U.U.D.M. Project Report 2018:19
Examensarbete i matematik, 15 hp
Handledare: Andreas Strömbergsson
Examinator: Martin Herschend
Juni 2018
Department of Mathematics
Uppsala University
Gauss and Jacobi Sums and the Congruence
Zeta Function
GAUSS AND JACOBI SUMS AND THE CONGRUENCE ZETA FUNCTION BACHELOR’S THESIS DEPARTMENT OF MATHEMATICS UPPSALA UNIVERSITY EINAR WAARA 1. Introduction 2 2. Prerequisite Theory 3 2.1. Multiplicative Characters 3
2.2. Basic Notions from Algebraic Geometry 5
2.3. Trace and Norm in Finite Fields 6
3. Gauss and Jacobi Sums 7
3.1. Gauss Sums 7
3.2. Jacobi Sums and Applications 8
4. The Zeta Function 15
4.1. The Zeta Analogy 17
4.2. The Rationality of the Zeta Function 19 Contents
1. Introduction
The German mathematician Carl Friedrich Gauss (1777-1855) was the first to introduce the notion of what is now called quadratic Gauss sums, which are sums of the form ga =
P
t∈Fp(t/p)e
2πiat/p, where F
p is the finite field of order p and (·/p) is the quadratic reciprocity
symbol. Among many other things, Gauss proved that g1 takes the values
√
p or i√p when p ≡ 1 (4) or p ≡ 3 (4) respectively. These sums can be, and have been, greatly generalized and appear very often in areas related to modern number theory, in the study of theta functions and the discrete Fourier transform for instance. In this exposition we present an introduction to the theory of Gauss and Jacobi sums (and their interrelations) and apply these notions to study the (congruence) zeta function. We give a proof of the Hasse-Davenport relation, which is a lifting relation relating the Gauss sums of two different finite fields. We outline a proof of a special case of the first part of the Weil conjectures (named after Andr´e Weil (1906-1998)) proved in 1959 by Bernard Dwork (1923-1998), which states that any algebraic set V has a rational zeta function ZV(u). We state (and prove) a characterization of
ratio-nal zeta functions Zf(u) and use this characterization to prove that Zf(u) is rational when
f (x0. . . , xn) = a0xm0 + a1x1m + . . . + anxmn where a0, . . . , an ∈ F∗ and F is a finite field of
order q ≡ 1 (m). We also further display the potency of Gauss and Jacobi sums by providing a short proof of the famous law of quadratic reciprocity, a theorem first proved by Gauss (who refereed to it as the ”golden theorem”) and who subsequently provided five additional proofs, the final of which was published in 1818. This exposition is greatly inspired by the excellent classic
IR. K. Ireland and M. Rosen, A Classical Introduction to Modern Num-ber Theory, Springer-Verlag, 1990.
2. Prerequisite Theory
Here we present some basic definitions and results from elementary number theory for future reference. We start with the concept of a quadratic residue.
Definition 1. Let a ∈ Z and suppose (a, m) = 1. Then a is called a quadratic residue modulo m if x2 ≡ a (m) has a solution, i.e. a is a perfect square modulo m. If a is not a quadratic residue, we say that a is a quadratic nonresidue.
For the remainder of this section, let p be an odd prime number. Next we define a convenient symbol for dealing with quadratic residues.
Definition 2. The Legendre symbol (·/p) is defined by
(a/p) =
1 if a is a quadratic residue modulo p −1 if a is a quadratic nonresidue modulo p 0 if p | a
Proposition 2.1. (IR, 5.1.2) (I) a(p−1)/2≡ (a/p) (p) (II) (ab/p) = (a/p)(b/p)
(III) a ≡ b (p) =⇒ (a/p) = (b/p)
Proof. Assume that p 6 | a and p 6 | b (if not, the proof is trivial). Since ap−1≡ 1 (p) we have (a(p−1)/2+ 1)(a(p−1)/2− 1) = ap−1− 1 ≡ 0 (p) and thus a(p−1)/2 ≡ ±1 (p). By a well-known
fact we have a(p−1)/2 ≡ 1 (p) if and only if a is a quadratic residue modulo p. The second part is easily proved using the first. The third part is trivial. Corollary 2.2. (IR, 5.1.2 corollary) Modulo p there is an equal number of quadratic residues as there are quadratic nonresidues.
Proof. The equation a(p−1)/2≡ 1 (p) has (p−1)/2 solutions, hence there are (p−1)/2 quadratic residues and p − 1 − (p − 1)/2 = (p − 1)/2 quadratic nonresidues. Corollary 2.3. (IR, 6.3 lemma 2)Pp−1
t=0(t/p) = 0.
Proof. p | 0 so (0/p) = 0 by definition. Thus there are p − 1 (p odd so that p − 1 even) terms of ±1’s left, and we know these terms cancel by the previous corollary. Let F be a finite field of order q.
