Gauss Sums, Quadratic Reciprocity, and the Jacobi Symbol
Our first goal in this chapter is to present Gauss’s sixth proof of his Law of Quadratic Reciprocity. The presentation here follows [32, §3.3] fairly closely, except that our Gauss sums are over the complex numbers, as opposed to ibid. where Gauss sums ar
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Gauss Sums, Quadratic Reciprocity, and the Jacobi Symbol
Our first goal in this chapter is to present Gauss’ sixth proof of his Law of Quadratic Reciprocity. The presentation here follows [32, §3.3] fairly closely, except that our Gauss sums are over the complex numbers, as opposed to ibid. where Gauss sums are considered over a finite field. Later in the chapter we introduce the Jacobi symbol and study its basic properties. We will also prove the Law of Quadratic Reciprocity for the Jacobi symbol. At the end of the chapter we will show examples that demonstrate how the Jacobi symbol can be used to compute the Legendre symbol efficiently. The Jacobi symbol will make an appearance in Chapter 10 when we give a proof of the Three Squares Theorem. In the Notes, we give some references for the various proofs of the Law of Quadratic Reciprocity.
7.1 Gauss sums and Quadratic Reciprocity For an odd prime p, let ζ = e
2πi p
and define the pth Gauss sum by p−1 k ζ k. τp = p k=1
We start with the following lemma: Lemma 7.1. For all odd primes p, τ p2
=
−1 p
p.
Proof. We have τ p2 =
p−1 p−1 k l k=1 l=1
p
p
ζ k+l =
p−1 p−1 kl k=1 l=1
p
ζ k+l .
© Springer Nature Switzerland AG 2018 R. Takloo-Bighash, A Pythagorean Introduction to Number Theory, Undergraduate Texts in Mathematics, https://doi.org/10.1007/978-3-030-02604-2_7
119
120
7 Gauss Sums, Quadratic Reciprocity, and the Jacobi Symbol
We make a change of variables by introducing a new variable m by l ≡ mk mod p. When k, l range over {1, . . . , p − 1}, m varies over the same set. So we get τ p2
=
p−1 p−1 mk 2
p
k=1 m=1
=
p−1 p−1 m
p
k=1 m=1
p−1
=
ζ
k+mk
m=1
m p
p−1
ζ k+mk
ζ k(m+1) .
k=1
The innermost sum is a geometric sum, and if ζ m+1 = 1, we get p−1
ζ k(m+1) =
k=1
1 − ζ m+1 (ζ p )m+1 − ζ m+1 = m+1 = −1. m+1 ζ −1 ζ −1
If on the other hand ζ m+1 = 1, we have e
2πi(m+1) p
= 1.
Consequently, p|m + 1, and, since 1 ≤ m ≤ p − 1, we conclude that m = p − 1. In this case, p−1 ζ k(m+1) = p − 1. k=1
Putting everything together, τ p2
=
p−1 p−1 m
p
m=1
=
p−2 p−1 m m=1
p
=−
m=1
=−
p−1 m m=1
p
k=1
p
m=1
=−
p
+
+p
k=1
ζ k(m+1) + ( p − 1)
p−2 m
p−1 m
ζ k(m+1)
+ ( p − 1)
p−1 p
p−1 p
+ ( p − 1)
=−
p−1 m m=1
So in order to prove the lemma it suffices to prove
m=1
p
p−1 p
p−1 m
p−1 p
= 0.
p
p−1 p
+p
−1 . p
7.1 Gauss sums and Quadratic Reciprocity
121
To see this, let
p−1 m
X=
p
m=1
.
Pick an integer b, e.g, a primitive root modulo p, such that (b/ p) = −1. Then p−1 p−1 m b bm b X= = . −X = p p p p m=1 m=1 But when m ranges over the numbers {1, . . . , p − 1}, the product mb ranges over the same set modulo p. Consequently, the last expression is equal to X as well. Hence −X = X. This implies X = 0, and we are done. Now we can proceed to prove the Quadratic Reciprocity, presenting a variation of Gauss’s extremely clever argument. This proof
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