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✓ 18 results · all verified · 6 also independently AI-judged

Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 12 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Dirichlets Unit Theorem Regulators and S Units

1 · Prerequisites

2 · Summary

The unit group of a number field is finite at rank zero and otherwise a finite torsion group times a free abelian group. This page proves that structure from the arithmetic of the maximal order: only finitely many roots of unity lie in $K$ , Kronecker's criterion converts bounded conjugates into torsion, and a unit of $O_{K}$ is exactly an element of norm $\pm 1$ . The product formula is stated in the normalization the argument needs, with finite absolute values $N p^{- v_{p}}$ , real ones $∣ σ (x) ∣$ and complex ones $∣ τ (x) ∣^{2}$ ; its ideal-factorization step assumes Choice. The full-lattice argument also states Choice for its Minkowski and measure inputs, and the unit and S-unit results carry the hypotheses of their dependencies.

The logarithmic embedding $λ$ doubles the complex coordinates, so that the product-formula identity becomes the literal coordinate-sum-zero hyperplane $H$ ; mixing the doubled and undoubled conventions silently rescales regulators by powers of two. On units, $λ$ has kernel exactly the roots of unity, its image is discrete, and the central theorem of the page, after Stein, shows that image is a full lattice in $H$ : the equality case of the Minkowski convex-body theorem produces a small element of $O_{K}$ , the boundedly many principal ideals of bounded norm reduce it to a unit, and the two-sided bound forces the span to fill $H$ . Dirichlet's unit theorem then reads $O_{K}^{\times} ≅ μ (K) \times Z^{r_{1} + r_{2} - 1}$ , with the rank $r_{1} + r_{2} - 1$ recorded as a signature corollary covering the rank-zero fields $Q$ and the imaginary quadratic fields, and the real quadratic case of rank one.

The regulator is defined by the absolute value of a deleted-row minor of the logarithmic matrix, with the empty determinant set to $1$ at rank zero. The deleted-row minors of a matrix with zero column sums are independent of the deleted row up to sign, and a unimodular change of generating system multiplies every minor by $\pm 1$ , so the regulator of a fundamental system is well defined; the definition and the well-definedness theorem are stated for the fundamental systems of the unit theorem, and the deletion normalization is made explicit. The page closes with $S$ -integers and $S$ -units for a finite set $S$ of finite primes. The valuation map to $Z^{S}$ has kernel $O_{K}^{\times}$ and finite-index image, since the class number kills the classes of primes in $S$ ; this gives $O_{K, S}^{\times} ≅ μ (K) \times Z^{r_{1} + r_{2} - 1 + ∣ S ∣}$ .

3 · Logical flowchart

4 · Definitions, theorems and proofs

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-02Open item page →

Finitely many roots of unity in a number field

Statement

Let $K$ be a number field. The group $μ (K)$ of roots of unity contained in $K$ is finite.

Facts & Assumptions

Given: A number field $K$ of degree $n = [K : Q]$ , and the set $μ (K) = ⋃_{N \geq 1} μ_{N} (K)$ of the elements of $K$ that satisfy $x^{N} = 1$ for some $N \geq 1$ .

[F1]

For every $N \geq 1$ the set $μ_{N} (K) = {x \in K : x^{N} = 1}$ is a subgroup of $K^{\times}$ , and an element $x \in K$ is a root of unity exactly when $x^{N} = 1$ for some $N \geq 1$ (The group $μ_{n} (K)$ of $n$ -th roots of unity in a field, and primitive $n$ -th roots of unity).

[F2]

An element $b$ of a commutative ring $B$ is integral over a subring $A$ when it is a root of a monic polynomial in $A [X]$ , and an algebraic integer is a complex number integral over $Z$ (Integral elements over a commutative ring and algebraic integers); the ring of integers $O_{K}$ is the integral closure of $Z$ in $K$ (Ring of integers).

[F3]

For $α \in K$ , one has $α \in O_{K}$ if and only if the monic minimal polynomial of $α$ over $Q$ lies in $Z [X]$ (Minimal-polynomial criterion for algebraic integers).

[F4]

For an algebraic element $a$ of an extension of $Q$ , the monic minimal polynomial $m_{a} \in Q [X]$ satisfies $f (a) = 0$ if and only if $m_{a} ∣ f$ in $Q [X]$ (The evaluation kernel and the unique monic irreducible minimal polynomial of an algebraic element).

[F5]

Complex modulus satisfies $∣ z w ∣ = ∣ z ∣ ∣ w ∣$ , $∣ z ∣ \geq 0$ and $∣ z ∣ = 0$ only for $z = 0$ (Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive); the $N$ -th roots of unity in $C$ are the numbers $e^{2 π i k / N}$ , $k = 0, \dots, N - 1$ , all of modulus one (The $n$ -th roots of a complex number and the $n$ distinct roots of unity for every $n \geq 1$ ).

[F6]

If $a$ is algebraic over $F$ with minimal polynomial of degree $d$ , then $[F (a) : F] = d$ (An element is algebraic over $F$ if and only if its simple extension $F (a) / F$ is finite).

[F7]

If $F \subseteq E \subseteq L$ and $L / F$ is finite, then $E / F$ and $L / E$ are finite and $[E : F]$ divides $[L : F]$ (The degree of an intermediate field divides the degree of a finite extension); for finite extensions the degrees multiply, $[L : F] = [L : E] [E : F]$ (Tower law for finite extensions: $[L : F] = [L : K] [K : F]$ , The degree $[K : F] = \dim_{F} K$ of a finite field extension).

[F8]

There are only finitely many monic integer polynomials of degree at most $n$ whose complex roots, counted with multiplicity, all have modulus at most $R = 1$ (Bounded roots give finitely many monic integer polynomials). This is the one batch-2 supplier consumed here, authored in this run; the exact obligation used is that the set of monic integer polynomials of degree at most $n$ all of whose complex roots have modulus at most $1$ is finite.

[F9]

A nonzero polynomial of degree $d$ over an integral domain has at most $d$ distinct roots in that domain (A nonzero polynomial of degree $n$ over an integral domain has at most $n$ distinct roots); $C$ is a field, hence an integral domain.

Proof

technique · every root of unity in $K$ is an algebraic integer whose minimal polynomial has degree at most $n$ and all of whose complex roots have modulus $1$; the bounded-conjugate polynomials of that degree form a finite box, and each polynomial has at most $n$ roots

1.1F1

The set $μ (K)$ is a subgroup of $K^{\times}$ : it contains $1$ ; if $ζ^{m} = 1$ and $η^{k} = 1$ then $(ζ η)^{m k} = ζ^{m k} η^{m k} = 1$ ; and if $ζ^{m} = 1$ then $(ζ^{- 1})^{m} = (ζ^{m})^{- 1} = 1$ .

1.2F2F3F4

Let $ζ \in K$ be a root of unity with $ζ^{N} = 1$ for some $N \geq 1$ . Then $ζ$ is a root of the monic polynomial $X^{N} - 1 \in Z [X]$ , so $ζ$ is integral over $Z$ and therefore lies in $O_{K}$ ; its monic minimal polynomial $m_{ζ} \in Q [X]$ has coefficients in $Z$ ; and $m_{ζ}$ divides $X^{N} - 1$ in $Q [X]$ , because the polynomial $X^{N} - 1$ vanishes at $ζ$ and $m_{ζ}$ is the minimal polynomial of $ζ$ .

1.3F5

Every complex root $w$ of the polynomial $X^{N} - 1$ satisfies $w^{N} = 1$ , hence $∣ w ∣^{N} = ∣ w^{N} ∣ = 1$ with $∣ w ∣ \geq 0$ , so $∣ w ∣ = 1$ ; equivalently the roots of $X^{N} - 1$ are the $N$ -th roots of unity, of modulus one.

1.4F6F7

The degree of $m_{ζ}$ equals $[Q (ζ) : Q]$ by [F6], and $Q \subseteq Q (ζ) \subseteq K$ with $K / Q$ finite, so $[Q (ζ) : Q]$ is finite and divides $[K : Q] = n$ by [F7]; in particular $d : = \deg m_{ζ} \leq n$ .

2.1step 1.2step 1.3step 1.4

Every complex root $w$ of $m_{ζ}$ is a complex root of $X^{N} - 1$ , since $m_{ζ} ∣ X^{N} - 1$ in $Q [X]$ and therefore in $C [X]$ ; by step 1.3 such a root has $∣ w ∣ = 1$ . Hence $m_{ζ}$ is a monic integer polynomial of degree $d \leq n$ all of whose complex roots have modulus at most $1$ , with the degree bound of step 1.4.

3.1F8F9step 2.1

By [F8] the monic integer polynomials of degree at most $n$ whose complex roots all have modulus at most $1$ are only finitely many; fix a list $g_{1}, \dots, g_{M}$ of them. Each $g_{j}$ has degree at most $n$ , hence at most $n$ distinct complex roots by [F9], so the union of their complex root sets has at most $n M$ elements.

4.1step 1.1step 3.1∎

Every root of unity $ζ \in K$ has $m_{ζ}$ of the form $g_{j}$ by step 2.1, so $ζ$ is a root of one of the finitely many polynomials $g_{1}, \dots, g_{M}$ ; therefore $μ (K)$ is contained in the finite union of their root sets, and $μ (K)$ is finite. By step 1.1 it is the group of roots of unity contained in $K$ .

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-02Open item page →

Kronecker root-of-unity criterion

Statement

Let $K$ be a number field and let $0 \neq α \in O_{K}$ be an algebraic integer all of whose complex conjugates satisfy $∣ σ (α) ∣ \leq 1$ . Then $α$ is a root of unity.

Facts & Assumptions

Given: A number field $K$ of degree $n = [K : Q]$ , the set $Σ = {Hom}_{Q} (K, C)$ of its $n$ embeddings into $C$ , and an element $0 \neq α \in O_{K}$ with $∣ σ (α) ∣ \leq 1$ for every $σ \in Σ$ .

[F1]

$K / Q$ is separable, because $Q$ has characteristic zero, hence is perfect, and algebraic extensions of perfect fields are separable (Fields of characteristic zero, finite fields, and algebraically closed fields are perfect, Every algebraic extension of a perfect field is separable); so the norm is the product over the distinct embeddings, $N_{K / Q} (α) = \prod_{σ \in Σ} σ (α)$ , with $∣ Σ ∣ = n = [K : Q]$ (Norm and trace from embeddings, with the inseparable exponent in the norm formula, Archimedean embeddings and signature).

[F2]

For $α \in O_{K}$ the norm $N_{K / Q} (α)$ is an integer (Trace and norm of an algebraic integer); if $α \neq 0$ then multiplication by $α$ is an invertible linear map, so $N_{K / Q} (α) \neq 0$ (The norm $N_{K / F}$ and trace ${Tr}_{K / F}$ of a finite field extension, Ring of integers).

[F3]

An element $β$ is conjugate to $α$ over $Q$ exactly when $β$ is a complex root of the minimal polynomial $m_{α} \in Q [X]$ (Conjugate algebraic elements over a field). Sending an embedding $τ : Q (α) \to C$ to $τ (α)$ is a bijection onto the set of distinct complex roots of $m_{α}$ ( $F$ -embeddings of $F (α)$ into an algebraically closed field correspond to the distinct roots of $m_{α}$ ); restriction $Σ \to {Hom}_{Q} (Q (α), C)$ is surjective (Restriction partitions embeddings in a finite tower into extension fibres); and for $σ \in Σ$ one has $m_{α} (σ (α)) = σ (m_{α} (α)) = 0$ . Hence the set of complex roots of $m_{α}$ is exactly ${σ (α) : σ \in Σ}$ .

[F4]

Complex modulus is multiplicative, $∣ z w ∣ = ∣ z ∣ ∣ w ∣$ , and $∣ z ∣ = 0$ only for $z = 0$ (Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive).

[F5]

Sums and products of elements integral over $Z$ are integral (Integral elements over a nonzero base ring form a subring), so for $m \geq 1$ the power $α^{m}$ is a nonzero element of $O_{K}$ , the integral closure of $Z$ in $K$ (Ring of integers).

[F6]

For every $m \geq 1$ the minimal polynomial $m_{α^{m}}$ of the algebraic element $α^{m}$ has coefficients in $Z$ , and $\deg m_{α^{m}} = [Q (α^{m}) : Q]$ divides $[K : Q] = n$ (Minimal-polynomial criterion for algebraic integers, An element is algebraic over $F$ if and only if its simple extension $F (a) / F$ is finite, The degree of an intermediate field divides the degree of a finite extension).

[F7]

Every complex root $w$ of $m_{α^{m}}$ is the image of $α^{m}$ under a $Q$ -embedding of $Q (α^{m})$ into $C$ ( $F$ -embeddings of $F (α)$ into an algebraically closed field correspond to the distinct roots of $m_{α}$ ). Such an embedding extends to a $Q$ -embedding $σ : K \to C$ (Restriction partitions embeddings in a finite tower into extension fibres), so $w = σ (α)^{m}$ .

[F8]

For fixed $n \geq 1$ and $R = 1$ there are only finitely many monic integer polynomials of degree at most $n$ whose complex roots, counted with multiplicity, all have modulus at most $1$ (Bounded roots give finitely many monic integer polynomials); this is the supplier consumed here, and the exact obligation used is this instance $R = 1$ .

[F9]

A nonzero polynomial of degree at most $n$ over the integral domain $C$ has at most $n$ distinct roots (A nonzero polynomial of degree $n$ over an integral domain has at most $n$ distinct roots).

[F10]

An element $ζ$ of $K$ is a root of unity exactly when $ζ^{N} = 1$ for some $N \geq 1$ (The group $μ_{n} (K)$ of $n$ -th roots of unity in a field, and primitive $n$ -th roots of unity).

Proof

technique · the norm bounds every conjugate of $\alpha$ above by $1$ and below by $1$ simultaneously; the powers $\alpha^{m}$ then have norm-controlled minimal polynomials of bounded degree, and only finitely many such polynomials exist

1.1F2

Since $0 \neq α \in O_{K}$ , the norm $N_{K / Q} (α)$ is a nonzero integer, so $∣ N_{K / Q} (α) ∣ \geq 1$ .

1.2F1

The set $Σ$ of $Q$ -embeddings $K \to C$ has $n$ elements and $N_{K / Q} (α) = \prod_{σ \in Σ} σ (α)$ .

1.3F3given

The complex roots of $m_{α}$ are exactly the numbers $σ (α)$ with $σ \in Σ$ ; each is a conjugate of $α$ , so by the hypothesis $∣ σ (α) ∣ \leq 1$ for every $σ \in Σ$ .

2.1F4step 1.2step 1.3

By multiplicativity of the modulus, $∣ N_{K / Q} (α) ∣ = \prod_{σ \in Σ} ∣ σ (α) ∣$ , and every factor is at most $1$ by step 1.3, so $∣ N_{K / Q} (α) ∣ \leq 1$ .

3.1step 1.1step 1.3step 2.1

Steps 1.1 and 2.1 give $1 \leq ∣ N_{K / Q} (α) ∣ \leq 1$ , so $∣ N_{K / Q} (α) ∣ = 1$ ; a product of finitely many real numbers in $[0, 1]$ equals $1$ only if every factor equals $1$ , so $∣ σ (α) ∣ = 1$ for every $σ \in Σ$ , and every complex root of $m_{α}$ has modulus exactly $1$ .

4.1F4F5F6F7step 3.1

Let $m \geq 1$ . Then $m_{α^{m}} \in Z [X]$ is monic of degree at most $n$ . By [F7], every complex root $w$ of $m_{α^{m}}$ equals $σ (α)^{m}$ for some $σ \in Σ$ ; hence $∣ w ∣ = ∣ σ (α) ∣^{m} = 1$ by step 3.1.

5.1F8F9step 4.1

By [F8] there are only finitely many monic integer polynomials of degree at most $n$ whose complex roots all have modulus at most $1$ , and each of them has at most $n$ distinct complex roots by [F9]; hence the union of the complex root sets of these finitely many polynomials is finite.

