number field $K$ the following are equivalent: (i) $h_{K} = 1$ ; (ii) $O_{K}$ is a principal ideal domain; (iii) $O_{K}$ is a unique factorisation domain.

Proof. (i) $\Leftrightarrow$ (ii). $h_{K} = 1$ means every fractional ideal is principal; restricting to integral ideals gives that $O_{K}$ is a PID, and conversely a fractional ideal $\frac{1}{d} b$ with $b = (β)$ equals $(β / d)$ .

(ii) $\Rightarrow$ (iii) is the standard fact that every PID is a UFD, proved in Polynomial Rings and Unique Factorization.

(iii) $\Rightarrow$ (ii). Let $O_{K}$ be a UFD and $p$ a nonzero prime ideal. Choose $0 \neq = α \in p$ and factor $α = π_{1} \dots π_{r}$ into irreducibles. Since $p$ is prime it contains some $π_{i}$ , so $(π_{i}) \subseteq p$ . In a UFD an irreducible element generates a prime ideal, and $(π_{i})$ is a nonzero prime, hence maximal by the theorem $O_{K}$ is a Dedekind domain. Therefore $(π_{i}) = p$ and every prime ideal is principal. By the theorem Unique factorisation of ideals every ideal is a product of primes, hence principal.∎

Intuition. The class number is the number of "kinds" of ideal. If $h_{K} = 3$ , then every ideal is, after multiplying by a suitable element, one of three shapes, and the shape multiplies like an element of $Z /3$ . The failure of unique factorisation of elements is then completely explained: an element factorisation can go wrong exactly when a non-principal ideal has a principal power, and $h_{K}$ is the order of the group of ways that can happen.

Exercises for §7.14 →

7.15Finiteness of the class group

Theorem 7.56 (Finiteness of the class group). For every number field $K$ , the class group $Cl (K)$ is a finite abelian group.

Proof. Labelled sketch — assumes Minkowski's convex body theorem, which is not proved here.

Assumed (M). Minkowski's lattice point theorem. If $Λ \subseteq R^{n}$ is a lattice of covolume $V$ and $S \subseteq R^{n}$ is convex, symmetric about the origin, and of volume $> 2^{n} V$ , then $S$ contains a nonzero point of $Λ$ .

Step 1 (the geometric setup, proved). Order the embeddings of $K$ as $r_{1}$ real ones and $r_{2}$ conjugate pairs of complex ones, $n = r_{1} + 2 r_{2}$ . The map

ι : K \to R^{r_{1}} \times C^{r_{2}} ≅ R^{n}, α \mapsto (σ_{1} α, \dots, σ_{r_{1}} α, σ_{r_{1} + 1} α, \dots)

sends any nonzero ideal $a$ to a full lattice, of covolume $2^{- r_{2}} N (a) ∣ d_{K} ∣$ ; this is a determinant computation from the definition Discriminant of a tuple, discriminant of a field.

Step 2 (the bound, assumes (M)). Applying (M) to a suitable convex symmetric region and optimising yields: every nonzero ideal $a$ contains a nonzero $α$ with

∣ N_{K / Q} (α) ∣ \leq M_{K} N (a), M_{K} = (\frac{4}{π})^{r_{2}} \frac{n !}{n ^{n}} ∣ d_{K} ∣ .

The constant $M_{K}$ is the Minkowski bound.

Step 3 (from the bound to finiteness, proved). Let $c$ be any ideal class and pick an integral ideal $b$ in the class $c^{- 1}$ , using the proposition The class group is well defined. By Step 2 there is $0 \neq = α \in b$ with $∣ N (α) ∣ \leq M_{K} N (b)$ . Since $α \in b$ , the corollary To contain is to divide gives $a$ with $(α) = ab$ , and then $a \sim b^{- 1}$ , so $a \in c$ . Its norm is

N (a) = \frac{∣ N ( α ) ∣}{N ( b )} \leq M_{K}

by the proposition The ideal norm is multiplicative and extends the element norm. So every class contains an integral ideal of norm at most $M_{K}$ .

Step 4 (finitely many such ideals, proved). An ideal of norm $m$ contains $m$ , hence divides $(m)$ , and $(m)$ has only finitely many divisors by unique factorisation. As there are finitely many $m \leq M_{K}$ , there are finitely many candidate ideals, hence finitely many classes.∎

Remark. Steps 1, 3 and 4 are complete arguments; only Step 2's constant needs the geometry of numbers. The practical content is Step 3 and Step 4 together: to compute a class group, factor every rational prime $p \leq M_{K}$ and find the relations among the resulting prime ideals. Every class is represented among those primes, so nothing else can occur.

Proposition 7.57 (Minkowski bound in the quadratic case). For $K = Q (d)$ with $d < 0$ the Minkowski bound is $M_{K} = \frac{2}{π} ∣ d_{K} ∣$ , and for $d > 0$ it is $M_{K} = \frac{1}{2} d_{K}$ .

Proof. Here $n = 2$ and $n! / n^{n} = 2/4 = 1/2$ . For $d < 0$ both embeddings are complex conjugates of each other, so $r_{1} = 0$ , $r_{2} = 1$ and $M_{K} = (4/ π) \cdot \frac{1}{2} ∣ d_{K} ∣ = \frac{2}{π} ∣ d_{K} ∣$ . For $d > 0$ both embeddings are real, $r_{2} = 0$ , and $M_{K} = \frac{1}{2} d_{K}$ .∎

Example 7.58 (Two fields with class number one, for free). Show $h = 1$ for $Q (i)$ and for $Q (5)$ without factoring anything.

Solution.

$Q (i)$ : $d_{K} = - 4$ , so $M_{K} = \frac{2}{π} 4 = \frac{4}{π} \approx 1.273$ .
Every class contains an integral ideal of norm $\leq 1.273$ , hence of norm $1$ , hence equal to $O_{K}$ . So there is one class: $h = 1$ .
$Q (5)$ : $5 \equiv 1 (mod 4)$ so $d_{K} = 5$ and $M_{K} = \frac{1}{2} 5 \approx 1.118 < 2$ . The same argument gives $h = 1$ .
The threshold is $M_{K} < 2$ : below it, only the unit ideal survives.

