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Dirichlets Unit Theorem Regulators and S Units — Examples

1 · Prerequisites

2 · Summary

The examples compute the unit group or the regulator in the three rank patterns the theorem allows. Rank zero is $Q$ and the imaginary quadratic fields: the groups ${\pm 1}$ , ${\pm 1, \pm i}$ and ${\pm 1, \pm ω, \pm ω^{2}}$ are found by solving the norm equation $N (u) = \pm 1$ with the quadratic norm formula. Rank one is the real quadratic field $Q (\sqrt{d})$ , where the units of the maximal order are governed by the negative Pell equation through the period of the continued fraction of $\sqrt{d}$ , the fundamental unit is $(1 + \sqrt{5}) / 2$ for $d = 5$ , and $R_{K} = \log ε$ . Rank two is the totally real cubic field of $2 \cos (2 π / 9)$ : the elements $α$ and $α - 1$ have norm $- 1$ and $1$ , their logarithmic vectors are independent, and the two independent units pin down the full rank.

Two computations make the regulator's conventions and invariance visible. The deleted-row minors of the cubic logarithmic matrix all have absolute value $0.849287 \dots$ , and the unimodular tuple $(α (α - 1), α - 1)$ reproduces them exactly, illustrating the ${G L}_{2} (Z)$ step of the well-definedness theorem; replacing the fundamental unit $(1 + \sqrt{5}) / 2$ of $Q (\sqrt{5})$ by the Pell generator $9 + 4 \sqrt{5}$ of the order $Z [\sqrt{5}]$ would multiply the regulator by six. The page ends with the $S$ -units of $Q$ , equal to ${\pm p_{1}^{n_{1}} \dots p_{m}^{n_{m}}}$ of rank $m$ , and with the counterexample showing that $Z [\sqrt{5}]^{\times} = \pm ⟨ (2 + \sqrt{5}) ⟩$ is a subgroup of index three in $O_{Q (\sqrt{5})}^{\times}$ .

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

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Units of $Q$ and the imaginary quadratic fields

Example

Assume the Axiom of Choice. For $K = Q$ one has $O_{K} = Z$ and $O_{K}^{\times} = {\pm 1}$ . For $K = Q (i)$ one has $O_{K} = Z [i]$ and $O_{K}^{\times} = {\pm 1, \pm i} = μ_{4} (K)$ . For $K = Q (\sqrt{- 3})$ , with $ω = (1 + \sqrt{- 3}) / 2$ , one has $O_{K} = Z [(1 + \sqrt{- 3}) / 2] = Z [ω]$ and $O_{K}^{\times} = μ_{6} (K) = {\pm 1, \pm ω, \pm ω^{2}}$ . All three unit groups are finite of rank $0$ .

Facts & Assumptions

Given: The Axiom of Choice and the three number fields $Q$ , $Q (i) = Q (\sqrt{- 1})$ and $Q (\sqrt{- 3})$ , with the element $ω : = (1 + \sqrt{- 3}) / 2$ (number field).

[F1]

The ring of integers $O_{Q}$ is the integral closure of $Z$ in $Q$ (Ring of integers); a rational number integral over $Z$ is an integer (The rational algebraic integers are exactly the integers), and conversely every integer $n$ is a root of the monic polynomial $X - n$ . Hence $O_{Q} = Z$ .

[F2]

For squarefree $d \neq 1$ one has $O_{Q (\sqrt{d})} = Z [(1 + \sqrt{d}) / 2]$ if $d \equiv 1 (m o d 4)$ and $O_{Q (\sqrt{d})} = Z [\sqrt{d}]$ otherwise (Integers in a quadratic field). Since $- 1 \equiv 3 (m o d 4)$ and $- 3 \equiv 1 (m o d 4)$ , this gives $O_{Q (i)} = Z [\sqrt{- 1}] = Z [i]$ and $O_{Q (\sqrt{- 3})} = Z [(1 + \sqrt{- 3}) / 2] = Z [ω]$ .

[F3]

For a number field $K$ and $u \in O_{K}$ , the element $u$ is a unit of $O_{K}$ 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]

Let $d \in Q$ be nonsquare and put $K = Q (\sqrt{d})$ . A degree- $2$ polynomial over $Q$ is irreducible exactly when it has no rational root (A polynomial of degree two or three over a field is irreducible exactly when it has no root in the field), and $d$ is not a square in $Q$ , so $X^{2} - d$ is the minimal polynomial of $\sqrt{d}$ over $Q$ (The evaluation kernel and the unique monic irreducible minimal polynomial of an algebraic element) and $[K : Q] = 2$ . By the correspondence between $F$ -embeddings and distinct roots of the minimal polynomial ( $F$ -embeddings of $F (α)$ into an algebraically closed field correspond to the distinct roots of $m_{α}$ ), the two $Q$ -embeddings of $K$ into $C$ send $\sqrt{d}$ to $\sqrt{d}$ and to $- \sqrt{d}$ ; since $char Q = 0$ , norm and trace are the product and sum over these embeddings (Norm and trace from embeddings, with the inseparable exponent in the norm formula). Hence for $α = a + b \sqrt{d}$ with $a, b \in Q$ , $N_{K / Q} (α) = (a + b \sqrt{d}) (a - b \sqrt{d}) = a^{2} - d b^{2} .$

[F5]

The units of $Z$ are exactly $1$ 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 $u = - 1$ ).

[F6]

For $n \geq 1$ , $μ_{n} (K) = {x \in K : x^{n} = 1}$ is the set of $n$ -th roots of unity in $K$ , and $μ (K)$ denotes the group of all roots of unity in $K$ , the union of the subgroups $μ_{n} (K)$ (The group $μ_{n} (K)$ of $n$ -th roots of unity in a field, and primitive $n$ -th roots of unity).

[F7]

If $x \in K$ satisfies $x^{n} = 1$ for some $n \geq 1$ , then $x$ is a root of the monic polynomial $T^{n} - 1 \in Z [T]$ , hence is integral over $Z$ (Integral elements over a commutative ring and algebraic integers) and lies in $O_{K}$ , the integral closure of $Z$ in $K$ (Ring of integers); also $x^{- 1} = x^{n - 1} \in O_{K}$ , so $x \in O_{K}^{\times}$ . In particular every root of unity in $K$ is a unit of $O_{K}$ , that is $μ (K) \subseteq O_{K}^{\times}$ .

[F8]

The unit rank $r_{1} + r_{2} - 1$ is $0$ exactly for $Q$ and for imaginary quadratic fields, and in the rank-zero cases $O_{K}^{\times} = μ (K)$ is finite (Unit ranks by signature); a quadratic field $Q (\sqrt{d})$ with $d < 0$ has no real embedding and two complex conjugate embeddings, hence signature $(0, 1)$ (Archimedean embeddings and signature).

[A1]

The Axiom of Choice is assumed; it is used only through the rank-zero unit structure [F8] (The Axiom of Choice).

Verification

technique · compute the unit group of each of the three rings of integers by solving the norm equation $N(u)=\pm1$ with the elementary norm formula for quadratic fields, and identify the results with the groups of fourth and sixth roots of unity

1.1F1F5F6F7

$O_{Q} = Z$ and $O_{Q}^{\times} = Z^{\times} = {\pm 1}$ : the rational units are the units of $Z$ , and $\pm 1$ are roots of unity while every element of $μ (Q)$ is a unit of $Z$ by [F7], so $μ (Q) = {\pm 1}$ as well.

1.2F2

By [F2] and the congruences $- 1 \equiv 3 (m o d 4)$ , $- 3 \equiv 1 (m o d 4)$ one has $O_{Q (i)} = Z [i] = {a + b i : a, b \in Z}$ and $O_{Q (\sqrt{- 3})} = Z [ω] = {a + b ω : a, b \in Z}$ , where $ω = (1 + \sqrt{- 3}) / 2$ .

1.3F4

Applying [F4] with the nonsquare rational $d = - 1$ to $a + b i = a + b \sqrt{- 1}$ with $a, b \in Z \subseteq Q$ gives $N_{Q (i) / Q} (a + b i) = a^{2} + b^{2}$ .

1.4F4algebra

Writing $a + b ω = (a + b / 2) + (b / 2) \sqrt{- 3}$ with $a, b \in Z$ and applying [F4] with the nonsquare rational $d = - 3$ gives $N_{Q (\sqrt{- 3}) / Q} (a + b ω) = (a + b / 2)^{2} + 3 (b / 2)^{2} = a^{2} + a b + b^{2}$ .

2.1F3step 1.3algebra

By [F3] and step 1.3, $a + b i \in Z [i]$ is a unit if and only if $a^{2} + b^{2} = \pm 1$ . Since $a^{2} + b^{2} \geq 0$ , this is the equation $a^{2} + b^{2} = 1$ with $a, b \in Z$ ; then $a^{2} \leq 1$ , $∣ a ∣ \leq 1$ , and $b^{2} = 1 - a^{2}$ , so either $a = 0$ and $b = \pm 1$ or $b = 0$ and $a = \pm 1$ . Hence $O_{Q (i)}^{\times} = {1, - 1, i, - i}$ .

2.2F3step 1.4algebra

By [F3] and step 1.4, $a + b ω \in Z [ω]$ is a unit if and only if $a^{2} + a b + b^{2} = \pm 1$ . Since $4 (a^{2} + a b + b^{2}) = (2 a + b)^{2} + 3 b^{2} \geq 0$ , the value $- 1$ cannot occur, and $a^{2} + a b + b^{2} = 1$ is equivalent to $(2 a + b)^{2} + 3 b^{2} = 4$ ; then $3 b^{2} \leq 4$ with $b \in Z$ forces $b \in {- 1, 0, 1}$ . If $b = 0$ then $(2 a)^{2} = 4$ gives $a = \pm 1$ ; if $b = 1$ then $(2 a + 1)^{2} = 1$ gives $a = 0$ or $a = - 1$ ; if $b = - 1$ then $(2 a - 1)^{2} = 1$ gives $a = 1$ or $a = 0$ .

3.1step 2.2algebra

Direct computation in $Z [ω]$ gives $ω^{2} = \frac{- 1 + \sqrt{- 3}}{2} = ω - 1$ and $ω^{3} = ω \cdot ω^{2} = ω^{2} - ω = (ω - 1) - ω = - 1$ , hence $ω^{6} = 1$ . Since $1, ω, ω - 1, - 1$ have pairwise different coordinates in the $Z$ -basis $(1, ω)$ of $Z [ω]$ , the elements $ω, ω^{2}, ω - 1$ are all different from $1$ , and the six units found in step 2.2 are exactly $ω^{0}, ω^{1}, ω^{2}, ω^{3} = - 1, ω^{4} = - ω, ω^{5} = - (ω - 1)$ .

3.2F6F7step 2.1

Each of the four elements $1, - 1, i, - i$ of $O_{Q (i)}^{\times}$ satisfies $x^{4} = 1$ , so $O_{Q (i)}^{\times} \subseteq μ_{4} (Q (i))$ by [F6], while $μ_{4} (Q (i)) \subseteq O_{Q (i)}^{\times}$ by [F7]; hence $O_{Q (i)}^{\times} = μ_{4} (Q (i)) = {\pm 1, \pm i}$ .

4.1F6F7step 2.2step 3.1

Each of the six units found in step 2.2 is a power of $ω$ by step 3.1, hence satisfies $x^{6} = 1$ ; therefore $O_{Q (\sqrt{- 3})}^{\times} \subseteq μ_{6} (Q (\sqrt{- 3}))$ by [F6], and $μ_{6} (Q (\sqrt{- 3})) \subseteq O_{Q (\sqrt{- 3})}^{\times}$ by [F7]; hence $O_{Q (\sqrt{- 3})}^{\times} = μ_{6} (Q (\sqrt{- 3})) = {\pm 1, \pm ω, \pm ω^{2}}$ , because $ω^{3} = - 1$ gives ${ω^{0}, \dots, ω^{5}} = {\pm 1, \pm ω, \pm ω^{2}}$ .