6.1step 4.1step 5.1

For every $m \geq 1$ , $α^{m}$ is a complex root of $m_{α^{m}}$ , so the set ${α^{m} : m \geq 1}$ is contained in the finite union of step 5.1 and is finite.

7.1F10step 6.1∎

Two distinct powers therefore coincide: $α^{k} = α^{m}$ for integers $k > m \geq 1$ , and since $α \neq 0$ this gives $α^{k - m} = 1$ , so $α$ is a root of unity.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-02Open item page →

A number-field unit is exactly an algebraic integer of norm plus or minus one

Statement

Let $K$ be a number field (Number field) with ring of integers $O_{K}$ (Ring of integers) and with field norm $N_{K / Q} (u) = \det (m_{u})$ of multiplication by $u$ (The norm $N_{K / F}$ and trace ${Tr}_{K / F}$ of a finite field extension). For $u \in O_{K}$ , the element $u$ is a unit of the ring $O_{K}$ (The units of a ring are the invertible elements of its multiplicative monoid, and $R^{\times}$ is a group under multiplication; $0 \in R^{\times}$ only in the zero ring) if and only if $N_{K / Q} (u) = \pm 1$ .

Facts & Assumptions

Given: A number field $K$ of degree $n$ , its ring of integers $O_{K}$ , and an element $u \in O_{K}$ .

[F1]

$O_{K}$ is a free $Z$ -module of rank $n = [K : Q]$ (The ring of integers has rank the degree).

[F2]

If $A \in M_{n} (R)$ is invertible over a commutative ring $R$ , then $\det A$ is a unit of $R$ , with $\det (A^{- 1})$ its inverse (An invertible square matrix over a commutative ring has unit determinant).

[F3]

If $\det (A)$ is a unit of $R$ , then $A^{- 1} = \det (A)^{- 1} adj (A)$ , and the adjugate of a matrix with entries in $R$ has entries in $R$ , its entries being cofactors (If $\det (A)$ is a unit, then $A^{- 1} = \det (A)^{- 1} adj (A)$ , Deleted-row-and-column minors, cofactors, the cofactor matrix and the adjugate over a commutative ring).

[F4]

The units of $Z$ are exactly $\pm 1$ ( $(Z, \cdot, 1)$ is a commutative monoid whose group of units is ${1, - 1}$ ; equivalently $u ∣ 1$ holds exactly for $u = 1$ and $u = - 1$ ).

Proof

Proof technique: read the norm as the determinant of multiplication by $u$ in an integral basis, and use the adjugate formula in one direction and the unit-determinant theorem in the other.

1.1F1given

Fix a $Z$ -basis of $O_{K}$ and let $M$ be the matrix of the $Q$ -linear map $m_{u} : K \to K$ , $m_{u} (x) = u x$ , in that basis (Coordinate columns $[v]_{B}$ and matrices $[T]_{B}^{C}$ of linear maps relative to ordered bases). Since $u \in O_{K}$ and $O_{K}$ is closed under multiplication, $u O_{K} \subseteq O_{K}$ ; hence every column of $M$ is the coordinate column of an element of $O_{K}$ , so $M$ has entries in $Z$ , and $N_{K / Q} (u) = \det M$ .

2.1F2F4step 1.1

Suppose first that $u$ is a unit of $O_{K}$ , so $u^{- 1} \in O_{K}$ . The inverse of $m_{u}$ is $m_{u^{- 1}}$ , and its matrix in the same basis is $M^{- 1}$ ; by the argument of step 1.1 with $u$ replaced by $u^{- 1}$ this matrix has integer entries. Thus $M$ is invertible over $Z$ , so by [F2] $\det M$ is a unit of $Z$ , and [F4] gives $\det M = \pm 1$ , that is, $N_{K / Q} (u) = \pm 1$ .

3.1F3step 1.1∎

Suppose conversely that $N_{K / Q} (u) = \det M = \pm 1$ . Then $M$ is invertible and $M^{- 1} = \det (M)^{- 1} adj (M)$ by [F3]; since $\det M = \pm 1$ is a unit of $Z$ and the adjugate of an integer matrix has integer entries, $M^{- 1}$ has integer entries. For every $x \in O_{K}$ the coordinate column of $u^{- 1} x$ is $M^{- 1}$ applied to the coordinate column of $x$ , hence is integral, so $u^{- 1} O_{K} \subseteq O_{K}$ ; taking $x = 1$ and using $1 \in O_{K}$ gives $u^{- 1} \in O_{K}$ . Therefore $u \cdot u^{- 1} = 1$ exhibits $u$ as a unit of $O_{K}$ together with its inverse $u^{- 1}$ .

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-02Open item page →

Product formula for a number field

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $K$ be a number field (Number field) with ring of integers $O_{K}$ (Ring of integers). Normalize the absolute values of $x \in K^{\times}$ as follows:

at a nonzero prime ideal $p$ of $O_{K}$ , set $∣ x ∣_{p} = N p^{- v_{p} (x)}$ , where $v_{p}$ is the prime-ideal valuation (Prime-ideal valuations on fractional ideals) and $N p$ is the absolute norm (The absolute norm of an integral ideal);
at a real embedding $σ$ , set $∣ x ∣_{σ} = ∣ σ (x) ∣$ ;
at a complex embedding $τ$ , one chosen from each complex conjugate pair, set $∣ x ∣_{τ} = ∣ τ (x) ∣^{2}$ .

Then

$\prod_{v} ∣ x ∣_{v} = 1 for every x \in K^{\times},$

the product being taken over the nonzero prime ideals and the chosen real and complex embeddings; only finitely many factors differ from $1$ .

Facts & Assumptions

Given: The Axiom of Choice, a number field $K$ with embeddings $σ_{1}, \dots, σ_{r_{1}}$ and $τ_{1}, \dots, τ_{r_{2}}$ as in the statement (Archimedean embeddings and signature), and an element $x \in K^{\times}$ .

[A1]

The Axiom of Choice is assumed for the whole argument; its single use is the Dedekind unique-factorisation route for fractional ideals (Unique factorization of nonzero fractional ideals into prime powers), whose statement assumes Choice, applied to ideals of $O_{K}$ (Rings of integers are Dedekind domains).

[F1]

Every nonzero fractional ideal of $O_{K}$ has a unique finite factorisation into prime ideals, and an integral ideal has only nonnegative exponents (Unique factorization of nonzero fractional ideals into prime powers); the valuation $v_{p} (I)$ is the exponent attached to $p$ (Prime-ideal valuations on fractional ideals).

[F2]

For nonzero integral ideals, $N (a b) = N (a) N (b)$ , and for $0 \neq α \in O_{K}$ one has $N ((α)) = ∣ N_{K / Q} (α) ∣$ (Ideal norm is multiplicative, The norm of a principal integral ideal).

[F3]

$N_{K / Q} (x) = \prod_{ψ} ψ (x)$ , the product being over the $[K : Q]$ embeddings $K \to C$ , and $N_{K / Q} (x y) = N_{K / Q} (x) N_{K / Q} (y)$ with $N_{K / Q} (1) = 1$ (Norm and trace from embeddings, with the inseparable exponent in the norm formula, Norm is multiplicative, trace is $F$ -linear, and both are transitive in towers).

[F4]

The modulus satisfies $∣ z w ∣ = ∣ z ∣ ∣ w ∣$ and $z \overline{z} = ∣ z ∣^{2}$ for complex numbers (Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive); in particular $∣ \overline{z} ∣ = ∣ z ∣$ , since $∣ \overline{z} ∣^{2} = \overline{z} \overline{\overline{z}} = \overline{z} z = ∣ z ∣^{2}$ and both sides are nonnegative.

Proof

technique · split the product into its finite and archimedean parts; the finite part is the reciprocal of $|N_{K/\mathbb Q}(x)|$ by unique factorisation and the ideal-norm formulas, and the archimedean part is $|N_{K/\mathbb Q}(x)|$ by the embedding formula for the norm

1.1givenalgebra

Write $x = a / b$ with $a, b \in O_{K} ∖ {0}$ . Indeed, $K / Q$ is finite so $x$ is algebraic over $Q$ and has a monic minimal polynomial $m (X) = X^{d} + c_{d - 1} X^{d - 1} + \dots + c_{0} \in Q [X]$ (The evaluation kernel and the unique monic irreducible minimal polynomial of an algebraic element); choose $M \geq 1$ with all $M^{j} c_{d - j} \in Z$ , and set $a = M x$ . Then $a^{d} + M c_{d - 1} a^{d - 1} + \dots + M^{d} c_{0} = M^{d} m (x) = 0$ , a monic integer polynomial relation, so $a \in O_{K}$ by the minimal-polynomial criterion (Minimal-polynomial criterion for algebraic integers); with $b = M \in Z ∖ {0} \subseteq O_{K}$ this gives $x = a / b$ .

1.2F3F4given

For the archimedean factors, [F3] gives $N_{K / Q} (x) = \prod_{ψ} ψ (x)$ over all $[K : Q]$ embeddings, and the embeddings consist of the $r_{1}$ real embeddings together with the $r_{2}$ conjugate pairs ${τ_{j}, \overline{τ_{j}}}$ ; taking absolute values and using $∣ z w ∣ = ∣ z ∣ ∣ w ∣$ and $∣ \overline{z} ∣ = ∣ z ∣$ from [F4], $∣ N_{K / Q} (x) ∣ = \prod_{ψ} ∣ ψ (x) ∣ = (\prod_{i = 1}^{r_{1}} ∣ σ_{i} x ∣) (\prod_{j = 1}^{r_{2}} ∣ τ_{j} x ∣ ∣ \overline{τ_{j} x} ∣) = (\prod_{i = 1}^{r_{1}} ∣ σ_{i} x ∣) (\prod_{j = 1}^{r_{2}} ∣ τ_{j} x ∣^{2})$ , which is exactly the product of the archimedean normalized absolute values.

2.1A1F1F2F3step 1.1

In the language of fractional ideals (Fractional ideals, The field of fractions $Frac (D) = (D ∖ {0})^{- 1} D$ of an integral domain) one has $(x) = (a) (b)^{- 1}$ , hence $v_{p} (x) = v_{p} ((a)) - v_{p} ((b))$ for every prime $p$ ; by [F1] write $(a) = \prod_{p} p^{e_{p}}$ and $(b) = \prod_{p} p^{f_{p}}$ with finite supports, so $v_{p} (x) = e_{p} - f_{p}$ vanishes outside the finite union of those supports and $\prod_{p} N p^{- v_{p} (x)} = (\prod_{p} N p^{f_{p}}) (\prod_{p} N p^{e_{p}})^{- 1} = N ((b)) / N ((a)) = ∣ N_{K / Q} (b) ∣ / ∣ N_{K / Q} (a) ∣ = 1 / ∣ N_{K / Q} (x) ∣$ , the third equality by [F2] applied to the two finite factorisations and the last by [F3], since $N_{K / Q} (a) = N_{K / Q} (x) N_{K / Q} (b)$ .

3.1A1step 2.1step 1.2∎

Multiplying the finite product of step 2.1 and the archimedean product of step 1.2 gives $\prod_{v} ∣ x ∣_{v} = (\prod_{i = 1}^{r_{1}} ∣ σ_{i} x ∣) (\prod_{j = 1}^{r_{2}} ∣ τ_{j} x ∣^{2}) (\prod_{p} N p^{- v_{p} (x)}) = ∣ N_{K / Q} (x) ∣ \cdot ∣ N_{K / Q} (x) ∣^{- 1} = 1$ , and only the finitely many primes in the supports of $(a)$ and $(b)$ contribute a finite factor different from $1$ , so the product is over a finite set of places; the only Choice in the argument is [A1], the norms, moduli and logarithms being computed without further selection.

DefinitionDefinition: Literature-sourcedProof: Not applicableprecheck passaudited 2026-10-02Open item page →

Logarithmic embedding of a number field

Definition

Let $K$ be a number field of signature $(r_{1}, r_{2})$ , with real embeddings $σ_{1}, \dots, σ_{r_{1}} : K \to R$ and one embedding $τ_{1}, \dots, τ_{r_{2}} : K \to C$ chosen from each complex conjugate pair (Archimedean embeddings and signature). The logarithmic embedding of $K$ is the map

$λ : K^{\times} ⟶ R^{r_{1} + r_{2}}, λ (x) = (\log ∣ σ_{1} x ∣, \dots, \log ∣ σ_{r_{1}} x ∣, 2 \log ∣ τ_{1} x ∣, \dots, 2 \log ∣ τ_{r_{2}} x ∣),$

where $K^{\times} = K ∖ {0}$ , $\log$ is the natural logarithm (The natural logarithm as the inverse of the exponential function), and $∣ \cdot ∣$ is the complex modulus (Real and imaginary parts, complex conjugation, and modulus).

The factor $2$ on the complex coordinates is part of the convention and is not optional. A real coordinate carries no factor, while a complex coordinate enters with weight $2$ , matching the squared modulus $∣ τ x ∣^{2}$ that is the normalized absolute value at a complex place. Taking the doubled coordinate is what makes the product-formula hyperplane the literal coordinate-sum-zero hyperplane and what fixes the determinant normalization of the regulator; for a fixed deleted row, doubling the retained complex rows multiplies the absolute determinant by $2$ for each such row. Euclidean covolumes in the corresponding hyperplanes need not change by a power of $2$ .

The map is well defined. If $x \neq 0$ then $σ (x) \neq 0$ and $τ (x) \neq 0$ for every embedding, since a field homomorphism has trivial kernel, so every modulus is a strictly positive real number and every logarithm is defined. Replacing a chosen $τ_{j}$ by its complex conjugate does not change $λ$ : conjugation fixes the real numbers and replaces $z = a + b i$ by $\overline{z} = a - b i$ , so $∣ \overline{τ_{j} x} ∣ = ∣ τ_{j} x ∣$ for every $x \in K^{\times}$ . The ordering of the coordinates is auxiliary: reordering them post-composes $λ$ with a linear isometry of $R^{r_{1} + r_{2}}$ , and every statement about $λ$ below is invariant under that reordering.

The coordinates are written in the display with the real embeddings first and the chosen complex embeddings after them; that ordering is the one used for the rest of this page.

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Unit logarithms lie in the trace-zero hyperplane

Statement

Assume the Axiom of Choice. Let $H = {(x_{i}) \in R^{r_{1} + r_{2}} : \sum_{i} x_{i} = 0}$ . Then $λ (O_{K}^{\times}) \subseteq H$ ; that is, the coordinate sum of $λ (u)$ is $\log ∣ N_{K / Q} (u) ∣$ , which vanishes for every unit.

Facts & Assumptions

Given: The Axiom of Choice, a number field $K$ of signature $(r_{1}, r_{2})$ with its logarithmic embedding $λ$ (Logarithmic embedding of a number field), and a unit $u \in O_{K}^{\times}$ .

[F1]

The logarithmic embedding is $λ (x) = (\log ∣ σ_{1} x ∣, \dots, \log ∣ σ_{r_{1}} x ∣, 2 \log ∣ τ_{1} x ∣, \dots, 2 \log ∣ τ_{r_{2}} x ∣)$ on $K^{\times}$ , where $σ_{1}, \dots, σ_{r_{1}}$ are the real embeddings and $τ_{1}, \dots, τ_{r_{2}}$ one embedding from each complex conjugate pair (Logarithmic embedding of a number field, Archimedean embeddings and signature).

[F2]

For $x \in K^{\times}$ the product formula reads $\prod_{v} ∣ x ∣_{v} = 1$ , where the finite absolute values are $∣ x ∣_{p} = N p^{- v_{p} (x)}$ with $v_{p} (x) = v_{p} ((x))$ the valuation of the principal fractional ideal, and the archimedean ones are $∣ x ∣_{σ} = ∣ σ (x) ∣$ and $∣ x ∣_{τ} = ∣ τ (x) ∣^{2}$ (Product formula for a number field, Prime-ideal valuations on fractional ideals, Fractional ideals).

[F3]

For $u \in O_{K}$ the element $u$ is a unit if and only if $N_{K / Q} (u) = \pm 1$ (A number-field unit is exactly an algebraic integer of norm plus or minus one).