Proposition 2.4. (IR, 7.1.2) The equation xn= α ∈ F∗ is solvable if and only if α(q−1)/d = 1, where d = (n, q − 1). If the equation is solvable, then there are d solutions.
Proof. The multiplicative group F∗ of F is cyclic. Let g be a generator of F∗. Set α = ga and x = gb for some a, b ∈ {1, 2, . . . , q − 1}. Substituting this into the equation yields gnb= ga if
and only if gnb−a= 1, thus q − 1 | nb − a i.e. nb ≡ a (q − 1), which proves the result by basic
facts about congruences.
2.1. Multiplicative Characters. The definitions of Gauss and Jacobi sums rely on the notion of a multiplicative character, we therefore provide a brief introduction of this topic here. We follow [IR, Ch. 8.1].
Definition 3. A multiplicative character on Fpis a group homomorphism from the
The Legendre symbol can therefore be regarded a special case of a character.
The set of characters on Fp constitute a group under the following definitions. Let λ and χ
be characters. We define λχ as the map a 7→ λ(a)χ(a), and χ−1 as a 7→ χ(a)−1. The identity will be denoted by and is characterized by (a) = 1 for all a ∈ Fp∗. It will be useful to extend the domain and let χ(0) = 0 for all χ 6= while (0) = 1. Notice that this is in accordance with the Legendre symbol. We shall denote the group of characters on Fp by Ωp. Since we are
now dealing with a group, we naturally define the order of a character as the smallest positive integer n such that χn= . The Legendre symbol is thus a character of order two.
It turns out that Ωp is a cyclic group of order p − 1. In order to prove this, we need a few
basic results easily proved from the definitions. Proposition 2.5. Let χ ∈ Ωp and a ∈ Fp∗, then
(I) χ(1) = 1
(II) χ(a) is a (p − 1)st root of unity (III) χ(a−1) = χ(a)−1 = χ(a)
Proof. χ(1) = χ(12) = χ(1)2 and since χ(1) 6= 0 we have χ(1) = 1. Also, since |Fp∗| = p − 1 we have ap−1 = 1 and thus χ(1) = χ(ap−1) = χ(a)p−1 = 1. Lastly χ(a−1)χ(a) = χ(a−1a) = χ(1) = 1 hence χ(a)−1= χ(a−1). The complex number χ(a) lies on the unit circle in C since |χ(1)| = |χ(ap−1)| = |χ(a)p−1| = |χ(a)|p−1 = 1 implies |χ(a)| = 1. Due to the definition of
complex multiplication we have χ(a)−1 = χ(a). Proposition 2.6. Let χ ∈ Ωp, then Pt∈Fpχ(t) =
(
0 if χ 6= p otherwise
Proof. The equality is trivial when χ = . If χ 6= , then there is some a ∈ Fp∗ such that χ(a) 6= 1. Then χ(a)P
t∈Fpχ(t) = P t∈Fpχ(at) = P t∈Fpχ(t), hence P t∈Fpχ(t) = 0.
Proposition 2.7. Ωp is a cyclic group of order p − 1.
Proof. We show that there exists a generator to Ωp. The multiplicative group Fp∗ is cyclic. Let
g ∈ Fp∗ be a generator. If a ∈ Fp∗ then a = gl for some l ∈ {0, 1, . . . , p − 1}. Thus any character χ ∈ Ωp is completely determined by its action on g since χ(a) = χ(gl) = χ(g)l. In general the
number of n’th roots of unity is n. Hence since χ(g) ∈ C \ {0} is a (p − 1)st root of unity, it follows that |Ωp| ≤ p − 1. We claim that a generator to Ωp is given by λ(gk) = e2πi(k/(p−1)).
Firstly λ is well defined since λ(gk+p−1) = λ(gk)e2πi= λ(gk). Since λ(gk1gk2) = λ(gk1)λ(gk2)
we also know that λ is character, hence λ ∈ Ωp. We now show that λ is a character of order
p − 1. Assume that λn = . Then λn(g) = λ(g)n = e2πin/(p−1) = (g) = 1 thus (p − 1) | n. Now, λp−1(a) = λ(a)p−1 = λ(ap−1) = λ(1) = 1 thus λp−1 = . There are no smaller positive integers which divide p − 1 than p − 1 itself. Thus p − 1 is the smallest positive integer n such that λn = and therefore the order of λ is p − 1. Hence we know that , λ, λ2, . . . , λp−2 are all distinct members of Ωp and since there are p − 1 of them and |Ωp| ≤ p − 1 we know that
Ωp= p − 1. Thus λ is indeed a generator, and hence Ωp is cyclic.