5.1F8step 1.1step 3.2step 4.1

Finally $Q$ has signature $(1, 0)$ and both imaginary quadratic fields have signature $(0, 1)$ , so [F8] gives unit rank $r_{1} + r_{2} - 1 = 0$ for all three fields and exhibits the unit groups as finite torsion groups $μ (K)$ ; the computed groups ${\pm 1}$ , ${\pm 1, \pm i}$ and ${\pm 1, \pm ω, \pm ω^{2}}$ have $2$ , $4$ and $6$ elements.

6.1A1F8∎

Choice accounting: AC is used only through the rank-zero structure [F8]; the norm computations, the enumeration of the norm-one solutions, and the powers of $ω$ are elementary computations in $Z$ and $Z [ω]$ and use no choice.

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Real quadratic units and Pell's equation

Example

Assume the Axiom of Choice. Let $d > 1$ be squarefree and $K = Q (\sqrt{d})$ . If $d \equiv 2, 3 (m o d 4)$ then $O_{K} = Z [\sqrt{d}]$ , and the unit group is $\pm ⟨ u_{d} ⟩$ , where $u_{d}$ is the least positive solution of the negative Pell equation $x^{2} - d y^{2} = - 1$ when that equation is solvable, and $u_{d} = ε_{d}$ is the fundamental Pell solution of $x^{2} - d y^{2} = 1$ otherwise; in the solvable case $ε_{d} = u_{d}^{2}$ and the norm-one Pell subgroup $\pm ⟨ ε_{d} ⟩$ has index $2$ in $O_{K}^{\times}$ . If $d \equiv 1 (m o d 4)$ then $O_{K} = Z [(1 + \sqrt{d}) / 2]$ strictly contains $Z [\sqrt{d}]$ ; the unit group of the order $Z [\sqrt{d}]$ is as above, while the fundamental unit of $O_{K}$ may be a half-integer element solving $x^{2} - d y^{2} = \pm 4$ that does not lie in $Z [\sqrt{d}]$ , in which case $Z [\sqrt{d}]^{\times}$ (in particular the norm-one Pell subgroup $\pm ⟨ ε_{d} ⟩$ ) is a proper subgroup of $O_{K}^{\times}$ . For $d = 5$ the fundamental unit of $O_{K}$ is $ε = (1 + \sqrt{5}) / 2$ (norm $- 1$ ), so $O_{K}^{\times} = {\pm ε^{n} : n \in Z}$ , while in $Z [\sqrt{5}]$ one has $2 + \sqrt{5} = ε^{3}$ , the fundamental Pell solution is $ε_{5} = 9 + 4 \sqrt{5} = ε^{6}$ , and $Z [\sqrt{5}]^{\times} = \pm ⟨ 2 + \sqrt{5} ⟩ = \pm ⟨ ε^{3} ⟩$ has index $3$ in $O_{K}^{\times}$ .

Facts & Assumptions

Given: The Axiom of Choice, a squarefree integer $d > 1$ , the field $K = Q (\sqrt{d})$ , the order $Z [\sqrt{d}]$ with its Pell norm $N_{d}$ (The norm on the explicit order $Z [\sqrt{D}]$ ), the fundamental Pell solution $ε_{d}$ of $x^{2} - d y^{2} = 1$ (The fundamental Pell solution), and, when the negative Pell equation $x^{2} - d y^{2} = - 1$ is solvable, its least positive solution $u_{d}$ .

[F1]

If $d \equiv 2, 3 (m o d 4)$ then $O_{K} = Z [\sqrt{d}]$ , and if $d \equiv 1 (m o d 4)$ then $O_{K} = Z [(1 + \sqrt{d}) / 2]$ , which strictly contains $Z [\sqrt{d}]$ (Integers in a quadratic field).

[F2]

For $u \in O_{K}$ , $u$ is a unit of $O_{K}$ if and only if $N_{K / Q} (u) = \pm 1$ , and for a real quadratic field the norm of $x + y \sqrt{d}$ is $x^{2} - d y^{2}$ (A number-field unit is exactly an algebraic integer of norm plus or minus one, Integers in a quadratic field).

[F3]

The Pell norm is multiplicative: $N_{d} (α β) = N_{d} (α) N_{d} (β)$ for $α, β \in Z [\sqrt{d}]$ (The Pell norm is multiplicative); consequently an element $α \in Z [\sqrt{d}]$ is a unit of the order $Z [\sqrt{d}]$ if and only if $N_{d} (α) = \pm 1$ (The norm on the explicit order $Z [\sqrt{D}]$ , 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).

[F4]

The integral solutions of $x^{2} - d y^{2} = 1$ are exactly the elements $\pm ε_{d}^{k}$ with $k \in Z$ , and the positive solutions are $ε_{d}^{k}$ with $k \geq 1$ ; also $ε_{d} > 1$ (All integral Pell solutions are $\pm ε_{D}^{k}$ , All positive Pell solutions are powers of the fundamental solution, The fundamental Pell solution). The equation $x^{2} - d y^{2} = 1$ has a positive nontrivial solution (Every Pell equation has a positive nontrivial integral solution).

[F5]

The negative Pell equation $x^{2} - d y^{2} = - 1$ is solvable if and only if the period length $ℓ$ of the continued fraction of $\sqrt{d}$ is odd; when it is solvable, the numerator-denominator pair $(p_{ℓ - 1}, q_{ℓ - 1})$ gives the least positive solution, and the least positive solution of $x^{2} - d y^{2} = 1$ is $(p_{2 ℓ - 1}, q_{2 ℓ - 1})$ (Negative Pell is soluble exactly for odd period length, Generalized and negative Pell equations).

[F6]

For a real quadratic field $K$ , $O_{K}^{\times} ≅ {\pm 1} \times Z$ ; in particular there is a unit $γ > 1$ with $O_{K}^{\times} = {\pm γ^{n} : n \in Z}$ (Unit ranks by signature).

[F7]

For $d = 5$ : $2 < \sqrt{5} < 3$ gives $⌊ \sqrt{5} ⌋ = 2$ , and $\sqrt{5} + 2 = 4 + \frac{1}{\sqrt{5} + 2}$ , so the complete quotients of $\sqrt{5}$ are $α_{1} = \sqrt{5} + 2$ and $α_{2} = α_{1}$ ; hence the digit sequence is $2, 4, 4, 4, \dots$ , that is $\sqrt{5} = [2; \overline{4}]$ , with period length $ℓ = 1$ (Complete quotients in the continued-fraction algorithm, Finite and infinite regular continued fractions, Eventually periodic regular continued fractions). The convergent recurrence gives $p_{0} / q_{0} = 2 / 1$ and $p_{1} / q_{1} = 9 / 4$ (Convergents of a regular continued fraction), and direct computation gives $2^{2} - 5 \cdot 1^{2} = - 1$ , $9^{2} - 5 \cdot 4^{2} = 1$ and $(2 + \sqrt{5})^{2} = 9 + 4 \sqrt{5}$ .

[A1]

The Axiom of Choice is assumed; it is used only through the rank-one structure of $O_{K}^{\times}$ in [F6]; the Pell solution theory quoted above is choice-free (The Axiom of Choice).

Verification

technique · identify units with Pell-type solutions of norm $\pm1$; in the solvable case compare the least negative solution with the fundamental positive solution to prove $\varepsilon_d=u_d^2$, and for $d=5$ enumerate the norm-$\pm4$ units of the maximal order and compare with the order's Pell units

1.1F1

The ring-of-integers formula gives $O_{K} = Z [\sqrt{d}]$ when $d \equiv 2, 3 (m o d 4)$ and $O_{K} = Z [(1 + \sqrt{d}) / 2] ⊋ Z [\sqrt{d}]$ when $d \equiv 1 (m o d 4)$ .

1.2F3

An element $α = x + y \sqrt{d} \in Z [\sqrt{d}]$ is a unit of the order $Z [\sqrt{d}]$ if and only if $N_{d} (α) = x^{2} - d y^{2} = \pm 1$ : if $N_{d} (α) = \pm 1$ then $α^{- 1} = \pm (x - y \sqrt{d}) \in Z [\sqrt{d}]$ , while for a unit $α$ one has $N_{d} (α) N_{d} (α^{- 1}) = N_{d} (1) = 1$ with both factors in $Z$ , so $N_{d} (α) = \pm 1$ .

1.3F2F3F4

Seen in $K$ , every unit of $Z [\sqrt{d}]$ has norm $\pm 1$ (its Pell norm is the field norm), and conversely a norm- $\pm 1$ element of $Z [\sqrt{d}]$ is a unit; thus the units of the order are exactly the integral solutions of $x^{2} - d y^{2} = \pm 1$ , with the norm-one solutions forming $\pm ⟨ ε_{d} ⟩$ .

1.4F1F2algebra

Take $d = 5$ and $ε : = (1 + \sqrt{5}) / 2 \in O_{K}$ . Its norm is $((1 + \sqrt{5}) / 2) ((1 - \sqrt{5}) / 2) = (1 - 5) / 4 = - 1$ , so $ε \in O_{K}^{\times}$ ; direct multiplication gives $ε^{2} = (3 + \sqrt{5}) / 2$ and $ε^{3} = 2 + \sqrt{5}$ .

2.1F4F5step 1.3

Suppose first that $x^{2} - d y^{2} = - 1$ is solvable, and let $u_{d} = x_{0} + y_{0} \sqrt{d}$ be its least positive solution, whose coordinates are the numerator and denominator of the convergent in [F5]. Then $u_{d}^{2}$ is a positive solution of $x^{2} - d y^{2} = 1$ , so $u_{d}^{2} = ε_{d}^{n}$ for a unique integer $n \geq 1$ .

2.2F1F2step 1.2

For $d \equiv 1 (m o d 4)$ the elements of $O_{K} = Z [(1 + \sqrt{d}) / 2]$ are the numbers $(x + y \sqrt{d}) / 2$ with $x \equiv y (m o d 2)$ , of norm $(x^{2} - d y^{2}) / 4$ ; such an element is a unit exactly when $x^{2} - d y^{2} = \pm 4$ , and it fails to lie in $Z [\sqrt{d}]$ exactly when $x$ and $y$ are both odd.

3.1F3step 2.1algebra

The exponent $n$ is odd: if $n = 2 m$ , then $u_{d}^{2} = (ε_{d}^{m})^{2}$ in the domain $Z [\sqrt{d}]$ , so $u_{d} = \pm ε_{d}^{m}$ and taking norms gives $N (u_{d}) = N (ε_{d}^{m}) = + 1$ , contradicting $N (u_{d}) = - 1$ .

3.2step 2.2step 1.4algebra

The units of $O_{K}$ for $d = 5$ are the elements $(x + y \sqrt{5}) / 2$ with $x \equiv y (m o d 2)$ and $x^{2} - 5 y^{2} = \pm 4$ ; for a unit $u > 1$ one has $y > 0$ , and $x = u \pm 1 / u > 0$ according as $N (u) = \pm 1$ , so $x, y$ are positive integers. Checking the admissible positive pairs in order of $y$ : for $y = 1$ the equation gives $x^{2} = 5 \pm 4$ , so $x = 1$ (the unit $ε$ ) or $x = 3$ (the unit $ε^{2}$ ); for $y = 2$ it gives $x^{2} = 20 \pm 4$ , so $x = 4$ (the unit $2 + \sqrt{5} = ε^{3}$ ); for $y \geq 3$ one has $x^{2} \geq 5 \cdot 9 - 4 = 41$ , hence $x \geq 7$ and $(x + y \sqrt{5}) / 2 \geq (7 + 3 \sqrt{5}) / 2 > ε^{3} > ε$ . Therefore $ε$ is the least unit $> 1$ of $O_{K}$ .