[F4]

The modulus of a complex number is nonnegative and vanishes only at $0$ , and satisfies $∣ z w ∣ = ∣ z ∣ ∣ w ∣$ (Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive).

[F5]

For $x, y > 0$ the natural logarithm satisfies $\log (x y) = \log x + \log y$ and $\log 1 = 0$ (Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm, The natural logarithm as the inverse of the exponential function).

[F6]

For a finite separable extension such as $K / Q$ the norm is the product of the images under the $[K : Q]$ embeddings $K \to C$ (Norm and trace from embeddings, with the inseparable exponent in the norm formula).

[A1]

The Axiom of Choice is assumed; its only use in this argument is the AC-qualified product formula [F2] (The Axiom of Choice).

Proof

Proof technique: a unit has valuation zero at every finite prime, so the product formula collapses to its archimedean part; taking logarithms turns that product into the coordinate sum of $λ (u)$ .

1.1F2given

The principal fractional ideal of the unit $u$ is $(u) = O_{K}$ , so $v_{p} (u) = v_{p} ((u)) = 0$ for every nonzero prime $p$ , and the normalized finite absolute value $∣ u ∣_{p} = N p^{- v_{p} (u)}$ equals $1$ at every finite place.

1.2F4given

Every modulus $∣ σ_{i} u ∣$ and $∣ τ_{j} u ∣$ is strictly positive: the embeddings are injective field homomorphisms, $u \neq 0$ , and a nonzero complex number has positive modulus.

1.3F3given

Since $u$ is a unit, $N_{K / Q} (u) = \pm 1$ and therefore $∣ N_{K / Q} (u) ∣ = 1$ .

2.1F2step 1.1

For $x = u$ the product formula gives $\prod_{v} ∣ u ∣_{v} = 1$ ; by step 1.1 every finite factor equals $1$ , so the archimedean factors satisfy $(\prod_{i = 1}^{r_{1}} ∣ σ_{i} u ∣) (\prod_{j = 1}^{r_{2}} ∣ τ_{j} u ∣^{2}) = 1$ .

3.1F1F5step 2.1step 1.2

Applying the logarithm to the identity of step 2.1 yields $\sum_{i = 1}^{r_{1}} \log ∣ σ_{i} u ∣ + \sum_{j = 1}^{r_{2}} \log ∣ τ_{j} u ∣^{2} = \log 1 = 0$ ; since $\log (a^{2}) = 2 \log a$ for $a > 0$ by step 1.2, the left-hand side equals $\sum_{k} λ (u)_{k}$ , the coordinate sum of $λ (u)$ ; hence this coordinate sum is $0$ and $λ (u) \in H$ .

4.1F5F6step 1.3step 3.1

The same coordinate sum equals $\log ∣ N_{K / Q} (u) ∣$ : the embedding formula [F6] gives $∣ N_{K / Q} (u) ∣ = (\prod_{i} ∣ σ_{i} u ∣) (\prod_{j} ∣ τ_{j} u ∣^{2})$ , whose logarithm is the sum of step 3.1, and by step 1.3 this is $\log 1 = 0$ .

5.1A1step 3.1step 4.1∎

As $u \in O_{K}^{\times}$ was arbitrary, $λ (O_{K}^{\times}) \subseteq H$ ; the only Choice used is [A1] through the AC-qualified product formula, the remaining computations being evaluations of norms, moduli and logarithms.

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Kernel of the unit logarithm is the roots of unity

Statement

$\ker (λ ∣_{O_{K}^{\times}}) = μ (K)$ , the group of roots of unity contained in $K$ ; in particular the kernel is finite and is the torsion subgroup of $O_{K}^{\times}$ .

Facts & Assumptions

Given: A number field $K$ with embeddings $σ_{1}, \dots, σ_{r_{1}}$ and $τ_{1}, \dots, τ_{r_{2}}$ as in the definition of the logarithmic embedding $λ$ (Logarithmic embedding of a number field, Archimedean embeddings and signature).

[F1]

For $x \in K^{\times}$ , $λ (x) = (\log ∣ σ_{1} x ∣, \dots, \log ∣ σ_{r_{1}} x ∣, 2 \log ∣ τ_{1} x ∣, \dots, 2 \log ∣ τ_{r_{2}} x ∣)$ , the $σ_{i}$ and $τ_{j}$ being the real and chosen complex embeddings of $K$ (Logarithmic embedding of a number field).

[F2]

$μ (K)$ is the set of $ζ \in K$ with $ζ^{N} = 1$ for some $N \geq 1$ ; for fixed $N$ the set $μ_{N} (K)$ is a finite cyclic subgroup of $K^{\times}$ , and an element of a field is a root of unity exactly when it has finite order in the multiplicative group (The group $μ_{n} (K)$ of $n$ -th roots of unity in a field, and primitive $n$ -th roots of unity, $μ_{n} (K)$ is cyclic of order dividing $n$ , and has a primitive $n$ -th root of unity exactly when its order is $n$ ).

[F3]

If $ζ \in K$ satisfies $ζ^{N} = 1$ , then $ζ$ is a root of the monic polynomial $X^{N} - 1 \in Z [X]$ , hence is integral over $Z$ and lies in $O_{K}$ ; its inverse $ζ^{N - 1}$ lies in $O_{K}$ as well, so $ζ \in O_{K}^{\times}$ (Integral elements over a commutative ring and algebraic integers, Ring of integers).

[F4]

Modulus is multiplicative, and $∣ \overline{z} ∣ = ∣ z ∣$ : writing $z = a + b i$ , the coordinate definition gives $∣ \overline{z} ∣ = \sqrt{a^{2} + (- b)^{2}} = \sqrt{a^{2} + b^{2}} = ∣ z ∣$ . Thus, for an embedding $ψ : K \to C$ and $ζ \in K$ with $ζ^{N} = 1$ , $ψ (ζ)^{N} = 1$ implies $∣ ψ (ζ) ∣^{N} = ∣ ψ (ζ)^{N} ∣ = 1$ and $∣ ψ (ζ) ∣ = 1$ (Real and imaginary parts, complex conjugation, and modulus, Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive).

[F5]

The natural logarithm satisfies $\log 1 = 0$ and is strictly increasing on $(0, \infty)$ (Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm); hence $\log a = 0$ for $a > 0$ forces $a = 1$ .

[F6]

If $0 \neq α \in O_{K}$ has all its complex conjugates of modulus at most $1$ , then $α$ is a root of unity (Kronecker root-of-unity criterion).

[F7]

The group $μ (K)$ of roots of unity in the number field $K$ is finite (Finitely many roots of unity in a number field).

[F8]

A unit $u$ of $O_{K}$ satisfies $N_{K / Q} (u) = \pm 1$ , so in particular $u \neq 0$ (A number-field unit is exactly an algebraic integer of norm plus or minus one).

Proof

Proof technique: compare the kernel of $λ$ with $μ (K)$ in both directions, the forward direction by Kronecker's criterion and the reverse by the multiplicativity of the embeddings.

1.1F2F3

Every root of unity $ζ \in μ (K)$ lies in $O_{K}^{\times}$ , and $μ (K)$ is a subgroup of $O_{K}^{\times}$ : each $ζ$ lies in $O_{K}$ with inverse $ζ^{N - 1}$ for $ζ^{N} = 1$ ; the product of roots of unity of orders $m$ and $k$ is a root of unity of order dividing $m k$ , and the inverse of a root of unity is a root of unity.

1.2F1F4F5

Let $ζ \in μ (K)$ with $ζ^{N} = 1$ . For every real embedding $σ$ and every chosen complex embedding $τ$ one has $σ (ζ)^{N} = 1 = τ (ζ)^{N}$ , so $∣ σ (ζ) ∣ = ∣ τ (ζ) ∣ = 1$ by [F4]; therefore $\log ∣ σ (ζ) ∣ = \log 1 = 0$ and $2 \log ∣ τ (ζ) ∣ = 0$ by [F5], and $λ (ζ) = 0$ . Hence $μ (K) \subseteq \ker (λ ∣_{O_{K}^{\times}})$ .

1.3F1F4F5F6F8

Conversely let $u \in O_{K}^{\times}$ with $λ (u) = 0$ . Every coordinate of $λ (u)$ vanishes: $\log ∣ σ_{i} u ∣ = 0$ for every real embedding and $2 \log ∣ τ_{j} u ∣ = 0$ for every chosen complex embedding; by [F5] this gives $∣ σ_{i} u ∣ = ∣ τ_{j} u ∣ = 1$ . Every complex conjugate of $u$ is a real embedding value, a chosen value $τ_{j} (u)$ , or its conjugate $\overline{τ_{j} (u)}$ ; by [F4], the latter also has modulus $1$ . Thus all conjugates have modulus at most $1$ . The unit $u$ is a nonzero algebraic integer by [F8], so Kronecker's criterion [F6] makes it a root of unity, that is, $u \in μ (K)$ . Hence $\ker (λ ∣_{O_{K}^{\times}}) \subseteq μ (K)$ .

2.1F2F7step 1.2step 1.3∎

Steps 1.2 and 1.3 give $\ker (λ ∣_{O_{K}^{\times}}) = μ (K)$ ; this kernel is finite by [F7]. Moreover an element $u \in O_{K}^{\times}$ has finite order in the group $O_{K}^{\times}$ exactly when $u^{N} = 1$ for some $N \geq 1$ , that is, exactly when $u$ is a root of unity in $K$ , by [F2]; hence $μ (K)$ is the torsion subgroup of $O_{K}^{\times}$ , and the kernel is both finite and the torsion subgroup.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-02Open item page →

Discrete subgroups of a real vector space are lattices

Statement

Let $V$ be a finite-dimensional real vector space (Vector space over a field, Finite-dimensional vector space, and its dimension $\dim_{F} V$ ; infinite-dimensional means having no finite basis) with the topology induced by a norm (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms, The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement), and let $Γ \leq V$ be a subgroup. The following are equivalent:

(a) $Γ$ is discrete in the subspace topology (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace);

(b) every bounded subset of $V$ (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space) meets $Γ$ in a finite set;

(c) $Γ = Z v_{1} \oplus \dots \oplus Z v_{r}$ for some $R$ -linearly independent $v_{1}, \dots, v_{r} \in V$ (Linear independence: a finite list $v : n \to V$ is independent when $\sum_{i < n} λ_{i} v_{i} = 0_{V}$ forces every $λ_{i} = 0_{F}$ , and a subset $S \subseteq V$ is independent when every injective finite list into $S$ is independent) with $r \leq \dim_{R} V$ ; that is, $Γ$ is a lattice in $V$ .

Facts & Assumptions

Given: A finite-dimensional real vector space $V$ of dimension $n$ with a norm, the induced metric and topology, and a subgroup $Γ \leq V$ .

[F1]

$\emptyset$ is a linearly independent list and every linearly independent subset of $V$ has at most $n$ elements, because $V$ has a basis of $n$ elements (If $V$ has a spanning set with $n$ elements, then every linearly independent subset of $V$ is finite with at most $n$ elements; in particular $V$ has no linearly independent subset equinumerous with $N$ , Linear independence: a finite list $v : n \to V$ is independent when $\sum_{i < n} λ_{i} v_{i} = 0_{V}$ forces every $λ_{i} = 0_{F}$ , and a subset $S \subseteq V$ is independent when every injective finite list into $S$ is independent, Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis).

[F2]

The induced metric is $d (x, y) = ∥ x - y ∥$ , the norm is homogeneous and satisfies the triangle inequality, a bounded set is contained in some ball, and balls are translation invariant; open sets contain a ball around each of their points (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms, The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, Open ball, closed ball and sphere in a metric space, Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space).

[F3]

Identifying $V$ with $R^{n}$ by one basis, the given norm corresponds to a norm on $R^{n}$ . Equivalence with the coordinate maximum norm gives a constant $c > 0$ such that $∥ x ∥ \geq c ∥ x ∥_{\infty}$ in these coordinates (For $n \geq 1$ all norms on $R^{n}$ are equivalent).

[F4]

Every subgroup of $(Z, +)$ is $d Z$ for a unique nonnegative integer $d$ (Every subgroup of $(Z, +)$ is $⟨ n ⟩ = n Z$ for exactly one natural number $n$ ). In particular, the image of a subgroup of $Z^{m}$ under projection to one coordinate is either ${0}$ or $d Z$ for some $d > 0$ .

[F5]

A finite group of order $N$ has $u^{N} = 1$ for every element $u$ , by Lagrange's theorem (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

[F6]

For every $ε > 0$ there is an integer $q \geq 1$ with $1 / q < ε$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ).

Proof

Proof technique: a direct chain (a) implies (b) implies (c) implies (a): isolation and a finite coordinate grid give bounded finiteness. A bounded fundamental parallelepiped gives a finite-index inclusion into the integer span of a maximal independent tuple. Scaling embeds the group in $Z^{r}$ ; a finite-rank subgroup induction then supplies a lattice basis using only the cyclic-subgroup-of- $Z$ result.

First handle $n = 0$ . Then $V = {0}$ and $Γ = {0}$ , so (a), (b), and (c) hold with $r = 0$ . Assume $n \geq 1$ below.

1.1F2

Suppose first that $Γ$ is discrete. Then ${0}$ is open in the subspace topology on $Γ$ , so ${0} = Γ \cap U$ for some open $U \subseteq V$ ; choosing a ball around $0$ inside $U$ gives an $ε > 0$ with $Γ \cap B (0, ε) = {0}$ .

1.2F2

Suppose next that (b) holds. The set $Γ \cap B (0, 1)$ is then finite; if it is ${0}$ take $ε = 1$ , and otherwise let $δ : = \min {∥ γ ∥ : 0 \neq γ \in Γ \cap B (0, 1)} > 0$ , a minimum of a nonempty finite set of positive reals, so that again $Γ \cap B (0, δ) = {0}$ . Since balls are translation invariant in the metric of a norm, $B (γ, δ) = γ + B (0, δ)$ for every $γ \in Γ$ , so $Γ \cap B (γ, δ) = {γ}$ : every point of $Γ$ is isolated in $Γ$ , that is, $Γ$ is discrete. Hence (b) implies (a).

1.3F1

Assume (b) from here on. By [F1] the lengths of the $R$ -linearly independent finite tuples of elements of $Γ$ form a nonempty subset of ${0, 1, \dots, n}$ , so a maximum $r$ exists; select an $R$ -linearly independent tuple $v_{1}, \dots, v_{r} \in Γ$ , and put $W : = {span}_{R} (v_{1}, \dots, v_{r})$ , so $\dim_{R} W = r$ .

1.4F2

Put $Γ_{0} : = Z v_{1} \oplus \dots \oplus Z v_{r} \subseteq Γ$ and $P : = {\sum_{i = 1}^{r} t_{i} v_{i} : 0 \leq t_{i} < 1}$ . Since $∥ \sum_{i} t_{i} v_{i} ∥ \leq \sum_{i} ∣ t_{i} ∣ ∥ v_{i} ∥ \leq \sum_{i} ∥ v_{i} ∥$ , the set $P$ is bounded, so $F : = Γ \cap P$ is finite by (b).

2.1F2F3F6step 1.1

This proves (a) $\Rightarrow$ (b) without selecting a sequence from a bounded set. Fix a basis $e_{1}, \dots, e_{n}$ of $V$ and put $S : = \sum_{i = 1}^{n} ∥ e_{i} ∥ > 0$ . Let $B \subseteq V$ be bounded; if $B = \emptyset$ the conclusion is immediate. Otherwise choose a ball $B (x_{0}, R)$ containing it. Let $a_{i}$ be the coordinates of $x_{0}$ . By [F3] there is $c > 0$ with $∥ v ∥ \geq c ∥ (v_{i}) ∥_{\infty}$ in these coordinates, so every $x \in B$ satisfies $\max_{i} ∣ x_{i} - a_{i} ∣ < R / c$ . Set $M : = \max {1, R / c}$ ; then the coordinate vectors of $B$ lie in the box $\prod_{i} [a_{i} - M, a_{i} + M]$ . Set $δ : = ε / (2 S) > 0$ and choose an integer $q \geq 1$ with $1 / q < δ / (2 M)$ by [F6]. Divide each coordinate interval into $q$ equal subintervals and take their finitely many product cells. In one cell, any two coordinate vectors differ by less than $δ$ in each coordinate, so the corresponding points $x, y$ satisfy $∥ x - y ∥ \leq \sum_{i} ∣ x_{i} - y_{i} ∣ ∥ e_{i} ∥ < δ S = ε / 2 < ε$ . By step 1.1, each cell therefore contains at most one point of $Γ$ . The finite collection of cells covers $B$ , so $B \cap Γ$ is finite.