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Exercises for §7.15 →

7.16What the class number counts, and the analytic formula

The class number is not merely finite; it is computed, in principle, by an analytic formula that ties it to the Dedekind zeta function. The formula is quoted, not proved, but its shape explains every phenomenon in the tables.

Definition 7.59 (Dedekind zeta function). For a number field $K$ and $ℜ (s) > 1$ ,

ζ_{K} (s) = a \neq = 0 \sum \frac{1}{N ( a ) ^{s}} = p \prod (1 - \frac{1}{N ( p ) ^{s}})^{- 1},

the sum over nonzero integral ideals and the product over nonzero primes. It continues meromorphically to $C$ with a simple pole at $s = 1$ .

The Euler product is exactly the theorem Unique factorisation of ideals written analytically: expanding each factor as a geometric series and multiplying gives each ideal once, because each ideal is one product of prime powers.

Theorem 7.60 (Analytic class number formula). With $r_{1}, r_{2}$ the numbers of real and complex places, $w_{K}$ the number of roots of unity in $K$ , and $R_{K}$ the regulator defined in Regulator and its meaning,

s \to 1 lim (s - 1) ζ_{K} (s) = \frac{2 ^{r_{1}} ( 2 π ) ^{r_{2}} h _{K} R _{K}}{w _{K} ∣ d _{K} ∣} .

Proof. Labelled sketch — assumes the lattice-point counting of the geometry of numbers and analytic continuation. Group the ideals by class. For a fixed class $c$ , counting integral ideals of norm $\leq X$ in $c$ is, via the correspondence $a \leftrightarrow (α) b^{- 1}$ used in Step 3 of the theorem Finiteness of the class group, the same as counting lattice points of bounded norm in a fundamental domain for the action of the units. That count is asymptotically $κ X$ with $κ = \frac{2 ^{r_{1}} ( 2 π ) ^{r_{2}} R _{K}}{w _{K} ∣ d _{K} ∣}$ , the volume constant of the region divided by the covolume of the lattice, where $R_{K}$ appears as the covolume of the unit lattice of The image is a lattice and $w_{K}$ accounts for the torsion that acts trivially. Summing over the $h_{K}$ classes gives $h_{K} κ$ , and a Tauberian argument converts the counting asymptotic into the residue at $s = 1$ .∎

Corollary 7.61 (Dirichlet's class number formula for imaginary quadratic fields). Let $K = Q (d)$ with $d < 0$ , discriminant $D = d_{K} < - 4$ . Then $w_{K} = 2$ , $R_{K} = 1$ , and

h_{K} = \frac{1}{∣ D ∣} a = 1 \sum ∣ D ∣ - 1 (\frac{D}{a}) \cdot (- a) = - \frac{1}{∣ D ∣} a = 1 \sum ∣ D ∣ - 1 a (\frac{D}{a}),

where $(\frac{D}{\cdot})$ is the Kronecker symbol.

Proof. Labelled sketch — assumes the theorem Analytic class number formula and the factorisation $ζ_{K} (s) = ζ (s) L (s, χ_{D})$ for the quadratic character $χ_{D} = (\frac{D}{\cdot})$ . With $r_{1} = 0$ , $r_{2} = 1$ , and unit rank $0$ so $R_{K} = 1$ , the formula reads $L (1, χ_{D}) = \frac{2 π h _{K}}{w _{K} ∣ D ∣}$ . Evaluating $L (1, χ_{D})$ by the classical closed form $L (1, χ_{D}) = \frac{π}{∣ D ∣ ^{3/2}} \sum_{a} χ_{D} (a) (- a) \cdot (- 1)$ and comparing gives the stated sum. For $D < - 4$ the only roots of unity are $\pm 1$ , so $w_{K} = 2$ .∎

Remark. The formula is why class numbers of imaginary quadratic fields grow roughly like $∣ D ∣$ — the sum is a character sum of size about $∣ D ∣ lo g ∣ D ∣$ — and why $h = 1$ can only happen for small $∣ D ∣$ . That it happens for exactly nine discriminants, ending at $D = - 163$ , is the Baker–Heegner–Stark theorem, one of the hardest results in the subject.

Exercises for §7.16 →

7.17Computing class groups

Method 7.62 (Computing a class group from the Minkowski bound).

Compute $d_{K}$ and $M_{K}$ .
For each rational prime $p \leq M_{K}$ , factor $p O_{K}$ into prime ideals. For $K = Q (d)$ with $O_{K} = Z [θ]$ , this follows the factorisation of $m_{θ} mod p$ : $p$ splits, is inert, or ramifies according as $d_{K}$ is a nonzero square, a non-square, or $0$ modulo $p$ .
Discard primes of norm $1$ ; the remaining primes generate $Cl (K)$ .
Decide which of them are principal, by searching for an element of the right norm — a finite search, since $∣ N (a + b θ) ∣$ grows in $a, b$ .
Read off the group from the relations, remembering that $p \overset{ˉ}{p} = (p)$ is always principal, so conjugate primes are mutually inverse in $Cl (K)$ .

Example 7.63 ( $Cl (Q (- 5)) ≅ Z /2$ ). Compute the class group of $K = Q (- 5)$ .

Solution.

$- 5 \equiv 3 (mod 4)$ , so $O_{K} = Z [- 5]$ and $d_{K} = - 20$ .
$M_{K} = \frac{2}{π} 20 \approx 2.847$ , so only $p = 2$ matters.
$x^{2} + 5 \equiv x^{2} + 1 \equiv (x + 1)^{2} (mod 2)$ , so $2$ ramifies: $(2) = p_{2}^{2}$ with $p_{2} = (2, 1 + - 5)$ , of norm $2$ .
Is $p_{2}$ principal? It would need $α = a + b - 5$ with $∣ N (α) ∣ = a^{2} + 5 b^{2} = 2$ . Then $b = 0$ and $a^{2} = 2$ , impossible. So $p_{2}$ is not principal.
Therefore $Cl (K)$ is generated by the class of $p_{2}$ , which is nontrivial and satisfies $[p_{2}]^{2} = [(2)] = 1$ . Hence $Cl (K) ≅ Z /2$ and $h_{K} = 2$ .

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Example 7.64 ( $Cl (Q (- 23)) ≅ Z /3$ ). Compute the class group of $K = Q (- 23)$ .

