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Quadratic subfield generated by the Gauss sum

Statement

For an odd prime $p$ , the unique intermediate field $Q \subseteq F \subseteq Q (ζ_{p})$ with $[F : Q] = 2$ is $Q (τ_{p}) = Q (\sqrt{p^{*}}), p^{*} = (- 1)^{(p - 1) / 2} p,$ where $τ_{p}$ is the quadratic Gauss sum attached to a chosen primitive $p$ -th root of unity $ζ_{p}$ . At $p = 3$ the field $Q (ζ_{3})$ itself has degree two over $Q$ , so the intermediate field of degree two may equal the full cyclotomic field.

Facts & Assumptions

Given: An odd prime $p$ , a fixed primitive $p$ -th root of unity $ζ$ 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]

$τ = \sum_{a = 1}^{p - 1} (a / p) ζ^{a} \in Z [ζ] \subseteq Q (ζ)$ , the term $a = 0$ vanishing, and the chosen root $ζ$ is part of the data (Quadratic Gauss sum in a prime cyclotomic field, The Legendre symbol, including its zero value).

[F2]

$τ^{2} = p^{*} = (- 1)^{(p - 1) / 2} p$ (Square of the quadratic Gauss sum).

[F3]

For every integer $b$ not divisible by $p$ , the automorphism $σ_{b} (ζ) = ζ^{b}$ satisfies $σ_{b} (τ) = (b / p) τ$ ; consequently the Gauss sum built from another primitive $p$ -th root $ζ^{b}$ equals $(b / p) τ$ and generates the same field as $τ$ (Galois action on the quadratic Gauss sum).

[F4]

$Q (ζ_{p})$ is a finite Galois extension of $Q$ with group $G : = Gal (Q (ζ_{p}) / Q)$ ; the embedding $G \to (Z / p)^{\times}$ taking $σ$ with $σ (ζ) = ζ^{b}$ to $b$ is an isomorphism, and $[Q (ζ_{p}) : Q] = p - 1$ . The group $(Z / p)^{\times}$ is cyclic of order $p - 1$ , so $G$ is cyclic of order $p - 1$ ( $[Q (ζ_{n}) : Q] = φ (n)$ and $Gal (Q (μ_{n}) / Q) ≅ (Z / n)^{\times}$ , For every prime $p$ , the multiplicative group $(Z / p Z)^{\times}$ is cyclic, The cyclotomic extension $K (μ_{n})$ as a splitting field of $t^{n} - 1$ ).

[F5]

In a finite cyclic group of order $n$ , every positive divisor $d$ of $n$ is the order of exactly one subgroup, and every subgroup is the unique subgroup of its own order; in particular the cyclic group $G$ of [F4] has exactly one subgroup of order $(p - 1) / 2$ (A finite cyclic group has exactly one subgroup of each order dividing its own).

[F6]

Fundamental theorem of finite Galois theory: for a finite Galois extension $K / F$ with group $G$ , the maps $H \mapsto K^{H}$ and $E \mapsto Gal (K / E)$ are mutually inverse bijections between subgroups of $G$ and intermediate fields, and $[K^{H} : F] = [G : H]$ ; in particular intermediate fields of degree two over $F$ correspond bijectively to subgroups of index two (The fundamental theorem of finite Galois theory).

[F7]

Lagrange's theorem: for a finite group $G$ and a subgroup $H \leq G$ one has $∣ G ∣ = [G : H] ∣ H ∣$ (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

[F8]

The Legendre symbol on $(Z / p)^{\times}$ is a homomorphism onto ${\pm 1}$ , so $(b / p)^{- 1} = (b / p)$ for every nonzero class $b$ (On the units, the Legendre symbol is the unique nontrivial homomorphism to ${\pm 1}$ ).

Proof

technique · direct

1.1F1F2

The sum $τ$ is an element of $Z [ζ]$ , hence of $Q (ζ)$ , and it satisfies $τ^{2} = p^{*} = (- 1)^{(p - 1) / 2} p$ .

