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

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Equivalent formulations of the Hilbert symbol

Statement

Let $a, b \in Q_{v}^{\times}$ . The following are equivalent:

$(a, b)_{v} = 1$ .
The ternary form $z^{2} - a x^{2} - b y^{2}$ is isotropic over $Q_{v}$ .
$b$ is a norm from the quadratic algebra $Q_{v} (\sqrt{a})$ , equivalently $b = u^{2} - a v^{2}$ for some $u, v \in Q_{v}$ .

Facts & Assumptions

Given: A place $v$ of $Q$ and nonzero elements $a, b \in Q_{v}^{\times}$ .

[L1]

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

[L2]

Put $A_{a} : = Q_{v} [T] / (T^{2} - a)$ . Relative to the basis $(1, T)$ , multiplication by $u + v T$ has matrix $(\begin{matrix} u & a v \\ v & u \end{matrix})$ and determinant $u^{2} - a v^{2}$ . When $a$ is nonsquare this is the field norm of The norm $N_{K / F}$ and trace ${Tr}_{K / F}$ of a finite field extension; the same determinant defines the norm in the split quadratic algebra when $a$ is square.

Proof

technique · direct

1.1

By [L1], condition 1 means that $a x^{2} + b y^{2} = 1$ has a solution, and then $(x, y, 1)$ is a nontrivial zero of $z^{2} - a x^{2} - b y^{2}$ . Conversely, let $(x, y, z)$ be a nontrivial zero of $z^{2} - a x^{2} - b y^{2}$ . If $z \neq 0$ , dividing by $z^{2}$ gives a solution of $a (x / z)^{2} + b (y / z)^{2} = 1$ . If $z = 0$ , then $y \neq 0$ and $r : = x / y$ satisfies $b = - a r^{2}$ . The explicit choice $X : = \frac{1 + a}{2 a}, Y : = \frac{1 - a}{2 a r}$ then gives $a X^{2} + b Y^{2} = 1$ . Thus conditions 1 and 2 are equivalent.

L1givenalgebra

2.1

If condition 2 holds and $y \neq 0$ , then dividing the identity $z^{2} - a x^{2} - b y^{2} = 0$ by $y^{2}$ gives $b = (z / y)^{2} - a (x / y)^{2}$ . If instead $y = 0$ , then $a = (z / x)^{2}$ is a square, say $a = r^{2}$ , and $b = {(\frac{b + 1}{2})}^{2} - a {(\frac{b - 1}{2 r})}^{2} .$ So condition 2 implies that $b$ has the form $u^{2} - a v^{2}$ , which is exactly the norm condition in [L2]. Conversely, if $b = u^{2} - a v^{2}$ , then $(v, 1, u)$ is a nontrivial zero of $z^{2} - a x^{2} - b y^{2}$ , so condition 2 holds and step 1.1 returns condition 1. Hence conditions 1 and 3 are equivalent.

L2step 1.1algebra

3.1

Steps 1.1 and 2.1 prove the three formulations equivalent.

step 1.1step 2.1∎

Depends on

Used by

Hilbert symbols over the real numbers Example
Binary quadratic representation via the Hilbert symbol Lemma
Hasse-Minkowski for ternary forms over Q Theorem
The Hilbert symbol is a symmetric bilinear nondegenerate pairing Theorem
The odd-prime Hilbert symbol formula Theorem
The two-adic Hilbert symbol formula Theorem

Dependency tree · two levels

6 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, Lemma 10.2 (standard reference, not scraped)
Sam Raskin, Introduction to the Arithmetic Theory of Quadratic Forms, section 4.2 (standard reference, not scraped)