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The odd-prime Hilbert symbol formula

Statement

Let $p$ be odd, and write $a = p^{α} u$ , $b = p^{β} v$ with $α, β \in Z$ and $u, v \in Z_{p}^{\times}$ . Then

for a $p$ -adic unit $w$ , write $(\frac{w}{p})$ for the Legendre symbol of any integer representative of its nonzero residue class modulo $p$ . With this convention,

$(a, b)_{p} = (- 1)^{α β (p - 1) / 2} {(\frac{u}{p})}^{β} {(\frac{v}{p})}^{α} .$

Facts & Assumptions

Given: An odd prime $p$ , elements $a = p^{α} u$ and $b = p^{β} v$ in $Q_{p}^{\times}$ , and unit parts $u, v \in Z_{p}^{\times}$ .

[L1]

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

[L2]

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

[L3]

The Legendre symbol of an integer detects whether its nonzero residue class is a square modulo $p$ (The Legendre symbol, including its zero value, Euler's criterion: $(a / p) \equiv a^{(p - 1) / 2} (m o d p)$ ); hence the notation $(\frac{w}{p})$ above is well defined for $w \in Z_{p}^{\times}$ .

[L4]

The square criterion in $Q_{p}$ for odd $p$ is parity of valuation plus a square residue unit (Square criterion in Q_p for odd p).

[L5]

A simple root modulo $p$ lifts to a $p$ -adic root (Simple roots lift uniquely in Z_p).

Proof

technique · direct

1.1

By [L2], only the parities of $α, β$ and the unit square classes of $u, v$ matter, so it is enough to treat $α, β \in {0, 1}$ . We also use three consequences of [L1]. First, the defining equation is symmetric in $a$ and $b$ , so $(a, b)_{p} = (b, a)_{p}$ . Second, if $c$ is a square then $(a, c)_{p} = 1$ . Third, if $(a, c)_{p} = 1$ then the norm subgroup from $Q_{p} (\sqrt{a})$ is multiplicative, so $(a, b c)_{p} = (a, b)_{p}$ ; by symmetry the same cancellation rule holds in the first argument.

L1L2givenalgebra

2.1

If $α = β = 0$ , both arguments are units. Consider the sets $U : = {u x^{2} m o d p : x \in F_{p}}, V : = {1 - v y^{2} m o d p : y \in F_{p}} .$ Each has $(p + 1) / 2$ elements, so they intersect. Hence there exist $x_{0}, y_{0} \in F_{p}$ with $u x_{0}^{2} + v y_{0}^{2} \equiv 1 (m o d p)$ . At least one of $x_{0}, y_{0}$ is nonzero, so one partial derivative of $u x^{2} + v y^{2} - 1$ is nonzero at $(x_{0}, y_{0})$ modulo $p$ ; [L5] lifts this solution to $Z_{p}$ . Therefore $(u, v)_{p} = 1$ , agreeing with the displayed formula when $α = β = 0$ .

L3L5step 1.1algebra

3.1

Suppose $α = 1$ and $β = 0$ . Step 2.1 gives $(u, v)_{p} = 1$ , so the cancellation rule from step 1.1 yields $(p u, v)_{p} = (p, v)_{p}$ . If $v$ is a square unit, then [L4] and step 1.1 give $(p, v)_{p} = 1$ . If $v$ is a nonsquare unit and $(p, v)_{p} = 1$ , then [L1] gives a primitive solution of $z^{2} - p x^{2} - v y^{2} = 0$ over $Z_{p}$ . The congruence $z^{2} \equiv v y^{2} (m o d p)$ forces $y$ to be divisible by $p$ , for otherwise [L4] would make $v$ a square in $Q_{p}$ . Then $z^{2} = p x^{2} + p^{2} (\dots)$ , so primitivity forces $x$ to be a unit and therefore $v_{p} (z^{2}) = 1$ , impossible. Hence $(p, v)_{p} = - 1$ in the nonsquare case. By [L3] and [L4], this is exactly $(v / p)$ , so $(p u, v)_{p} = (\frac{v}{p}) .$

L1L3L4step 1.1step 2.1algebra

4.1

By symmetry, the case $α = 0, β = 1$ gives $(u, p v)_{p} = (\frac{u}{p}) .$

step 1.1step 3.1algebra

4.2

When $α = β = 1$ , step 1.1 gives $(- p v, p v)_{p} = 1$ , so $(p u, p v)_{p} = (p u, p v)_{p} (- p v, p v)_{p} = (- p^{2} u v, p v)_{p} = (- u v, p v)_{p} = (p v, - u v)_{p} .$ Now apply step 3.1 with $- u v$ in place of $v$ : $(p v, - u v)_{p} = (\frac{- u v}{p}) = (\frac{- 1}{p}) (\frac{u}{p}) (\frac{v}{p}) = (- 1)^{(p - 1) / 2} (\frac{u}{p}) (\frac{v}{p}),$ where the last identity is Euler's criterion from [L3]. This matches the displayed formula for $α = β = 1$ .

L3step 1.1step 3.1algebra

5.1

Steps 2.1 through 4.2 settle all four parity cases, so the claimed formula holds for all $a = p^{α} u$ and $b = p^{β} v$ .

step 2.1step 3.1step 4.1step 4.2∎

Depends on

Used by

A Hilbert-symbol computation at an odd prime Example
Hasse-Minkowski theorem over Q Theorem
Hilbert reciprocity over the rationals Theorem
The Hilbert symbol is a symmetric bilinear nondegenerate pairing Theorem

Dependency tree · two levels

19 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 10.7 (standard reference, not scraped)
Sam Raskin, Introduction to the Arithmetic Theory of Quadratic Forms, section 4.3 (standard reference, not scraped)