PropositionStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-16

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

On the units, the Legendre symbol is the unique nontrivial homomorphism to ${\pm 1}$

Statement

Let $p$ be an odd prime. Restricted to $(Z / p)^{\times}$ , the Legendre symbol is the unique nontrivial homomorphism

$χ_{p} : (Z / p)^{\times} ⟶ {\pm 1} .$

It is surjective, and its kernel is the subgroup of nonzero square classes.

Facts & Assumptions

Given: An odd prime $p$ , the unit group $G = (Z / p)^{\times}$ , and the multiplicative group ${\pm 1}$ .

[L1]

On unit classes, the Legendre symbol takes values in ${\pm 1}$ and is representative-independent (The Legendre symbol is well defined on residue classes).

[L2]

The nonzero square classes form an index-two subgroup $Q$ of $G$ (The nonzero squares modulo an odd prime form an index-two subgroup).

[L3]

For a group homomorphism $f : G \to H$ , $\ker f = {g : f (g) = e_{H}}$ and $f$ is surjective exactly when its image is $H$ (The kernel and image of a group homomorphism).

[L4]

A group homomorphism satisfies $f (x y) = f (x) f (y)$ for every $x, y$ in its domain (Monoid homomorphism and group homomorphism).

[L5]

The group $G$ is cyclic of order $p - 1$ (For every prime $p$ , the multiplicative group $(Z / p Z)^{\times}$ is cyclic).

[L6]

Every group homomorphism preserves integer powers: $f (g^{n}) = f (g)^{n}$ (A group homomorphism automatically satisfies $f (e) = e^{'}$ and $f (g^{- 1}) = f (g)^{- 1}$ , and $f (g^{n}) = f (g)^{n}$ for every $n \in Z$ ; for monoid homomorphisms preservation of the identity must be assumed).

[L7]

For an odd prime $p$ , $(\frac{a}{p}) = 1$ when $p ∤ a$ and $a$ is a quadratic residue modulo $p$ , and $(\frac{a}{p}) = - 1$ when $p ∤ a$ and $a$ is a quadratic nonresidue modulo $p$ (The Legendre symbol, including its zero value).

Proof

technique · direct

1.1

By [L2] the quadratic-residue classes are exactly the members of $Q$ , so [L7] assigns the value $1$ to every class in $Q$ and, every unit class outside $Q$ being a nonresidue, the value $- 1$ to every class in the other coset; [L1] makes this independent of representatives. Thus its value-one set is exactly $Q$ , and both values occur.

L1L2L3L7given

2.1

Choose $h \in G ∖ Q$ , which is possible because [L2] gives index two. The two left cosets are $Q$ and $h Q$ , so every element outside $Q$ has the form $h q$ with $q \in Q$ . The cyclic group $G$ from [L5] is abelian, and $h^{2}$ is a square, hence belongs to $Q$ by [L2]. It follows that products from $Q \cdot Q$ , $Q \cdot h Q$ , and $h Q \cdot h Q$ lie respectively in $Q$ , $h Q$ , and $Q$ . The corresponding signs multiply as $1 \cdot 1 = 1$ , $1 \cdot (- 1) = - 1$ , and $(- 1)^{2} = 1$ . Hence $χ_{p} (x y) = χ_{p} (x) χ_{p} (y)$ , so [L4] makes $χ_{p}$ a homomorphism; step 1.1 and [L3] give its kernel and surjectivity.

L2L3L4L5step 1.1choose

3.1

Choose a generator $g$ of $G$ by [L5]. Were $g \in Q$ , every power of $g$ would lie in the subgroup $Q$ and so $G = ⟨ g ⟩ \subseteq Q$ , contradicting the index two of [L2]; hence $g \notin Q$ and step 1.1 gives $χ_{p} (g) = - 1$ , so $χ_{p} (g^{n}) = (- 1)^{n}$ by [L6]. If a homomorphism $ψ : G \to {\pm 1}$ sent $g$ to $1$ , then [L6] would make $ψ (g^{n}) = 1$ for every integer $n$ , so $ψ$ would be trivial. Every nontrivial $ψ$ therefore sends $g$ to $- 1$ , and [L6] gives $ψ (g^{n}) = (- 1)^{n} = χ_{p} (g^{n})$ for every $n$ . Since $g$ generates $G$ , $ψ = χ_{p}$ .

L2L5L6step 1.1step 2.1choose∎

Depends on

Used by

The Legendre symbol is multiplicative for all integer numerators Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 94 results over 20 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

W. Stein, Elementary Number Theory, Section 4.1 (standard reference, not scraped)

On the units, the Legendre symbol is the unique nontrivial homomorphism to {±1}{±1}