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Hilbert reciprocity over the rationals

Statement

For all $a, b \in Q^{\times}$ ,

\prod_{v \leq \infty} (a, b)_{v} = 1,

and all but finitely many factors are $1$ .

Facts & Assumptions

Given: Two nonzero rational numbers $a$ and $b$ .

[L1]

The Hilbert symbol is bilinear on the square-class group (The Hilbert symbol is a symmetric bilinear nondegenerate pairing).

[L2]

The explicit local formulas are known at $\infty$ , odd $p$ , and $2$ (The real Hilbert symbol formula, The odd-prime Hilbert symbol formula, The two-adic Hilbert symbol formula).

[L3]

Quadratic reciprocity and its two supplements are already proved (Quadratic reciprocity for distinct odd primes, First supplement: $(- 1 / p) = (- 1)^{(p - 1) / 2}$ , Second supplement: $(2 / p) = (- 1)^{(p^{2} - 1) / 8}$ ).

Proof

technique · direct

1.1

By [L1], the map $(a, b) \mapsto \prod_{v} (a, b)_{v}$ is bilinear on $Q^{\times} / (Q^{\times})^{2}$ . This square-class group is generated by the classes of $- 1$ , $2$ , and the odd primes. Hence it is enough to check the product formula on pairs of generators.

L1givenalgebra

1.2

For the pair $(- 1, - 1)$ , [L2] gives $(- 1, - 1)_{\infty} = - 1$ and $(- 1, - 1)_{2} = - 1$ , while $(- 1, - 1)_{p} = 1$ for every odd prime $p$ , so the global product is $1$ . For the pair $(- 1, 2)$ , the identity $- 1 \cdot 1^{2} + 2 \cdot 1^{2} = 1$ shows $(- 1, 2)_{v} = 1$ at every place $v$ , so the global product is again $1$ .

L2givenalgebra

2.1

Let $p$ be an odd prime. Then [L2] gives $(- 1, p)_{v} = 1$ for $v \notin {2, p}$ and $(- 1, p)_{2} = (- 1)^{(p - 1) / 2}, (- 1, p)_{p} = (\frac{- 1}{p}) = (- 1)^{(p - 1) / 2}$ by the first supplement from [L3], so the product for $(- 1, p)$ is $1$ . The same local formulas imply $(p, p)_{v} = (- 1, p)_{v}$ for every place $v$ , so the pair $(p, p)$ also has global product $1$ .

L2L3step 1.1algebra

2.2

If $p$ is an odd prime, then $(2, p)_{v} = 1$ for $v \notin {2, p}$ and $(2, p)_{2} = (- 1)^{(p^{2} - 1) / 8}, (2, p)_{p} = (\frac{2}{p}) = (- 1)^{(p^{2} - 1) / 8}$ by the second supplement from [L3], so the pair $(2, p)$ has global product $1$ .

L2L3step 1.1algebra

3.1

If $p$ and $q$ are distinct odd primes, then $(p, q)_{v} = 1$ for $v \notin {2, p, q}$ , while $(p, q)_{2} = (- 1)^{(p - 1) (q - 1) / 4}, (p, q)_{p} = (\frac{q}{p}), (p, q)_{q} = (\frac{p}{q}) .$ Quadratic reciprocity [L3] says that the product of these three terms is $1$ . Every generator pair therefore has global product $1$ , and bilinearity from step 1.1 gives the reciprocity law for all $a, b \in Q^{\times}$ .

L2L3step 1.1algebra∎

Depends on

Used by

One local place is determined by the others for ternary forms Corollary

Dependency tree · two levels

20 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 10.11 (standard reference, not scraped)
Sam Raskin, Introduction to the Arithmetic Theory of Quadratic Forms, section 4.4 (standard reference, not scraped)