2.2step 1.3

If $γ \in Γ ∖ W$ , then any relation $λ_{1} v_{1} + \dots + λ_{r} v_{r} + μ γ = 0$ must have $μ = 0$ , since otherwise it would express $γ$ as an element of $W$ . The independence of $v_{1}, \dots, v_{r}$ then forces every $λ_{i} = 0$ , so adjoining $γ$ would give $r + 1$ independent elements of $Γ$ , contradicting maximality in step 1.3. Thus $Γ \subseteq W$ , and since the $v_{i}$ lie in $Γ$ , ${span}_{R} Γ = W$ .

3.1step 1.4step 2.2

Every $γ \in Γ$ differs from an element of $Γ_{0}$ by an element of $F$ : by step 2.2 write $γ = \sum_{i} s_{i} v_{i}$ with $s_{i} \in R$ and write $s_{i} = m_{i} + t_{i}$ with $m_{i} \in Z$ and $0 \leq t_{i} < 1$ ; then $γ - \sum_{i} m_{i} v_{i} \in Γ \cap P = F$ .

4.1F5step 1.4step 3.1

The map $F \to Γ / Γ_{0}$ , $f \mapsto f + Γ_{0}$ , is surjective by step 3.1. Thus $Γ / Γ_{0}$ is finite; let its order be $N \geq 1$ . By [F5], $N γ \in Γ_{0}$ for every $γ \in Γ$ . The map $T : Γ \to Γ_{0}$ , $T (γ) = N γ$ , is an injective homomorphism: if $N γ = 0$ , then $γ = 0$ because $V$ is a real vector space and $N > 0$ . Consequently its image is a subgroup of $Γ_{0} ≅ Z^{r}$ .

5.1F4choosestep 4.1

By step 4.1, $T (Γ)$ is a subgroup of $Γ_{0} ≅ Z^{r}$ . For this use, every subgroup $H \leq Z^{m}$ has a finite $Z$ -basis of length at most $m$ , by induction on $m$ . For $m = 0$ the subgroup is zero and the empty list is a basis. For $m \geq 1$ , project $H$ onto its first coordinate. By [F4] the image is $d Z$ for some $d \geq 0$ . If $d = 0$ , identify $H$ with a subgroup of the last $m - 1$ coordinates and apply induction. If $d > 0$ , choose $h \in H$ with first coordinate $d$ ; the kernel $H_{0}$ of that projection is a subgroup of $Z^{m - 1}$ , so induction gives it a basis of length at most $m - 1$ . Every $x \in H$ has first coordinate $k d$ for some $k \in Z$ ; then $x - k h \in H_{0}$ , so $h$ together with a basis of $H_{0}$ generates $H$ . They are independent because projecting any integer relation to the first coordinate forces the coefficient of $h$ to be zero, after which independence in $H_{0}$ forces all remaining coefficients to vanish. This proves the claim, including that the basis has at most $m$ elements.

6.1F1step 2.2step 5.1

Apply step 5.1 to $T (Γ) \leq Γ_{0} ≅ Z^{r}$ and pull its $Z$ -basis back through the isomorphism $T : Γ \to T (Γ)$ . This gives a $Z$ -basis $b_{1}, \dots, b_{s}$ of $Γ$ with $s \leq r$ . By step 2.2, $v_{1}, \dots, v_{r}$ are linearly independent and span $W$ , while ${span}_{R} (b_{1}, \dots, b_{s}) = W$ because they generate $Γ$ . The finite-dimensional independent-set bound [F1] gives $r \leq s$ , so $s = r$ . If $r > 0$ and these $r$ spanning vectors were linearly dependent, one could remove a vector and still span $W$ , contradicting [F1] applied to $v_{1}, \dots, v_{r}$ ; when $r = 0$ , the empty list is independent. Hence they are $R$ -linearly independent and $Γ = Z b_{1} \oplus \dots \oplus Z b_{r}$ , proving (c).

7.1F1F3∎

Finally assume (c): $Γ = Z v_{1} \oplus \dots \oplus Z v_{r}$ with $v_{1}, \dots, v_{r}$ $R$ -linearly independent. Extend this tuple to a basis $v_{1}, \dots, v_{r}, w_{1}, \dots, w_{n - r}$ of $V$ and let $φ : R^{n} \to V$ send the standard basis to it. The function $t \mapsto ∥ φ (t) ∥$ is a norm on $R^{n}$ , hence equivalent to the coordinate norm $∣ t ∣_{\infty} = \max_{i} ∣ t_{i} ∣$ by [F3], so there is $c > 0$ with $∥ φ (t) ∥ \geq c ∣ t ∣_{\infty}$ for all $t$ . A nonzero element of $Γ$ has coordinates $(m_{1}, \dots, m_{r}, 0, \dots, 0)$ with some $m_{i} \in Z ∖ {0}$ , so $∣ m ∣_{\infty} \geq 1$ and $∥ γ ∥ \geq c$ ; thus $Γ \cap B (0, c) = {0}$ and $Γ$ is discrete, proving (a). This closes the cycle (a) $\Rightarrow$ (b) $\Rightarrow$ (c) $\Rightarrow$ (a).

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-10-02Open item page →

The logarithmic unit image is discrete

Statement

Assume the Axiom of Choice. The subgroup $λ (O_{K}^{\times}) \subset H$ is discrete; equivalently, every bounded subset of $H$ meets $λ (O_{K}^{\times})$ in finitely many points.

Facts & Assumptions

Given: The Axiom of Choice, a number field $K$ of degree $n = [K : Q]$ with logarithmic embedding $λ$ and hyperplane $H$ (Logarithmic embedding of a number field), and a bounded subset $C \subseteq H$ .

[F1]

The map $λ$ is given by $λ (x) = (\log ∣ σ_{1} x ∣, \dots, \log ∣ σ_{r_{1}} x ∣, 2 \log ∣ τ_{1} x ∣, \dots, 2 \log ∣ τ_{r_{2}} x ∣)$ , with $σ_{1}, \dots, σ_{r_{1}}$ the real embeddings and $τ_{1}, \dots, τ_{r_{2}}$ one embedding from each complex conjugate pair (Logarithmic embedding of a number field, Archimedean embeddings and signature).

[F2]

For every unit $u \in O_{K}^{\times}$ one has $λ (u) \in H$ , so $λ (O_{K}^{\times})$ is a subgroup of the finite-dimensional real vector space $H$ (Unit logarithms lie in the trace-zero hyperplane).

[F3]

For a subgroup $Γ$ of a finite-dimensional real vector space with the topology induced by a norm, $Γ$ is discrete if and only if every bounded subset of the space meets $Γ$ in a finite set (Discrete subgroups of a real vector space are lattices).

[F4]

The natural logarithm is strictly increasing with inverse the exponential function on $(0, \infty)$ (Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm, The exponential function is strictly increasing, The natural logarithm as the inverse of the exponential function); hence for real $a > 0$ and real $t$ , $a \leq e^{t}$ if and only if $\log a \leq t$ , and similarly $\log a \geq - t$ if and only if $a \geq e^{- t}$ .

[F5]

For fixed $n \geq 1$ and $R \geq 1$ , only finitely many monic integer polynomials of degree at most $n$ have all their complex roots of modulus at most $R$ (Bounded roots give finitely many monic integer polynomials). This batch-2 supplier is authored in this run, and the exact obligation used is the instance for the fixed real $R = e^{M} \geq 1$ of the argument.

[F6]

For $u \in O_{K}^{\times}$ the minimal polynomial $m_{u} \in Z [X]$ is monic of degree $[Q (u) : Q]$ , which divides $n$ ; its complex roots are exactly the numbers $ψ (u)$ , where $ψ$ ranges over the $Q$ -embeddings $K \to C$ (Minimal-polynomial criterion for algebraic integers, The degree of an intermediate field divides the degree of a finite extension, $F$ -embeddings of $F (α)$ into an algebraically closed field correspond to the distinct roots of $m_{α}$ , Restriction partitions embeddings in a finite tower into extension fibres).

[F7]

A nonzero polynomial of degree at most $n$ over $C$ has at most $n$ distinct roots (A nonzero polynomial of degree $n$ over an integral domain has at most $n$ distinct roots).

[F8]

The kernel of $λ ∣_{O_{K}^{\times}}$ is finite (Kernel of the unit logarithm is the roots of unity).

[A1]

The Axiom of Choice is assumed; it is used only through the AC-qualified hyperplane lemma [F2] (The Axiom of Choice).

Proof

technique · boundedness of the log image bounds all conjugates of a unit between two positive constants, and the bounded-conjugate polynomials of bounded degree are finite in number

1.1F2

The image $λ (O_{K}^{\times})$ lies in $H$ and is a subgroup of $H$ .

1.2givenalgebra

Since $C \subseteq H$ is bounded, there is a real $M \geq 0$ with $∣ x_{i} ∣ \leq M$ for every $x = (x_{i}) \in C$ and every coordinate $i$ ; fix such an $M$ and put $R = e^{M} \geq 1$ .

2.1F1step 1.2

Let $u \in O_{K}^{\times}$ with $λ (u) \in C$ . Then $∣ \log ∣ σ_{i} u ∣ ∣ \leq M$ for every real embedding and $∣ 2 \log ∣ τ_{j} u ∣ ∣ \leq M$ , that is, $- M \leq \log ∣ σ_{i} u ∣ \leq M$ and $- M / 2 \leq \log ∣ τ_{j} u ∣ \leq M / 2$ .

3.1F4step 2.1

Exponentiating the inequalities of step 2.1, using that the exponential is strictly increasing and inverse to the logarithm, gives $e^{- M} \leq ∣ σ_{i} u ∣ \leq e^{M} = R$ for every real embedding and $e^{- M / 2} \leq ∣ τ_{j} u ∣ \leq e^{M / 2}$ for every complex embedding; in particular every conjugate of $u$ has modulus at most $R$ .

4.1F5F6F7step 3.1

Consequently the minimal polynomial $m_{u}$ of such a unit $u$ is a monic integer polynomial of degree at most $n$ all of whose complex roots have modulus at most $R$ ; by [F5] there are only finitely many such polynomials, and each of them has at most $n$ distinct complex roots by [F7], so the set $S : = {u \in O_{K}^{\times} : λ (u) \in C}$ is finite.

5.1F3step 1.1step 4.1step 1.2

The intersection $C \cap λ (O_{K}^{\times})$ is the image under $λ$ of $S$ , hence is finite; therefore every bounded subset of $H$ meets the subgroup $λ (O_{K}^{\times})$ in a finite set, and by the lattice criterion [F3] the subgroup $λ (O_{K}^{\times})$ is discrete.

6.1A1F8step 5.1∎

The single Choice use is [A1] through the AC-qualified product formula behind the hyperplane lemma; the bounded-conjugate and root-bound arguments select nothing, and the kernel [F8] is finite by the choice-free finiteness of the roots of unity.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-10-02Open item page →

The logarithmic unit image is a full lattice

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $K$ be a number field of signature $(r_{1}, r_{2})$ with logarithmic embedding $λ$ and hyperplane $H = {(x_{i}) : \sum_{i} x_{i} = 0}$ (Logarithmic embedding of a number field). The discrete subgroup $λ (O_{K}^{\times}) \subset H$ spans $H$ over $R$ ; hence it is a full lattice in $H$ of rank $r_{1} + r_{2} - 1$ .

Facts & Assumptions

Given: The Axiom of Choice, a number field $K$ of degree $n = r_{1} + 2 r_{2}$ and signature $(r_{1}, r_{2})$ , its logarithmic embedding $λ$ and hyperplane $H$ (Logarithmic embedding of a number field), the subspace $W : = {span}_{R} λ (O_{K}^{\times}) \subseteq H$ , and the fixed constant $A : = \sqrt{∣ d_{K} ∣} (2 / π)^{r_{2}}$ .

[F1]

$λ (x) = (\log ∣ σ_{1} x ∣, \dots, \log ∣ σ_{r_{1}} x ∣, 2 \log ∣ τ_{1} x ∣, \dots, 2 \log ∣ τ_{r_{2}} x ∣)$ for the real embeddings $σ_{1}, \dots, σ_{r_{1}}$ and one chosen embedding $τ_{1}, \dots, τ_{r_{2}}$ from each complex conjugate pair, and $H = {(x_{i}) \in R^{r_{1} + r_{2}} : \sum_{i} x_{i} = 0}$ is a hyperplane of dimension $r_{1} + r_{2} - 1$ (Logarithmic embedding of a number field, Archimedean embeddings and signature).

[F2]

$λ (u) \in H$ for every unit $u \in O_{K}^{\times}$ (Unit logarithms lie in the trace-zero hyperplane), so both $λ (O_{K}^{\times})$ and $W$ lie in $H$ .

[F3]

$λ (O_{K}^{\times})$ is discrete (The logarithmic unit image is discrete).

[F4]

For a subgroup $Γ$ of a finite-dimensional real vector space: if every bounded set meets $Γ$ in a finite set, then there are $R$ -linearly independent $v_{1}, \dots, v_{s} \in Γ$ with $Γ = Z v_{1} \oplus \dots \oplus Z v_{s}$ and ${span}_{R} Γ = R v_{1} \oplus \dots \oplus R v_{s}$ ; conversely such a subgroup is discrete (Discrete subgroups of a real vector space are lattices).

[F5]

Orthogonal complements are taken in the standard inner product on $R^{r_{1} + r_{2}}$ : $U^{⊥} = {v : ⟨ v, u ⟩ = 0$ for all $u \in U}$ , one has $W^{⊥ ⊥} = W$ for every subspace $W$ , and $U_{1} \subseteq U_{2}$ implies $U_{2}^{⊥} \subseteq U_{1}^{⊥}$ . Since $H = {x : ⟨ x, (1, \dots, 1) ⟩ = 0}$ , we have $H^{⊥} = R (1, \dots, 1)$ ; hence $z \notin H^{⊥}$ exactly when the coordinates of $z$ are not all equal (The orthogonal complement $W^{⊥} = {v : ⟨ v, w ⟩ = 0 for all w \in W}$ , In finite dimension, $W^{⊥ ⊥} = W$ and $\dim W + \dim W^{⊥} = \dim V$ ).

[F6]

The Minkowski embedding $σ : K \to R^{r_{1}} \times R^{2 r_{2}} = R^{n}$ is injective, sends $x$ to $(σ_{1} x, \dots, σ_{r_{1}} x, Re τ_{1} x, Im τ_{1} x, \dots, Re τ_{r_{2}} x, Im τ_{r_{2}} x)$ , is additive, and in the complex coordinate pairs $∣ τ_{j} x ∣^{2} = (Re τ_{j} x)^{2} + (Im τ_{j} x)^{2}$ (Unscaled Minkowski embedding).

[F7]

The image $σ (O_{K})$ is a full lattice in $R^{n}$ , and its covolume, for the Lebesgue volume on $R^{r_{1}} \times R^{2 r_{2}}$ induced by the identification $C ≅ R^{2}$ just fixed, is $covol (σ (O_{K})) = 2^{- r_{2}} \sqrt{∣ d_{K} ∣}$ , where $d_{K} = disc (O_{K})$ is the nonzero field discriminant (Number-field integer rings and ideals are full lattices, Covolume of an integral ideal lattice, Discriminant of a basis and order, Number-field discriminant is well-defined and nonzero).

[F8]

Equality case of the lattice point principle: if $Λ \subseteq R^{n}$ is a full lattice and $S \subseteq R^{n}$ is closed, bounded, convex and centrally symmetric with $Vol (S) \geq 2^{n} covol (Λ)$ , then $S \cap Λ$ contains a nonzero point (Minkowski convex-body theorem at equality).

