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The two-adic Hilbert symbol formula

Statement

Write $a = 2^{α} u$ and $b = 2^{β} v$ with $α, β \in Z$ and odd units $u, v \in Z_{2}^{\times}$ . Put

ϵ (w) : = \frac{w - 1}{2} m o d 2, ω (w) : = \frac{w^{2} - 1}{8} m o d 2

for odd $w$ . Then

(a, b)_{2} = (- 1)^{ϵ (u) ϵ (v) + α ω (v) + β ω (u)} .

Facts & Assumptions

Given: Elements $a = 2^{α} u$ and $b = 2^{β} v$ in $Q_{2}^{\times}$ with odd units $u, v$ .

[L1]

The Hilbert symbol is equivalent to solvability of $z^{2} - a x^{2} - b y^{2} = 0$ and to the norm condition from $Q_{2} (\sqrt{a})$ (Equivalent formulations of the Hilbert symbol).

[L2]

The Hilbert symbol depends only on square classes (The Hilbert symbol depends only on square classes).

[L3]

An element of $Q_{2}^{\times}$ is a square exactly when its valuation is even and its odd unit part is $1$ modulo $8$ (Square criterion in Q_2).

Proof

technique · direct

1.1

By [L2], only the parities of $α, β$ and the odd unit classes modulo $8$ matter, so it is enough to treat $α, β \in {0, 1}$ and $u, v \in {\pm 1, \pm 5}$ . As in the odd-prime proof, [L1] gives three useful identities: the symbol is symmetric; $(- c, c)_{2} = 1$ for every $c$ ; and, if $(d, c)_{2} = 1$ , then $(b, c)_{2} = (b d, c)_{2}$ . In particular, $(2 u, 2 v)_{2} = (2 u, 2 v)_{2} (- 2 v, 2 v)_{2} = (- 4 u v, 2 v)_{2} = (- u v, 2 v)_{2} = (2 v, - u v)_{2},$ so the case $α = β = 1$ reduces to the case $α = 1, β = 0$ .

L1L2givenalgebra

2.1

First suppose $α = β = 0$ , so both arguments are odd units. If one of $u, v$ is $1$ , then the symbol is $1$ . The remaining positive cases are $(5, - 1)_{2}$ , $(5, 5)_{2}$ , and $(5, - 5)_{2}$ up to symmetry: they are witnessed respectively by $5 \cdot 1^{2} - 2^{2} = 1, 5 \cdot 1^{2} + 5 \cdot 2^{2} = 5^{2}, (- 5, 5)_{2} = 1.$ For the negative cases $u, v \in {- 1, - 5}$ , any primitive solution of $z^{2} = u x^{2} + v y^{2}$ would have at least one of $x, y$ odd. If exactly one of $x, y$ were odd, then the right-hand side would be congruent to $3$ or $7$ modulo $8$ ; if both were odd, it would be congruent to $2$ or $6$ modulo $8$ . None of these is a $2$ -adic square by [L3], so these pairs have symbol $- 1$ . Thus $(u, v)_{2} = (- 1)^{ϵ (u) ϵ (v)} .$

L1L3step 1.1algebra

3.1

Next suppose $α = 1$ and $β = 0$ . If $v = 1$ , then $(2 u, 1)_{2} = 1$ . If $v = - 1$ , the formula predicts $(- 1)^{ϵ (u)}$ : for $u = 1$ and $u = 5$ the identities $2 \cdot 1^{2} - 1^{2} = 1, 10 \cdot 1^{2} - 3^{2} = 1$ show that the symbol is $1$ , while for $u = - 1$ and $u = - 5$ every primitive value of $- 2 x^{2} - y^{2}$ or $- 10 x^{2} - y^{2}$ is congruent to $2$ , $5$ , $6$ , or $7$ modulo $8$ , so the symbol is $- 1$ by [L3]. If $v = 5$ , then for any primitive solution of $z^{2} = 2 u x^{2} + 5 y^{2}$ the right-hand side is congruent modulo $8$ to one of $5$ , $2 u$ , or $2 u + 5$ , namely to $5$ , $2$ , $6$ , $7$ , or $3$ ; none is a square, so $(2 u, 5)_{2} = - 1$ . If $v = - 5$ , the formula predicts $(- 1)^{ϵ (u) + 1}$ : for $u = - 1$ and $u = - 5$ the choice $x = y = 1$ gives right-hand sides $- 7$ and $- 15$ , both congruent to $1$ modulo $8$ and therefore square by [L3], so the symbol is $1$ ; for $u = 1$ and $u = 5$ , the same parity check as above shows that $2 u x^{2} - 5 y^{2}$ is never a square modulo $8$ , so the symbol is $- 1$ . Therefore $(2 u, v)_{2} = (- 1)^{ϵ (u) ϵ (v) + ω (v)}$ for every odd-unit representative $v$ .

L3step 1.1step 2.1algebra

4.1

Step 3.1 and symmetry give the case $α = 0, β = 1$ . For odd units modulo $8$ , direct calculation gives $ϵ (x y) = ϵ (x) + ϵ (y), ω (x y) = ω (x) + ω (y), ϵ (- 1) = 1, ω (- 1) = 0$ in $Z / 2 Z$ . In the remaining case $α = β = 1$ , step 1.1 and step 3.1 therefore give the exponent $ϵ (v) ϵ (- u v) + ω (- u v) = ϵ (u) ϵ (v) + ω (u) + ω (v),$ which is exactly the displayed formula. Hence the formula holds for all $a = 2^{α} u$ and $b = 2^{β} v$ in $Q_{2}^{\times}$ .

step 1.1step 2.1step 3.1algebra∎

Depends on

Used by

The two-adic Hilbert symbol table Example
Hilbert reciprocity over the rationals Theorem
The Hilbert symbol is a symmetric bilinear nondegenerate pairing Theorem

Dependency tree · two levels

7 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 10.9 (standard reference, not scraped)
Sam Raskin, Introduction to the Arithmetic Theory of Quadratic Forms, Proposition 3.16.3 (standard reference, not scraped)