4.1F3step 2.1step 3.1

In fact $n = 1$ . If $n \geq 3$ , write $n = 2 m + 1$ and put $w : = u_{d} ε_{d}^{- m}$ . Both $u_{d}$ and $ε_{d}$ are units of $Z [\sqrt{d}]$ by [F3], so $w$ is a unit of that order as well; in particular $w = x + y \sqrt{d}$ for integers $x, y$ . From step 2.1, $u_{d}^{2} = ε_{d}^{n}$ , and therefore $w^{2} = u_{d}^{2} ε_{d}^{- 2 m} = ε_{d}$ and $u_{d} = w ε_{d}^{m} = w (w^{2})^{m} = w^{n}$ . Since $u_{d} > 0$ and $ε_{d} > 1$ , the definition of $w$ gives $w > 0$ , and $w^{2} = ε_{d} > 1$ gives $w > 1$ . By [F3] the order norm $N_{d} (w)$ is $+ 1$ or $- 1$ . If $N_{d} (w) = + 1$ , multiplicativity in [F3] and $u_{d} = w^{n}$ give $N_{d} (u_{d}) = N_{d} (w)^{n} = + 1$ , contradicting $N_{d} (u_{d}) = - 1$ ; hence $N_{d} (w) = - 1$ . Thus $w = x + y \sqrt{d}$ is a positive integral solution of $x^{2} - d y^{2} = - 1$ : its conjugate is $- 1 / w$ , so $x = (w - 1 / w) / 2 > 0$ and $y = (w + 1 / w) / (2 \sqrt{d}) > 0$ . As $n \geq 3$ and $w > 1$ , $u_{d} = w^{n} > w$ ; moreover $x = (w - 1 / w) / 2 < (u_{d} - 1 / u_{d}) / 2$ , so this solution has smaller first coordinate than the least positive solution $u_{d}$ , a contradiction. Therefore $n = 1$ and $ε_{d} = u_{d}^{2}$ .

4.2F6step 3.2

Since $O_{K}^{\times} = {\pm γ^{n}}$ for some $γ > 1$ and the least unit $> 1$ in such a group is $γ$ , step 3.2 gives $γ = ε$ , so $O_{K}^{\times} = {\pm ε^{n} : n \in Z}$ .

5.1step 1.3step 4.1

Consequently, in the solvable case every norm-one unit is $\pm ε_{d}^{m} = \pm u_{d}^{2 m}$ and every norm- $(- 1)$ unit is $\pm u_{d}^{2 m + 1}$ , because multiplying it by $u_{d}^{- 1}$ gives norm $1$ ; thus $Z [\sqrt{d}]^{\times} = \pm ⟨ u_{d} ⟩$ , and the norm-one subgroup $\pm ⟨ ε_{d} ⟩ = \pm ⟨ u_{d}^{2} ⟩$ consists of the even powers of $u_{d}$ , of index $2$ .

6.1F3F4step 1.3step 5.1

If instead $x^{2} - d y^{2} = - 1$ is unsolvable, every unit of $Z [\sqrt{d}]$ has norm $+ 1$ , so $Z [\sqrt{d}]^{\times} = \pm ⟨ ε_{d} ⟩$ ; setting $u_{d} : = ε_{d}$ gives $Z [\sqrt{d}]^{\times} = \pm ⟨ u_{d} ⟩$ in both cases.

7.1F1step 6.1step 2.2

The order $Z [\sqrt{d}]$ is a subring of $O_{K}$ , so its unit group is a subgroup of $O_{K}^{\times}$ and equals the units computed in steps 5.1 and 6.1; if the fundamental unit of $O_{K}$ (its least unit $> 1$ ) is an element with $x, y$ both odd, then it is not in $Z [\sqrt{d}]$ , so $Z [\sqrt{d}]^{\times} ⊊ O_{K}^{\times}$ , and since $\pm ⟨ ε_{d} ⟩ \subseteq Z [\sqrt{d}]^{\times}$ also $\pm ⟨ ε_{d} ⟩ ⊊ O_{K}^{\times}$ .

7.2F5F7step 6.1step 1.4

In the order, [F5] applied with the period length $ℓ = 1$ computed in [F7] makes the pair $(p_{0}, q_{0}) = (2, 1)$ the least positive solution of $x^{2} - 5 y^{2} = - 1$ and the pair $(p_{1}, q_{1}) = (9, 4)$ the least positive solution of $x^{2} - 5 y^{2} = 1$ , whose associated element is the fundamental Pell unit $ε_{5} = 9 + 4 \sqrt{5}$ ; by steps 5.1 and 6.1 applied to $d = 5$ , $Z [\sqrt{5}]^{\times} = \pm ⟨ 2 + \sqrt{5} ⟩$ , and by step 1.4 this is $\pm ⟨ ε^{3} ⟩$ , while the identity $(2 + \sqrt{5})^{2} = 9 + 4 \sqrt{5}$ of [F7] gives $9 + 4 \sqrt{5} = ε^{6} = ε_{5}$ .

8.1step 6.1step 4.2step 7.2

Finally $Z [\sqrt{5}]^{\times} = \pm ⟨ ε^{3} ⟩ \subseteq \pm ⟨ ε ⟩ = O_{K}^{\times}$ with index $[⟨ ε ⟩ : ⟨ ε^{3} ⟩] = 3$ , because the multiples of $3$ in $Z$ have index $3$ ; a generator of the larger group, for instance $ε$ , is not in the smaller order, so the two unit groups are not equal and the order's norm-one Pell subgroup is proper in $O_{K}^{\times}$ .

9.1A1F6step 7.1step 8.1∎

Scope and choice accounting: the general statements of the example are steps 5.1, 6.1 and 7.1, and the failure of equality is witnessed by $d = 5$ in steps 7.2 and 8.1; AC is used only through the rank-one structure [F6], all computations here being elementary arithmetic in $Z [\sqrt{5}]$ .

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-10-02Open item page →

Two independent units in a real cubic field

Example

Assume the Axiom of Choice. Let $α = 2 \cos (2 π / 9)$ , the root in $(1, 2)$ of $f (X) = X^{3} - 3 X + 1$ . Then $K = Q (α)$ is a totally real cubic field with signature $(r_{1}, r_{2}) = (3, 0)$ , so the unit rank is $r_{1} + r_{2} - 1 = 2$ . The elements $α$ (norm $- 1$ ) and $α - 1$ (norm $1$ ) are units of $O_{K}$ whose logarithmic vectors $λ (α), λ (α - 1) \in H$ are linearly independent over $R$ ; hence $⟨ α, α - 1 ⟩$ is a rank- $2$ subgroup of $O_{K}^{\times}$ of finite index, confirming the rank.

Facts & Assumptions

Given: The Axiom of Choice, the real number $θ : = 2 π / 9$ , the element $α : = 2 \cos θ$ , and the polynomial $f (X) = X^{3} - 3 X + 1$ (cosine, number field).

[F1]

$π > 0$ and $π / 2$ is the smallest positive zero of cosine (Pi as twice the smallest positive zero of cosine).

[F2]

For every real $x$ one has $\cos (x + π) = - \cos x$ and $\cos π = - 1$ (Quarter-turn values and shifts by pi/2 and pi).

[F3]

For every real $x$ one has $\cos (- x) = \cos x$ (Parity and the Pythagorean identity for sine and cosine).

[F4]

For every real $x$ one has $\cos 3 x = 4 \cos^{3} x - 3 \cos x$ (Triple-angle identities for sine, cosine, and tangent).

[F5]

For every real $x$ one has $\cos 2 x = 2 \cos^{2} x - 1$ (Double-angle and quadratic power-reduction identities).

[F6]

$\cos 0 = 1$ (The derivatives of sine and cosine are cosine and minus sine).

[F7]

Cosine is strictly decreasing on $[0, π]$ , with range $[- 1, 1]$ (Signs, monotonicity intervals, and ranges of sine and cosine).

[F8]

If a rational number $p / q$ in lowest terms is a root of a polynomial with integer coefficients $a_{n} X^{n} + \dots + a_{0}$ , then $p$ divides $a_{0}$ and $q$ divides $a_{n}$ (Rational root theorem).

[F9]

A polynomial of degree $2$ or $3$ over a field is irreducible if and only if it has no root in that field (A polynomial of degree two or three over a field is irreducible exactly when it has no root in the field).

[F10]

A continuous real function on a closed bounded interval attains every value between its endpoint values (Intermediate value theorem, by bisection with a canonical left-half rule: a continuous function on $[a, b]$ takes every value between $f (a)$ and $f (b)$ ).

[F11]

Sending an embedding of $F (α)$ into an algebraically closed field to the image of $α$ is a bijection onto the set of distinct roots of the minimal polynomial of $α$ ; in particular the number of embeddings equals the number of distinct roots ( $F$ -embeddings of $F (α)$ into an algebraically closed field correspond to the distinct roots of $m_{α}$ ).

[F12]

For a separable finite extension the field norm is the product of the images under the distinct embeddings: $N_{K / Q} (u) = \prod_{σ} σ (u)$ (Norm and trace from embeddings, with the inseparable exponent in the norm formula).

[F13]

If $f (t) = t^{n} + a_{1} t^{n - 1} + \dots + a_{n}$ splits over a commutative ring as $f (t) = \prod_{i = 1}^{n} (t - α_{i})$ , then $a_{k} = (- 1)^{k} e_{k} (α_{1}, \dots, α_{n})$ for each $k$ ; in particular $a_{n} = (- 1)^{n} α_{1} \dots α_{n}$ (Vieta's formulas identify the coefficients of a split monic polynomial with elementary symmetric functions of its roots).

[F14]

For $u \in O_{K}$ , the element $u$ is a unit of $O_{K}$ 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).

[F15]

$O_{K}$ is the integral closure of $Z$ in $K$ ; an element of $K$ that is a root of a monic polynomial in $Z [X]$ is integral over $Z$ and hence lies in $O_{K}$ , and $O_{K}$ is a subring of $K$ containing $1$ (Integral elements over a commutative ring and algebraic integers, Ring of integers).

[F16]

The unit rank of $K$ is $r_{1} + r_{2} - 1$ , where $(r_{1}, r_{2})$ is the signature; a totally real cubic field has signature $(3, 0)$ and unit rank $2$ (Unit ranks by signature, Archimedean embeddings and signature).

[F17]

The logarithmic embedding is the map $λ (x) = (\log ∣ σ_{1} x ∣, \dots, 2 \log ∣ τ x ∣, \dots)$ on $K^{\times}$ , and $λ (u v) = λ (u) + λ (v)$ for $u, v \in K^{\times}$ (Logarithmic embedding of a number field, Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm); for a totally real field its values are the vectors of the logarithms of the absolute values of the conjugates.

[F18]

The image $λ (O_{K}^{\times})$ is contained in the hyperplane $H = {x : \sum_{i} x_{i} = 0}$ (Unit logarithms lie in the trace-zero hyperplane).

[F19]

$O_{K}^{\times} ≅ μ (K) \times Z^{r_{1} + r_{2} - 1}$ ; in particular the unit group is finitely generated of rank $r_{1} + r_{2} - 1$ with finite torsion subgroup $μ (K)$ (Dirichlet unit theorem).

[F20]

A system of fundamental units $(ε_{1}, \dots, ε_{r})$ of $K$ exists, its logarithms $λ (ε_{i})$ form a $Z$ -basis of $λ (O_{K}^{\times})$ , and every unit has a unique expression $u = ζ ε_{1}^{m_{1}} \dots ε_{r}^{m_{r}}$ with $ζ \in μ (K)$ and $m_{i} \in Z$ (System of fundamental units).

[F21]

For $n \geq 1$ and $A \in M_{n} (Z)$ with $\det A \neq 0$ , the subgroup $L = A Z^{n}$ generated by the columns has finite index $∣ \det A ∣$ in $Z^{n}$ (The index of a full-rank subgroup of $Z^{n}$ is the absolute determinant of a generating matrix).

[A1]

The Axiom of Choice is assumed; it is used only through the unit theorem [F19] and the existence of a system of fundamental units [F20] (The Axiom of Choice).

