Tools
2.2 The Quadratic Reciprocity Law
2.2.5 The Sign of the Quadratic Gauss Sum
Corollary 2.1.47 gives the square ofτ(χ) whenχis a real character, in other words by the preceding section, whenχis the Legendre–Kronecker symbol. A more difficult result due to Gauss is that one can give the value ofτ(χ) itself (Proposition 2.2.24). Before proving it, we need some results of independent interest.
Proposition 2.2.16 (Poisson summation formula). Letf be a continu- ous function and locally of bounded variation on some not necessarily bounded interval[A, B]. Then
AnB
f(n) =
m∈Z
B A
f(t) exp(2iπmt)dt ,
where
means that the terms for n =A and n =B, if present, must be counted with coefficient1/2.
Proof. Letf1be a piecewise continuous function locally of bounded vari- ation, that tends to zero sufficiently rapidly (we will in fact have f1 with compact support, so this is no problem). Setg(x) =
n∈Zf1(n+x). Then g(x) is an absolutely convergent series that converges normally in any com- pact subset ofR, and clearlyg(x) is periodic of period dividing 1. Thus we may apply the standard theorem on Fourier series that tells us that for allx we have
g(x+) +g(x−)
2 =
m∈Z
cmexp(2iπmx), where as usual
g(x±) = lim
ε→0,sign(ε)=±g(x+ε), and the Fourier coefficientscm are given by
cm= 1
0
g(t) exp(−2iπmt)dt=
n∈Z
1 0
f1(n+t) exp(−2iπmt)dt
=
n∈Z
n+1
n
f1(t) exp(−2iπmt)dt=f1(m), where the Fourier transformf1(y) is defined as usual by
f1(y) = +∞
−∞ f1(t) exp(−2iπyt)dt . Setting in particulary= 0, we obtain
n∈Z
f1(n+) +f1(n−)
2 =
m∈Z
f1(m).
Choose nowf1(t) =f(t) fort∈[A, B] andf1(t) = 0 elsewhere. Then f1(y) =
B
A
f(t) exp(−2iπyt)dt .
Furthermore, since f is continuous on ]A, B[ , when A < n < B we have (f1(n+) +f1(n−))/2 = f(n), while if n = A (of course only when A ∈ Z) then (f1(n+) +f1(n−))/2 = f(n+)/2 = f(n)/2, and similarly if n = B (whenB ∈Z), then (f1(n+) +f1(n−))/2 =f(n−)/2 =f(n)/2, proving the
proposition after changingminto −m.
Corollary 2.2.17. Let f be a continuous function and locally of bounded variation onR. Then for allx∈Rwe have
n∈Z
f(x+n) =
m∈Z
f(m) exp(2iπmx),
where as abovef(m)is the Fourier transform off. In particular
n∈Zf(n) =
m∈Zf(m).
Proof. Apply the proposition to [A, B] =R, and note that by an evident change of variable the Fourier transform off(x+t) aty isf(y)e2iπyx. Lemma 2.2.18. Let pbe an odd prime number, and let χ(n) =n
p
be the Legendre symbol. Then
τ(χ) =
xmodp
ζpx2 .
Proof.This immediately follows from the trivial observation that the num- ber of solutions moduloptox2≡n(modp) is equal to 1 +χ(n) and the fact that
nmodpχ(n) = 0.
We can now obtain the fundamental result on the sign of the Gauss sum.
Theorem 2.2.19. Letpbe an odd prime number, and let χ(n) =n
p
be the Legendre symbol. Then
τ(χ) =
p1/2 if p≡1 (mod 4), p1/2i if p≡3 (mod 4). Proof. By the above lemma, we have τ(χ) =
0xp−1exp(2iπx2/p).
We apply the Poisson summation formula proved above to [A, B] = [0, p] and f(x) = exp(2iπx2/p). Sincef(0) =f(p), we have
0np
f(n) =
0np−1
f(n) =τ(χ). On the other hand,
p 0
f(t) exp(2iπmt)dt= p
0
exp(2iπ(t2+pmt)/p)dt
= exp(−2iπpm2/4) p
0
exp(2iπ(t+pm/2)2/p)dt
= exp(−2iπpm2/4)
p(m+2)/2 pm/2
exp(2iπt2/p)dt . Changingtinto p1/2tit follows that
m∈Z,2|m
p
0
f(t) exp(2iπmt)dt= +∞
−∞
exp(2iπt2/p)dt=p1/2I , where
I= +∞
−∞ exp(2iπt2)dt .
The value of this integral is well known, but we do not need it since it will follow from the proof. Note that we know in advance that it converges, but this can be checked directly for example by settingt2=xand integrating by parts.