Remark. Let a ∈ Fp∗\ {1} then a = gl for some l ∈ {1, 2, . . . , p − 2}, so p − 1 - l. Thus
λ(a) = λ(g)l = e2πi(l/(p−1)) 6= 1. We conclude that given a ∈ Fp∗\ {1} there is always some character χ such that χ(a) 6= 1. This is crucial for the following corollary.
Corollary 2.8. If a ∈ Fp∗\ {1} thenP
Proof. Let z be the complex numberP
χ∈Ωpχ(a). Since a 6= 1 there is (by the remark above)
some character χ such that χ(a) 6= 1, thus λ(a)z = X
χ∈Ωp
λ(a)χ(a) = X
χ∈Ωp
λχ(a) = z.
The last equality holds since {λχ}χ∈Ωp = Ωp. This follows from the implication λχ1 =
λχ2 =⇒ χ1 = χ2. Thus, since λ(a) 6= 1, we see that z = 0.
We shall see how characters can be used in the study of equations in finite fields. The following proposition is an early tool in this direction.
Proposition 2.9. If a ∈ Fp∗, n|p − 1 and the equation xn = a is unsolvable, then there is a character χ such that χn= and χ(a) 6= 1.
Proof. Let g and λ be as in the proof of proposition 2.7. We know that a = gl for some l and by assumption xn= a is not solvable, thus n - l (otherwise gl/n would be a solution). Let χ = λ(p−1)/n. Then χ(g) = λ(g)(p−1)/n= e2πi/n and thus χ(a) = χ(g)l= e2πi(l/n) 6= 1. Lastly χn= λp−1= since λ was determined to be a generator to Ωp.
Let N (xn = a) denote the number of solutions to the equation xn = a in Fp. The following
proposition will be a useful tool for counting the number of solutions to more complicated equations in Fp and will be used often.
Proposition 2.10. If n | p − 1, then N (xn= a) =P
χn=χ(a) where the sum is taken over
all characters χ such that χn= .
Proof. If χ is a character such that χn = , then χ(g) ∈ C is a n’th root of unity since χ(g)n= χn(g) = (g) = 1. Thus there are at most n such characters in Ωp. But in the previous
proposition we defined χ = λ(p−1)/nwhich meant that χ(g) = e2πi/n, thus , χ, χ2, . . . , χn−1are n distinct characters, all of order n. Thus, these are the characters in the sum of consideration. To prove the equality we consider three cases. First, if a = 0, clearly there is only one solution to xn= a, namely x = 0. Since χ(0) = 0 except for when χ = which maps 0 to 1 the equality
holds in this case. Next, assume a 6= 0 and that the equation xn = a is solvable. Let b be a solution, then χ(a) = χ(bn) = χ(b)n = (b) = 1 and thus P
χn=χ(a) = n = N (xn = a).
Lastly, assume that a 6= 0 and that the equation xn = a is unsolvable. By proposition 2.9, there is a ρ ∈ Ωp such that ρn = and ρ(a) 6= 1. We have {χ : χn = } < Ωp, thus
ρ(a)P
χn=χ(a) =
P
χn=χρ(a) =
P
χn=χ(a), hence ρ(a) − 1
P
χn=χ(a)
= 0 which
proves the result.
2.2. Basic Notions from Algebraic Geometry. In order to discuss the congruence zeta function, we need some fundamental definitions from algebraic geometry. We follow [IR, Ch. 10.1]. For the remainder of this section, let F be a field of order q.
Definition 4. An(F ) = {(a1, a2, . . . , an) : ai ∈ F for i = 1, 2, . . . , n} is called the affine
n-space over F .
An(F ) can be considered to be a vector space over F with the usual definitions of scalar multiplication and vector addition. Let us consider An+1(F ) \ {(0, 0, . . . , 0)} and define an
equivalence relation ∼ on this set by
(a0, a1, . . . , an) ∼ (b0, b1, . . . , bn) ⇐⇒ ∃γ ∈ F∗ such that ai = γbi for i = 0, 1, . . . , n.
The size of the affine n-space is qn, while the size of the projective n-space is qn+1q−1−1 = qn+ qn−1+ . . . + q + 1.
A homogeneous polynomial f (x1, x2, . . . xn) =P(i1,i2,...,in)ai1i2...inx
i1
1 x i2
2 . . . xinnin F [x1, x2, . . . , xn]
is one where each nonzero term have the same degree.
For the rest of this section, let F ⊆ K be a field extension. Given some nonzero (not necessarily homogeneous) polynomial f ∈ F [x1, x2, . . . xn] we can view it as a function f : An(K) −→ K
by evaluating f at points in An(K) in the usual way.
Definition 6. Hf(K) = {a ∈ An(K) : f (a) = 0} is called the affine hypersurface of f in
An(K).
Let g ∈ F [x0, x1, . . . , xn] be some homogeneous polynomial of degree d.
