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Quadratic reciprocity as a Frobenius restriction identity

Statement

Let $p \neq q$ be odd primes, let $ζ_{p}$ be a fixed primitive $p$ -th root of unity, let $τ_{p} = \sum_{a m o d p} (a / p) ζ_{p}^{a}$ be the quadratic Gauss sum attached to it, and put $p^{*} = (- 1)^{(p - 1) / 2} p$ . The arithmetic Frobenius ${Frob}_{q}$ of $q$ in $Q (ζ_{p})$ sends $ζ_{p} ⟼ ζ_{p}^{q}, τ_{p} ⟼ (\frac{q}{p}) τ_{p} .$ Its restriction to the quadratic subfield $Q (\sqrt{p^{*}})$ acts by $τ_{p} ⟼ (\frac{p^{*}}{q}) τ_{p},$ and consequently $(\frac{p^{*}}{q}) = (\frac{q}{p}) .$

Facts & Assumptions

Given: Distinct odd primes $p$ and $q$ , a fixed primitive $p$ -th root of unity $ζ = ζ_{p}$ , the Gauss sum $τ = τ_{p}$ attached to it, the field $K = Q (ζ)$ , the element $p^{*} = (- 1)^{(p - 1) / 2} p$ , and the automorphism $σ_{q} \in Gal (K / Q)$ with $σ_{q} (ζ) = ζ^{q}$ .

[F1]

The index $p$ is reduced and $q ∤ p$ ; hence $σ_{q}$ is the arithmetic Frobenius at every prime $P$ of $O_{K}$ above $q$ : it is the unique element of $Gal (K / Q)$ with $σ_{q} (x) \equiv x^{q} (m o d P)$ for all $x \in O_{K}$ , and it does not depend on the choice of $P$ (Arithmetic Frobenius is the power map in an unramified cyclotomic field, Arithmetic frobenius coset).

[F2]

For every integer $b$ not divisible by $p$ one has $σ_{b} (τ) = (b / p) τ$ , where $σ_{b} (ζ) = ζ^{b}$ ; in particular $σ_{q} (τ) = (q / p) τ$ (Galois action on the quadratic Gauss sum).

[F3]

$τ^{2} = p^{*}$ and $Q (τ) = Q (\sqrt{p^{*}})$ is the unique intermediate field $Q \subseteq F \subseteq K$ with $[F : Q] = 2$ ; moreover $τ \in O_{K}$ and $τ \notin Q$ (Square of the quadratic Gauss sum, Quadratic subfield generated by the Gauss sum, Quadratic Gauss sum in a prime cyclotomic field).

[F4]

Euler's criterion: for every integer $a$ and the odd prime $q$ , $(a / q) \equiv a^{(q - 1) / 2} (m o d q)$ , and $(a / q) \in {- 1, 0, 1}$ (Euler's criterion: $(a / p) \equiv a^{(p - 1) / 2} (m o d p)$ , The Legendre symbol, including its zero value).

[F5]

$q ∤ p^{*}$ : indeed $p^{*} = \pm p$ with $q \neq p$ , so $p^{*} \equiv̸ 0 (m o d q)$ . [given, arithmetic]

Proof

technique · direct

1.1F1F2F3

By [F1] the automorphism $σ_{q}$ sends $ζ$ to $ζ^{q}$ and is the arithmetic Frobenius at each prime above $q$ , while by [F2] it sends $σ_{q} (τ) = (q / p) τ$ ; since $(q / p) = \pm 1$ , it maps $τ$ to $\pm τ$ and therefore preserves $F = Q (τ)$ .

1.2F3F5

Let $P$ be a prime of $O_{K}$ above $q$ and let $κ = O_{K} / P$ be its residue field. Since $τ^{2} = p^{*}$ and $q ∤ p^{*}$ by [F5], we have $τ^{2} \equiv p^{*} \equiv̸ 0 (m o d P)$ , so the residue class of $τ$ in the field $κ$ is nonzero.

1.3F4

Euler's criterion [F4] with $a = p^{*}$ gives $(p^{*})^{(q - 1) / 2} \equiv (p^{*} / q) (m o d q)$ .

1.4F1F3

By the congruence property of [F1] applied to $x = τ \in O_{K}$ , for every prime $P$ above $q$ one has $σ_{q} (τ) \equiv τ^{q} (m o d P)$ .

2.1F3step 1.3

In $κ$ one has $τ^{q} = τ \cdot (τ^{2})^{(q - 1) / 2} = τ \cdot (p^{*})^{(q - 1) / 2} = (p^{*} / q) τ$ , the last equality because $(p^{*})^{(q - 1) / 2} \equiv (p^{*} / q) (m o d q)$ by step 1.3 and $P$ contains $q$ .

3.1step 1.1step 1.2step 1.4step 2.1

Fix $P$ above $q$ . Steps 1.1, 1.4 and 2.1 compare the same element in the field $κ$ and give $(q / p) τ = σ_{q} (τ) \equiv τ^{q} = (p^{*} / q) τ (m o d P)$ ; the residue class of $τ$ is nonzero by step 1.2, so cancellation gives $(\frac{q}{p}) \equiv (\frac{p^{*}}{q}) (m o d P)$ .

4.1step 3.1F4

Both $(\frac{q}{p})$ and $(\frac{p^{*}}{q})$ lie in ${- 1, 1}$ ; their difference lies in $P \cap Z$ , which is the prime ideal $(q)$ because $P$ lies above $q$ and $q Z$ is maximal. A difference of two elements of ${- 1, 1}$ that is divisible by the odd prime $q$ must be $0$ ; hence $(\frac{p^{*}}{q}) = (\frac{q}{p})$ .

5.1step 1.1step 4.1F3∎

The element $τ$ generates $F = Q (\sqrt{p^{*}})$ over $Q$ by [F3], and step 1.1 together with step 4.1 gives $σ_{q} (τ) = (\frac{q}{p}) τ = (\frac{p^{*}}{q}) τ$ ; therefore the restriction of the arithmetic Frobenius to the quadratic subfield $Q (\sqrt{p^{*}})$ acts on $τ$ by multiplication by $(\frac{p^{*}}{q})$ , that is, it is the identity if $(\frac{p^{*}}{q}) = 1$ and the nontrivial automorphism if $(\frac{p^{*}}{q}) = - 1$ .

Remarks

Arithmetic convention. The result uses the arithmetic Frobenius $ζ \mapsto ζ^{q}$ ; the geometric inverse would send $ζ$ to $ζ^{q^{- 1}}$ , whose exponent is a different nonzero class modulo $p$ in general, and the identity with $(\frac{q}{p})$ would then read with the inverse symbol.
No prior reciprocity. Neither this theorem nor its supplier Square of the quadratic Gauss sum uses quadratic reciprocity: the comparison is between two independently computed signs for the same residue class.

Depends on

Used by

Quadratic reciprocity via Frobenius Corollary
Frobenius restriction for p=5 and q=3 Example

Dependency tree · two levels

44 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

J. S. Milne, Algebraic Number Theory, Ch. 8, Example 8.19 (standard reference, not scraped)
Jerry Shurman, Math 361 Ninth Lecture, section 4 (standard reference, not scraped)
Conrad-Landesman, Math 154 Algebraic Number Theory, Ch. 12 (standard reference, not scraped)