CorollaryStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-02

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Quadratic reciprocity via Frobenius

Statement

For distinct odd primes $p$ and $q$ , $(\frac{p}{q}) (\frac{q}{p}) = (- 1)^{(p - 1) (q - 1) / 4} .$

Facts & Assumptions

Given: Distinct odd primes $p$ and $q$ , and $p^{*} = (- 1)^{(p - 1) / 2} p$ .

[F1]

Quadratic Frobenius restriction identity: for distinct odd primes $p, q$ , $(\frac{p^{*}}{q}) = (\frac{q}{p})$ (Quadratic reciprocity as a Frobenius restriction identity).

[F2]

Legendre symbol multiplicativity: $(a b / q) = (a / q) (b / q)$ for all integers $a, b$ ; consequently $(((- 1)^{k} p) / q) = ((- 1) / q)^{k} (p / q)$ for every integer $k \geq 0$ , and $(a / q) \in {- 1, 0, 1}$ with $(a / q) = 0$ exactly when $q ∣ a$ (The Legendre symbol is multiplicative for all integer numerators, The Legendre symbol, including its zero value).

[F3]

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

[F4]

$q ∤ p$ , so $(p / q) \neq 0$ and hence $(p / q)^{2} = 1$ (The Legendre symbol, including its zero value).

Proof

technique · direct

1.1F2

By [F2], $p^{*} = (- 1)^{(p - 1) / 2} p$ gives $(\frac{p^{*}}{q}) = {(\frac{- 1}{q})}^{(p - 1) / 2} (\frac{p}{q})$ .

1.2F3algebra

By [F3], ${(\frac{- 1}{q})}^{(p - 1) / 2} = (- 1)^{((p - 1) / 2) ((q - 1) / 2)} = (- 1)^{(p - 1) (q - 1) / 4}$ , the exponent $(p - 1) (q - 1) / 4$ being an integer because $p - 1$ and $q - 1$ are even.

1.3F4

Since $p \neq q$ , the symbol $(p / q)$ is $\pm 1$ , so $(p / q)^{2} = 1$ .

2.1step 1.1step 1.2

Substituting step 1.2 into step 1.1 gives $(\frac{p^{*}}{q}) = (- 1)^{(p - 1) (q - 1) / 4} (\frac{p}{q})$ .

3.1F1step 2.1

Combining with [F1], $(\frac{q}{p}) = (\frac{p^{*}}{q}) = (- 1)^{(p - 1) (q - 1) / 4} (\frac{p}{q})$ .

4.1step 1.3step 3.1∎

Multiplying both sides of step 3.1 by $(\frac{p}{q})$ and using $(p / q)^{2} = 1$ from step 1.3 yields $(\frac{p}{q}) (\frac{q}{p}) = (- 1)^{(p - 1) (q - 1) / 4} {(\frac{p}{q})}^{2} = (- 1)^{(p - 1) (q - 1) / 4}$ .

Remarks

The earlier reciprocity theorem is not used. The only inputs are the Frobenius restriction identity, Legendre multiplicativity and the first supplement; in particular neither the published quadratic reciprocity theorem nor an analytic Gauss-sum sign is a supplier.
Symmetry check. $(p - 1) (q - 1) / 4$ is symmetric in $p$ and $q$ , as the product form must be; the two special values $p = q$ and the case $p = 2$ are excluded, the latter being exactly the case covered by the second supplement Second supplement from Frobenius on Q(zeta_8).

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

18 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)