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# Law of quadratic reciprocity: Statement and demonstration

## Definition: Symbol of Legendre

For p odd prime and a ⩾ 1, we define the Legendre symbol by;

\left (\dfrac{a}{p} \right) = \left \{ \begin{array}{lll} 1& \text{if a is a square in }& \mathbb{F}_p^*\\ - 1& \text{if a is not a square in}& \mathbb{F}_p^*\\ 0 & \text{if } p | a& \mathbb{F}_p^*\\ \end{array}\right.

Moreover, for p prime, we denote

\mathbb{F}_p = \Z / p\Z

## Statement of the law of quadratic reciprocity

If p and q are two prime numbers odd distinct, then

\left (\dfrac{p}{q} \right)\left (\dfrac{q}{p} \right) = (-1)^{\frac{p-1}{2}\frac{q- 1}{2}}

## Demonstration of the law of quadratic reciprocity

To prove this result, we start by stating and proving a lemma

### Lemma

If p is odd prime and

a \in \mathbb{F}_p^*

then

|\{ x \in \mathbb{F}_p|ax^2 = 1 \}| = 1 + \left (\dfrac{a}{p} \right)

### Proof of the lemma

a is a square if and only if a-1 is also a square. So in terms of solutions, the equation ax2 = 1 which is equivalent to

x^2 = a^{-1}

has as many solutions as the equation

x^2 = a

Furthermore :

• If b is a square, the polynomial X2 – ba two distinct root roots denoted c and -c.
• If b is not a square, this equation has no solution

We then verify the desired equality! End of the lemma.

### 1st step: Study of a first set

Consider the quadratic form defined by

\forall x = (x_1, \ldots, x_p) \in \mathbb{F}_q^p, \varphi(x) = x_1^2 + \ldots + x_p^2

Note also

X = \{ x \in \mathbb{F}_q^p, \varphi(x) = 1\}

the unit ball.
We're going to use a group action, swapping the coordinates. We define Φ by

\Phi: \mathbb{F}_p \times X \to X

by

\Phi(\alpha, (x_1, \ldots, x_p)) = (x_{1+ \alpha [p]}, \ldots, x_{p + \alpha [p]})

The orbit-stabilizer relationship gives us:

\forall x \in \mathbb{F}_p ^q, |Stab_x|.|Orb_x| = |\mathbb{F}_p | =p

p being prime, the orbits have two possible sizes:

• Of size p, in this case, the stabilizer is {e}.
• Of size 1, the stabilizers are therefore of the form
\{(x, \ldots,x), x \in \mathbb{F}_q^p \}

and therefore verifying px2 = 1. There are therefore, according to the lemma demonstrated above,

1 + \left ( \dfrac{p}{q} \right)

Since the orbits are disjoint, we have

\begin{array}{ll} |X| &= \displaystyle \sum |\Omega_p| + \sum |\Omega_1|\\ & = \left( 1 + \left( \dfrac{p}{q} \right) \right) \times 1 + k \times p \\ & \equiv1 + \left( \dfrac{p}{q} \right) [p] \end{array}

### 2nd step: Study of a second set

We define the quadratic form φ' by

\forall x \in \mathbb{F}_q^p, \varphi '(x_1, \ldots,x_p ) = 2(x_1x_2 + \ldots + x_{p-2}x_{p-1})+(-1 )^{\frac{p-1}{2} }x_p^2

We also note here X' the set

\{ x \in \mathbb{F}_q^p, \varphi'(x) = 1 \}

the ball for φ′. We pose

d = \frac{p-1}{2}, a = (-1)^d

If we write the matrix of φ′ in the canonical basis, we get:

A = \begin{pmatrix} 0 & 1 & &&&&\\ 1 & 0 & & &&&\\ & & 0 & 1 && &\\ & & 1 & 0 & & & \\ & & & & \ddots & & \ \ & & & & & \ddots & \\ & & & & & & 0 & 1&\\ & & & & & & 1 & 0&\\ & & & & & & & & &a\\ \end{pmatrix}

She

\det( \varphi') = \det(A) = (-1)^da = 1 =\det(\varphi)

So, by the classification theorem of, quadratic forms over finite fields, φ et φ′ are equivalent. So there exists a linear isomorphism u such that

\varphi' = \varphi \circ u

This isomorphism gives us a bijection between X and X', so that

|X|= |X'|

### Step 3: Let's put the good pieces together to conclude

Let us count X' modulo p.
First caseand

(x_1,x_3 \ldots,x_{p-2}) = (0, \ldots, 0)

Then we choose either the vector

(x_2, x_4, \ldots, x_{p-1})

which leaves us qd possible choices. We then choose xp and there we are reduced to the equation

ax_p^2 = 1

which leaves us by the lemma

1 + \left( \dfrac{a}{q}\right)

possibilities.
Second case, we choose the vector

(x_1,x_3 \ldots,x_{p-2})

with qd – 1 choice. Then, we have a choice on xp. We then choose the vector

(x_2, x_4, \ldots, x_{p-1})

As such manner that

2(x_1x_2 + \ldots + x_{p-2}x_{p-1}) + ax_p^2 - 1 = 0

Which leaves us qd-1 choice.

Finally, we have:

\begin{array}{ll} |X'| &= q^d \left( 1 + \left( \dfrac{a}{q} \right) \right) + (q^d -1) qq^{d-1}\\ &= q^{2d } + q^d\left( \dfrac{a}{q} \right) \end{array}

As |X| = |X'|, we obtain modulo p:

1 + \left( \dfrac{p}{q} \right) = \left( \dfrac{a}{q} \right)q^d + q^{2d}

Urban artist

q^d = q^{\frac{p-1}{2}} = \left( \dfrac{p}{q} \right)

And also, using Fermat's little theorem

q^{2d} = q^{p-1}\equiv 1 [p]

Urban artist

\left( \dfrac{a}{q} \right) = a ^{\frac{p-1}{2}}

We are then reduced to the equality

1 + \left( \dfrac{p}{q} \right) = \left( \dfrac{q}{p} \right)a ^{\frac{q-1}{2}} +1

Which gives good (remember the value of a)

\left( \dfrac{p}{q} \right) = \left( \dfrac{q}{p} \right)(-1) ^{\frac{p-1}{2}\frac{q- 1}{2}}

Which can therefore be rewritten as:

\left( \dfrac{q}{p} \right)\left( \dfrac{p}{q} \right) = (-1) ^{\frac{p-1}{2}\frac{q- 1}{2}}

Which gives us the law of quadratic reciprocity that we were trying to demonstrate! 