Verification

technique · verify by the triple-angle identity that $2\cos(2\pi/9)$ is the root of $X^3-3X+1$ in $(1,2)$, locate the three real conjugates by the intermediate value theorem, read the norms off Vieta's formulas, and detect the linear independence of the two logarithmic vectors through exact signs of the logarithms of the conjugates; finite index follows from the integrality of the coordinates in a fundamental system

1.1F1algebra

$π > 0$ , hence $0 < θ < π / 3 < π$ for $θ = 2 π / 9$ .

1.2F2F3F5F7algebra

With $c : = \cos (π / 3)$ one has $\cos (2 π / 3) = \cos (π - π / 3) = \cos ((- π / 3) + π) = - \cos (- π / 3) = - \cos (π / 3) = - c$ by the shift and parity identities, while the double-angle identity gives $\cos (2 π / 3) = 2 c^{2} - 1$ ; hence $2 c^{2} - 1 = - c$ , that is $(2 c - 1) (c + 1) = 0$ . Since $π / 3 < π$ and cosine is strictly decreasing on $[0, π]$ with $\cos π = - 1$ , one has $c > - 1$ , so $c = 1 / 2$ .

1.3algebra

Values of $f$ : $f (2) = 3$ , $f (1) = - 1$ , $f (0) = 1$ , $f (- 1) = 3$ , $f (- 2) = - 1$ .

2.1step 1.2

Hence $\cos (2 π / 3) = - 1 / 2$ .

2.2F6F7step 1.1step 1.2algebra

Moreover $1 < α < 2$ : from $0 < θ < π / 3$ and strict decrease of cosine on $[0, π]$ together with $\cos 0 = 1$ and $\cos (π / 3) = 1 / 2$ one gets $1 / 2 < \cos θ < 1$ , hence $1 < α < 2$ .

2.3F8F9step 1.3

The polynomial $f$ is irreducible over $Q$ : by [F8] a rational root of the monic integer polynomial $f$ must be an integer dividing the constant term $1$ , hence equal to $1$ or $- 1$ ; but $f (1) = - 1 \neq 0$ and $f (- 1) = 3 \neq 0$ , so $f$ has no rational root, and since $\deg f = 3$ it is irreducible by [F9].

3.1F4step 2.1algebra

The triple-angle identity gives $α^{3} - 3 α = (2 \cos θ)^{3} - 3 (2 \cos θ) = 2 (4 \cos^{3} θ - 3 \cos θ) = 2 \cos (3 θ) = 2 \cos (2 π / 3) = - 1$ , so $f (α) = α^{3} - 3 α + 1 = 0$ .

4.1F10step 3.1step 2.2step 1.3algebra

The polynomial $f$ has exactly one root in $(1, 2)$ , namely $α$ : by step 1.3 and [F10] the sign change from $f (1) < 0$ to $f (2) > 0$ gives a root in $(1, 2)$ , and for $1 < x < y$ one has $f (y) - f (x) = (y - x) (x^{2} + x y + y^{2} - 3) > 0$ , so $f$ is strictly increasing on $(1, \infty)$ and the root is unique; steps 3.1 and 2.2 show that $α$ is such a root.

5.1F10step 4.1algebra

By step 1.3 and [F10] there are also roots in $(- 2, - 1)$ and in $(0, 1)$ , and these three roots are distinct because the intervals are disjoint; a cubic has at most three roots, so these are all the roots of $f$ and all of them are real.

6.1F11F16step 3.1step 5.1step 2.3

Therefore $f$ is the minimal polynomial of $α$ over $Q$ (monic, irreducible, with $f (α) = 0$ by step 3.1), so $K = Q (α)$ has degree $3$ over $Q$ ; by [F11] the three embeddings $K \to C$ send $α$ to the three roots of $f$ , which are all real by step 5.1, so $K$ is totally real with signature $(3, 0)$ and unit rank $r_{1} + r_{2} - 1 = 2$ by [F16].

7.1F12F13step 6.1step 1.3algebra

By [F12] and [F13] applied to $f (t) = \prod_{i = 1}^{3} (t - α_{i})$ , where $α_{1}, α_{2}, α_{3}$ are the three conjugates of $α$ given by the embeddings of step 6.1, one has $N_{K / Q} (α) = α_{1} α_{2} α_{3} = (- 1)^{3} a_{3} = - 1$ because the constant coefficient of $f$ is $a_{3} = 1$ ; and $N_{K / Q} (α - 1) = \prod_{i = 1}^{3} (α_{i} - 1) = (- 1)^{3} \prod_{i = 1}^{3} (1 - α_{i}) = - f (1) = - (- 1) = 1$ , using $f (1) = 1 - 3 + 1 = - 1$ .

7.2step 2.2step 5.1step 6.1algebra

Writing the three real embeddings as $σ_{1} = i d, σ_{2}, σ_{3}$ , the conjugates satisfy $σ_{1} (α) = α \in (1, 2)$ , $0 < σ_{2} (α) < 1$ and $- 2 < σ_{3} (α) < - 1$ by steps 2.2 and 5.1; hence $\log ∣ σ_{1} α ∣ = \log α > 0$ , $\log ∣ σ_{1} (α - 1) ∣ = \log (α - 1) < 0$ , $\log ∣ σ_{2} α ∣ < 0$ , $\log ∣ σ_{2} (α - 1) ∣ = \log (1 - σ_{2} α) < 0$ , $\log ∣ σ_{3} α ∣ > 0$ and $\log ∣ σ_{3} (α - 1) ∣ > 0$ .

8.1F14F15step 7.1

The elements $α$ and $α - 1$ lie in $O_{K}$ : $α$ is a root of the monic polynomial $f \in Z [X]$ , hence integral over $Z$ and in $O_{K}$ by [F15], and $α - 1 \in O_{K}$ because $O_{K}$ is a subring of $K$ ; by [F14] with the norms of step 7.1, both are units of $O_{K}$ .

8.2step 7.2algebra

The vectors $λ (α)$ and $λ (α - 1)$ are linearly independent over $R$ : if $a λ (α) + b λ (α - 1) = 0$ , then reading the first coordinate and dividing by $\log α \neq 0$ gives $a + b r_{1} = 0$ with $r_{1} : = \log (α - 1) / \log α < 0$ by step 7.2, while reading the second coordinate and dividing by $\log ∣ σ_{2} α ∣ \neq 0$ gives $a + b r_{2} = 0$ with $r_{2} : = \log ∣ σ_{2} (α - 1) ∣ / \log ∣ σ_{2} α ∣ > 0$ , a quotient of two negative numbers; subtracting the two equations gives $b (r_{1} - r_{2}) = 0$ , and $r_{1} \neq r_{2}$ because their signs differ, so $b = 0$ and then $a = 0$ from the first equation.

9.1F17F18step 8.1

By [F17] the map $λ$ turns products into sums, and by [F18] both $λ (α)$ and $λ (α - 1)$ lie in $H$ , since $α$ and $α - 1$ are units of $O_{K}$ by step 8.1.

9.2F17step 8.2

Hence the subgroup $⟨ α, α - 1 ⟩$ is free abelian of rank $2$ : if $α^{m} (α - 1)^{n} = 1$ for integers $m, n$ , then $m λ (α) + n λ (α - 1) = λ (1) = 0$ by [F17] and step 8.2 forces $m = n = 0$ ; thus the homomorphism $Z^{2} \to O_{K}^{\times}$ , $(m, n) \mapsto α^{m} (α - 1)^{n}$ , has trivial kernel, and its image is exactly $⟨ α, α - 1 ⟩$ .

10.1F20F21step 8.2step 9.2

The image has finite index in $λ (O_{K}^{\times})$ : by [F20] fix a system of fundamental units $ε_{1}, ε_{2}$ and write $α = ζ ε_{1}^{m_{1}} ε_{2}^{m_{2}}$ and $α - 1 = ζ^{'} ε_{1}^{n_{1}} ε_{2}^{n_{2}}$ with $ζ, ζ^{'} \in μ (K)$ and integers $m_{i}, n_{i}$ ; since $λ$ kills $μ (K)$ , $λ (α) = m_{1} λ (ε_{1}) + m_{2} λ (ε_{2})$ and $λ (α - 1) = n_{1} λ (ε_{1}) + n_{2} λ (ε_{2})$ , so with respect to the $Z$ -basis $(λ (ε_{1}), λ (ε_{2}))$ of $λ (O_{K}^{\times})$ the two vectors have the integer coordinate columns $(m_{1}, m_{2})^{T}$ and $(n_{1}, n_{2})^{T}$ , and the matrix $A = (\begin{array}{cc} m_{1} & n_{1} \\ m_{2} & n_{2} \end{array})$ has $\det A \neq 0$ because a zero determinant would make the two coordinate columns, hence $λ (α)$ and $λ (α - 1)$ , linearly dependent over $R$ , contradicting step 8.2; therefore the subgroup $Z λ (α) + Z λ (α - 1) = A Z^{2}$ has finite index in $λ (O_{K}^{\times})$ by [F21].

11.1F19step 10.1

Consequently $⟨ α, α - 1 ⟩$ has finite index in $O_{K}^{\times}$ : the canonical map $O_{K}^{\times} / ⟨ α, α - 1 ⟩ \to λ (O_{K}^{\times}) / (Z λ (α) + Z λ (α - 1))$ is surjective onto a finite group by step 10.1, and its kernel $(μ (K) ⟨ α, α - 1 ⟩) / ⟨ α, α - 1 ⟩ ≅ μ (K) / (μ (K) \cap ⟨ α, α - 1 ⟩)$ is a quotient of the finite group $μ (K)$ of [F19]; hence the quotient is finite, as claimed.

12.1A1F19F20∎

Choice accounting: AC is used only through the unit theorem [F19], which supplies the finite generation and rank, and through the existence of the system of fundamental units [F20]; the trigonometric, polynomial, norm and logarithm computations, and the independence argument via signs, are elementary and use no further choice.

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Regulator of a real quadratic field

Example

Assume the Axiom of Choice. Let $K = Q (\sqrt{d})$ be a real quadratic field with fundamental unit $ε > 1$ (the least unit of $O_{K}$ greater than $1$ ). Then $K$ has signature $(2, 0)$ , so its unit rank is $r_{1} + r_{2} - 1 = 1$ and $ε$ is a system of fundamental units; its logarithmic vector is $λ (ε) = (\log ε, \log ∣ σ_{2} ε ∣) = (\log ε, - \log ε),$ and the absolute deleted-row determinant of the $2 \times 1$ logarithmic matrix is $\log ε$ , so $R_{K} = \log ε$ . The factor $2$ of the doubled-complex convention never enters, since a real quadratic field has no complex place. For $K = Q (\sqrt{5})$ the fundamental unit is $ε = (1 + \sqrt{5}) / 2$ and $R_{K} = \log ((1 + \sqrt{5}) / 2) \approx 0.4812118251$ ; the positive generator $9 + 4 \sqrt{5} = ε^{6}$ of the norm-one Pell subgroup of the order $Z [\sqrt{5}]$ has $\log (9 + 4 \sqrt{5}) = 6 R_{K} \approx 2.8872709504$ , so using that generator of the nonmaximal order as if it were the fundamental unit of the maximal order would multiply the regulator by six.

Facts & Assumptions

Given: The Axiom of Choice, a squarefree integer $d > 1$ , the field $K = Q (\sqrt{d})$ , its fundamental unit $ε$ (the least unit of $O_{K}$ greater than $1$ ), and its two real embeddings $σ_{1}, σ_{2}$ (Real quadratic units and Pell's equation, Logarithmic embedding of a number field).

[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 $K^{\times}$ , with one coordinate for each real embedding and one doubled coordinate for each complex place, and it is well defined (Logarithmic embedding of a number field).