[F9]

For every real $B \geq 1$ only finitely many nonzero integral ideals $a \subseteq O_{K}$ satisfy $N a \leq B$ (Finitely many ideals of bounded norm).

[F10]

For $0 \neq a \in O_{K}$ the principal ideal $(a) = a O_{K} = {c a : c \in O_{K}}$ satisfies $N ((a)) = ∣ N_{K / Q} (a) ∣$ ; if $(a) \subseteq (b)$ then $a = c b$ for some $c \in O_{K}$ , so $(a) = (b)$ gives $a = u b$ and $b = v a$ with $u, v \in O_{K}$ and $u v = 1$ , that is, $u \in O_{K}^{\times}$ (The norm of a principal integral ideal, The ideal generated by a subset and principal ideals).

[F11]

If $0 \neq a \in O_{K}$ then $N_{K / Q} (a) = \prod_{i = 1}^{r_{1}} σ_{i} (a) \cdot \prod_{j = 1}^{r_{2}} ∣ τ_{j} (a) ∣^{2}$ and $N_{K / Q} (a) \in Z$ ; consequently $N_{K / Q} (a) \neq 0$ and $∣ N_{K / Q} (a) ∣ \geq 1$ (Norm and trace from embeddings, with the inseparable exponent in the norm formula, Trace and norm of an algebraic integer, Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive).

[F12]

AC implies Countable Choice (AC implies DC implies countable choice). Under Countable Choice, each closed real interval $[- c, c]$ has Lebesgue measure $2 c$ (A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included). Each closed disc of radius $c > 0$ is a bounded Jordan-measurable region between continuous graphs (Riemann area between continuous graphs equals Jordan content) with Jordan content $π c^{2}$ (A closed disc of radius $r \geq 0$ has Jordan content $π r^{2}$ ), so its Lebesgue measure is $π c^{2}$ (Lebesgue outer measure is at most Jordan outer content, and a bounded Jordan measurable set is Lebesgue measurable with Lebesgue measure equal to its Jordan content). Each Euclidean Lebesgue measure $λ_{d}$ is sigma-finite, since the cubes $[- N, N]^{d}$ exhaust $R^{d}$ and have finite measure by the box formula. For Borel sets $E_{i} \subseteq R^{d_{i}}$ with $d_{i} \in {1, 2}$ , the measure of a finite Cartesian product is the product of the factor measures: iterate the rectangle formula for sigma-finite product measures and the agreement of product measure with Euclidean Lebesgue measure on Borel sets (For sigma-finite factors, the product measure exists, has the rectangle formula, is sigma-finite, and is unique, On Borel subsets of R^{m+n}, the product lambda_m times lambda_n agrees with lambda_{m+n}). Therefore the Euclidean volume of the product of these intervals and discs is the product of their Lebesgue measures.

[F13]

The natural logarithm is strictly increasing, maps $(0, \infty)$ onto $R$ , satisfies $\log (x y) = \log x + \log y$ and $\log 1 = 0$ , so $\log t \to \infty$ as $t \to \infty$ and $\log (s / t) = \log s - \log t$ for $s, t > 0$ (Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm).

[F14]

Since $d_{K} \neq 0$ by [F7], $∣ d_{K} ∣ > 0$ and its nonnegative square root is positive; also $π > 0$ . Thus the fixed constant $A : = \sqrt{∣ d_{K} ∣} (2 / π)^{r_{2}}$ is positive (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ , Pi as twice the smallest positive zero of cosine).

[A1]

The Axiom of Choice is assumed; it is used through the equality-case lattice point principle [F8], the discreteness input [F3] whose own AC use is inherited, and the Countable Choice measure interfaces in [F12]. AC supplies Countable Choice by [F12] (The Axiom of Choice).

Proof

technique · contraposition on a linear functional. If some $z$ is not orthogonal to $H$, products of intervals and discs of fixed volume produce, through the equality case of the lattice point principle, an algebraic integer of bounded norm; the finite list of principal ideals of bounded norm converts it to a unit $u$ with $f(u)$ bounded away from a tunable number $t_c$, and rescaled logarithms make $|t_c|$ exceed that bound, forcing $f(u)\ne0$

1.1F1F2F5

The image $λ (O_{K}^{\times})$ is a subgroup of $R^{r_{1} + r_{2}}$ contained in $H$ , so $W$ is a subspace of $H$ ; also $H^{⊥} = R (1, \dots, 1)$ .

1.2F1F2F4

If $r_{1} + r_{2} = 1$ then $H = {0}$ by [F1], so $λ (O_{K}^{\times}) = {0} = H$ by [F2], and the image is a full lattice of rank $0$ in $H$ ; hence assume $r_{1} + r_{2} \geq 2$ from now on.

1.3F5

Fix $z \in R^{r_{1} + r_{2}}$ with $z \notin H^{⊥}$ ; then the coordinates $z_{1}, \dots, z_{r_{1} + r_{2}}$ of $z$ are not all equal, by [F5].

1.4F1algebra

Define $f (x) : = ⟨ z, λ (x) ⟩ = \sum_{i = 1}^{r_{1}} z_{i} \log ∣ σ_{i} x ∣ + \sum_{j = 1}^{r_{2}} 2 z_{r_{1} + j} \log ∣ τ_{j} x ∣$ for $x \in K^{\times}$ , a group homomorphism $K^{\times} \to (R, +)$ ; since $λ (O_{K}^{\times})$ spans $W$ and $⟨ z, \cdot ⟩$ is linear, $f (u) = 0$ for every $u \in O_{K}^{\times}$ exactly when $z \in W^{⊥}$ , so it suffices to exhibit one unit $u$ with $f (u) \neq 0$ .

1.5F12F14algebra

Let $c_{1}, \dots, c_{r_{1} + r_{2}}$ be positive reals with $\prod_{i = 1}^{r_{1}} c_{i} \cdot \prod_{j = 1}^{r_{2}} c_{r_{1} + j}^{2} = A$ , and let $S_{c} \subseteq R^{n}$ be the set of points whose first $r_{1}$ coordinates $y_{i}$ satisfy $∣ y_{i} ∣ \leq c_{i}$ and whose $j$ -th complex pair $(u_{j}, v_{j})$ satisfies $u_{j}^{2} + v_{j}^{2} \leq c_{r_{1} + j}^{2}$ ; then $S_{c}$ is closed, bounded, convex and centrally symmetric. Positivity of $A$ is established in [F14]. By [F12], its volume is $\prod_{i = 1}^{r_{1}} (2 c_{i}) \cdot \prod_{j = 1}^{r_{2}} π c_{r_{1} + j}^{2} = 2^{r_{1}} π^{r_{2}} A = 2^{n} \cdot 2^{- r_{2}} \sqrt{∣ d_{K} ∣}$ .

1.6F14

The fixed positive constant $A$ depends only on $K$ , as recorded in [F14].

2.1F7F8step 1.5

By [F7] we have $covol (σ (O_{K})) = 2^{- r_{2}} \sqrt{∣ d_{K} ∣}$ , so $Vol (S_{c}) = 2^{n} covol (σ (O_{K}))$ ; applying [F8] to the full lattice $σ (O_{K})$ and the set $S_{c}$ produces a nonzero point $x \in S_{c} \cap σ (O_{K})$ .

3.1F6step 2.1

Write $x = σ (a)$ with $a \in O_{K}$ ; then $a \neq 0$ and the coordinates of $a$ satisfy $∣ σ_{i} (a) ∣ \leq c_{i}$ for $i \leq r_{1}$ and $(Re τ_{j} a)^{2} + (Im τ_{j} a)^{2} \leq c_{r_{1} + j}^{2}$ , that is $∣ τ_{j} (a) ∣^{2} \leq c_{r_{1} + j}^{2}$ .

4.1F11step 3.1

By [F11], $∣ N_{K / Q} (a) ∣ = \prod_{i = 1}^{r_{1}} ∣ σ_{i} (a) ∣ \cdot \prod_{j = 1}^{r_{2}} ∣ τ_{j} (a) ∣^{2} \leq \prod_{i = 1}^{r_{1}} c_{i} \cdot \prod_{j = 1}^{r_{2}} c_{r_{1} + j}^{2} = A$ ; since also $a \neq 0$ and $N_{K / Q} (a) \in Z ∖ {0}$ , we get $1 \leq ∣ N (a) ∣ \leq A$ , and in particular $A \geq 1$ .

5.1step 4.1algebra

If for some $i \leq r_{1}$ one had $∣ σ_{i} (a) ∣ < c_{i} / A$ , then the remaining factors of $∣ N (a) ∣$ being bounded by $A / c_{i}$ would give $∣ N (a) ∣ < (c_{i} / A) (A / c_{i}) = 1$ , contradicting $∣ N (a) ∣ \geq 1$ ; hence $∣ σ_{i} (a) ∣ \geq c_{i} / A$ . Likewise, if $∣ τ_{j} (a) ∣^{2} < c_{r_{1} + j}^{2} / A$ for some $j \leq r_{2}$ , then $∣ N (a) ∣ < (c_{r_{1} + j}^{2} / A) (A / c_{r_{1} + j}^{2}) = 1$ , again a contradiction, so $∣ τ_{j} (a) ∣^{2} \geq c_{r_{1} + j}^{2} / A$ .

5.2F9F10step 4.1

Let $B_{0} : = \max {1, A} \geq 1$ ; by [F9] only finitely many nonzero integral ideals of $O_{K}$ have norm at most $B_{0}$ , hence only finitely many have norm at most $A$ . The finite subcollection of principal ideals is nonempty because $(1)$ has norm $1 \leq A$ by [F10] and step 4.1. Choose generators $0 \neq b_{1}, \dots, b_{m} \in O_{K}$ for these principal ideals, so that $(b_{1}), \dots, (b_{m})$ are exactly the principal ideals of norm at most $A$ ; choosing these $m$ generators is a selection from finitely many nonempty sets. Since $N ((a)) = ∣ N (a) ∣ \leq A$ by [F10], $(a) = (b_{j})$ for some $j$ , and then $a = u b_{j}$ with $u \in O_{K}^{\times}$ .

6.1F13step 4.1step 5.2

Put $t_{c} : = \sum_{i = 1}^{r_{1}} z_{i} \log c_{i} + \sum_{j = 1}^{r_{2}} z_{r_{1} + j} \log (c_{r_{1} + j}^{2})$ ; the finite numbers $f (b_{j}) = ⟨ z, λ (b_{j}) ⟩$ being fixed, $B : = \max_{j} ∣ f (b_{j}) ∣ + \log A \cdot \sum_{i = 1}^{r_{1} + r_{2}} ∣ z_{i} ∣$ is a real number depending only on $z$ , on $K$ and on the chosen list, not on $c$ .

7.1F13step 5.1step 5.2step 6.1

For the unit $u$ of step 5.2 we have $∣ f (u) - t_{c} ∣ = ∣ f (a) - f (b_{j}) - t_{c} ∣ \leq ∣ f (b_{j}) ∣ + ∣ f (a) - t_{c} ∣$ , and expanding $f (a) - t_{c} = \sum_{i = 1}^{r_{1}} z_{i} \log (∣ σ_{i} (a) ∣ / c_{i}) + \sum_{j = 1}^{r_{2}} z_{r_{1} + j} \log (∣ τ_{j} (a) ∣^{2} / c_{r_{1} + j}^{2})$ shows, by step 5.1 and by $∣ σ_{i} (a) ∣ \leq c_{i}$ , $∣ τ_{j} (a) ∣^{2} \leq c_{r_{1} + j}^{2}$ , that each logarithm lies in $[- \log A, 0]$ ; hence $∣ f (a) - t_{c} ∣ \leq \log A \cdot \sum_{i} ∣ z_{i} ∣$ and $∣ f (u) - t_{c} ∣ \leq B$ .

7.2F13step 1.3step 6.1

Choose indices $p < q$ with $z_{p} \neq z_{q}$ , possible by step 1.3. Since $\log d \to \infty$ as $d \to \infty$ and $z_{p} - z_{q} \neq 0$ , the absolute value of $(z_{p} - z_{q}) \log d + z_{q} \log A$ tends to infinity; hence choose $d > 0$ with $∣ (z_{p} - z_{q}) \log d + z_{q} \log A ∣ > B$ .

8.1F13step 7.2

Set $d_{p} : = d$ , $d_{q} : = A / d$ and $d_{i} : = 1$ for the remaining indices $i$ ; then $\prod_{i = 1}^{r_{1} + r_{2}} d_{i} = A$ and $\sum_{i} z_{i} \log d_{i} = z_{p} \log d + z_{q} \log (A / d) = (z_{p} - z_{q}) \log d + z_{q} \log A$ , so this $d$ can be chosen with $∣ \sum_{i} z_{i} \log d_{i} ∣ > B$ .

9.1step 8.1algebra

Define the admissible tuple $c$ by $c_{i} : = d_{i}$ for $i \leq r_{1}$ and $c_{r_{1} + j} : = \sqrt{d_{r_{1} + j}}$ for $1 \leq j \leq r_{2}$ ; then $\prod_{i = 1}^{r_{1}} c_{i} \cdot \prod_{j = 1}^{r_{2}} c_{r_{1} + j}^{2} = \prod_{i} d_{i} = A$ and $t_{c} = \sum_{i} z_{i} \log d_{i}$ , so $∣ t_{c} ∣ > B$ .

10.1step 1.4step 1.5step 5.2step 7.1step 9.1

Applying the fixed-product construction and bounded-norm argument of steps 1.5 through 5.2 to the admissible tuple $c$ from step 9.1 yields a unit $u \in O_{K}^{\times}$ with $∣ f (u) - t_{c} ∣ \leq B$ ; since $∣ t_{c} ∣ > B$ , the triangle inequality gives $∣ f (u) ∣ \geq ∣ t_{c} ∣ - ∣ f (u) - t_{c} ∣ > 0$ , so $f (u) \neq 0$ and therefore $z \notin W^{⊥}$ by step 1.4.

11.1step 1.3step 10.1

As $z \notin H^{⊥}$ was arbitrary, every $z$ outside $H^{⊥}$ lies outside $W^{⊥}$ , which means $W^{⊥} \subseteq H^{⊥}$ .

12.1F5step 1.1step 11.1

Since $W \subseteq H$ we have $H^{⊥} \subseteq W^{⊥}$ , and with step 11.1 this gives $H^{⊥} = W^{⊥}$ ; taking orthogonal complements and using $W^{⊥ ⊥} = W$ and $H^{⊥ ⊥} = H$ yields $W = H$ , so $λ (O_{K}^{\times})$ spans $H$ over $R$ .

13.1F1F3F4step 12.1

By the discreteness of $λ (O_{K}^{\times})$ and [F4], there are $R$ -linearly independent $v_{1}, \dots, v_{s} \in λ (O_{K}^{\times})$ with $λ (O_{K}^{\times}) = Z v_{1} \oplus \dots \oplus Z v_{s}$ and ${span}_{R} λ (O_{K}^{\times}) = R v_{1} \oplus \dots \oplus R v_{s}$ ; this span is $H$ by step 12.1, so $s = \dim_{R} H = r_{1} + r_{2} - 1$ and $λ (O_{K}^{\times})$ is a full lattice in $H$ of rank $r_{1} + r_{2} - 1$ .

14.1A1F3F8F9F12step 1.2step 5.2step 7.2∎

Choice accounting: AC is invoked through [F8], the discreteness input [F3], and the Countable Choice measure interfaces in [F12]. The only other selections are the finitely many generators $b_{1}, \dots, b_{m}$ of step 5.2 and the single positive real $d$ of step 7.2; the rank-zero case of step 1.2 is choice-free.

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Dirichlet unit theorem

Statement

Assume the Axiom of Choice (The Axiom of Choice). For a number field $K$ of signature $(r_{1}, r_{2})$ there is an isomorphism $O_{K}^{\times} ≅ μ (K) \times Z^{r_{1} + r_{2} - 1}$ . In particular $O_{K}^{\times}$ is finitely generated of rank $r_{1} + r_{2} - 1$ and its torsion subgroup is $μ (K)$ .