Similarly we find that
m∈Z,2m
p 0
f(t) exp(2iπmt)dt= exp(−2iπp/4)p1/2I . Putting everything together, we thus obtain
τ(χ) = (1 +i−p)p1/2I .
We can first deduce from this the value ofI: indeed, we simply choose a small value ofp, for examplep= 3. Then
τ(χ) = exp(2iπ/3)−exp(4iπ/3) =i31/2 henceI=i/(1 +i) = (1 +i)/2.
Thus
τ(χ) = (1 +i)(1 +i−p) 2 p1/2,
proving the theorem after separation of cases.
For simplicity of notation, whenD is a fundamental discriminant we de- note byχD the character such that χD(n) =D
n
. By quadratic reciprocity, we note that the above theorem can be reformulated as τ(χD) = D1/2 for D= (−1)(p−1)/2p, wherepis an odd prime, choosing the principal branch of the square root, i.e., such that−π/2<Arg(z1/2)π/2. We are now going to show that this is true for any fundamental discriminant D by proving a few lemmas.
Lemma 2.2.20. We haveτ(χD) =D1/2 forD=−4,D=−8, andD= 8.
Proof. Clear by direct computation.
Lemma 2.2.21. LetD1 andD2be two coprime fundamental discriminants.
Ifτ(χD1) =D11/2 andτ(χD2) =D1/22 , thenτ(χD1D2) = (D1D2)1/2.
Proof.First note the important fact that it isnot true that (D1D2)1/2= D1/21 D1/22 (exampleD1=−3,D2=−7).
SinceD1andD2are coprime, by the Chinese remainder theorem a residue moduloD1D2 can be written uniquely in the form n2D1+n1D2, wheren2
is moduloD2andn1 is moduloD1. Thus, τ(χD1D2) =
nmodD1D2
D1D2
n
ζ|nD1D2|
=
n1modD1
n2modD2
D1D2 n2D1+n1D2
ζ|Dn21DD12+n| 1D2
= D1
D2 D2
D1
n1modD1
D1
n1
ζ|nD1
1|
n2modD2
D2
n2
ζ|nD2
2|
= (−1)(sign(D1)−1)(sign(D2)−1)/4τ(χD1)τ(χD2)
by Proposition 2.2.6. It is clear that (D1D2)1/2 =D11/2D1/22 except if both D1 andD2 are negative, in which case (D1D2)1/2=−D1/21 D1/22 , and this is exactly compensated by (−1)(sign(D1)−1)(sign(D2)−1)/4, proving the lemma.
Definition 2.2.22. A fundamental discriminant D is said to be a prime discriminant if it is either equal to−4,−8, or8, or equal to(−1)(p−1)/2pfor pan odd prime.
Note that all these expressions are indeed fundamental discriminants.
Lemma 2.2.23. Any fundamental discriminantDcan be written in a unique way as a product of prime fundamental discriminants.
Proof. Since D is fundamental, no odd prime can divide D to a power larger than 1. Thus, we may write D = 2u
p∈Sp, where S is a finite set of odd primes. It follows thatD =ε2u
p∈S(−1)(p−1)/2pfor some ε=±1.
Note that the product overp∈S is congruent to 1 modulo 4. Thus, either u= 0, in which case we must have ε= 1 (sinceD ≡1 (mod 4)); oru= 2, in which case we must haveε=−1 (otherwise D/4 is also a discriminant), so the factor in front of the product is −4; or finally u= 3, in which case εcan be±1, giving the two factors ±8. Uniqueness of the decomposition is
clear.
The proof of the result that we are after is now immediate.
Proposition 2.2.24. Letχ be a real primitive character modulo m, so that χ(n) =D
n
forD=χ(−1)ma fundamental discriminant. Then
τ(χ) =
m1/2 ifχ(−1) = 1, m1/2i ifχ(−1) =−1.
Proof. By Theorem 2.2.15, we know that χ = χD with D = χ(−1)m a fundamental discriminant. By Lemma 2.2.23,D is equal to a product of prime fundamental discriminants that are necessarily coprime. By Lemma 2.2.21, it is thus sufficient to prove the proposition for prime fundamental discriminants, and this is exactly the content of Theorem 2.2.19 and Lemma
2.2.20.
In view of the functional equation for DirichletL-functions that we will study in Chapter 10 we make the following definition:
Definition 2.2.25. Let χ be any primitive character modulom. We define the root numberW(χ)by the formula
W(χ) =
τ(χ)
m1/2 if χ(−1) = 1, τ(χ)
m1/2i if χ(−1) =−1.
Thus a restatement of Proposition 2.2.24 is that whenχ is real we have W(χ) = 1. In the general case, since|τ(χ)|=m1/2 we have|W(χ)|= 1, and one can show thatW(χ) is a root of unity if and only ifχ is real, in which caseW(χ) = 1 (see Exercise 17).