TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05

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The real Hilbert symbol formula

Statement

For $a, b \in R^{\times}$ ,

$(a, b)_{\infty} = {\begin{cases} 1, & if a > 0 or b > 0, \\ - 1, & if a < 0 and b < 0. \end{cases}$

Facts & Assumptions

Given: Nonzero real numbers $a$ and $b$ .

[L1]

By definition, $(a, b)_{\infty} = 1$ exactly when $a x^{2} + b y^{2} = 1$ has a real solution (The Hilbert symbol over a rational completion).

Proof

technique · direct

1.1

If $a > 0$ , then $x = 1 / \sqrt{a}$ and $y = 0$ solve $a x^{2} + b y^{2} = 1$ , so [L1] gives $(a, b)_{\infty} = 1$ ; the same argument works when $b > 0$ .

L1givenalgebra

2.1

If $a < 0$ and $b < 0$ , then $a x^{2} \leq 0$ and $b y^{2} \leq 0$ for all real $x, y$ , so $a x^{2} + b y^{2}$ can never equal $1$ . Therefore [L1] gives $(a, b)_{\infty} = - 1$ . Combining with step 1.1 proves the formula.

L1step 1.1algebra∎

Depends on

The Hilbert symbol over a rational completion

Used by

Hilbert symbols over the real numbers Example
Hilbert reciprocity over the rationals Theorem
The Hilbert symbol is a symmetric bilinear nondegenerate pairing Theorem

Dependency tree · two levels

2 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

Andrew V. Sutherland, 18.782 Lecture 10, Theorem 10.3 (standard reference, not scraped)
Sam Raskin, Introduction to the Arithmetic Theory of Quadratic Forms, section 4.1 (standard reference, not scraped)