1.2F1F3F8

If $ζ^{'} = ζ^{b}$ is any primitive $p$ -th root of unity, then the Gauss sum built from $ζ^{'}$ is $\sum_{a = 1}^{p - 1} (a / p) ζ^{a b} = σ_{b} (τ) = (b / p) τ$ , a nonzero multiple of $τ$ ; hence it generates the same subfield $Q (τ)$ .

1.3F2

The rational number $p^{*} = \pm p$ is not a square in $Q$ : if $x = a / b$ in lowest terms had $x^{2} = p^{*}$ , then $a^{2} = p^{*} b^{2}$ ; for $p^{*} = p$ the exponent of $p$ on the left side of $a^{2} = p b^{2}$ is $2 v_{p} (a)$ , while on the right it is $1 + 2 v_{p} (b)$ , which is odd, a contradiction, and for $p^{*} = - p$ the left side $a^{2} / b^{2}$ is positive while $- p$ is negative.

2.1step 1.1step 1.3

Since $τ^{2} = p^{*} \neq 0$ , we have $τ \neq 0$ , and $τ \notin Q$ by step 1.3; as $τ$ is a root of the degree-two polynomial $X^{2} - p^{*} \in Q [X]$ , the degree $[Q (τ) : Q]$ divides $2$ and is not $1$ , so it equals $2$ , and $Q (τ) \subseteq Q (ζ)$ by step 1.1. Moreover $τ$ is a square root of $p^{*}$ , so $τ = \pm \sqrt{p^{*}}$ and $Q (τ) = Q (\sqrt{p^{*}})$ .

3.1F4F5F6F7step 2.1

Let $F \subseteq Q (ζ_{p})$ be any intermediate field with $[F : Q] = 2$ and put $H : = Gal (Q (ζ_{p}) / F)$ . By [F6] the subgroup $H$ has index $[G : H] = [F : Q] = 2$ in $G$ , so [F7] together with $∣ G ∣ = p - 1$ from [F4] gives $∣ H ∣ = (p - 1) / 2$ . As $(p - 1) / 2$ is a positive divisor of $p - 1$ , [F5] shows that the cyclic group $G$ has exactly one subgroup of order $(p - 1) / 2$ , so $H$ is that same subgroup for every such $F$ ; the bijection [F6] then gives $F = Q (ζ_{p})^{H}$ , the same field for every such $F$ . Since step 2.1 exhibits $Q (τ)$ as one of them, $Q (τ)$ is the unique degree-two intermediate field.

4.1F4step 1.2step 2.1step 3.1∎

Combining steps 1.2 and 3.1 with the identification of step 2.1: the unique degree-two intermediate field is $Q (τ_{p}) = Q (\sqrt{p^{*}})$ , independently of the chosen primitive $p$ -th root, and for $p = 3$ one has $[Q (ζ_{3}) : Q] = p - 1 = 2$ by [F4], so the unique degree-two intermediate field there is $Q (ζ_{3})$ itself.

Remarks

Nonvanishing is the decisive input. Without $τ_{p}^{2} = p^{*} \neq 0$ the element $τ_{p}$ might generate only $Q$ ; the square computation of Square of the quadratic Gauss sum is what makes the generated field quadratic.
The sign of $τ_{p}$ is not needed. Replacing $ζ_{p}$ by $ζ_{p}^{b}$ multiplies $τ_{p}$ by the Legendre sign $(b / p)$ , which leaves both the square and the generated field unchanged; this is the content of the sign counterexample on the companion examples page.

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 subfield of Q(zeta₇) Example
Quadratic reciprocity as a Frobenius restriction identity Theorem

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Sources

Jerry Shurman, Math 361 Ninth Lecture, sections 2-3 (standard reference, not scraped)
J. S. Milne, Algebraic Number Theory, Ch. 8, Example 8.19 (standard reference, not scraped)