LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05

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Character values on units are roots of unity

Statement

Let $χ$ be a Dirichlet character modulo $q$ . If $(n, q) = 1$ , then $χ (n)$ is a root of unity and $\overline{χ (n)} = χ (n)^{- 1}$ . If $(n, q) > 1$ , then $χ (n) = 0$ .

Facts & Assumptions

Given: A Dirichlet character $χ$ modulo $q$ and an integer $n$ .

[L1]

A Dirichlet character modulo $q$ is a homomorphism on $(Z / q Z)^{\times}$ , extended by zero on nonunits (Dirichlet characters modulo q).

[L2]

The extension vanishes exactly when $(n, q) > 1$ (Extension by zero is well defined and periodic).

[A1]

The finite group $(Z / q Z)^{\times}$ has finite order, so every element of it has finite order.

Proof

technique · direct

1.1

If $(n, q) > 1$ , then [L2] gives $χ (n) = 0$ . Assume now that $(n, q) = 1$ . By [A1], the unit class $\overset{ˉ}{n}$ has some positive order $m$ , so ${\overset{ˉ}{n}}^{m} = \overset{ˉ}{1}$ . Applying the homomorphism of [L1] gives $χ (n)^{m} = \overset{ˉ}{χ} (\overset{ˉ}{n})^{m} = \overset{ˉ}{χ} (\overset{ˉ}{1}) = 1$ . Thus $χ (n)$ is a root of unity, hence nonzero.

L1L2A1givenalgebra

2.1

For a nonzero complex number on the unit circle, complex conjugation equals reciprocal. Since step 1.1 gives $χ (n)^{m} = 1$ , the value $χ (n)$ lies on the unit circle, so $\overline{χ (n)} = χ (n)^{- 1}$ . Together with the nonunit case from step 1.1, this proves the statement.

step 1.1algebra∎

Depends on

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Sources

Kiran S. Kedlaya, Notes on Analytic Number Theory, section 3.1 (standard reference, not scraped)
Andrew V. Sutherland, Number Theory I, section 18.2 (standard reference, not scraped)