[F2]

$\log : (0, \infty) \to R$ is strictly increasing with $\log 1 = 0$ , and $\log (x y) = \log x + \log y$ , $\log (1 / x) = - \log x$ for $x, y > 0$ (Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm).

[F3]

A quadratic field $K = Q (\sqrt{d})$ with $d > 0$ has two real embeddings $a + b \sqrt{d} \mapsto a \pm b \sqrt{d}$ and no complex place, so its signature is $(2, 0)$ ; its unit rank is $1$ , and $O_{K}^{\times} ≅ {\pm 1} \times Z$ with torsion subgroup $μ (K) = {\pm 1}$ (Unit ranks by signature).

[F4]

For $u \in O_{K}$ , $u$ is a unit of $O_{K}$ if and only if $N_{K / Q} (u) = \pm 1$ ; for the real quadratic field $N_{K / Q} (x) = x σ_{2} (x)$ (A number-field unit is exactly an algebraic integer of norm plus or minus one, Real quadratic units and Pell's equation).

[F5]

A system of fundamental units of $K$ is a tuple $(ε_{1}, \dots, ε_{r})$ of units with $λ (O_{K}^{\times}) = Z λ (ε_{1}) \oplus \dots \oplus Z λ (ε_{r})$ (System of fundamental units).

[F6]

The regulator is $R_{K} = ∣ \det A_{k} ∣$ , where the columns of $A$ are the logarithmic vectors of a system of fundamental units and $A_{k}$ is obtained by deleting row $k$ ; $R_{K}$ is independent of the deleted row and of the chosen system and is positive. The definition records that for a real quadratic field the regulator is $\log ε$ for its fundamental unit $ε > 1$ (Regulator of a number field, The regulator is well defined).

[F7]

For $K = Q (\sqrt{5})$ : the element $ε = (1 + \sqrt{5}) / 2$ has norm $- 1$ , so it is a unit; it is the least unit $> 1$ of $O_{K}$ and $O_{K}^{\times} = {\pm ε^{n} : n \in Z}$ ; moreover $2 + \sqrt{5} = ε^{3}$ , the fundamental Pell solution is $9 + 4 \sqrt{5} = ε^{6}$ , and $Z [\sqrt{5}]^{\times} = \pm ⟨ 2 + \sqrt{5} ⟩ = \pm ⟨ ε^{3} ⟩$ (Real quadratic units and Pell's equation).

[A1]

The Axiom of Choice is assumed; it is used only through the rank-one structure [F3] and the unit-theoretic inputs of [F7] (The Axiom of Choice).

Verification

technique · identify the fundamental unit as a system of fundamental units for the rank-one real quadratic field, compute its logarithmic vector from $|N(\varepsilon)|=1$, read off the deleted-row determinant, and specialize to $\mathbb Q(\sqrt5)$ where the maximal-order fundamental unit and the Pell generator of $\mathbb Z[\sqrt5]$ differ by a sixth power

1.1F1F3

$K$ has signature $(2, 0)$ and $λ (x) = (\log ∣ σ_{1} x ∣, \log ∣ σ_{2} x ∣)$ for $x \in K^{\times}$ : there are two real embeddings and no complex place by [F3], so [F1] has no doubled coordinate, and no logarithm of a nonzero element is undefined.

1.2F1F2F3F5algebra

The fundamental unit generates the unit group: by [F3] the group $O_{K}^{\times}$ has torsion subgroup ${\pm 1}$ and free part of rank $1$ , so there is a unit $γ > 1$ with $O_{K}^{\times} = {\pm γ^{n} : n \in Z}$ ; every unit $v > 1$ is then $γ^{n}$ with $n \geq 1$ , and $γ^{n} = γ \cdot γ^{n - 1} \geq γ$ because $γ^{n - 1} \geq 1$ for $n \geq 1$ , so $γ$ is the least unit $> 1$ and hence $γ = ε$ ; therefore $O_{K}^{\times} = {\pm ε^{n}}$ and, by [F2] and [F1], $λ (- 1) = 0$ and $λ (ε^{n}) = n λ (ε)$ for every $n \in Z$ , so $λ (O_{K}^{\times}) = Z λ (ε) .$ Thus $(ε)$ is a system of fundamental units of $K$ in the sense of [F5].

2.1F1F2F4step 1.1algebra

Logarithmic vector: $σ_{1}$ may be taken to be the identity, so $∣ σ_{1} ε ∣ = ε > 0$ ; and $∣ σ_{2} ε ∣ = 1 / ε$ , because $ε$ is a unit and $N_{K / Q} (ε) = ε σ_{2} (ε) = \pm 1$ by [F4], so $σ_{2} (ε) = \pm 1 / ε$ . Hence $λ (ε) = (\log ε, \log (1 / ε)) = (\log ε, - \log ε)$ by [F2], a nonzero vector in the hyperplane ${(ξ_{1}, ξ_{2}) : ξ_{1} + ξ_{2} = 0}$ .

3.1F2F6step 1.2step 2.1algebra

Regulator: the logarithmic matrix of the system $(ε)$ is the $2 \times 1$ matrix $A$ with entries $\log ε$ and $- \log ε$ , so deleting row $1$ gives the $1 \times 1$ determinant $- \log ε$ and deleting row $2$ gives $\log ε$ ; by [F6] and step 1.2, $R_{K} = ∣ \det A_{k} ∣ = \log ε,$ which is positive because $ε > 1$ and $\log$ is strictly increasing with $\log 1 = 0$ by [F2]. In particular the two deleted rows give the same absolute value, and the factor $2$ of the complex coordinates of [F1] is absent.

4.1F7step 3.1algebra

For $K = Q (\sqrt{5})$ : by [F7] the fundamental unit is $ε = (1 + \sqrt{5}) / 2 \approx 1.6180339887$ , so $R_{K} = \log ((1 + \sqrt{5}) / 2) \approx 0.4812118251.$

5.1F2F5F7step 3.1step 4.1algebra

By [F7], the full order unit group is $Z [\sqrt{5}]^{\times} = \pm ⟨ ε^{3} ⟩$ , and its norm-one Pell subgroup is $\pm ⟨ ε^{6} ⟩ = \pm ⟨ 9 + 4 \sqrt{5} ⟩$ . The Pell generator has logarithmic coordinate $\log (9 + 4 \sqrt{5}) = \log (ε^{6}) = 6 \log ε = 6 R_{K} \approx 2.8872709504,$ so its rank-one deleted-row determinant is six times the field regulator, which is defined using the maximal-order fundamental unit $ε$ .

6.1A1F3F6F7step 3.1step 5.1∎

Conclusion and choice accounting: for every real quadratic field the fundamental unit $ε > 1$ is a system of fundamental units, its logarithmic vector is $(\log ε, - \log ε)$ , and $R_{K} = \log ε$ ; the doubled-complex normalization is vacuous here, and for $K = Q (\sqrt{5})$ the value is $R_{K} = \log ((1 + \sqrt{5}) / 2) \approx 0.4812118251$ , six times smaller than the determinant $6 R_{K}$ obtained from the Pell generator $9 + 4 \sqrt{5}$ of the order $Z [\sqrt{5}]$ . Choice enters only through the rank-one unit structure [F3] and the $d = 5$ input [F7]; the logarithm computations and the determinant of the $2 \times 1$ matrix use no choice.

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A unimodular change of generators preserves the regulator determinants

Example

Assume the Axiom of Choice. Let $α = 2 \cos (2 π / 9)$ , the root in $(1, 2)$ of $X^{3} - 3 X + 1$ , with conjugates $x = σ_{1} α \in (1, 2)$ , $y = σ_{2} α \in (0, 1)$ and $z = σ_{3} α \in (- 2, - 1)$ , and let $u_{1} : = α$ , $u_{2} : = α - 1$ be the two independent units of the cubic field example. Then:

the tuples $(u_{1}, u_{2})$ and $(u_{1} u_{2}, u_{2})$ generate the same lattice $Z λ (u_{1}) + Z λ (u_{2})$ in the hyperplane $H$ ; the second logarithmic matrix is $A (\begin{array}{cc} 1 & 0 \\ 1 & 1 \end{array})$ with the matrix acting on columns, a unimodular integer matrix of determinant $1$ ;
every deleted-row $2 \times 2$ determinant of the logarithmic matrix has the same value for the two tuples, namely $\pm (a^{2} + a b + b^{2})$ with $a = \log x$ , $b = \log y$ , and its absolute value is $0.849287 \dots$ in both cases (the three deleted rows have the signs $-, +, -$ );
interchanging $u_{1}$ and $u_{2}$ , that is right multiplication by the unimodular integer matrix $(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array})$ of determinant $- 1$ , reverses the sign of every deleted-row determinant and leaves its absolute value unchanged, which is why the regulator is defined from an absolute determinant.

Facts & Assumptions

Given: The Axiom of Choice, the element $α = 2 \cos (2 π / 9)$ , the field $K = Q (α)$ , its three real embeddings $σ_{1}, σ_{2}, σ_{3}$ with conjugates $x = σ_{1} α$ , $y = σ_{2} α$ , $z = σ_{3} α$ , and the units $u_{1} = α$ , $u_{2} = α - 1$ (Two independent units in a real cubic field, Logarithmic embedding of a number field).

[F1]

The cubic example gives: $f (X) = X^{3} - 3 X + 1$ is irreducible with $α \in (1, 2)$ one of its three real roots; $K$ is totally real of signature $(3, 0)$ , so $r_{1} + r_{2} = 3$ and the logarithms of the three embeddings are the coordinates of $λ$ ; the conjugates satisfy $x \in (1, 2)$ , $y \in (0, 1)$ , $z \in (- 2, - 1)$ ; $f$ is strictly increasing on $(1, \infty)$ , so $x$ is its only root there; $N_{K / Q} (α) = - 1$ and $N_{K / Q} (α - 1) = 1$ ; $α$ and $α - 1$ are units of $O_{K}$ ; and $λ (α)$ , $λ (α - 1)$ are $R$ -linearly independent (Two independent units in a real cubic field).

[F2]

The logarithmic embedding is $λ (w) = (\log ∣ σ_{1} w ∣, \dots, \log ∣ σ_{r_{1}} w ∣, 2 \log ∣ τ_{1} w ∣, \dots)$ on $K^{\times}$ ; for the totally real $K$ of [F1] it is $λ (w) = (\log ∣ σ_{1} w ∣, \log ∣ σ_{2} w ∣, \log ∣ σ_{3} w ∣)$ , and it is well defined because nonzero elements have nonzero images under every embedding (Logarithmic embedding of a number field).

[F3]

$\log$ is additive over products, so for $u, v \in K^{\times}$ the coordinatewise additivity $λ (u v) = λ (u) + λ (v)$ holds, the absolute values of the conjugates being multiplicative (Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm).

[F4]

Every value $λ (u)$ of a unit $u \in O_{K}^{\times}$ lies in the hyperplane $H = {ξ : \sum_{i = 1}^{3} ξ_{i} = 0}$ (Unit logarithms lie in the trace-zero hyperplane).

[F5]

The regulator of $K$ is $R_{K} = ∣ \det A_{k} ∣$ , where $A$ has the columns $λ (ε_{1}), \dots, λ (ε_{r})$ for a system of fundamental units and $A_{k}$ is obtained by deleting row $k$ ; deleting rows may be done before or after finite matrix products. The definition records that 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" (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).

[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. Then for the deleted-row determinants $Δ_{k}$ one has $Δ_{k} \neq 0$ 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 over a commutative ring, $\det (B C) = \det (B) \det (C)$ (For same-sized finite square matrices over a commutative ring, $\det (A B) = \det (A) \det (B)$ ).

[F8]

An invertible square matrix over a commutative ring has unit determinant (An invertible square matrix over a commutative ring has unit determinant), and the units of $Z$ are $\pm 1$ ( $(Z, \cdot, 1)$ is a commutative monoid whose group of units is ${1, - 1}$ ; equivalently $u ∣ 1$ holds exactly for $u = 1$ and $u = - 1$ ).

