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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

1.1L1L2L3L7given

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.

2.1L2L3L4L5step 1.1choose

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.

3.1L2L5L6step 1.1step 2.1choose∎

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 $G$ , $ψ = χ_{p}$ .

Depends on

Used by

The Legendre symbol is multiplicative for all integer numerators Theorem

Dependency tree · two levels

36 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

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}