Solution.

$- 23 \equiv 1 (mod 4)$ , so $O_{K} = Z [ω]$ with $ω = \frac{1 + - 23}{2}$ and $d_{K} = - 23$ . Norms take the form $N (\frac{a + b - 23}{2}) = \frac{a ^{2} + 23 b ^{2}}{4}$ with $a \equiv b (mod 2)$ .
$M_{K} = \frac{2}{π} 23 \approx 3.053$ , so $p = 2$ and $p = 3$ must be examined.
$m_{ω} (x) = x^{2} - x + 6$ . Modulo $2$ : $x^{2} + x = x (x + 1)$ , two distinct factors, so $(2) = p_{2} \overset{ˉ}{p}_{2}$ with norms $2$ . Modulo $3$ : $x^{2} - x + 6 \equiv x^{2} + 2 x = x (x + 2)$ , so $(3) = p_{3} \overset{ˉ}{p}_{3}$ with norms $3$ .
Principality: norm $2$ needs $a^{2} + 23 b^{2} = 8$ , impossible; norm $3$ needs $a^{2} + 23 b^{2} = 12$ , impossible. So none of the four primes is principal.
Relations: norm $8$ is achieved by $\frac{3 + - 23}{2}$ , since $\frac{9 + 23}{4} = 8$ . Its ideal has norm $8 = 2^{3}$ and is not divisible by both $p_{2}$ and $\overset{ˉ}{p}_{2}$ (else it would be divisible by $(2)$ , forcing $\frac{3 + - 23}{4} \in O_{K}$ , which is false). Hence $p_{2}^{3}$ is principal.
So $[p_{2}]$ has order dividing $3$ and is nontrivial, hence order exactly $3$ . Since $[p_{3}]$ must also lie in this group and $[\overset{ˉ}{p}_{2}] = [p_{2}]^{- 1}$ , the group is generated by $[p_{2}]$ : $Cl (K) ≅ Z /3$ , $h_{K} = 3$ .

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Pitfall. Step 4 in both examples is a norm obstruction, and it only proves non-principality because the norm form is positive definite, so the equation $a^{2} + ∣ d ∣ b^{2} = m$ has finitely many candidates. For real quadratic fields the norm form is indefinite and takes negative values, so "no solution of small size" proves nothing — one must work with the fundamental unit and continued fractions instead. Do not transplant the imaginary-quadratic argument to a real field.

Exercises for §7.17 →

7.18Number fields and their rings of integers: the picture so far

It is worth collecting what has been built before pushing on to the units, because the four objects fit together in one diagram: a number field $K$ , its ring of integers $O_{K}$ , the group of fractional ideals $I_{K}$ , and the finite quotient $Cl (K)$ .

Summary.

$K$ is a finite extension of $Q$ of degree $n$ , with $n$ embeddings into $C$ .
$O_{K}$ is the integral closure of $Z$ in $K$ : a free $Z$ -module of rank $n$ , with fraction field $K$ and discriminant $d_{K}$ .
$O_{K}$ is a Dedekind domain, so its nonzero ideals factor uniquely into primes, and its fractional ideals form the free abelian group $I_{K}$ on those primes.
$Cl (K) = I_{K} / P_{K}$ is finite of order $h_{K}$ , and $h_{K} = 1$ exactly when $O_{K}$ is a UFD.
Each rational prime $p$ factors as $\prod p_{i}^{e_{i}}$ with $\sum e_{i} f_{i} = n$ , and ramifies exactly when $p ∣ d_{K}$ .

Example 7.65 (A complete dossier on one field). Assemble $n$ , $O_{K}$ , $d_{K}$ , the ramified primes, the factorisation of $2$ and $3$ , and $h_{K}$ , for $K = Q (- 6)$ .

Solution.

$n = 2$ ; $- 6 \equiv 2 (mod 4)$ , so $O_{K} = Z [- 6]$ and $d_{K} = - 24$ .
Ramified primes are those dividing $- 24 = - 2^{3} \cdot 3$ : namely $2$ and $3$ .
$x^{2} + 6 \equiv x^{2} (mod 2)$ so $(2) = p_{2}^{2}$ with $p_{2} = (2, - 6)$ ; likewise $x^{2} + 6 \equiv x^{2} (mod 3)$ so $(3) = p_{3}^{2}$ with $p_{3} = (3, - 6)$ . In both cases $e = 2$ , $f = 1$ , $g = 1$ , and $e f = 2 = n$ . ✓
$M_{K} = \frac{2}{π} 24 \approx 3.119$ , so $Cl (K)$ is generated by $[p_{2}]$ and $[p_{3}]$ .
Norm $2$ would need $a^{2} + 6 b^{2} = 2$ and norm $3$ would need $a^{2} + 6 b^{2} = 3$ ; neither has a solution, so both primes are non-principal. But $p_{2} p_{3} = (- 6)$ is principal of norm $6$ , so $[p_{3}] = [p_{2}]^{- 1} = [p_{2}]$ (as $[p_{2}]^{2} = [(2)] = 1$ ).
Hence $Cl (K) = {1, [p_{2}]} ≅ Z /2$ and $h_{K} = 2$ .

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Exercises for §7.18 →

7.19The failure of unique factorisation

Here is the example that started the subject, worked in full.

Theorem 7.66 (Unique factorisation fails in $Z [- 5]$ ). In $R = Z [- 5]$ the four elements $2$ , $3$ , $1 + - 5$ , $1 - - 5$ are irreducible and pairwise non-associate, and

6 = 2 \cdot 3 = (1 + - 5) (1 - - 5)

are two essentially different factorisations of $6$ into irreducibles. Hence $R$ is not a UFD.

Proof. The norm here is $N (a + b - 5) = a^{2} + 5 b^{2}$ , multiplicative by the proposition Norm detects units setup, and always a non-negative integer. Note which values it can take: $b = 0$ gives squares $0, 1, 4, 9, \dots$ , and $b \neq = 0$ gives at least $5$ . So no element has norm $2$ or $3$ .

Units. By the proposition Norm detects units, $α$ is a unit iff $N (α) = 1$ , i.e. $a^{2} + 5 b^{2} = 1$ , i.e. $α = \pm 1$ . So $R^{\times} = {\pm 1}$ .

