Prime Numbers

4.4 Gauss and Jacobi sums 195
ζq−1 p = 1. And, as χp,q(m) p = 1 for every nonzero residue m mod q, and
χp,q(gq) = 1, it follows that χp,q has order p. Let
G(p, q) =τ(χp,q) =
q−1
m=1
χp,q(m)ζ m q−1
q =
k=1
ζ k p ζ gk q−1
q
q = ζ
k mod p
p q
k=1
ζ gk
q
mod q
(That this definition in the case p = 2 is equivalent to that in Definition 2.3.6
is the subject of Exercise 4.20.)
We are interested in the Gauss sums G(p, q) for their arithmetic properties,
though it may not be clear what a sum of lots of complex numbers has to do
with arithmetic! The Gauss sum G(p, q) is an element of the ring Z[ζp,ζq].
Elements of the ring can be expressed uniquely as sums p−2 q−2 j=0 k=0 aj,kζ j pζ k q
where each aj,k ∈ Z. We thus can say what it means for two elements
of Z[ζp,ζq] to be congruent modulo n; namely, the corresponding integer
coefficients are congruent modulo n. Also note that if α is in Z[ζp,ζq], then
so is its complex conjugate α.
It is very important in actual ring computations to treat ζp,ζq symbolically.
As with Lucas sequences, where we work symbolically with the roots
of quadratic polynomials, we treat ζp,ζq as symbols x, y, say, which obey the
rules
x p−1 + x p−2 + ···+1=0, y q−1 + y q−2 + ···+1=0.
In particular, one may avoid complex-floating-point methods.
We begin with a well-known result about Gauss sums.
Lemma 4.4.1. If p, q are primes with p | q − 1, then G(p, q)G(p, q) =q.
Proof. Let χ = χp,q. Wehave
Let m −1
2
G(p, q)G(p, q) =
q−1
q−1
χ(m1)χ(m2)ζ
m1=1 m2=1
m1−m2
q .
denote a multiplicative inverse of m2 modulo q, sothatχ(m2) =
χ(m −1
−1
2 ). Note that if m1m2 ≡ a (mod q), then χ(m1)χ(m2) =χ(a) and
m1 − m2 ≡ (a − 1)m2 (mod q). Thus,
q−1
G(p, q)G(p, q) = χ(a)
a=1
q−1
m=1
ζ (a−1)m
q .
The inner sum is q − 1 in the case a = 1 and is −1 in the cases a>1. Thus,
q−1
q−1
G(p, q)G(p, q) =q − 1 − χ(a) =q − χ(a).
Finally, by (1.28), this last sum is 0, which proves the lemma. ✷
a=2
a=1
.