[F9]

Assume the Axiom of Choice. For two systems of fundamental units of a number field $K$ the regulator $R_{K}$ is the same number: the absolute deleted-row determinant does not depend on the deleted row nor on the chosen system, and it is positive (The regulator is well defined).

[A1]

The Axiom of Choice is assumed; it is used only through the unit-theoretic inputs quoted in [F1] and [F9] and the hyperplane input [F4] (The Axiom of Choice).

Verification

technique · derive the exact algebraic relations among the three real conjugates from the polynomial $X^3-3X+1$, translate them into relations among the logarithmic coordinates, and compare the deleted-row determinants of the two tuples through the explicit unimodular matrices

1.1F1F2F3F4

The setting is as in [F1]-[F4]: $λ (α) = (a, b, c)$ and $λ (α - 1)$ are vectors in the hyperplane $H$ of $R^{3}$ , where $a = \log x$ , $b = \log y$ , $c = \log ∣ z ∣$ ; these logarithms are defined because $x > 1$ , $0 < y < 1$ and $∣ z ∣ > 1$ ; and $λ$ is additive, $λ (u_{1} u_{2}) = λ (u_{1}) + λ (u_{2})$ .

1.2F1algebra

Conjugate relations I: if $r$ is any root of $f$ , then $r^{2} - 2$ is again a root: from $r^{3} = 3 r - 1$ one computes $r^{4} = 3 r^{2} - r$ , $r^{5} = - r^{2} + 9 r - 3$ , $r^{6} = 9 r^{2} - 6 r + 1$ , hence $(r^{2} - 2)^{3} = 3 r^{2} - 7$ and $(r^{2} - 2)^{3} - 3 (r^{2} - 2) + 1 = (3 r^{2} - 7) - 3 r^{2} + 6 + 1 = 0$ . Thus $φ (r) : = r^{2} - 2$ maps the set ${x, y, z}$ of the three distinct roots into itself. Now $x > \sqrt{2}$ , because $f$ is strictly increasing on $(1, \infty)$ by [F1] and $f (\sqrt{2}) = 2 \sqrt{2} - 3 \sqrt{2} + 1 = 1 - \sqrt{2} < 0 < f (x)$ ; hence $x^{2} - 2 \in (0, 2)$ , and the only roots in $(0, 2)$ are $y \in (0, 1)$ and $x \in (1, 2)$ , with $x^{2} - 2 \neq x$ because $x^{2} - x - 2 = 0$ would give $x \in {- 1, 2}$ ; therefore $φ (x) = y$ . Similarly $φ (y) = y^{2} - 2 \in (- 2, - 1)$ is the root in $(- 2, - 1)$ , namely $z$ . Finally $z \in (- 2, - \sqrt{3})$ , since for $s < t < - 1$ one has $f (t) - f (s) = (t - s) (s^{2} + s t + t^{2} - 3) > 0$ , so $f$ is strictly increasing on $(- \infty, - 1)$ , and $f (- 2) = - 1 < 0 < f (- \sqrt{3}) = 1$ , so $φ (z) = z^{2} - 2 \in (1, 2)$ , and the only root in $(1, 2)$ is $x$ ; hence $φ (z) = x$ .

2.1F1step 1.2algebra

Conjugate relations II: for every root $r$ of $f$ one has $r (r^{2} - 3) = - 1$ , hence $1 / r = 3 - r^{2}$ . Applying this to $r = x$ and using $x^{2} - 2 = y$ gives $1 / x = 3 - x^{2} = 1 - y$ ; applied to $r = y$ and using $y^{2} - 2 = z$ it gives $1 / y = 1 - z$ ; applied to $r = z$ and using $z^{2} - 2 = x$ it gives $1 / z = 1 - x$ .

3.1F1F2F3step 2.1algebra

Logarithmic coordinates: since $1 - y > 0$ is the inverse of $x$ , $\log (1 - y) = - a$ , that is $\log ∣ y - 1 ∣ = - a$ ; since $1 - z > 0$ is the inverse of $y$ , $\log ∣ z - 1 ∣ = - b$ ; and since $1 / z = 1 - x$ is negative with $x - 1 > 0$ , $\log ∣ z ∣ = - \log (x - 1)$ , that is $\log (x - 1) = - c$ . Moreover $a + b + c = \log ∣ x y z ∣ = \log ∣ N_{K / Q} (α) ∣ = \log 1 = 0$ because $N_{K / Q} (α) = - 1$ . Therefore $λ (α) = (a, b, c)$ and $λ (α - 1) = (\log (x - 1), \log ∣ y - 1 ∣, \log ∣ z - 1 ∣) = (- c, - a, - b)$ with $a + b + c = 0$ .

4.1F5F6step 3.1algebra

The logarithmic matrix of the tuple $(u_{1}, u_{2})$ is $A = (\begin{array}{cc} a & - c \\ b & - a \\ c & - b \end{array}), a + b + c = 0.$ Deleting row $1$ gives $Δ_{1} = b (- b) - (- a) c = - b^{2} + a c = - (a^{2} + a b + b^{2})$ ; deleting row $2$ gives $Δ_{2} = a (- b) - (- c) c = - a b + c^{2} = a^{2} + a b + b^{2}$ ; and deleting row $3$ gives $Δ_{3} = a (- a) - (- c) b = - a^{2} + b c = - (a^{2} + a b + b^{2})$ , where $c = - a - b$ is used in each reduction. In particular all three deleted-row determinants are nonzero and have absolute value $Q : = a^{2} + a b + b^{2}$ .

5.1F3F5F7step 4.1algebra

The second tuple has the same lattice and the same determinants: by additivity $λ (u_{1} u_{2}) = λ (u_{1}) + λ (u_{2})$ , so $Z λ (u_{1}) + Z λ (u_{2}) = Z (λ (u_{1}) + λ (u_{2})) + Z λ (u_{2})$ , an equality of subgroups of $H$ . In coordinates, the logarithmic matrix of $(u_{1} u_{2}, u_{2})$ is $A^{'} = A C$ with $C = (\begin{array}{cc} 1 & 0 \\ 1 & 1 \end{array})$ acting on columns, whose inverse $C^{- 1} = (\begin{array}{cc} 1 & 0 \\ - 1 & 1 \end{array})$ is integral and whose determinant is $1$ ; deleting row $k$ commutes with right multiplication, so $A_{k}^{'} = A_{k} C$ and $\det A_{k}^{'} = \det (A_{k}) \det (C) = \det A_{k}$ by [F7]. Thus the determinants of the two tuples are equal, not merely equal in absolute value, and $∣ \det A_{k}^{'} ∣ = Q$ as well.

6.1F7F8step 4.1step 5.1algebra

Swapping the two units, that is passing to $(u_{2}, u_{1})$ , replaces $A$ by $A^{''} = A P$ with $P = (\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array})$ , whose inverse is itself and whose determinant is $- 1$ ; then $\det A_{k}^{''} = \det (A_{k}) \det (P) = - \det A_{k}$ , so the sign of every deleted-row determinant is reversed and the absolute value $Q$ is unchanged. Both $C$ and $P$ are invertible over $Z$ , so by [F8] their determinants are units of $Z$ , that is $\pm 1$ , in agreement with the direct computations $\det C = 1$ and $\det P = - 1$ .

6.2step 3.1step 4.1step 5.1algebra

Numerical value: evaluating cosine gives $\cos (2 π / 9) \approx 0.7660444431$ , so $x = α = 2 \cos (2 π / 9) \approx 1.5320888862$ , $y = x^{2} - 2 \approx 0.3472963553$ , and $z = y^{2} - 2$ satisfies $∣ z ∣ \approx 1.8793852416$ because $z \in (- 2, - 1)$ by step 1.2; hence $a = \log x \approx 0.4266320894$ , $b = \log y$ satisfies $∣ b ∣ \approx 1.0575768136$ , and $Q = a^{2} + a b + b^{2} \approx 0.8492874506.$ So every deleted-row determinant of the two tuples has absolute value the number $Q$ of the display, the signs for the tuple $(u_{1}, u_{2})$ being $-, +, -$ as computed in step 4.1.

7.1A1F1F4F5F9step 5.1step 6.2∎

Relation to the regulator definition and choice accounting: the computation is the explicit ${GL}_{2} (Z)$ step that [F5] and [F9] single out — right multiplication by an integral matrix of determinant $\pm 1$ changes the deleted-row determinants by that same factor, so the absolute value is the invariant and the regulator is defined from it. Nothing here asserts that $(u_{1}, u_{2})$ or $(u_{1} u_{2}, u_{2})$ is a system of fundamental units: the common number $Q$ is the absolute deleted-row determinant of the rank-two subgroup lattice generated by the tuple, and it coincides with the field regulator only when the tuple generates all of $λ (O_{K}^{\times})$ . AC enters only through the unit-theoretic inputs quoted in [F1] and [F9] and the hyperplane input [F4]; the algebraic relations among conjugates, the logarithmic matrix computations and the determinant identities use no choice.

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S-units of Q

Example

Assume the Axiom of Choice. Let $S = {p_{1}, \dots, p_{m}}$ be a finite set of primes, let $M = {p_{1}^{e_{1}} \dots p_{m}^{e_{m}} : e_{1}, \dots, e_{m} \in N}$ be the multiplicative subset of $Z$ generated by $S$ (the empty product gives $1$ , so that $M = {1}$ when $m = 0$ ), and let $Z [S^{- 1}] = M^{- 1} Z$ be the localisation of $Z$ at $M$ . Then, under the canonical embedding of $Z [S^{- 1}]$ in $Q$ , $Z [S^{- 1}]^{\times} = {\pm p_{1}^{n_{1}} \dots p_{m}^{n_{m}} : n_{1}, \dots, n_{m} \in Z} ≅ {\pm 1} \times Z^{m},$ the group law on the right being addition of the exponent vector. This agrees with the $S$ -unit theorem for $K = Q$ : applied to the finite set $S_{Q} = {(p_{1}), \dots, (p_{m})}$ of nonzero prime ideals of $O_{Q} = Z$ it gives $O_{Q, S_{Q}}^{\times} ≅ μ (Q) \times Z^{m}$ , of rank $r_{1} + r_{2} - 1 + ∣ S_{Q} ∣ = 1 + 0 - 1 + m = m$ .

Facts & Assumptions

Given: The Axiom of Choice, a finite set $S = {p_{1}, \dots, p_{m}}$ of primes (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ), the multiplicative subset $M \subseteq Z$ generated by $S$ , and the localisation $R : = Z [S^{- 1}] = M^{- 1} Z$ (Multiplicative subsets and the localisation $S^{- 1} R$ as equivalence classes of fractions).

[F1]

$M$ is multiplicative and its elements are exactly the products $p_{1}^{e_{1}} \dots p_{m}^{e_{m}}$ with all $e_{i} \in N$ , including the empty product $1$ . Elements of $R$ are classes $a / b$ with $a \in Z$ , $b \in M$ , with $a / b + a^{'} / b^{'} = (a b^{'} + a^{'} b) / (b b^{'})$ and $(a / b) (a^{'} / b^{'}) = (a a^{'}) / (b b^{'})$ ; two classes are equal, $a / b = a^{'} / b^{'}$ , exactly when $u (a b^{'} - a^{'} b) = 0$ for some $u \in M$ ; the localisation map $a \mapsto a / 1$ is a ring homomorphism, and every $s \in M$ maps to a unit, $(s / 1)^{- 1} = 1 / s$ (Multiplicative subsets and the localisation $S^{- 1} R$ as equivalence classes of fractions).

[F2]

A product of two nonzero integers is nonzero, so $u t = 0$ with $u, t \in Z$ and $u \neq 0$ forces $t = 0$ ; since $0 \notin M$ , the equality criterion of [F1] reduces to $a b^{'} = a^{'} b$ (The integers have no zero divisors; multiplicative cancellation).