Irreducibility of $2$ . $N (2) = 4$ . If $2 = α β$ with both non-units, then $N (α) N (β) = 4$ with both factors $> 1$ , so $N (α) = N (β) = 2$ — impossible. So $2$ is irreducible.

Irreducibility of $3$ . $N (3) = 9$ ; a non-trivial splitting needs two factors of norm $3$ , impossible.

Irreducibility of $1 \pm - 5$ . $N (1 \pm - 5) = 1 + 5 = 6$ ; a non-trivial splitting needs norms $2$ and $3$ , impossible.

Non-associate. Associates differ by a unit $\pm 1$ . Now $2$ and $3$ have different norms from $1 \pm - 5$ , and $\pm (1 + - 5) \neq = 1 - - 5$ , so no two of the four are associates. In particular $2$ does not appear, even up to sign, among ${1 + - 5, 1 - - 5}$ .

Both products equal $6$ : $(1 + - 5) (1 - - 5) = 1 + 5 = 6$ . So $6$ has two factorisations into irreducibles whose multisets of associate-classes differ, which is exactly the failure of uniqueness.∎

Figure 7.67. The two element-factorisations of

6

Z [- 5]

are incompatible: no irreducible on the left is an associate of one on the right, and the norms

4 \cdot 9 = 6 \cdot 6

agree only in their product.

Example 7.68 (Why the gcd breaks down). Show that $2$ and $1 + - 5$ have no greatest common divisor in $Z [- 5]$ , and identify the ideal that should have been it.

Solution.

Common divisors of $2$ and $1 + - 5$ : a divisor has norm dividing $g cd (4, 6) = 2$ in the sense that it divides both $4$ and $6$ ; possible norms are $1$ and $2$ , and norm $2$ is impossible. So the only common divisors are the units.
If the gcd existed it would therefore be $1$ , i.e. $2$ and $1 + - 5$ would be coprime.
But then $2$ would divide $(1 + - 5) (1 - - 5) = 6$ "away from" $1 + - 5$ , forcing $2 ∣ 1 - - 5$ , which is false since $\frac{1 - - 5}{2} \in / R$ . Coprime elements cannot behave this way in a UFD.
The missing common divisor is the ideal $p_{2} = (2, 1 + - 5)$ , of norm $2$ . It divides both $(2)$ and $(1 + - 5)$ , and it is not principal — precisely because no element of norm $2$ exists.

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Exercises for §7.19 →

7.20Ideals restore what elements lost

Theorem 7.69 (The factorisations of $6$ reconcile at the ideal level). In $O_{K} = Z [- 5]$ set

p_{2} = (2, 1 + - 5), p_{3} = (3, 1 + - 5), q_{3} = (3, 1 - - 5) .

These are prime, of norms $2, 3, 3$ , and

(2) = p_{2}^{2}, (3) = p_{3} q_{3}, (1 + - 5) = p_{2} p_{3}, (1 - - 5) = p_{2} q_{3},

so that both element factorisations refine to the single ideal factorisation $(6) = p_{2}^{2} p_{3} q_{3}$ .

Proof. Primality and norms. $O_{K} / p_{2}$ : modulo $2$ and $1 + - 5$ we get $- 5 \equiv - 1 \equiv 1$ , so the quotient is $F_{2}$ , a field; hence $p_{2}$ is prime of norm $2$ . Modulo $p_{3}$ : $3 \equiv 0$ and $- 5 \equiv - 1$ , so the quotient is $F_{3}$ and $N (p_{3}) = 3$ ; similarly $N (q_{3}) = 3$ with $- 5 \equiv 1$ .

$(2) = p_{2}^{2}$ . Shown in the example Norm of a prime above 2 in $Z [- 5]$ .

$(3) = p_{3} q_{3}$ . The product is generated by $9$ , $3 (1 - - 5)$ , $3 (1 + - 5)$ and $(1 + - 5) (1 - - 5) = 6$ , all divisible by $3$ ; and $9 - 6 = 3$ lies in it. So the product is $(3)$ . Norms confirm: $3 \cdot 3 = 9 = ∣ N (3) ∣$ .

$(1 + - 5) = p_{2} p_{3}$ . Both $p_{2}$ and $p_{3}$ contain $1 + - 5$ , so by the corollary To contain is to divide each divides $(1 + - 5)$ , and being distinct primes their product divides it too. Norms: $N (p_{2} p_{3}) = 6 = ∣ N (1 + - 5) ∣$ , and two ideals with one dividing the other and equal norms are equal. The same argument with $q_{3}$ gives the last identity.

Reconciliation. $(6) = (2) (3) = p_{2}^{2} \cdot p_{3} q_{3}$ and $(6) = (1 + - 5) (1 - - 5) = (p_{2} p_{3}) (p_{2} q_{3}) = p_{2}^{2} p_{3} q_{3}$ . The two agree, as the theorem Unique factorisation of ideals demands.∎

Intuition. The element factorisations were two different ways of grouping the same four prime ideals into pairs that happen to be principal. ${p_{2} p_{2}}, {p_{3} q_{3}}$ is one grouping; ${p_{2} p_{3}}, {p_{2} q_{3}}$ is another. Neither $p_{2}$ nor $p_{3}$ is principal on its own, so neither grouping can be refined at the element level — and the ambiguity is exactly the non-triviality of $[p_{2}]$ in $Cl (K) ≅ Z /2$ .

Exercises for §7.20 →

7.21The class group as the exact obstruction

Proposition 7.70 (Non-principal primes produce non-unique factorisations). Let $K$ be a number field with $h_{K} > 1$ . Then $O_{K}$ has an element with two essentially different factorisations into irreducibles.

Proof. Pick a non-principal prime ideal $p$ — one exists, since if every prime ideal were principal then by unique factorisation every ideal would be, giving $h_{K} = 1$ by the theorem The class number detects unique factorisation. Let $k > 1$ be the order of $[p]$ in $Cl (K)$ , so $p^{k} = (π)$ is principal while no smaller power is.

Then $π$ is irreducible: a factorisation $π = α β$ into non-units gives $(α) (β) = p^{k}$ , so by unique factorisation $(α) = p^{i}$ with $0 < i < k$ , contradicting minimality of $k$ .