Facts & Assumptions

Given: The Axiom of Choice, a number field $K$ of signature $(r_{1}, r_{2})$ with logarithmic embedding $λ$ (Logarithmic embedding of a number field), and the image $Λ : = λ (O_{K}^{\times})$ .

[F1]

For $x, y \in K^{\times}$ the multiplicativity of every embedding and of the complex modulus gives $∣ σ_{i} (x y) ∣ = ∣ σ_{i} x ∣ ∣ σ_{i} y ∣$ and $∣ τ_{j} (x y) ∣ = ∣ τ_{j} x ∣ ∣ τ_{j} y ∣$ , and $\log (x y) = \log x + \log y$ for positive reals; hence $λ (x y) = λ (x) + λ (y)$ , so $λ$ restricts to a group homomorphism $O_{K}^{\times} \to R^{r_{1} + r_{2}}$ (Logarithmic embedding of a number field, Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm, Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive).

[F2]

$\ker (λ ∣_{O_{K}^{\times}}) = μ (K)$ ; this kernel is finite and is exactly the torsion subgroup of $O_{K}^{\times}$ (Kernel of the unit logarithm is the roots of unity, The group $μ_{n} (K)$ of $n$ -th roots of unity in a field, and primitive $n$ -th roots of unity).

[F3]

$Λ$ is a full lattice in the hyperplane $H$ of the definition of $λ$ , of rank $r_{1} + r_{2} - 1$ ; hence $Λ = Z v_{1} \oplus \dots \oplus Z v_{r}$ for $R$ -linearly independent $v_{1}, \dots, v_{r} \in Λ$ and $r = r_{1} + r_{2} - 1$ (The logarithmic unit image is a full lattice).

[F4]

Every finitely generated abelian group $G$ has a decomposition $G ≅ Z^{s} \oplus T$ with $T$ its finite torsion subgroup, and the integer $s$ (the free rank) and the torsion subgroup are determined by $G$ (The fundamental theorem of finitely generated abelian groups from PID modules).

[F5]

The units of a commutative ring form an abelian group under multiplication; in particular $O_{K}^{\times}$ is an abelian group (The units of a ring are the invertible elements of its multiplicative monoid, and $R^{\times}$ is a group under multiplication; $0 \in R^{\times}$ only in the zero ring).

Proof

Proof technique: split the unit group by the logarithmic embedding. The image is a free abelian group whose rank is computed by the full-lattice theorem, and the kernel is the finite roots-of-unity group, so lifting a $Z$ -basis of the image exhibits $O_{K}^{\times}$ as $μ (K) \times Z^{r_{1} + r_{2} - 1}$ .

1.1F1F2

The restriction $λ ∣_{O_{K}^{\times}}$ is a group homomorphism onto $Λ$ with kernel $μ (K)$ , a finite group that coincides with the torsion subgroup of $O_{K}^{\times}$ .

1.2F3

By [F3] the image is $Λ = Z v_{1} \oplus \dots \oplus Z v_{r}$ with $v_{1}, \dots, v_{r} \in Λ$ $R$ -linearly independent and $r = r_{1} + r_{2} - 1$ ; choose once and for all units $u_{1}, \dots, u_{r} \in O_{K}^{\times}$ with $λ (u_{i}) = v_{i}$ , a selection of finitely many elements.

2.1F1step 1.1step 1.2

Let $u \in O_{K}^{\times}$ ; since $Λ$ is generated by $v_{1}, \dots, v_{r}$ there are unique integers $m_{1}, \dots, m_{r}$ with $λ (u) = \sum_{i} m_{i} v_{i} = λ (u_{1}^{m_{1}} \dots u_{r}^{m_{r}})$ , so $λ (u (u_{1}^{m_{1}} \dots u_{r}^{m_{r}})^{- 1}) = 0$ and hence $u (u_{1}^{m_{1}} \dots u_{r}^{m_{r}})^{- 1} = ζ$ for some $ζ \in μ (K)$ ; therefore $u = ζ u_{1}^{m_{1}} \dots u_{r}^{m_{r}}$ .

3.1F1step 1.2

The expression in step 2.1 is unique: if $ζ u_{1}^{m_{1}} \dots u_{r}^{m_{r}} = ζ^{'} u_{1}^{m_{1}^{'}} \dots u_{r}^{m_{r}^{'}}$ with $ζ, ζ^{'} \in μ (K)$ , then applying the homomorphism and using $λ (ζ) = λ (ζ^{'}) = 0$ gives $\sum_{i} (m_{i} - m_{i}^{'}) v_{i} = 0$ , hence $m_{i} = m_{i}^{'}$ for every $i$ by linear independence of the $v_{i}$ , and then $ζ = ζ^{'}$ .

4.1F5step 1.2step 2.1step 3.1

Consequently the map $μ (K) \times Z^{r} \to O_{K}^{\times}$ , $(ζ, m_{1}, \dots, m_{r}) \mapsto ζ u_{1}^{m_{1}} \dots u_{r}^{m_{r}}$ , is a bijective group homomorphism, because $O_{K}^{\times}$ is abelian and $u_{1}^{m_{1}} \dots u_{r}^{m_{r}} u_{1}^{n_{1}} \dots u_{r}^{n_{r}} = u_{1}^{m_{1} + n_{1}} \dots u_{r}^{m_{r} + n_{r}}$ ; hence $O_{K}^{\times} ≅ μ (K) \times Z^{r}$ with $r = r_{1} + r_{2} - 1$ , and $O_{K}^{\times}$ is finitely generated.

5.1F2F4step 4.1

Since $O_{K}^{\times}$ is finitely generated abelian, [F4] exhibits it as $Z^{s} \oplus T$ with $T$ its finite torsion subgroup; the displayed isomorphism of step 4.1 has free part $Z^{r_{1} + r_{2} - 1}$ and torsion factor $μ (K)$ , so by the uniqueness in [F4] the free rank is $s = r_{1} + r_{2} - 1$ and the torsion subgroup is $T = μ (K)$ .

6.1F3step 1.2∎

Choice accounting: the assumption AC enters only through the full-lattice theorem [F3]; the only selections made here are the finitely many units $u_{1}, \dots, u_{r}$ lifting a finite basis, and no choice over an infinite family occurs.

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System of fundamental units

Definition

Assume the Axiom of Choice (The Axiom of Choice). Let $K$ be a number field of signature $(r_{1}, r_{2})$ and unit rank $r = r_{1} + r_{2} - 1$ (Dirichlet unit theorem). By the full-lattice theorem the image $λ (O_{K}^{\times})$ is a free abelian subgroup of the hyperplane $H$ of rank $r$ (The logarithmic unit image is a full lattice, Free abelian group on a set). A system of fundamental units of $K$ is a tuple $(ε_{1}, \dots, ε_{r})$ of units of $O_{K}$ such that $λ (ε_{1}), \dots, λ (ε_{r})$ is a $Z$ -basis of $λ (O_{K}^{\times})$ , that is

$λ (O_{K}^{\times}) = Z λ (ε_{1}) \oplus \dots \oplus Z λ (ε_{r}) .$

Existence. The unit theorem gives $O_{K}^{\times} ≅ μ (K) \times Z^{r}$ (Dirichlet unit theorem), so $λ (O_{K}^{\times})$ is a free abelian group of rank $r$ and has a $Z$ -basis; since $λ$ maps $O_{K}^{\times}$ onto its image, each basis vector is $λ (ε_{i})$ for some unit $ε_{i}$ . This exhibits a system of fundamental units, and the construction makes only the finitely many choices of preimages of a finite basis, so the Axiom of Choice is used here only through the unit theorem. In rank $r = 0$ the empty tuple is the unique system of fundamental units.

The equivalent product description. A tuple $(ε_{1}, \dots, ε_{r})$ is a system of fundamental units if and only if every unit $u \in O_{K}^{\times}$ admits a unique expression

$u = ζ ε_{1}^{m_{1}} \dots ε_{r}^{m_{r}}, ζ \in μ (K), m_{1}, \dots, m_{r} \in Z .$

Indeed, if the $λ (ε_{i})$ form a $Z$ -basis and $u \in O_{K}^{\times}$ , then $λ (u) = \sum_{i} m_{i} λ (ε_{i}) = λ (ε_{1}^{m_{1}} \dots ε_{r}^{m_{r}})$ for unique integers $m_{i}$ , so $u (ε_{1}^{m_{1}} \dots ε_{r}^{m_{r}})^{- 1}$ lies in the kernel of $λ$ on $O_{K}^{\times}$ , which is $μ (K)$ (Kernel of the unit logarithm is the roots of unity); uniqueness of the exponents follows from the $Z$ -independence of the basis and then uniqueness of $ζ$ from cancellation in the group $O_{K}^{\times}$ (The units of a ring are the invertible elements of its multiplicative monoid, and $R^{\times}$ is a group under multiplication; $0 \in R^{\times}$ only in the zero ring). Conversely, if every unit has such a unique expression, then $\sum_{i} m_{i} λ (ε_{i}) = 0$ forces the unit $ε_{1}^{m_{1}} \dots ε_{r}^{m_{r}}$ to lie in $μ (K)$ , hence by uniqueness all $m_{i} = 0$ , so the $λ (ε_{i})$ are $Z$ -independent; and applying $λ$ to the expression of an arbitrary unit shows that they generate $λ (O_{K}^{\times})$ . Thus the two descriptions of the definition agree.

A system is auxiliary data, not canonical field data. Different systems of fundamental units are related by a unimodular integer change of coordinates: the tuples $(λ (ε_{i}))$ and $(λ (ε_{i}^{'}))$ are two $Z$ -bases of the same free abelian group, so $λ (ε_{i}^{'}) = \sum_{j} c_{i j} λ (ε_{j})$ with $(c_{i j}) \in {GL}_{r} (Z)$ . No system is singled out by the field, and the definition introduces no sign or ordering convention: the regulator constructed from these units is independent of the system, a fact proved separately.

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Regulator of a number field

Definition

Assume the Axiom of Choice (The Axiom of Choice). Let $K$ be a number field of signature $(r_{1}, r_{2})$ (Number field, Archimedean embeddings and signature) and unit rank $r = r_{1} + r_{2} - 1$ , and let $(ε_{1}, \dots, ε_{r})$ be a system of fundamental units of $K$ (System of fundamental units). Let $A$ be the $(r_{1} + r_{2}) \times r$ real matrix

$A = (λ (ε_{1}) \dots λ (ε_{r}))$

whose columns are the logarithmic vectors $λ (ε_{i})$ in the doubled convention of the logarithmic embedding (Logarithmic embedding of a number field). For $k = 1, \dots, r_{1} + r_{2}$ let $A_{k}$ be the $r \times r$ matrix obtained from $A$ by deleting row $k$ , and let $\det A_{k}$ be its determinant (For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix). The regulator of $K$ is

$R_{K} : = ∣ \det A_{k} ∣ .$

For $r = 0$ the matrices $A$ and $A_{k}$ are empty, and the empty determinant is defined to be $1$ ; thus $R_{Q} = 1$ and $R_{K} = 1$ for imaginary quadratic $K$ .

Why a deleted row gives a well-defined number. The columns of $A$ are linearly independent over $R$ : the full-lattice theorem says that $λ (O_{K}^{\times})$ is a full lattice in $H$ of rank $r = r_{1} + r_{2} - 1 = \dim_{R} H$ . Thus a $Z$ -basis has $r$ vectors spanning $H$ , hence is also an $R$ -basis of $H$ . The vectors $λ (ε_{1}), \dots, λ (ε_{r})$ are another $Z$ -basis of the same group, and their change-of-basis matrix lies in ${GL}_{r} (Z)$ , hence is invertible over $R$ . Therefore these columns are $R$ -linearly independent and $A$ has rank $r$ , so $r_{1} + r_{2} = r + 1 \geq 1$ (Row space, column space, nullspace, row rank, column rank and matrix rank, The logarithmic unit image is a full lattice). Each column lies in the hyperplane $H$ , so its coordinates sum to zero (Unit logarithms lie in the trace-zero hyperplane). The deleted-row minors of a rank- $r$ real matrix with $r + 1$ rows and zero column sums therefore satisfy $\det A_{k} = (- 1)^{k - 1} \det A_{1}$ and are all nonzero (Deleted-row minors of a zero-column-sum matrix agree up to sign); in particular $∣ \det A_{k} ∣$ does not depend on the deleted row $k$ .

Independence of the fundamental system. The number $R_{K}$ above is defined from one chosen system of fundamental units; that the absolute deleted-row determinant does not depend on this auxiliary choice, so that $R_{K}$ depends on $K$ alone, is the content of The regulator is well defined ↗, which also shows $R_{K} > 0$ in the rank- $r$ case. A change of fundamental system multiplies $A$ on the right by a matrix in ${GL}_{r} (Z)$ , which is why the absolute determinant, and not the signed one, is the invariant.

Normalization. The factor $2$ on the complex coordinates of $λ$ is part of the doubled convention fixed in the definition of the logarithmic embedding; with it, the regulator of a real quadratic field is $\log ε$ for its fundamental unit $ε > 1$ , and mixing conventions changes the value. Ordering the coordinates differently permutes rows of $A$ , which leaves every $∣ \det A_{k} ∣$ unchanged, so the definition is insensitive to the ordering of the embeddings.

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Deleted-row minors of a zero-column-sum matrix agree up to sign

Statement

Let $m \geq 1$ and let $A = (a_{k j})$ be an $(m + 1) \times m$ real matrix of rank $m$ (Row space, column space, nullspace, row rank, column rank and matrix rank) each of whose columns has coordinate sum zero, that is $\sum_{k = 1}^{m + 1} a_{k j} = 0$ for every column index $j$ . For $k = 1, \dots, m + 1$ let $Δ_{k}$ be the determinant (For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix) of the $m \times m$ matrix obtained by deleting row $k$ . Then $Δ_{k} \neq 0$ for every $k$ and

$Δ_{k} = (- 1)^{k - 1} Δ_{1} (k = 1, \dots, m + 1);$

in particular all $∣ Δ_{k} ∣$ are equal.

Facts & Assumptions

Given: An integer $m \geq 1$ and an $(m + 1) \times m$ real matrix $A = (a_{k j})$ of rank $m$ whose columns each have coordinate sum zero.

[F1]

Expanding a determinant along its last column, with $M_{k m}$ the determinant of the matrix obtained by deleting row $k$ and the last column and $C_{k m} = (- 1)^{k - 1 + m} M_{k m}$ the corresponding cofactor, gives $\det B = \sum_{k = 1}^{m + 1} b_{k, m + 1} C_{k, m + 1}$ ; a matrix with two equal columns has determinant zero (Laplace expansion computes the determinant along every row and every column over a commutative ring, Deleted-row-and-column minors, cofactors, the cofactor matrix and the adjugate over a commutative ring, A square matrix with a zero column or two equal columns has determinant zero).

[F2]

For a real $m \times (m + 1)$ matrix the rank and the dimension of the kernel satisfy $rank + \dim N = m + 1$ ; transposition does not change the rank (For an $m \times n$ matrix $A$ , $rank (A) + \dim N (A) = n$ , Row rank equals column rank, and both equal the number of pivots, The transpose $A^{T}$ of a matrix).

[F3]

If $A$ is $(m + 1) \times m$ of rank $m$ , then some $m$ -rowed minor of $A$ is nonzero (A matrix has rank at least $r$ exactly when it has a nonzero $r$ -rowed minor).

Proof

Proof technique: build a linear relation among the minors from the equality of two columns of an augmented matrix, then identify the resulting kernel with the all-ones line.

1.1F1given

Fix a column index $j$ and let $B$ be the $(m + 1) \times (m + 1)$ real matrix whose first $m$ columns are the columns of $A$ and whose last column is the $j$ -th column of $A$ ; its entries in the last column are $b_{k, m + 1} = a_{k j}$ . The last column of $B$ equals column $j$ , so $\det B = 0$ , and expanding along the last column as in [F1] gives $\sum_{k = 1}^{m + 1} (- 1)^{k - 1 + m} a_{k j} Δ_{k} = 0$ , because deleting row $k$ and the last column of $B$ leaves exactly the matrix whose determinant is $Δ_{k}$ .

