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The sign of a quadratic Gauss sum needs a chosen primitive root

Statement refuted

"For a fixed odd prime $p$ the quadratic Gauss sum $τ_{p}$ is independent of the choice of the primitive $p$ -th root of unity used to define it, so that its sign is a function of $p$ alone."

Facts & Assumptions

Given: The prime $p = 3$ , a primitive third root of unity $ζ$ with $ζ \neq 1$ , the second primitive third root $ζ^{'} : = ζ^{2}$ , and the Gauss sums $τ = \sum_{a m o d 3} (a / 3) ζ^{a}$ and $τ^{'} = \sum_{a m o d 3} (a / 3) (ζ^{'})^{a}$ attached to them (Quadratic Gauss sum in a prime cyclotomic field).

[F1]

In the cyclic group $μ_{3}$ of third roots of unity the elements $ζ$ and $ζ^{2}$ are the two primitive third roots of unity, and $Q (ζ^{'}) = Q (ζ)$ (The $n$ -th roots of a complex number and the $n$ distinct roots of unity for every $n \geq 1$ , The cyclotomic extension $K (μ_{n})$ as a splitting field of $t^{n} - 1$ ).

[F2]

$Gal (Q (ζ) / Q) ≅ (Z / 3)^{\times}$ acts by $σ_{b} (ζ) = ζ^{b}$ for $b = 1, 2$ , so $σ_{2}$ is an automorphism of $Q (ζ)$ and $σ_{2} (ζ) = ζ^{2} = ζ^{'}$ ( $[Q (ζ_{n}) : Q] = φ (n)$ and $Gal (Q (μ_{n}) / Q) ≅ (Z / n)^{\times}$ ).

[F3]

For every integer $b$ not divisible by $3$ , the automorphism $σ_{b}$ of $Q (ζ)$ satisfies $σ_{b} (τ) = (b / 3) τ$ (Galois action on the quadratic Gauss sum); and $(2 / 3) = - 1$ while $(1 / 3) = 1$ (The Legendre symbol, including its zero value).

[F4]

For every odd prime $p$ , $τ_{p}^{2} = p^{*} = (- 1)^{(p - 1) / 2} p$ ; at $p = 3$ this is $τ^{2} = - 3$ , and the corresponding statement holds for $τ^{'}$ because $ζ^{'}$ is again a primitive third root of unity (Square of the quadratic Gauss sum).

[F5]

For an odd prime $p$ , $Q (τ_{p}) = Q (\sqrt{p^{*}})$ is the unique intermediate field of $Q (ζ_{p}) / Q$ of degree $2$ (Quadratic subfield generated by the Gauss sum); at $p = 3$ this is $Q (τ) = Q (τ^{'}) = Q (\sqrt{- 3}) = Q (ζ)$ .

[F6]

For the standard complex root $ζ_{3} = e^{2 π i / 3}$ one has $τ_{3} = ζ_{3} - ζ_{3}^{2} = i \sqrt{3}$ (Quadratic Gauss sum at p=3).

Counterexample

technique · direct

1.1F1F2

Replacing the primitive root $ζ$ by $ζ^{'} = ζ^{2}$ rewrites the attached Gauss sum as $τ^{'} = \sum_{a m o d 3} (a / 3) (ζ^{2})^{a} = \sum_{a m o d 3} (a / 3) σ_{2} (ζ)^{a} = σ_{2} (τ)$ , where $σ_{2} \in Gal (Q (ζ) / Q)$ ; the two sums are formed from two different primitive third roots of unity of the same field.

1.2F3

$(2 / 3) = - 1$ : the nonzero square class modulo $3$ is $1$ , and $2 \equiv - 1$ is a nonresidue modulo $3$ .

1.3F4

By [F4], $τ^{2} = (τ^{'})^{2} = - 3 \neq 0$ , so $τ \neq 0$ and $τ^{'} \neq 0$ ; in particular $τ^{'} = - τ$ will mean $τ^{'} \neq τ$ .

2.1F3step 1.1step 1.2

Applying [F3] with $b = 2$ gives $τ^{'} = σ_{2} (τ) = (2 / 3) τ = - τ$ , so the Gauss sum changes sign when the primitive root is replaced by its square.

3.1F5F6step 1.3step 2.1∎

Steps 1.3 and 2.1 show $τ^{'} \neq τ$ , yet $τ^{' 2} = τ^{2} = - 3$ and $Q (τ^{'}) = Q (τ) = Q (\sqrt{- 3})$ ; thus the square of the Gauss sum and the field it generates are unchanged, while the sign of the sum depends on the chosen primitive root. In the standard complex embedding, $ζ = e^{2 π i / 3}$ gives $τ = i \sqrt{3}$ by [F6], while the same computation with $ζ^{'} = ζ^{2}$ gives $τ^{'} = - i \sqrt{3}$ ; the value differs and no root-independent sign is well defined.

Remarks

What remains canonical. Only the square $τ_{p}^{2} = p^{*}$ and the field $Q (τ_{p}) = Q (\sqrt{p^{*}})$ are independent of the chosen primitive root; the element itself is determined only up to the sign $(b / p)$ of the automorphism relating two chosen roots.
No contradiction with the analytic sign theorem. Statements that fix a standard complex root $ζ_{p} = e^{2 π i / p}$ (or an embedding together with a root) do determine the sign, e.g. $τ_{3} = i \sqrt{3}$ here; the counterexample only refutes root-independence.

Depends on

Used by

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Sources

Jerry Shurman, Math 361 Ninth Lecture, sections 2-4 (standard reference, not scraped)
J. S. Milne, Algebraic Number Theory, Ch. 8, Example 8.19 (standard reference, not scraped)