Now choose any prime $q \neq = p$ with $[q] = [p]^{- 1}$ ; then $pq = (γ)$ and $q^{k} = (ρ)$ are principal, and the same argument makes $γ$ and $ρ$ irreducible. Comparing ideals,

(π ρ) = p^{k} q^{k} = (pq)^{k} = (γ^{k}),

so $π ρ$ and $γ^{k}$ generate the same ideal and differ by a unit. That is one element written as a product of $2$ irreducibles and as a product of $k \geq 2$ irreducibles, with $π$ not an associate of $γ$ since $(π) = p^{k} \neq = pq = (γ)$ . The factorisations are essentially different.∎

The proposition and the theorem The class number detects unique factorisation together say the obstruction is exactly $Cl (K)$ : trivial group, unique factorisation; non-trivial group, a concrete counterexample built from any element of non-trivial class. With $k = 2$ , $p = p_{2}$ , $q = p_{3}$ in $Z [- 5]$ , the recipe produces $2 \cdot 3 = (1 + - 5) (1 - - 5)$ — the classical example is the general construction specialised.

Example 7.71 (Reading a class group off a table). The imaginary quadratic fields $Q (d)$ for $d = - 1, - 2, - 3, - 7, - 11, - 19, - 43, - 67, - 163$ all have $h = 1$ . What does that say about $Z [i]$ , and what fails for $d = - 14$ , where $h = 4$ ?

Solution.

$h = 1$ means $O_{K}$ is a UFD by the theorem The class number detects unique factorisation; for $d = - 1$ that is the classical unique factorisation in the Gaussian integers $Z [i]$ .
These nine are the complete list (the Baker–Heegner–Stark theorem), so no other imaginary quadratic field has unique factorisation.
For $d = - 14$ : $d_{K} = - 56$ , $h_{K} = 4$ , and $Cl (K) ≅ Z /4$ . There is a prime of order $4$ , so the construction of the proposition Non-principal primes produce non-unique factorisations gives an element that is a product of $2$ irreducibles in one way and of $4$ in another.

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Exercises for §7.21 →

7.22Units: the shape of the answer

The last invariant is the unit group $O_{K}^{\times}$ . For $Z$ it is ${\pm 1}$ ; for $Z [2]$ it is infinite, containing all powers of $1 + 2$ . Dirichlet's theorem says the difference is entirely accounted for by how many embeddings are real.

Definition 7.72 (Real and complex places). Let $K$ be a number field of degree $n$ . Among its $n$ embeddings into $C$ , let $r_{1}$ be the number landing in $R$ and $2 r_{2}$ the number that do not, these occurring in complex-conjugate pairs. Then $n = r_{1} + 2 r_{2}$ , and $(r_{1}, r_{2})$ is the signature of $K$ .

Theorem 7.73 (Dirichlet's unit theorem). Let $K$ have signature $(r_{1}, r_{2})$ and put $r = r_{1} + r_{2} - 1$ . Then

O_{K}^{\times} ≅ μ_{K} \times Z^{r},

where $μ_{K}$ is the finite cyclic group of roots of unity in $K$ . Equivalently, there are fundamental units $ε_{1}, \dots, ε_{r}$ such that every unit is uniquely $ζ ε_{1}^{a_{1}} \dots ε_{r}^{a_{r}}$ with $ζ \in μ_{K}$ and $a_{i} \in Z$ .

The proof occupies the remaining sections: the logarithmic embedding turns the multiplicative group into a lattice, and the rank is computed from the geometry. The rank formula itself is immediate to apply.

Proposition 7.74 (Unit rank in low degree).

Imaginary quadratic $Q (d)$ , $d < 0$ : $(r_{1}, r_{2}) = (0, 1)$ , so $r = 0$ and $O_{K}^{\times} = μ_{K}$ is finite.
Real quadratic $Q (d)$ , $d > 0$ : $(r_{1}, r_{2}) = (2, 0)$ , so $r = 1$ and $O_{K}^{\times} = {\pm ε^{a}}$ for a single fundamental unit $ε$ .
$Q (3 m)$ for $m$ a non-cube positive integer: $(r_{1}, r_{2}) = (1, 1)$ , so $r = 1$ .
$Q (ζ_{p})$ for an odd prime $p$ : degree $p - 1$ , all embeddings complex, so $(r_{1}, r_{2}) = (0, \frac{p - 1}{2})$ and $r = \frac{p - 3}{2}$ .

Proof. In each case count embeddings. For $d < 0$ the two embeddings send $d$ to $\pm i ∣ d ∣$ , neither real, and they are conjugate: $r_{1} = 0$ , $r_{2} = 1$ , $r = 0 + 1 - 1 = 0$ . For $d > 0$ both embeddings are real: $r_{1} = 2$ , $r_{2} = 0$ , $r = 1$ . For $θ = 3 m$ the three embeddings send $θ$ to $θ$ , $ω θ$ , $ω^{2} θ$ with $ω$ a primitive cube root of unity; the first is real and the other two are a conjugate pair, so $(1, 1)$ and $r = 1$ . For $Q (ζ_{p})$ the embeddings send $ζ_{p}$ to the $p - 1$ primitive $p$ -th roots of unity, none real for $p$ odd, pairing as $ζ \leftrightarrow \overset{ˉ}{ζ}$ : $r_{2} = \frac{p - 1}{2}$ and $r = \frac{p - 1}{2} - 1 = \frac{p - 3}{2}$ .∎

Example 7.75 (Units of $Z [2]$ and of $Z [i]$ ). Determine $O_{K}^{\times}$ for $K = Q (2)$ and for $K = Q (i)$ .

Solution.