2.1F2step 1.1

Define $c_{k} : = (- 1)^{k - 1 + m} Δ_{k}$ for $k = 1, \dots, m + 1$ . Step 1.1 says $\sum_{k = 1}^{m + 1} a_{k j} c_{k} = 0$ for every column index $j$ , that is $A^{T} c = 0$ for the transpose $A^{T}$ ; and the column-sum hypothesis says $\sum_{k = 1}^{m + 1} a_{k j} \cdot 1 = 0$ for every $j$ , that is $A^{T} 1 = 0$ with $1 = (1, \dots, 1) \neq 0$ .

3.1F2step 2.1

Since $A$ has rank $m$ , its transpose $A^{T}$ has rank $m$ , so by rank-nullity its kernel has dimension $(m + 1) - m = 1$ and is therefore a line. Both $c$ and $1$ lie in that kernel and $1 \neq 0$ , so $c = λ 1$ for some $λ \in R$ .

4.1F3step 3.1∎

By [F3] some $m$ -rowed minor of $A$ is nonzero, and the $m$ -rowed minors of $A$ are exactly the determinants $Δ_{1}, \dots, Δ_{m + 1}$ ; since $c_{k} = \pm Δ_{k}$ , this makes $c \neq 0$ , hence $λ \neq 0$ and $Δ_{k} = (- 1)^{k - 1 + m} λ \neq 0$ for every $k$ . In particular $Δ_{1} = (- 1)^{m} λ$ , so $Δ_{k} = (- 1)^{k - 1 + m} λ = (- 1)^{k - 1} \cdot (- 1)^{m} λ = (- 1)^{k - 1} Δ_{1}$ , which is the claimed sign pattern and nonvanishing.

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The regulator is well defined

Statement

Assume the Axiom of Choice. The regulator $R_{K}$ of a number field $K$ is independent of the deleted row and of the chosen system of fundamental units, and $R_{K} > 0$ . Consequently $R_{K}$ is an invariant of $K$ (in the doubled logarithmic normalization), with $R_{K} = 1$ for rank zero.

Facts & Assumptions

Given: The Axiom of Choice, a number field $K$ of signature $(r_{1}, r_{2})$ with unit rank $r = r_{1} + r_{2} - 1$ , a system of fundamental units $(ε_{1}, \dots, ε_{r})$ , the matrix $A$ with columns $λ (ε_{1}), \dots, λ (ε_{r})$ , and the deleted-row matrices $A_{k}$ of the regulator definition (Regulator of a number field, Logarithmic embedding of a number field).

[F1]

The regulator is $R_{K} = ∣ \det A_{k} ∣$ for the $r \times r$ matrix $A_{k}$ obtained from $A$ by deleting row $k$ ; every square matrix has a determinant, and for $r = 0$ the matrices $A$ and $A_{k}$ are empty and the empty determinant is $1$ (Regulator of a number field, For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix).

[F2]

The logarithms $λ (ε_{1}), \dots, λ (ε_{r})$ form a $Z$ -basis of the free abelian group $λ (O_{K}^{\times})$ , that is $λ (O_{K}^{\times}) = Z λ (ε_{1}) \oplus \dots \oplus Z λ (ε_{r})$ ; in rank $r = 0$ the empty tuple is the unique system of fundamental units; and for two systems $(ε_{i})$ and $(ε_{i}^{'})$ there is a matrix $C = (c_{i j}) \in {GL}_{r} (Z)$ with $λ (ε_{i}^{'}) = \sum_{j} c_{i j} λ (ε_{j})$ (System of fundamental units).

[F3]

Every column of $A$ lies in $H = {x : \sum_{i} x_{i} = 0}$ , because $λ (u) \in H$ for every unit $u$ (Unit logarithms lie in the trace-zero hyperplane, Logarithmic embedding of a number field).

[F4]

$λ (O_{K}^{\times})$ is discrete in $H$ and spans $H$ over $R$ ; it is a full lattice in $H$ of rank $r_{1} + r_{2} - 1$ (The logarithmic unit image is a full lattice).

[F5]

If $Γ$ is a discrete subgroup of a finite-dimensional real vector space, then there are $R$ -linearly independent $v_{1}, \dots, v_{s} \in Γ$ with $Γ = Z v_{1} \oplus \dots \oplus Z v_{s}$ and ${span}_{R} Γ = R v_{1} \oplus \dots \oplus R v_{s}$ (Discrete subgroups of a real vector space are lattices).

[F6]

Let $m \geq 1$ and let $B$ be an $(m + 1) \times m$ real matrix of rank $m$ whose columns have coordinate sum zero. For the determinant $Δ_{k}$ of the matrix obtained by deleting row $k$ one has $Δ_{k} \neq 0$ for every $k$ and $Δ_{k} = (- 1)^{k - 1} Δ_{1}$ ; in particular all $∣ Δ_{k} ∣$ are equal (Deleted-row minors of a zero-column-sum matrix agree up to sign).

[F7]

For square matrices of the same size, $\det (B C) = \det (B) \det (C)$ (For same-sized finite square matrices over a commutative ring, $\det (A B) = \det (A) \det (B)$ ), the rank of a matrix is the dimension of its column space (Row space, column space, nullspace, row rank, column rank and matrix rank), and an invertible square matrix over a commutative ring has unit determinant (An invertible square matrix over a commutative ring has unit determinant); the units of $Z$ are $1$ and $- 1$ ( $(Z, \cdot, 1)$ is a commutative monoid whose group of units is ${1, - 1}$ ; equivalently $u ∣ 1$ holds exactly for $u = 1$ and $u = - 1$ ).

[A1]

The Axiom of Choice is assumed; it is used only through the existence of a system of fundamental units supplied by the AC-qualified unit theorem [F2] (The Axiom of Choice).

Proof

technique · prove the two invariance statements of the definition separately. The deleted-row statement reduces to the sign pattern of the cofactors of a rank-$r$ matrix with zero column sums. For a change of fundamental system, form the integer matrix whose columns give the new basis vectors in the old basis; it is unimodular, so every deleted-row determinant is multiplied by $\pm1$. Positivity follows because the common determinant is nonzero

1.1F1F2

Rank zero: if $r = 0$ then by [F1] the matrix $A$ has no columns and each $A_{k}$ is the empty matrix with determinant $1$ , while by [F2] the empty tuple is the unique system of fundamental units; hence $R_{K} = 1$ is independent of the deleted row and of the fundamental system, and $R_{K} > 0$ .

1.2F2F4F5F7

Assume $r \geq 1$ from here on, so $A$ is an $(r + 1) \times r$ real matrix with $r_{1} + r_{2} = r + 1 \geq 2$ rows. Its columns are $R$ -linearly independent, that is $A$ has rank $r$ : by [F4] the group $λ (O_{K}^{\times})$ is a discrete subgroup of the finite-dimensional real vector space $H$ , so by [F5] it equals $Z v_{1} \oplus \dots \oplus Z v_{s}$ with $v_{1}, \dots, v_{s}$ $R$ -linearly independent and ${span}_{R} λ (O_{K}^{\times}) = R v_{1} \oplus \dots \oplus R v_{s}$ of dimension $s$ ; by [F2] the elements $λ (ε_{1}), \dots, λ (ε_{r})$ form a $Z$ -basis of that same group, so $s = r$ (equal free rank) and their $R$ -span is all of ${span}_{R} λ (O_{K}^{\times})$ , of dimension $r$ ; a spanning set of $r$ vectors in a space of dimension $r$ is a basis, so the columns of $A$ are $R$ -linearly independent and the rank of $A$ is $r$ by [F7].

1.3F3

Every column of $A$ lies in $H$ , hence has coordinate sum zero.

1.4F2

Independence of the fundamental system: let $(ε_{1}^{'}, \dots, ε_{r}^{'})$ be a second system of fundamental units and let $A^{'}$ be the matrix with columns $λ (ε_{1}^{'}), \dots, λ (ε_{r}^{'})$ . By [F2] both logarithm lists are $Z$ -bases of the same group. For each $i$ , write $λ (ε_{i}^{'}) = \sum_{j} b_{j i} λ (ε_{j})$ using unique integers $b_{j i}$ , and let $B = (b_{j i})$ , so the $i$ -th column of $B$ records the coordinates of $λ (ε_{i}^{'})$ in the old basis. The reverse basis change also has integer coefficients, so $B \in {GL}_{r} (Z)$ . With these column coordinates, $A^{'} = A B$ ; deleting row $k$ gives $A_{k}^{'} = A_{k} B$ for every $k$ .

2.1F7step 1.4

The matrix $B$ of step 1.4 has an integer inverse, so [F7] makes $\det B$ a unit of $Z$ , and the description of the units of $Z$ in [F7] gives $\det B = \pm 1$ .

2.2F6step 1.2step 1.3

Independence of the deleted row: by step 1.2 and step 1.3 the matrix $A$ satisfies the hypotheses of [F6] with $m = r$ , so for the determinants $Δ_{k} = \det A_{k}$ one has $Δ_{k} \neq 0$ and $Δ_{k} = (- 1)^{k - 1} Δ_{1}$ for every $k$ ; therefore $∣ \det A_{k} ∣ = ∣ Δ_{1} ∣$ is the same number for every deleted row, and it is positive because $Δ_{1} \neq 0$ .

3.1F7step 1.4step 2.1

For every $k$ , multiplicativity of the determinant [F7] applied to $A_{k}^{'} = A_{k} B$ of step 1.4 gives $\det A_{k}^{'} = \det A_{k} \det B$ , so $∣ \det A_{k}^{'} ∣ = ∣ \det A_{k} ∣ ∣ \det B ∣ = ∣ \det A_{k} ∣$ by step 2.1; hence every deleted-row determinant of the second system has the same absolute value as the first.

4.1F1step 1.1step 2.2step 3.1

Combining steps 2.2 and 3.1: in the case $r \geq 1$ the number $R_{K} = ∣ \det A_{k} ∣$ depends neither on the deleted row nor on the chosen system of fundamental units, and it is positive; in the case $r = 0$ step 1.1 gives $R_{K} = 1$ . Therefore $R_{K}$ is well defined and is an invariant of $K$ alone, equal to $1$ in rank zero.

5.1A1F2∎

Choice accounting: the only place where AC enters is the existence of the systems of fundamental units in [F2], inherited from the AC-qualified unit theorem; the linear algebra of ranks, determinants and unimodular change of basis is elementary and choice-free.

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Unit ranks by signature

Statement

Assume the Axiom of Choice. The unit rank $r_{1} + r_{2} - 1$ is $0$ exactly for $K = Q$ and for imaginary quadratic fields, and it is $1$ for every real quadratic field; in general it equals $r_{1} + r_{2} - 1$ . The rank- $0$ cases have $O_{K}^{\times} = μ (K)$ finite and the rank- $1$ real quadratic case has $O_{K}^{\times} ≅ μ (K) \times Z$ with $μ (K) = {\pm 1}$ .

Facts & Assumptions

Given: The Axiom of Choice and a number field $K$ of signature $(r_{1}, r_{2})$ (Number field, Archimedean embeddings and signature).

[F1]

The signature satisfies $r_{1} + 2 r_{2} = [K : Q]$ , with $r_{1}$ the number of real embeddings and $r_{2}$ the number of complex conjugate pairs (Archimedean embeddings and signature).

[F2]

$O_{K}^{\times} ≅ μ (K) \times Z^{r_{1} + r_{2} - 1}$ ; the rank of $O_{K}^{\times}$ is $r_{1} + r_{2} - 1$ , and when $r_{1} + r_{2} - 1 = 0$ one has $O_{K}^{\times} = μ (K)$ (Dirichlet unit theorem).

[F3]

For a finite field extension $K / Q$ one has $[K : Q] = 1$ if and only if $K = Q$ (A finite extension has degree one if and only if the two fields are equal). Moreover $O_{Q}$ , the integral closure of $Z$ in $Q$ (Ring of integers), equals $Z$ : a rational number integral over $Z$ is an integer (The rational algebraic integers are exactly the integers), and every integer $n$ is a root of the monic polynomial $X - n$ , so $O_{Q} = Z$ with unit group $O_{Q}^{\times} = Z^{\times} = {\pm 1}$ ( $(Z, \cdot, 1)$ is a commutative monoid whose group of units is ${1, - 1}$ ; equivalently $u ∣ 1$ holds exactly for $u = 1$ and $u = - 1$ ).

[F4]

If $x \in R$ and $x^{n} = 1$ for some integer $n \geq 1$ , then $x = \pm 1$ : the inequalities $x > 1$ , $0 < x < 1$ are preserved by taking $n$ -th powers, and $x < 0$ with $x^{n} = 1$ forces $n$ even and $(- x)^{n} = 1$ , so $- x = 1$ (Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ , Sign rules for products and monotonicity of multiplication, The group $μ_{n} (K)$ of $n$ -th roots of unity in a field, and primitive $n$ -th roots of unity).

[F5]

For a nonzero squarefree integer $d \neq 1$ , $d$ is not a rational square by unique prime factorisation (The fundamental theorem of arithmetic: every integer $n \geq 1$ is a product of primes, and the factorisation is unique up to order — if $\prod_{i < r} p_{i} = \prod_{j < s} q_{j}$ with every $p_{i}$ and $q_{j}$ prime, then $r = s$ and $q_{i} = p_{π (i)}$ for some $π \in Sym (r)$ ), so $X^{2} - d$ is irreducible over $Q$ and $Q (\sqrt{d})$ has degree $2$ (The evaluation kernel and the unique monic irreducible minimal polynomial of an algebraic element, An element is algebraic over $F$ if and only if its simple extension $F (a) / F$ is finite). Its embeddings correspond exactly to the two roots $\pm \sqrt{d}$ ( $F$ -embeddings of $F (α)$ into an algebraically closed field correspond to the distinct roots of $m_{α}$ ): if $d > 0$ both embeddings are real, and if $d < 0$ neither is real and they are complex conjugates. The signatures are therefore $(2, 0)$ and $(0, 1)$ respectively (Archimedean embeddings and signature).

[F6]

In a finite tower, the intermediate degree divides the total degree (The degree of an intermediate field divides the degree of a finite extension). For an algebraic element $α$ , $[Q (α) : Q]$ is the degree of its monic irreducible minimal polynomial (An element is algebraic over $F$ if and only if its simple extension $F (a) / F$ is finite, The evaluation kernel and the unique monic irreducible minimal polynomial of an algebraic element).

[F7]

Every positive integer has a unique prime factorisation (The fundamental theorem of arithmetic: every integer $n \geq 1$ is a product of primes, and the factorisation is unique up to order — if $\prod_{i < r} p_{i} = \prod_{j < s} q_{j}$ with every $p_{i}$ and $q_{j}$ prime, then $r = s$ and $q_{i} = p_{π (i)}$ for some $π \in Sym (r)$ ); consequently every nonzero rational $Δ$ can be written as $q^{2} d$ with $q \in Q^{\times}$ and $d$ a nonzero squarefree integer, by writing each prime exponent of $Δ$ as $2 k + e$ with $e \in {0, 1}$ and retaining the sign in $d$ .

[A1]

The Axiom of Choice is assumed; it is used only through the AC-qualified unit theorem [F2] (The Axiom of Choice).

Proof

technique · read the rank $r_1+r_2-1$ off $\mathcal O_K^\times\cong\mu(K)\times\mathbb Z^{r_1+r_2-1}$, enumerate the signatures with $r_1+r_2=1$ to identify the rank-zero fields, and compute the real quadratic torsion from the real roots of unity

1.1F1F2F5

By [F2] the rank of $O_{K}^{\times}$ equals $r_{1} + r_{2} - 1$ ; in particular rank $0$ means $O_{K}^{\times} = μ (K)$ , and a real quadratic field has signature $(2, 0)$ and rank $1$ .

1.2F1F2F3

$Q$ has signature $(1, 0)$ , so its unit rank is $0$ ; by [F3] its ring of integers is $Z$ with unit group ${\pm 1} = μ (Q)$ , a finite group.

