CS681
Computational Number Theory
Lecture 19: Quadratic Reciprocity (contd.)
Instructor: Piyush P Kurur
Scribe: Ramprasad Saptharishi
Overview
Last class we proved one part of the Quadratic Reciprocity Theorem. We
shall first finish the proof of the other part and then get to a generalization
of the Legendre symbol - the Jacobi symbol.
1
Proof of Reciprocity Theorem (contd.)
Recall the statement of the theorem. If p 6= q are odd primes, then
p−1 q−1
p
q
= (−1) 2 2
q
p
The idea of the proof is just the same. We choses τ = 1 + i last time and
p−1
we got a 2 2 term while computing τ p . It would be the same here as well.
Let ζ be a principle q-th root of unity. We shall work with the field Fp (ζ).
Let
X a
τ=
ζa
q
?
a∈Fq
Just as before, we compute τ 2 .



X a
X b
τ2 = 
ζ a 
ζ b
q
q
a∈F?q
b∈F?q
X a
b
=
ζ a+b
q
q
a,b∈F?q
X a b−1 =
ζ a+b
q
q
a,b∈F?q
X ab−1 =
ζ a+b
q
?
a,b∈Fq
1
Let us do a change of variable, by putting c = ab−1
X c
2
τ =
ζ bc+b
q
c,b∈F?q
X c X
X −1 c+1 b
=
(ζ ) +
q
q
?
?
?
−16=c∈Fq
b∈Fq
b∈Fq
Since both p and q are primes and −1 6= c ∈ F?q , ζ c+1 is also a principle q-th
P
root of unity. And therefore, b (ζ c+1 )b = −1. Therefore,
X c
−1
2
τ =
+ (q − 1)
q
q
?
−16=c∈Fq
Look at the first term. If one were to sum over all elements of F?q , then half
of the qc would be 1 and the other would be −1 thus fully cancelling off.
Since we are just excluding c = −1, the first term is just −1
.
q
−1
−1
−1
2
+ (q − 1)
=q
τ =
q
q
q
Now to evaluate τ p .
X a
ζ ap
q
a∈F?q
X cp−1 =
ζc
(c = ap)
q
c∈F?q
−1 X p
c
=
ζc
q
q
c∈F?q
p
=
τ
q
τp =
p−1
τ p = τ (τ 2 ) 2
p−1
p−1
p
−1 2
=⇒
τ = τq 2
q
q
p−1 q−1
q
= τ
(−1) 2 2
p
p−1 q−1
p
q
=⇒
= (−1) 2 2
q
p
2
2
Jacobi Symbol
The legendre symbol m
n can be naturally generalized to the case when m
and n are odd and coprime numbers.
αk
α1 α2
Definition 1.
Suppose n = p1 p2 · · · pk . Then the Jacobi symbol, also reprem
sented as n , is defined as follows
m
n
(
0
= Qk
i=1
α i
m
pi
if (m, n) 6= 1
otherwise
The Jacobi symbol also satisfies some nice multiplicative properties
• m1nm2 = mn1 mn2
m
• n1mn2 = nm1
n2
Using the above properties, we can get a generalize the theorem on the
legendre symbol as well.
Theorem 1. If m, n are odd numbers such that (m, n) = 1, then
n2 −1
2
= (−1) 8
n
m n m−1 n−1
= (−1) 2 2
n
m
We shall prove this theorem in the next class. The proof will just be using induction on the factors of m and n.
WARNING: Please note that the Jacobi symbol m
n doesn’t say anything about whether or not m is a square modulo n. For example, if n =
p1 p2 and m was chosen
such that m is not a square modulo p1 or p2 . Then
the Jacobi symbol
m
p1 p2
= (−1)(−1) = 1 but m is not a square in p1 p2 .
3