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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.

Depends on

Used by

System of fundamental units Definition
The logarithmic unit image is discrete Lemma
Dirichlet unit theorem Theorem

Dependency tree · two levels

61 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. S. Milne, Algebraic Number Theory v3.08 (standard reference, not scraped)
William A. Stein, Algebraic Number Theory: A Computational Approach (standard reference, not scraped)
Andrew V. Sutherland, MIT 18.785 Lecture 15: Dirichlet's Unit Theorem (Fall 2021) (standard reference, not scraped)