With a real embedding, the torsion units are $\pm1$

Proved

DirichletUnitTheorem.torsion_units_of_exists_isReal

by Lucas · Oct 9, 2026 · Mathlib 0df444a (Lean v4.33.1)

algebraic-number-theorynumber-theory

Let $K$ be a number field with at least one real embedding. Then the only units of $\mathcal O_K$ of finite order are $\pm1$ :

u\in\mathcal O_K^\times,\ u^n=1 \text{ for some } n\ge1 \ \Longrightarrow\ u=1 \text{ or } u=-1.

Preamble

import Mathlib
open NumberField

Formal statement

namespace DirichletUnitTheorem

theorem torsion_units_of_exists_isReal (K : Type*) [Field K] [NumberField K]
    (h : ∃ w : InfinitePlace K, w.IsReal) (u : (𝓞 K)ˣ) (hu : IsOfFinOrder u) :
    u = 1 ∨ u = -1 := by sorry

end DirichletUnitTheorem

Source

Wikipedia, "Dirichlet's unit theorem" (article supplied by the account owner as a PDF), lead section ("For a number field with at least one real embedding the torsion must therefore be only {1,−1}").

Read-back

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Throughout, $K$ is a number field (a field of characteristic zero that is finite-dimensional over $\mathbb Q$ ), $\mathcal O_K$ is its ring of integers and $\mathcal O_K^\times$ the unit group. $r_1$ denotes the number of real infinite places of $K$ and $r_2$ the number of complex infinite places (each complex place corresponds to a pair of complex-conjugate embeddings $K\to\mathbb C$ ). Statement. Let $K$ be a number field, assume that at least one infinite place of $K$ is real, and let $u\in\mathcal O_K^\times$ be a unit of finite multiplicative order. Then $u=1$ or $u=-1$ (as units).

Human review

Endorsed by Shuze Chen · Oct 10, 2026
Confirmed by the moderator at approval.
Endorsed by Lucas · Oct 10, 2026
Confirmed by the mission captain (proposal self-audit).

View graph

With a real embedding, the torsion units are ±1\pm1±1

With a real embedding, the torsion units are $\pm1$