[F3]

Every positive integer $n \geq 1$ is a finite product of primes, and the factorisation is unique up to order; a prime is an integer $p > 1$ whose only positive divisors are $1$ and $p$ (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 $π \in Sym (r)$ , Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ).

[F4]

The units of $Z$ are exactly $1$ 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 $u = - 1$ ).

[F5]

For every prime $q$ the principal ideal $(q) = q Z$ is a nonzero prime ideal of $Z$ , indeed a maximal one: $Z / (q)$ is a field (For every prime $p$ , the two operations on $Z / p$ make it a field) and a quotient ring is a field exactly when the ideal is maximal ( $R / M$ is a field if and only if $M$ is a maximal ideal), while a maximal ideal is prime (Every maximal ideal of a commutative ring is prime); the ideal is nonzero because $q \neq 0$ and proper because $q > 1$ . Distinct primes give distinct ideals, since $q \in (q^{'})$ forces $q^{'} = q$ . The localisation $Z_{(q)}$ of $Z$ at the prime ideal $(q)$ consists of the fractions $m / n$ with $q ∤ n$ , and every element of $Z$ outside $(q)$ becomes a unit there (Localisation at a prime ideal: $R_{p} = (R ∖ p)^{- 1} R$ , Multiplicative subsets and the localisation $S^{- 1} R$ as equivalence classes of fractions). Moreover, assuming the Axiom of Choice, $Z = O_{Q}$ is a Dedekind domain (Rings of integers are Dedekind domains), each localisation $Z_{p}$ at a nonzero prime is a discrete valuation ring (Localizing a Dedekind domain at a nonzero prime gives a DVR), and every nonzero fractional ideal $I$ carries prime-ideal valuations defined by $I_{p} = p^{v_{p} (I)} Z_{p}$ (Fractional ideals, Prime-ideal valuations on fractional ideals); the nonzero ideals of the discrete valuation ring $Z_{p}$ are the powers $p^{n} Z_{p}$ with $n \geq 0$ (Ideals in a DVR are powers of the maximal ideal), and for a nonzero rational the principal fractional ideal $(x) = x Z$ localises to $(x)_{p} = x Z_{p}$ (Localisation of a module at a multiplicative subset).

[F6]

For a number field $K$ (Number field) and a finite set $S$ of nonzero prime ideals of $O_{K}$ , $O_{K, S} = {0} \cup {x \in K^{\times} : v_{p} (x) \geq 0 for every nonzero prime p \notin S}$ consists of zero and the nonzero elements whose principal fractional ideal involves no prime outside $S$ in a denominator, and $O_{K, S}^{\times} = {x \in K^{\times} : v_{p} (x) = 0$ for every nonzero prime $p \notin S}$ ; no infinite place belongs to $S$ , and the rank formula is $r_{1} + r_{2} - 1 + ∣ S ∣$ (S-integers and S-units of a number field).

[F7]

Assume the Axiom of Choice. For a number field $K$ of signature $(r_{1}, r_{2})$ and a finite set $S$ of nonzero prime ideals, $O_{K, S}^{\times} ≅ μ (K) \times Z^{r_{1} + r_{2} - 1 + ∣ S ∣}$ (S-unit theorem).

[F8]

$O_{Q} = Z$ : the ring of integers $O_{Q}$ is the integral closure of $Z$ in $Q$ (Ring of integers), 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 $X - n$ . Also $Q$ has a single archimedean place, which is real, so its signature is $(1, 0)$ (Archimedean embeddings and signature).

[F9]

For a field $K$ and $n \geq 1$ , $μ_{n} (K) = {x \in K : x^{n} = 1}$ and $μ (K)$ is the group of all roots of unity in $K$ , the union of the $μ_{n} (K)$ (The group $μ_{n} (K)$ of $n$ -th roots of unity in a field, and primitive $n$ -th roots of unity).

[A1]

The Axiom of Choice is assumed. It is used through the $S$ -unit theorem [F7] and through the Dedekind structure of $Z$ quoted in [F5], where the ring-of-integers corollary is itself AC-qualified; the localisation, exponent and sign computations of the verification use no choice (The Axiom of Choice).

Verification

technique · identify the localisation with a subring of $\mathbb Q$, express membership in it and in its unit group through the exponents of the prime factorisation of a rational, and then compare the resulting monomial group with the $S$-unit theorem applied to $K=\mathbb Q$

1.1F1F2algebra

The localisation $R$ embeds in $Q$ by $φ (a / b) = a / b$ : the map is well defined because if $a / b = a^{'} / b^{'}$ in $R$ , then $u (a b^{'} - a^{'} b) = 0$ for some $u \in M$ by [F1] and hence $a b^{'} = a^{'} b$ by [F2], so the two rationals agree; it respects the fraction arithmetic of [F1], which is the arithmetic of the field $Q$ ; and it is injective, since $φ (a / b) = 0$ forces $a = 0$ , and then $1 \cdot (a \cdot 1 - 0 \cdot b) = 0$ gives $a / b = 0 / 1$ by [F1]. Hence $R$ is identified with the subring ${a / b \in Q : a \in Z, b \in M}$ of $Q$ , in which $1 / s$ is the inverse of $s$ for every $s \in M$ .

1.2F3algebra

Fix $x \in Q^{\times}$ and write its prime factorisation as $x = \pm \prod_{q p r i m e} q^{e_{q} (x)}, e_{q} (x) \in Z,$ with all but finitely many exponents zero; existence and uniqueness of the triple (sign, exponent vector) is [F3], applied to the numerator and denominator of $x$ in lowest terms. Writing $x = a / b$ in lowest terms with $b > 0$ and fixing a prime $q$ , one has $e_{q} (x) = v_{q} (a) - v_{q} (b)$ ; if $q ∤ b$ this is the nonnegative integer $v_{q} (a)$ , while if $q ∣ b$ then $v_{q} (b) > 0$ and $v_{q} (a) = 0$ by coprimality, so $e_{q} (x) < 0$ . Thus $e_{q} (x) \geq 0$ if and only if $q ∤ b$ .

1.3F5algebra

For a prime $q$ , the localisation $Z_{(q)}$ consists of the fractions $m / n$ with $q ∤ n$ by [F5], so for $x = a / b \in Q^{\times}$ in lowest terms with $b > 0$ one has $x \in Z_{(q)}$ if and only if $q ∤ b$ : if $q ∤ b$ then $x = a / b$ has the required form, while if $x = a / b = m / n$ with $q ∤ n$ , then $a n = b m$ , and $q ∣ b$ would give $q ∣ a n$ and hence $q ∣ a$ , contradicting coprimality.

1.4F5algebra

For a nonzero prime ideal $p$ of $Z$ and $x \in Q^{\times}$ , [F5] writes $v_{p} (x) = v_{p} ((x))$ as the unique integer $n$ with $x Z_{p} = (x)_{p} = p^{n} Z_{p}$ . The powers $p^{n} Z_{p}$ with $n \geq 0$ are exactly the ideals of the discrete valuation ring $Z_{p}$ , and for $n < 0$ the fractional ideal $p^{n} Z_{p}$ properly contains $Z_{p}$ ; hence $v_{p} (x) \geq 0$ if and only if $x Z_{p} \subseteq Z_{p}$ , that is, if and only if $x \in Z_{p}$ .

2.1F1F3F5step 1.1step 1.2algebra

An element $x \in Q^{\times}$ lies in $R$ if and only if $e_{q} (x) \geq 0$ for every prime $q \notin S$ . If $x = a / b$ with $a \in Z$ and $b \in M$ , then $b$ has no prime divisor outside $S$ , so for $q \notin S$ one has $e_{q} (x) = v_{q} (a) - v_{q} (b) = v_{q} (a) - 0 \geq 0$ , where the exponents are those of the factorisations of $a$ and $b$ ; conversely, if $e_{q} (x) \geq 0$ for all $q \notin S$ and $x = a / b$ is in lowest terms with $b > 0$ , then a prime $q ∣ b$ satisfies $v_{q} (a) = 0$ and $v_{q} (x) = - v_{q} (b) < 0$ , so $q \in S$ ; hence every prime divisor of $b$ lies in $S$ , that is $b \in M$ , and $x = a / b \in R$ .

2.2F3F4F8F9step 1.2algebra

For the comparison with the $S$ -unit theorem, evaluate its torsion factor and its signature input. $Q$ is a number field with signature $(1, 0)$ and $O_{Q} = Z$ by [F8]; the roots of unity in $Q$ are $\pm 1$ , since $x = a / b \in Q^{\times}$ in lowest terms with $x^{n} = 1$ gives $a^{n} = b^{n}$ , whence $∣ a ∣^{n} = ∣ b ∣^{n}$ and, by uniqueness of prime factorisation, $∣ a ∣ = ∣ b ∣$ , while coprimality forces $∣ a ∣ = ∣ b ∣ = 1$ ; conversely $\pm 1$ are roots of unity. Thus $μ (Q) = {\pm 1}$ by [F9].

2.3F5F6F8step 1.3step 1.4

The same group is obtained from the $S$ -unit theorem. By [F5] the ideals $(p_{1}), \dots, (p_{m})$ are nonzero prime ideals of $O_{Q} = Z$ by [F8], so $S_{Q} : = {(p_{1}), \dots, (p_{m})}$ is a finite set of nonzero prime ideals. The two rings agree on $Q^{\times}$ , and both contain $0$ . Indeed, let $0 \neq x = a / b \in R$ with $a \in Z$ and $b \in M$ , and let $p \notin S_{Q}$ be a nonzero prime ideal: then $b \notin p$ , because otherwise some prime $p_{i}$ with $p_{i} ∣ b$ lies in $p$ , so $(p_{i}) \subseteq p$ , and maximality of $(p_{i})$ with $p$ proper forces $p = (p_{i}) \in S_{Q}$ , a contradiction. Hence $b / 1$ is a unit of $Z_{p}$ and $x = (a / 1) (b / 1)^{- 1} \in Z_{p}$ , so $v_{p} (x) \geq 0$ by step 1.4 and $x \in O_{Q, S_{Q}}$ . Conversely, let $0 \neq x = a / b \in O_{Q, S_{Q}}$ be in lowest terms with $b > 0$ , and suppose a prime $q \notin S$ divides $b$ ; then $(q)$ is a nonzero prime ideal of $Z$ and $(q) \notin S_{Q}$ by the distinctness in [F5], so $v_{(q)} (x) \geq 0$ and step 1.4 gives $x \in Z_{(q)}$ , contradicting step 1.3. Therefore every prime divisor of $b$ lies in $S$ , that is $b \in M$ and $x = a / b \in R$ . Thus $O_{Q, S_{Q}} = R$ as subrings of $Q$ , and their unit groups inside $Q$ coincide: $O_{Q, S_{Q}}^{\times} = R^{\times}$ .

3.1F1step 2.1algebra

Consequently an element $x \in Q^{\times}$ is a unit of $R$ if and only if $e_{q} (x) = 0$ for every prime $q \notin S$ : a unit of $R$ lies in $R$ together with its inverse, so step 2.1 gives $e_{q} (x) \geq 0$ and $e_{q} (x^{- 1}) = - e_{q} (x) \geq 0$ for $q \notin S$ , and conversely the two inequalities $e_{q} (x) \geq 0$ , $- e_{q} (x) \geq 0$ for $q \notin S$ place both $x$ and $x^{- 1}$ in $R$ .

4.1F3step 3.1algebra

Therefore $R^{\times} = {x \in Q^{\times} : e_{q} (x) = 0$ for all $q \notin S} = {\pm p_{1}^{n_{1}} \dots p_{m}^{n_{m}} : n_{i} \in Z}$ : for the first description, an element with vanishing exponents outside $S$ has prime factorisation involving only primes of $S$ and a sign, and conversely a monomial $\pm p_{1}^{n_{1}} \dots p_{m}^{n_{m}}$ has all exponents outside $S$ equal to zero; the sign and the exponents in such an expression are unique by the uniqueness clause of [F3].