$Q (i)$ : signature $(0, 1)$ , so $r = 0$ and the unit group is just $μ_{K} = {\pm 1, \pm i}$ , cyclic of order $4$ . Check: $N (a + bi) = a^{2} + b^{2} = 1$ has exactly those four solutions.
$Q (2)$ : signature $(2, 0)$ , so $r = 1$ and $O_{K}^{\times} = {\pm ε^{a} : a \in Z}$ , since the only roots of unity in a real field are $\pm 1$ .
Finding $ε$ : $N (a + b 2) = a^{2} - 2 b^{2} = \pm 1$ . The smallest solution with $a, b > 0$ is $(a, b) = (1, 1)$ , giving $ε = 1 + 2$ with $N (ε) = - 1$ .
It is fundamental because any unit $> 1$ is $\geq ε$ : a unit $η = a + b 2 > 1$ has $∣ η^{- 1} ∣ = ∣ a - b 2 ∣ < 1$ , forcing $a, b \geq 1$ and hence $η \geq 1 + 2$ .
So $O_{K}^{\times} = {\pm (1 + 2)^{a}}$ , infinite — in bijection with ${\pm 1} \times Z$ .

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Exercises for §7.22 →

7.23The logarithmic embedding

Multiplicative structure is hard to see; additive structure is easy. Logarithms convert one into the other, and the whole proof of Dirichlet's theorem is an exercise in that conversion.

Definition 7.76 (Logarithmic embedding). Let $K$ have signature $(r_{1}, r_{2})$ , with real embeddings $σ_{1}, \dots, σ_{r_{1}}$ and one from each complex conjugate pair $τ_{1}, \dots, τ_{r_{2}}$ . The logarithmic embedding is

λ : K^{\times} \to R^{r_{1} + r_{2}}, λ (α) = (lo g ∣ σ_{1} α ∣, \dots, lo g ∣ σ_{r_{1}} α ∣, 2 lo g ∣ τ_{1} α ∣, \dots, 2 lo g ∣ τ_{r_{2}} α ∣),

where the factor $2$ on the complex coordinates records that each such embedding occurs twice among the $n$ embeddings of $K$ .

Proposition 7.77 ( $λ$ is a homomorphism sending units into a hyperplane). $λ (α β) = λ (α) + λ (β)$ , and the sum of the coordinates of $λ (α)$ is $lo g ∣ N_{K / Q} (α) ∣$ . Consequently $λ (O_{K}^{\times})$ lies in the hyperplane

H = {x \in R^{r_{1} + r_{2}} : \sum_{i} x_{i} = 0},

of dimension $r = r_{1} + r_{2} - 1$ .

Proof. Each $σ_{i}$ and $τ_{j}$ is multiplicative and $lo g ∣ \cdot ∣$ turns products into sums, giving the homomorphism property. For the coordinate sum, the $n$ embeddings of $K$ consist of the $r_{1}$ real ones and, for each $j$ , the pair $τ_{j}, \overset{τ}{ˉ}_{j}$ with $∣ τ_{j} α ∣ = ∣ \overset{τ}{ˉ}_{j} α ∣$ . Hence

i \sum λ (α)_{i} = i \leq r_{1} \sum lo g ∣ σ_{i} α ∣ + j \sum 2 lo g ∣ τ_{j} α ∣ = lo g k = 1 \prod n ∣ σ_{k} α ∣ = lo g ∣ N_{K / Q} (α) ∣.

If $α \in O_{K}^{\times}$ then $∣ N (α) ∣ = 1$ by the proposition Norm detects units, so the sum is $lo g 1 = 0$ .∎

Proposition 7.78 (The kernel is the roots of unity). $ker (λ ∣_{O_{K}^{\times}}) = μ_{K}$ , the group of roots of unity in $K$ , which is finite and cyclic.

Proof. A root of unity $ζ$ has $∣ σ ζ ∣ = 1$ for every embedding, since $σ ζ$ is again a root of unity; so $λ (ζ) = 0$ .

Conversely, suppose $λ (α) = 0$ for $α \in O_{K}^{\times}$ : then $∣ σ α ∣ = 1$ for every embedding $σ$ . The coefficients of the characteristic polynomial $χ_{α}$ are, up to sign, elementary symmetric functions of the $σ_{i} α$ , hence bounded in absolute value by $(k n)$ ; and they are rational integers by the corollary Norms and traces of algebraic integers are rational integers. So there are only finitely many possible $χ_{α}$ , and each has at most $n$ roots: the kernel is a finite set. A finite subgroup of $K^{\times}$ consists of roots of unity (each element has finite order) and is cyclic, since a finite subgroup of the multiplicative group of a field is cyclic.∎

Intuition. The picture to carry: a unit is a point of $O_{K}$ whose "size" is $1$ overall, but which can be enormous under one embedding provided it is correspondingly tiny under another. The logarithmic embedding plots those trade-offs as a point on the hyperplane $\sum x_{i} = 0$ , and Dirichlet's theorem says the trade-offs available form a lattice of full rank in that hyperplane — as many independent trade-offs as there are places, minus the one constraint.

Exercises for §7.23 →

7.24The image is a full lattice

Theorem 7.79 (The unit lattice). $Λ = λ (O_{K}^{\times})$ is a full lattice in $H$ : a discrete subgroup spanning $H$ over $R$ , hence free abelian of rank $r = r_{1} + r_{2} - 1$ . Together with the proposition The kernel is the roots of unity this proves the theorem Dirichlet's unit theorem.

Proof. Discreteness, proved. Let $B \subseteq H$ be bounded, say $∣ x_{i} ∣ \leq C$ for all $x \in B$ . If $λ (α) \in B$ then $∣ σ α ∣ \leq e^{C}$ for every embedding, so as in the proof of the proposition The kernel is the roots of unity the coefficients of $χ_{α}$ are bounded rational integers and only finitely many $α$ occur. Hence $Λ \cap B$ is finite for every bounded $B$ : $Λ$ is discrete, and a discrete subgroup of $R^{m}$ is free abelian of rank at most $m$ , generated by $R$ -independent vectors.

Full rank — labelled sketch, assumes Minkowski's convex body theorem (M) as in the theorem Finiteness of the class group. One must show $Λ$ spans $H$ , i.e. has rank exactly $r$ . The standard argument fixes an index $k$ and uses (M) to produce, for each $k$ , a unit $η_{k}$ whose $k$ -th logarithmic coordinate is positive while every other coordinate is negative. Concretely, (M) supplies infinitely many elements of $O_{K}$ of bounded norm with prescribed embedding sizes; boundedly many ideals have a given norm, so two of these elements generate the same ideal and their ratio is a unit with the required lopsided profile. The matrix whose rows are $λ (η_{1}), \dots, λ (η_{r})$ then has negative off-diagonal entries and positive row sums in absolute value, and such a matrix is nonsingular — an elementary determinant lemma. So the $λ (η_{k})$ are independent and $Λ$ has rank $r$ .

