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Square of the quadratic Gauss sum

Statement

For every odd prime $p$ and every primitive $p$ -th root of unity $ζ_{p}$ , with $τ_{p} = \sum_{a m o d p} (a / p) ζ_{p}^{a}$ , $τ_{p}^{2} = p^{*} = (- 1)^{(p - 1) / 2} p .$

Facts & Assumptions

Given: An odd prime $p$ , a fixed primitive $p$ -th root of unity $ζ = ζ_{p}$ in a fixed algebraic closure of $Q$ , and the Gauss sum $τ : = τ_{p} = \sum_{a m o d p} (a / p) ζ^{a}$ .

[F1]

$τ = \sum_{s = 1}^{p - 1} (s / p) ζ^{s} \in Z [ζ]$ , the term $a = 0$ vanishing because $(0 / p) = 0$ (Quadratic Gauss sum in a prime cyclotomic field, The Legendre symbol, including its zero value).

[F2]

The Legendre symbol is multiplicative for all integer numerators: $(u v / p) = (u / p) (v / p)$ for all $u, v \in Z$ (The Legendre symbol is multiplicative for all integer numerators).

[F3]

Restricted to $(Z / p)^{\times}$ the Legendre symbol is a surjective homomorphism onto ${\pm 1}$ with kernel the nonzero square classes (On the units, the Legendre symbol is the unique nontrivial homomorphism to ${\pm 1}$ ); hence $\sum_{u = 1}^{p - 1} (u / p) = 0$ , because choosing $v$ with $(v / p) = - 1$ and substituting $u \mapsto v u$ permutes the nonzero classes and multiplies the sum by $- 1$ , forcing it to be $0$ .

[F4]

First supplement: $(- 1 / p) = (- 1)^{(p - 1) / 2}$ (First supplement: $(- 1 / p) = (- 1)^{(p - 1) / 2}$ ).

[F5]

For an element $x$ with $x^{p} = 1$ and $x \neq 1$ one has $\sum_{s = 0}^{p - 1} x^{s} = \frac{x^{p} - 1}{x - 1} = 0$ , so $\sum_{s = 1}^{p - 1} x^{s} = - 1$ ; and $ζ$ has order $p$ , so $ζ^{1 + u} \neq 1$ exactly when $u \equiv̸ - 1 (m o d p)$ (The group $μ_{n} (K)$ of $n$ -th roots of unity in a field, and primitive $n$ -th roots of unity).

Proof

technique · direct

1.1F1

Squaring the sum of [F1] and multiplying out the finite product of sums gives $τ^{2} = \sum_{s = 1}^{p - 1} \sum_{t = 1}^{p - 1} (s / p) (t / p) ζ^{s + t}$ .

1.2F5

For every integer $u \equiv̸ 0 (m o d p)$ the inner geometric sum over $s$ is $\sum_{s = 1}^{p - 1} ζ^{s (1 + u)} = - 1$ when $u \equiv̸ - 1 (m o d p)$ , while for $u \equiv - 1 (m o d p)$ it equals $\sum_{s = 1}^{p - 1} 1 = p - 1$ .

2.1F2step 1.1

Multiplication by $u$ permutes the nonzero classes modulo $p$ , so substituting $t \equiv s u (m o d p)$ in the double sum of step 1.1 rewrites it as $τ^{2} = \sum_{u = 1}^{p - 1} \sum_{s = 1}^{p - 1} (s / p) (s u / p) ζ^{s (1 + u)} = \sum_{u = 1}^{p - 1} (u / p) \sum_{s = 1}^{p - 1} ζ^{s (1 + u)}$ , using multiplicativity of the Legendre symbol in the last equality.

3.1F3step 1.2step 2.1

Substituting the two evaluations of step 1.2, the contribution of $u \equiv - 1 (m o d p)$ is $(- 1 / p) (p - 1)$ and every other $u$ contributes $- (u / p)$ ; hence $τ^{2} = (- 1 / p) (p - 1) - (\sum_{u = 1}^{p - 1} (u / p) - (- 1 / p)) = p (- 1 / p) - \sum_{u = 1}^{p - 1} (u / p) = p (- 1 / p)$ , since the character sum vanishes by [F3].

4.1F4step 3.1∎

By the first supplement [F4], $(- 1 / p) = (- 1)^{(p - 1) / 2}$ , so $τ^{2} = (- 1)^{(p - 1) / 2} p = p^{*}$ .

Remarks

Only finite sums and the first supplement are used. Neither quadratic reciprocity nor any analytic determination of the sign of $τ_{p}$ enters; the square $p^{*}$ is insensitive to replacing $ζ_{p}$ by another primitive $p$ -th root, since that multiplies $τ_{p}$ by $\pm 1$ .
Nonvanishing. Since $p^{*} = \pm p \neq 0$ in the domain $Z [ζ_{p}]$ , the identity shows that $τ_{p} \neq 0$ , which is what makes the fixed field computation in Quadratic subfield generated by the Gauss sum possible.

Depends on

Used by

The sign of a quadratic Gauss sum needs a chosen primitive root Counterexample
Frobenius restriction for p=5 and q=3 Example
Quadratic Gauss sum at p=3 Example
Quadratic Gauss sum at p=5 Example
Quadratic subfield of Q(zeta₇) Example
Quadratic reciprocity as a Frobenius restriction identity Theorem
Quadratic subfield generated by the Gauss sum Theorem

Dependency tree · two levels

24 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

Jerry Shurman, Math 361 Ninth Lecture, section 2 (standard reference, not scraped)
J. S. Milne, Algebraic Number Theory, Ch. 8, Example 8.19 (standard reference, not scraped)