5.1F3step 4.1algebra

The assignment $(ε, n_{1}, \dots, n_{m}) \mapsto ε p_{1}^{n_{1}} \dots p_{m}^{n_{m}}$ is an isomorphism ${\pm 1} \times Z^{m} \to R^{\times}$ : it is a group homomorphism because the exponents add, it is surjective by step 4.1, and it is injective because $ε p_{1}^{n_{1}} \dots p_{m}^{n_{m}} = 1$ forces $ε = 1$ and $n_{1} = \dots = n_{m} = 0$ by the uniqueness of the factorisation of the positive integer obtained after moving negative exponents to the other side.

6.1F7step 5.1step 2.3step 2.2

By the $S$ -unit theorem [F7] applied to $K = Q$ and $S_{Q}$ , $O_{Q, S_{Q}}^{\times} ≅ μ (Q) \times Z^{r_{1} + r_{2} - 1 + ∣ S_{Q} ∣} = {\pm 1} \times Z^{1 + 0 - 1 + m} = {\pm 1} \times Z^{m},$ of rank $m$ . Combined with step 2.3 this agrees with the explicit computation of steps 4.1 and 5.1, and the $m$ generators are the classes of $p_{1}, \dots, p_{m}$ .

7.1A1F5F7step 1.1step 5.1∎

Scope and boundary cases: for $m = 0$ the set $M = {1}$ and $R = Z$ , and the computation returns $Z^{\times} = {\pm 1} ≅ {\pm 1} \times Z^{0}$ by [F4], while $S_{Q} = \emptyset$ has rank $0$ ; for $m \geq 1$ the exponents $n_{i}$ range over $Z$ and negative exponents are allowed, the generators being units because $p_{i} / 1 \in M$ has inverse $1 / p_{i}$ in $R$ . Choice enters through the $S$ -unit theorem [F7] and through the AC-qualified Dedekind interface of [F5]; the identification of $R$ with a subring of $Q$ , the exponent bookkeeping and the enumeration of signs use no choice.

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

Units of Z[√5] are a proper subgroup of the units of its maximal order

Statement refuted

Let $d > 1$ be squarefree. Whenever $d \equiv 1 (m o d 4)$ the Pell order $Z [\sqrt{d}]$ is a proper subring of the maximal order $O_{Q (\sqrt{d})}$ , and one might expect that this inclusion of rings is the only difference between them, so that their unit groups still coincide: $Z [\sqrt{d}]^{\times} = O_{Q (\sqrt{d})}^{\times},$ with the fundamental Pell solution $ε_{d}$ of $x^{2} - d y^{2} = 1$ generating the full unit group of the maximal order. This is false. At $d = 5$ , with $ε = (1 + \sqrt{5}) / 2$ , one has $O_{Q (\sqrt{5})} = Z [ε]$ , the element $2 + \sqrt{5} = ε^{3}$ generates the unit group of the Pell order, $Z [\sqrt{5}]^{\times} = {\pm (2 + \sqrt{5})^{n} : n \in Z} = {\pm ε^{3 n} : n \in Z}$ , while $O_{Q (\sqrt{5})}^{\times} = {\pm ε^{n} : n \in Z}$ ; the Pell order's unit group is a proper subgroup of index $3$ , and $ε$ itself is a unit of the maximal order that is not in $Z [\sqrt{5}]$ .

Facts & Assumptions

Given: The Axiom of Choice, the field $K = Q (\sqrt{5})$ , the Pell order $Z [\sqrt{5}]$ with its Pell norm $N_{5}$ (The norm on the explicit order $Z [\sqrt{D}]$ ), the fundamental Pell solution $ε_{5}$ of $x^{2} - 5 y^{2} = 1$ (The fundamental Pell solution), and the element $ε : = (1 + \sqrt{5}) / 2$ .

[F1]

$O_{K} = Z [(1 + \sqrt{5}) / 2] = Z [ε]$ , which strictly contains $Z [\sqrt{5}]$ , and its elements are the numbers $(x + y \sqrt{5}) / 2$ with $x, y \in Z$ and $x \equiv y (m o d 2)$ , of field norm $N_{K / Q} ((x + y \sqrt{5}) / 2) = (x^{2} - 5 y^{2}) / 4$ (Integers in a quadratic field, Real quadratic units and Pell's equation).

[F2]

For $u \in O_{K}$ , the element $u$ is a unit of the ring $O_{K}$ 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).

[F3]

The Pell norm on $Z [\sqrt{5}]$ is multiplicative, and $α = x + y \sqrt{5} \in Z [\sqrt{5}]$ is a unit of the order $Z [\sqrt{5}]$ if and only if $N_{5} (α) = x^{2} - 5 y^{2} = \pm 1$ (The norm on the explicit order $Z [\sqrt{D}]$ ). The norm-one integral solutions are exactly the elements $\pm ε_{5}^{k}$ , $k \in Z$ , and the positive ones are $ε_{5}^{k}$ for $k \geq 1$ (All integral Pell solutions are $\pm ε_{D}^{k}$ , All positive Pell solutions are powers of the fundamental solution, Integral Pell solutions form an abelian group).

[F4]

For $d = 5$ : the element $ε = (1 + \sqrt{5}) / 2 \in O_{K}$ has $N_{K / Q} (ε) = - 1$ , so it is a unit of $O_{K}$ ; direct multiplication gives $ε^{2} = (3 + \sqrt{5}) / 2$ and $ε^{3} = 2 + \sqrt{5}$ ; $2 + \sqrt{5}$ is the least positive solution of $x^{2} - 5 y^{2} = - 1$ and $ε_{5} = 9 + 4 \sqrt{5} = (2 + \sqrt{5})^{2}$ ; the unit group of the Pell order is $Z [\sqrt{5}]^{\times} = \pm ⟨ 2 + \sqrt{5} ⟩ = \pm ⟨ ε^{3} ⟩$ ; and $ε$ is the least unit $> 1$ of $O_{K}$ (Real quadratic units and Pell's equation, The fundamental Pell solution).

[F5]

Assume the Axiom of Choice. The maximal order has $O_{K}^{\times} ≅ {\pm 1} \times Z$ with torsion subgroup ${\pm 1}$ ; explicitly there is a unit $γ > 1$ with $O_{K}^{\times} = {\pm γ^{n} : n \in Z}$ , and then $γ$ is the least unit $> 1$ of $O_{K}$ (Unit ranks by signature).

[A1]

The Axiom of Choice is assumed; it is used only through the rank-one unit structure [F5] (The Axiom of Choice).

Counterexample

technique · direct; identify the units of the two orders by solving the norm equations $x^2-5y^2=\pm1$ and $x^2-5y^2=\pm4$, and compare the two cyclic groups through the common generator $\varepsilon$

1.1F1

The maximal order is $O_{K} = Z [ε]$ with $ε = (1 + \sqrt{5}) / 2$ , and $Z [\sqrt{5}] ⊊ O_{K}$ ; its elements are the $(x + y \sqrt{5}) / 2$ with $x \equiv y (m o d 2)$ , of nonzero norm when the element is nonzero because the norm is a product of the two embeddings.

1.2F2F4algebra

The element $ε$ is a unit of $O_{K}$ : its norm is $N_{K / Q} (ε) = \frac{1 + \sqrt{5}}{2} \cdot \frac{1 - \sqrt{5}}{2} = \frac{1 - 5}{4} = - 1,$ so [F2] applies; also $ε > 1$ because $\sqrt{5} > 1$ , and direct multiplication gives $ε^{2} = \frac{(1 + \sqrt{5})^{2}}{4} = \frac{3 + \sqrt{5}}{2}, ε^{3} = ε \cdot ε^{2} = \frac{(1 + \sqrt{5}) (3 + \sqrt{5})}{4} = \frac{8 + 4 \sqrt{5}}{4} = 2 + \sqrt{5} .$

2.1F3F4step 1.2

The units of the Pell order are $Z [\sqrt{5}]^{\times} = \pm ⟨ 2 + \sqrt{5} ⟩ = \pm ⟨ ε^{3} ⟩$ : an element $x + y \sqrt{5}$ of $Z [\sqrt{5}]$ is a unit of that order exactly when $x^{2} - 5 y^{2} = \pm 1$ by [F3], the norm-one solutions are $\pm ε_{5}^{k}$ , and $2 + \sqrt{5}$ is the least positive norm- $(- 1)$ solution, so every norm- $\pm 1$ solution is $\pm (2 + \sqrt{5})^{n} = \pm ε^{3 n}$ .

2.2F4F5step 1.2

The maximal order has $O_{K}^{\times} = {\pm ε^{n} : n \in Z}$ : by [F5] there is a unit $γ > 1$ with $O_{K}^{\times} = {\pm γ^{n} : n \in Z}$ , and then every unit $v > 1$ is $γ^{n}$ with $n \geq 1$ , hence $v \geq γ$ , so $γ$ is the least unit $> 1$ of $O_{K}$ ; by [F4] the element $ε > 1$ is the least unit $> 1$ of $O_{K}$ , so $γ = ε$ .

3.1step 1.1step 2.1step 2.2algebra

The inclusion of unit groups is proper: the element $ε$ lies in $O_{K}^{\times}$ by step 1.2, while $ε = (1 + \sqrt{5}) / 2$ is not of the form $x + y \sqrt{5}$ with $x, y \in Z$ and so does not lie in $Z [\sqrt{5}]$ , hence not in $Z [\sqrt{5}]^{\times}$ ; therefore $Z [\sqrt{5}]^{\times} = \pm ⟨ ε^{3} ⟩ ⊊ \pm ⟨ ε ⟩ = O_{K}^{\times}$ .

3.2step 2.1step 2.2algebra

The index is $3$ : the assignment $n \mapsto ε^{n}$ is an isomorphism $Z \to ⟨ ε ⟩ \subseteq O_{K}^{\times}$ , since $ε > 1$ forces $ε^{n - m} = 1$ only for $n = m$ . It maps $3 Z$ onto $⟨ ε^{3} ⟩$ , so $[⟨ ε ⟩ : ⟨ ε^{3} ⟩] = [Z : 3 Z] = 3$ ; multiplying by the common sign group ${\pm 1}$ does not change the index, hence $[O_{K}^{\times} : Z [\sqrt{5}]^{\times}] = 3$ .

4.1A1F4F5step 2.1step 2.2step 3.2∎

Conclusion: the maximal order $O_{K}$ of $K = Q (\sqrt{5})$ has unit group ${\pm ε^{n}}$ generated modulo its sign subgroup by $ε = (1 + \sqrt{5}) / 2$ , while the Pell order $Z [\sqrt{5}]$ has unit group ${\pm (2 + \sqrt{5})^{n}} = {\pm ε^{3 n}}$ , a proper subgroup of index $3$ . The fundamental Pell solution $ε_{5} = 9 + 4 \sqrt{5} = ε^{6}$ generates the norm-one Pell subgroup of the order up to sign; the maximal-order fundamental unit is $ε$ . The counterexample is the sharpened form of the design's warning for this pair: it identifies both groups and the exact index rather than merely exhibiting one missing unit. Choice is used only through the rank-one structure [F5]; all arithmetic in $Z [\sqrt{5}]$ and the comparisons of the two generators are elementary.

Dirichlets Unit Theorem Regulators and S Units — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

Units of QQ and the imaginary quadratic fields

Example

Facts & Assumptions

Verification

Real quadratic units and Pell's equation

Example

Facts & Assumptions

Verification

Two independent units in a real cubic field

Example

Facts & Assumptions

Verification

Regulator of a real quadratic field

Example

Facts & Assumptions

Verification

A unimodular change of generators preserves the regulator determinants

Example

Facts & Assumptions

Verification

S-units of Q

Example

Facts & Assumptions

Verification

Units of Z[√5] are a proper subgroup of the units of its maximal order

Statement refuted

Facts & Assumptions

Counterexample

Sources

Units of $Q$ and the imaginary quadratic fields