Granting that, $λ$ gives a short exact sequence $1 \to μ_{K} \to O_{K}^{\times} \to Λ \to 0$ with $Λ ≅ Z^{r}$ free, so the sequence splits and $O_{K}^{\times} ≅ μ_{K} \times Z^{r}$ .∎

Figure 7.80. For a real quadratic field the hyperplane

H

is the line

x_{1} + x_{2} = 0

, and the unit lattice is the set of multiples of

λ (ε) = (lo g ε, - lo g ε)

. Here

K = Q (2)

ε = 1 + 2

and

lo g ε \approx 0.8814

, which is the regulator.

Example 7.81 (The lattice for $Q (3)$ ). Compute $λ (ε)$ for the fundamental unit of $Q (3)$ and describe the lattice.

Solution.

$3 \equiv 3 (mod 4)$ , so $O_{K} = Z [3]$ and units solve $a^{2} - 3 b^{2} = \pm 1$ .
$a^{2} - 3 b^{2} = - 1$ has no solution modulo $3$ (it would need $a^{2} \equiv - 1 \equiv 2$ , and squares mod $3$ are $0, 1$ ). The smallest solution of $a^{2} - 3 b^{2} = 1$ is $(2, 1)$ , so $ε = 2 + 3$ .
The two real embeddings send $ε$ to $2 + 3 \approx 3.732$ and $2 - 3 \approx 0.268 = ε^{- 1}$ .
So $λ (ε) = (lo g (2 + 3), - lo g (2 + 3)) \approx (1.3170, - 1.3170)$ , and $Λ = Z λ (ε)$ is the set of points $(k \cdot 1.3170, - k \cdot 1.3170)$ on the line $x_{1} + x_{2} = 0$ .

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Exercises for §7.24 →

7.25The regulator

Definition 7.82 (Regulator). Let $ε_{1}, \dots, ε_{r}$ be fundamental units and delete any one column from the $r \times (r_{1} + r_{2})$ matrix $(λ (ε_{i})_{j})$ . The regulator $R_{K}$ is the absolute value of the determinant of the resulting $r \times r$ matrix. For $r = 0$ one sets $R_{K} = 1$ .

Proposition 7.83 (The regulator is well defined). $R_{K}$ does not depend on which column is deleted, nor on the choice of fundamental units. It equals the covolume of $Λ$ inside $H$ up to the fixed factor $r_{1} + r_{2}$ , and $R_{K} > 0$ whenever $r \geq 1$ .

Proof. Independence of the units. Two systems of fundamental units differ by a matrix in $G L_{r} (Z)$ , of determinant $\pm 1$ , which does not change the absolute determinant.

Independence of the column. Each row of the full matrix sums to $0$ by the proposition $λ$ is a homomorphism sending units into a hyperplane. Adding all other columns to column $j$ therefore turns column $j$ into $0$ ; equivalently, the $r \times r$ minor obtained by deleting column $j$ equals, up to sign, the minor obtained by deleting column $k$ , since both are computed from the same rank- $r$ row space subject to one linear relation among the columns. Hence all such minors have the same absolute value.

Positivity and covolume. By the theorem The unit lattice, $Λ$ has rank $r = dim H$ , so its generators are independent and the determinant is nonzero. The covolume statement is the standard fact that projecting $H$ orthogonally onto the coordinate hyperplane obtained by deleting one axis scales volume by $1/ r_{1} + r_{2}$ .∎

Example 7.84 (Regulators of three real quadratic fields). Compute $R_{K}$ for $Q (2)$ , $Q (5)$ and $Q (6)$ .

Solution.

$r = 1$ , so the matrix is $1 \times 2$ and $R_{K} = ∣ lo g ∣ σ_{1} ε ∣∣ = lo g ε$ for the fundamental unit $ε > 1$ .
$Q (2)$ : $ε = 1 + 2$ , $R_{K} = lo g (1 + 2) \approx 0.8814$ .
$Q (5)$ : $5 \equiv 1 (mod 4)$ so $O_{K} = Z [\frac{1 + 5}{2}]$ and $ε = \frac{1 + 5}{2}$ , of norm $\frac{1 - 5}{4} = - 1$ . So $R_{K} = lo g \frac{1 + 5}{2} \approx 0.4812$ — smaller than for $2$ , precisely because the larger ring of integers contains a smaller unit.
$Q (6)$ : $6 \equiv 2 (mod 4)$ , $O_{K} = Z [6]$ , and $a^{2} - 6 b^{2} = \pm 1$ has smallest solution $(5, 2)$ : $25 - 24 = 1$ . So $ε = 5 + 26$ and $R_{K} = lo g (5 + 26) \approx 2.2924$ .

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Remark. The regulator is the analytic counterweight to the class number. In the theorem Analytic class number formula only the product $h_{K} R_{K}$ is determined by the zeta function, which is why real quadratic class numbers are hard to compute: a large regulator and small class number are analytically indistinguishable from the reverse without further work.

Exercises for §7.25 →

7.26Assembling the proof of Dirichlet's theorem

Summary. The proof of the theorem Dirichlet's unit theorem has exactly four moving parts, three of which are proved above.

$λ$ is a homomorphism into the hyperplane $H$ — proved, from multiplicativity of the norm.
The kernel is $μ_{K}$ , finite and cyclic — proved, by bounding the coefficients of characteristic polynomials.
The image is discrete — proved, by the same bounding argument applied to a bounded region.
The image spans $H$ — assumed, via Minkowski's convex body theorem; this is the only genuinely analytic input.

Proposition 7.85 (Why the rank cannot exceed $r_{1} + r_{2} - 1$ ). Without any appeal to the geometry of numbers, $rank Λ \leq r_{1} + r_{2} - 1$ .