1.3F1F2F5

An imaginary quadratic field has signature $(0, 1)$ , so its unit rank is $0$ and hence $O_{K}^{\times} = μ (K)$ , a finite group.

1.4F1F3F5F6F7algebra

Conversely, rank $0$ gives $r_{1} + r_{2} = 1$ . If $r_{2} = 0$ , then $[K : Q] = 1$ and $K = Q$ . Otherwise $(r_{1}, r_{2}) = (0, 1)$ and $[K : Q] = 2$ . Choose $α \in K ∖ Q$ . The degree $[Q (α) : Q]$ divides $2$ and is not $1$ , so $K = Q (α)$ and the monic minimal polynomial is $X^{2} + b X + c$ with $b, c \in Q$ . Thus $β : = 2 α + b$ satisfies $β^{2} = Δ : = b^{2} - 4 c$ , where $Δ \neq 0$ is not a rational square, since otherwise $α$ would be rational. Factoring the numerator and denominator of $Δ$ into primes and removing even exponents gives $Δ = q^{2} d$ with $q \in Q^{\times}$ and $d \neq 1$ a nonzero squarefree integer. Hence $K = Q (β / q) = Q (\sqrt{d})$ . Since $K$ has no real embedding, [F5] forces $d < 0$ , so $K$ is imaginary quadratic.

1.5F1F2F4F5

For a real quadratic field, signature $(2, 0)$ gives rank $1$ and $O_{K}^{\times} ≅ μ (K) \times Z$ ; moreover every $ζ \in μ (K)$ lies in the image of one of the two real embeddings, so $ζ$ is a real root of unity and hence $ζ = \pm 1$ by [F4]; thus $μ (K) = {\pm 1}$ and $O_{K}^{\times} ≅ {\pm 1} \times Z$ .

1.6F1F2

The rank formula $r_{1} + r_{2} - 1$ applies to every signature; a field of signature $(1, 1)$ , such as a complex cubic field, also has rank $1$ , so the real quadratic case is a computed example of rank one rather than a characterisation of it.

2.1A1F2∎

Choice accounting: the corollary uses only the AC-qualified unit theorem [F2]; the signature enumeration and the real-roots-of-unity computation are choice-free.

DefinitionDefinition: Literature-sourcedProof: Not applicableprecheck passaudited 2026-10-02Open item page →

S-integers and S-units of a number field

Definition

Let $K$ be a number field (Number field), and let $S$ be a finite set of nonzero prime ideals of $O_{K}$ (Ring of integers); only finite primes belong to $S$ . Write $v_{p}$ for the prime-ideal valuation of a nonzero fractional ideal (Prime-ideal valuations on fractional ideals) and, for $x \in K^{\times}$ , write $v_{p} (x) : = v_{p} ((x))$ for the principal fractional ideal $(x)$ (Fractional ideals).

The ring of $S$ -integers of $K$ is

$O_{K, S} = {0} \cup {x \in K^{\times} : v_{p} (x) \geq 0 for every nonzero prime p \notin S},$

and the group of $S$ -units is

$O_{K, S}^{\times} = {x \in K^{\times} : v_{p} (x) = 0 for every nonzero prime p \notin S}$

with the group structure inherited from $K^{\times}$ (The units of a ring are the invertible elements of its multiplicative monoid, and $R^{\times}$ is a group under multiplication; $0 \in R^{\times}$ only in the zero ring).

No infinite place belongs to $S$ . Only nonzero prime ideals, that is finite places, are admitted into $S$ ; a real or complex place is never a member. The archimedean places are already carried by the logarithmic embedding, so admitting them into $S$ would count them twice. Consequently the rank formula of the $S$ -unit theorem is $r_{1} + r_{2} - 1 + ∣ S ∣$ , with $∣ S ∣$ the number of finite primes chosen.

The description via the fractional ideal is the one used below. $O_{K, S}$ consists of $0$ and the nonzero elements of $K$ whose principal fractional ideal involves no prime outside $S$ in a denominator, and an element $x$ of $O_{K, S}^{\times}$ is precisely an element of $K^{\times}$ whose principal fractional ideal involves no prime outside $S$ at all, in numerator or denominator. This intrinsic description, and not a localisation, is the one the S-unit theorem consumes; it also makes visible why $S$ is a set of finite primes only.

Well-definedness

Put $R = O_{K}$ . By Clearing denominators for an algebraic number, every $x \in K$ has $m x \in R$ for a positive integer $m$ , so $K = Frac (R)$ . For $x \neq 0$ , $(x) = x R$ is consequently a nonzero fractional ideal: it is an $R$ -submodule of $K$ and $m (x) \subseteq R$ .

The Dedekind property can also be established without Choice. By The ring of integers has rank the degree, $R ≅ Z^{n}$ additively, where $n = [K : Q] \geq 1$ . Every subgroup of $Z^{n}$ has a finite integer basis (the finite induction in that theorem, steps 1.2, 2.3 and 3.2). Thus every ideal of $R$ is finitely generated over $R$ , so $R$ is Noetherian. It is an integrally closed domain by The integral closure of a domain in a field extension is integrally closed. For every nonzero prime $p$ , $R / p$ is a finite domain by A nonzero number-field ideal has finite quotient, hence a field: multiplication by any nonzero element is injective on this finite set and therefore surjective. Thus every nonzero prime is maximal. Moreover $R / 2 R$ has $2^{n} > 1$ elements; among its proper ideals one of largest cardinality is maximal, and its inverse image is a nonzero prime of $R$ . Together with the prime $(0)$ this proves $\dim R = 1$ , so $R$ is Dedekind (Dedekind domains).

To justify the local valuation without the Choice-qualified general invertibility theorem, use the local calculation in Integral ideal factorisation in a number field, in ZF, steps 2.1--6.1, with the nonzero integral ideal $a = p$ . It proves that $R_{p}$ has maximal ideal $(π)$ and each nonzero element is uniquely $u π^{k}$ , with $u$ a unit and $k \geq 0$ . Its fraction field is $K$ , so each $x \in K^{\times}$ is uniquely $u π^{j}$ with $j \in Z$ . Hence $(x)_{p} = π^{j} R_{p}$ , exactly the valuation used in the Definition. Changing $π$ by a unit does not change $j$ . Multiplication adds these exponents, inversion negates them, and $v_{p} (x) \geq 0$ is equivalent to $x \in R_{p}$ . Consequently $O_{K, S}$ is the intersection of these local subrings over $p \notin S$ , and $x$ is a unit of that intersection precisely when both $x$ and $x^{- 1}$ belong to it, equivalently all those exponents vanish. This verifies the ring and group assertions without selecting uniformizers simultaneously; the construction uses no Choice.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-10-02Open item page →

S-unit theorem

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $K$ be a number field of signature $(r_{1}, r_{2})$ and let $S$ be a finite set of nonzero prime ideals of $O_{K}$ . Then $O_{K, S}^{\times} ≅ μ (K) \times Z^{r_{1} + r_{2} - 1 + ∣ S ∣}$ ; in particular the $S$ -unit group is finitely generated of rank $r_{1} + r_{2} - 1 + ∣ S ∣$ with torsion subgroup $μ (K)$ .

Facts & Assumptions

Given: The Axiom of Choice, a number field $K$ of signature $(r_{1}, r_{2})$ , and a finite set $S$ of nonzero prime ideals of $O_{K}$ with $G : = O_{K, S}^{\times}$ (S-integers and S-units of a number field).

[F1]

$G = {x \in K^{\times} : v_{p} (x) = 0$ for every nonzero prime $p \notin S}$ , where $v_{p} (x) : = v_{p} ((x))$ for the principal fractional ideal $(x)$ ; every $v_{p} (x)$ is an integer (S-integers and S-units of a number field, Prime-ideal valuations on fractional ideals).

[F2]

The unit group satisfies $O_{K}^{\times} ≅ μ (K) \times Z^{r_{1} + r_{2} - 1}$ , is finitely generated of rank $r_{1} + r_{2} - 1$ , and has torsion subgroup $μ (K)$ (Dirichlet unit theorem).

[F3]

The ideal class group $Cl (O_{K})$ is finite of order $h \geq 1$ , and the class of a nonzero fractional ideal is its image under $cl$ (Finiteness of the number-field class group, The ideal class group, The ideal class group quotient is well defined).

[F4]

The principal-divisor sequence $0 \to O_{K}^{\times} \to K^{\times} \overset{div}{\to} Div (O_{K}) \overset{cl}{\to} Cl (O_{K}) \to 0$ is exact, where $div (x) = \sum_{p} v_{p} ((x)) [p]$ ; thus $\ker div = O_{K}^{\times}$ and the principal divisors are exactly the kernel of $cl$ (The principal-divisor exact sequence for a Dedekind domain).

[F5]

Every nonzero fractional ideal has the unique factorisation $I = \prod_{p} p^{v_{p} (I)}$ , and $v_{p} (I J) = v_{p} (I) + v_{p} (J)$ (Unique factorization of nonzero fractional ideals into prime powers, Prime-ideal valuations of a fractional ideal have finite support and add under products); in particular $v_{p} (p) = 1$ and $v_{q} (p) = 0$ for $q \neq p$ .

[F6]

Every subgroup of $Z^{n}$ is free of rank at most $n$ , and a finitely generated abelian group has an intrinsic free rank and finite torsion subgroup (Integer abelian structure and rank by finite reduction).

[F7]

An element of $K^{\times}$ of finite order is a root of unity, and every root of unity in $K$ lies in $O_{K}^{\times}$ ; thus $μ (K)$ is exactly the torsion subgroup of $O_{K}^{\times}$ and is finite (The group $μ_{n} (K)$ of $n$ -th roots of unity in a field, and primitive $n$ -th roots of unity, Dirichlet unit theorem).

[F8]

If $G$ is a finite group of order $h$ , then $g^{h} = 1$ for every $g \in G$ (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

Proof

technique · measure an $S$-unit by its valuations at the primes in $S$. The kernel of that map is the ordinary unit group, and the image has finite index in $\mathbb Z^S$ because taking $h$-th powers of the prime ideals turns them principal; the image is therefore free of rank $|S|$, so the surjection onto it splits and exposes $G$ as $\mu(K)$ times a free abelian group of rank $r_1+r_2-1+|S|$

1.1F1F5

Define $φ : G \to Z^{S}$ by $φ (u) = (v_{p} (u))_{p \in S}$ ; it is a group homomorphism because $v_{p} (x y) = v_{p} (x) + v_{p} (y)$ for all nonzero fractional ideals.

1.2F1F4

The kernel of $φ$ is $O_{K}^{\times}$ : a unit $u \in G$ has $φ (u) = 0$ exactly when $v_{p} (u) = 0$ for every $p \in S$ , and by definition $v_{p} (u) = 0$ for every $p \notin S$ , so $φ (u) = 0$ exactly when $div (u) = 0$ , that is $u \in \ker div = O_{K}^{\times}$ .

1.3F3F4F8

Put $h : = ∣ Cl (O_{K}) ∣ \geq 1$ , a finite number; for $p \in S$ let $[p]$ be its class, so $[p]^{h} = 1$ by [F8], and hence the divisor $h [p]$ lies in the kernel of $cl$ , i.e. is principal: there is $π_{p} \in K^{\times}$ with $(π_{p}) = p^{h}$ ; since $p^{h}$ is an integral ideal, $π_{p} \in O_{K} ∖ {0}$ .

2.1F5step 1.3

For $p, q \in S$ one has $v_{q} (π_{p}) = v_{q} (p^{h}) = h v_{q} (p) = h δ_{p q}$ , using additivity of valuations and $v_{q} (p) = 1$ for $q = p$ and $0$ otherwise.

3.1step 1.1step 2.1

Consequently $h Z^{S} \subseteq φ (G)$ , because $φ (π_{p}) = h e_{p}$ for each $p \in S$ and the elements $h e_{p}$ generate $h Z^{S}$ .

4.1F6step 3.1

The image $M : = φ (G)$ is a subgroup of $Z^{S}$ , hence free of rank $s \leq ∣ S ∣$ ; since $h Z^{S}$ is a subgroup of $M$ isomorphic to $Z^{∣ S ∣}$ , the same rank bound gives $∣ S ∣ \leq s$ , so $s = ∣ S ∣$ and $M ≅ Z^{∣ S ∣}$ ; also $M$ has finite index in $Z^{S}$ , because it contains $h Z^{S}$ .

5.1step 1.2step 4.1

Since $M$ is free abelian, the surjection $φ : G ↠ M$ splits: choose a $Z$ -basis $m_{1}, \dots, m_{∣ S ∣}$ of $M$ and preimages $u_{1}, \dots, u_{∣ S ∣} \in G$ with $φ (u_{i}) = m_{i}$ , and define the homomorphism $σ : M \to G$ by $σ (m_{i}) = u_{i}$ ; then $φ \circ σ = {i d}_{M}$ , and every $g \in G$ is written as $g = σ (φ (g)) \cdot (g σ (φ (g))^{- 1})$ with $g σ (φ (g))^{- 1} \in \ker φ$ and $σ (φ (g)) \in im σ$ , while $\ker φ \cap im σ = {1}$ because $φ (σ (m)) = m$ ; hence $G ≅ \ker φ \times M$ .

6.1F2step 5.1

By [F2] and step 5.1, $G ≅ (μ (K) \times Z^{r_{1} + r_{2} - 1}) \times Z^{∣ S ∣} ≅ μ (K) \times Z^{r_{1} + r_{2} - 1 + ∣ S ∣}$ , so $G$ is finitely generated, its free rank is $r_{1} + r_{2} - 1 + ∣ S ∣$ , and the torsion subgroup of $G$ is the torsion subgroup of $μ (K) \times Z^{r_{1} + r_{2} - 1 + ∣ S ∣}$ , namely $μ (K)$ .

7.1F2F7step 6.1

Directly: if $g \in G \subseteq K^{\times}$ has finite order then $g$ is a root of unity and so lies in $μ (K)$ ; conversely every root of unity lies in $μ (K) \subseteq O_{K}^{\times} \subseteq G$ ; hence the torsion subgroup of $G$ is $μ (K)$ , finite, in agreement with step 6.1.

8.1F2F3F4F5step 1.3step 5.1∎

Choice accounting: the unit theorem [F2], class-group finiteness input [F3], principal-divisor exact sequence [F4], and ideal-factorisation and valuation inputs [F5] assume AC. The selections performed here are finitely many (the elements $π_{p}$ for $p \in S$ and the lifts $u_{i}$ of a basis of $M$ ), so these selections need only finite choice; the splitting is noncanonical because it depends on those lifts.

5 · Examples, counterexamples and false statements

None yet.

Dirichlets Unit Theorem Regulators and S Units

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

Finitely many roots of unity in a number field

Statement

Facts & Assumptions

Proof

Kronecker root-of-unity criterion

Statement

Facts & Assumptions

Proof

A number-field unit is exactly an algebraic integer of norm plus or minus one

Statement

Facts & Assumptions

Proof

Product formula for a number field

Statement

Facts & Assumptions

Proof

Logarithmic embedding of a number field

Definition

Unit logarithms lie in the trace-zero hyperplane

Statement

Facts & Assumptions

Proof

Kernel of the unit logarithm is the roots of unity

Statement

Facts & Assumptions

Proof

Discrete subgroups of a real vector space are lattices

Statement

Facts & Assumptions

Proof

The logarithmic unit image is discrete

Statement

Facts & Assumptions

Proof

The logarithmic unit image is a full lattice

Statement

Facts & Assumptions

Proof

Dirichlet unit theorem

Statement

Facts & Assumptions

Proof

System of fundamental units

Definition

Regulator of a number field

Definition

Deleted-row minors of a zero-column-sum matrix agree up to sign

Statement

Facts & Assumptions

Proof

The regulator is well defined

Statement

Facts & Assumptions

Proof

Unit ranks by signature

Statement

Facts & Assumptions

Proof

S-integers and S-units of a number field

Definition

Well-definedness

S-unit theorem

Statement

Facts & Assumptions

Proof

5 · Examples, counterexamples and false statements

Sources