Proof. $Λ \subseteq H$ by the proposition $λ$ is a homomorphism sending units into a hyperplane, and $dim_{R} H = r_{1} + r_{2} - 1$ because $H$ is the kernel of the nonzero linear functional $x \mapsto \sum x_{i}$ on $R^{r_{1} + r_{2}}$ . A discrete subgroup of a real vector space has rank at most its dimension, and $Λ$ is discrete by the theorem The unit lattice.∎

So the upper bound on the rank is elementary, and only the matching lower bound needs (M). This is a useful thing to know: if a specific field is at hand and one can exhibit $r$ multiplicatively independent units, the rank is settled without invoking Dirichlet's theorem at all.

Example 7.86 (Confirming the rank by hand for $Q (32)$ ). Show that $O_{K}^{\times}$ for $K = Q (32)$ has rank exactly $1$ , using only the proposition Why the rank cannot exceed $r_{1} + r_{2} - 1$ and one explicit unit.

Solution.

Signature: one real embedding $θ \mapsto 32$ and one conjugate pair, so $(r_{1}, r_{2}) = (1, 1)$ and the upper bound is $1 + 1 - 1 = 1$ .
A unit: let $η = θ - 1$ where $θ = 32$ . From $N_{K / Q} (c - θ) = m_{θ} (c)$ with $m_{θ} = x^{3} - 2$ we get $N (1 - θ) = 1 - 2 = - 1$ , and $N (θ - 1) = (- 1)^{3} N (1 - θ) = 1$ .
Since $∣ N (η) ∣ = 1$ and $η \in O_{K}$ , the proposition Norm detects units makes $η$ a unit.
$η = 32 - 1 \approx 0.2599$ is not a root of unity (a real root of unity is $\pm 1$ ), so it has infinite order and $Λ \neq = 0$ .
Rank is at least $1$ and at most $1$ , hence exactly $1$ : $O_{K}^{\times} = {\pm η^{a}}$ up to a possible finite-index subtlety in the choice of fundamental unit, and in fact $η$ is fundamental here.

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Example 7.87 (Rank for a cyclotomic field). Compute the unit rank of $Q (ζ_{7})$ and of $Q (ζ_{5})$ .

Solution.

$[Q (ζ_{7}) : Q] = φ (7) = 6$ , all embeddings complex, so $(r_{1}, r_{2}) = (0, 3)$ and $r = 0 + 3 - 1 = 2$ .
$[Q (ζ_{5}) : Q] = φ (5) = 4$ , so $(r_{1}, r_{2}) = (0, 2)$ and $r = 1$ .
Consistent with the formula $r = \frac{p - 3}{2}$ of the proposition Unit rank in low degree: $\frac{7 - 3}{2} = 2$ and $\frac{5 - 3}{2} = 1$ .
For $Q (ζ_{5})$ an explicit fundamental unit is $\frac{1 + 5}{2} = - ζ_{5}^{2} - ζ_{5}^{3}$ , which lies in the real subfield $Q (5)$ .

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Exercises for §7.26 →

Common Mistakes

Forgetting "monic". $\frac{1}{2}$ is a root of $2 x - 1$ , an integer polynomial, and is not an algebraic integer. Only monic integer polynomials count; the reliable test is the theorem Integrality is visible in the minimal polynomial.
Writing $O_{K} = Z [d]$ for every $d$ . When $d \equiv 1 (mod 4)$ the ring of integers is $Z [\frac{1 + d}{2}]$ , strictly larger, with discriminant $d$ rather than $4 d$ . Check $d mod 4$ before anything else.
Computing a discriminant from a non-integral basis. $disc (1, 5) = 20$ , but $d_{K} = 5$ . The difference is the square of the index, and using the wrong one invents ramified primes that do not exist.
Expecting unique factorisation of elements. It holds only when $h_{K} = 1$ . In $Z [- 5]$ the element $6$ has two genuinely different factorisations; what is unique is the factorisation of the ideal $(6)$ .
Confusing irreducible with prime. In $Z [- 5]$ the element $2$ is irreducible but not prime: it divides $(1 + - 5) (1 - - 5) = 6$ without dividing either factor. The two notions coincide only in a UFD.
Thinking a non-principal ideal has no factorisation. Every nonzero ideal factors into primes, principal or not. Non-principality is about generators, not about factorisation.
Using a norm obstruction in a real quadratic field. "There is no element of norm $2$ , so the ideal is not principal" is valid when the norm form is positive definite. For $d > 0$ the form is indefinite and the argument collapses.
Misremembering the unit rank. It is $r_{1} + r_{2} - 1$ , not $r_{1} + r_{2}$ and not $n - 1$ . Imaginary quadratic fields have rank $0$ — finitely many units — while real quadratic fields have rank $1$ and therefore infinitely many.
Treating $R_{K} = 1$ as a special case to memorise. It is the empty determinant, which arises exactly when $r = 0$ , i.e. for $Q$ and for imaginary quadratic fields.
Assuming a power basis exists. $O_{K} = Z [θ]$ can fail for every $θ$ . Compute an integral basis by the recipe Enlarging an order to the maximal one rather than guessing.

About the practice questions. They check that you can carry out this chapter's computations correctly, and each one is graded on a single answer. They are not proof exercises: working through them confirms the mechanics, not that you could prove the results yourself. For that, re-read the statements above and try to reconstruct their proofs with the page closed.

Practice this chapter

Introduction

7.1What it means to be an algebraic integer

7.2Algebraic numbers, algebraic integers, and the rational case

7.3The algebraic integers form a ring

7.4Minimal polynomial, characteristic polynomial, norm and trace

7.5The ring of integers as a Z -module

7.6Rings of integers of quadratic fields: power basis or not

7.7Discriminant and ramification

7.8Computing an integral basis

7.9Dedekind domains

7.10Unique factorisation of ideals

7.11Localisation: every Dedekind domain is locally a DVR

7.12Fractional ideals and the ideal group

7.13Rings of integers are Dedekind domains

7.14The ideal class group

7.15Finiteness of the class group

7.16What the class number counts, and the analytic formula

7.17Computing class groups

7.18Number fields and their rings of integers: the picture so far

7.19The failure of unique factorisation

7.20Ideals restore what elements lost

7.21The class group as the exact obstruction

7.22Units: the shape of the answer

7.23The logarithmic embedding

7.24The image is a full lattice

7.25The regulator

7.26Assembling the proof of Dirichlet's theorem

7.5The ring of integers as a $Z$ -module