Definition 7. Hg(K) = {[a] ∈ Pn(K) : g(a) = 0} is called the projective hypersurface of
g in Pn(K).
The set Hg(K) is well defined since if γ ∈ F∗ we have g(γx) = γdg(x) . This is because g is
homogeneous of degree d. Thus if g(a) = 0 for some representative a in [a], then g maps all elements in that equivalence class to 0.
We define a similar set as above, but for several polynomials instead of only one. In words, it is the set of common zeros to a finite set of polynomials. Let f1, f2, . . . , fm ∈ F [x1, x2, . . . , xn].
Definition 8. V = {a ∈ An(K) : fj(a) = 0 for j = 1, 2, . . . , m} is called an algebraic set
defined over K.
2.3. Trace and Norm in Finite Fields. In this short section we provide some basic defini-tions and results which we will need later. We follow [IR, Ch. 11.2]. Let F be a field of order q = pn and let E ⊇ F be a field of order qs.
Definition 9. The trace of α ∈ E from E to F is defined as trE/F(α) =
Ps−1
i=0αq
i
and the norm of α from E to F is defined as NE/F(α) =
Qs−1
i=0αq
i
. Proposition 2.11. Let α, β ∈ E and a ∈ F , then
(I) trE/F(α) ∈ F
(II) trE/F(α + β) = trE/F(α) + trE/F(β) (III) trE/F(aα) = a trE:F(α)
(IV) trE/F : E −→ F is a surjective mapping. (V) NE/F(α) ∈ F
(VI) NE/F(αβ) = NE/F(α)NE/F(β)
(VII) NE/F(aα) = asNE/F(α)
(VIII) NE/F : E∗ −→ F∗ is a surjective mapping.
Proof. We prove the last part. Note that NE/F is a group homomorphism, so that α ∈
ker(NE/F) if and only if NE/F(α) =
Qs−1
i=0αq
i
= α(qs−1)/(q−1) = 1. By proposition 2.4 we know that the equation x(qs−1)/(q−1) = 1 has (qs − 1)/(q − 1) solutions in E∗. The first
isomorphism theorem gives |im(NE/F)| = |E∗|/|ker(NE/F)| = q − 1 = |F∗|, hence NE/F is
3. Gauss and Jacobi Sums
We start by introducing the notion of Gauss and Jacobi sums over Fp, after which we widen
our view and consider Gauss and Jacobi sums over arbitrary finite fields of order q = pr. We then give a short proof of the law of quadratic reciprocity using Gauss and Jacobi sums. We follow [IR, Ch. 8.2-7].
3.1. Gauss Sums. Let χ ∈ Ωp, a ∈ Fp and ζ = e2πi/p.
Definition 10. A sum of the form P
t∈Fpχ(t)ζ
at is called a Gauss sum belonging to the
character χ and will be denoted ga(χ), and g1(χ) in particular will be denoted g(χ).
Lemma 3.1. Pp−1
t=0 ζat =
(
p if a ≡ 0 (p) 0 otherwise
Proof. If a ≡ 0 (p) then a/p is an integer so ζa = 1, hence Pp−1
t=0ζat = p. Otherwise, if
a 6≡ 0 (p), then ζa6= 1 and Pp−1
t=0 ζat = ζap−1
ζa−1 = 0.
Notice that, as suggested by the notation, the Gauss sum depends on two objects: a and χ. The following proposition gives some insight in how the Gauss sum behaves with respect to these in extreme cases.
Proposition 3.2.
(I) a 6= 0 and χ 6= =⇒ ga(χ) = χ(a−1)g1(χ)
(II) a 6= 0 and χ = =⇒ ga() = 0
(III) a = 0 and χ 6= =⇒ g0(χ) = 0
(IV) a = 0 and χ = =⇒ g0() = p
Proof. We prove each statement in turn. Notice that χ(a)ga(χ) = χ(a)Pt∈Fpχ(t)ζ
at =
P
t∈Fpχ(at)ζ
at = g
1(χ). Multiplying both sides by χ(a)−1= χ(a−1) proves the first statement.
Next we see that ga() =Pt∈Fp(t)ζ
at=P
t∈Fpζ
at = 0 by lemma 3.1, this proves the second
statement. Lastly, g0(χ) =
P
t∈Fpχ(t), hence the third and fourth statement both follow from
proposition X.
Proposition 3.3. If χ 6= , then |g(χ)| =√p. Proof. Our strategy will be to evaluate the sum
S = X
a∈Fp
ga(χ)ga(χ)
in two different ways. First, suppose that a 6= 0, then we have ga(χ)ga(χ) = χ(a−1)g(χ)χ(a−1)g(χ)
= χ(a−1)g(χ)χ(a)g(χ) = |g(χ)|2.
By proposition 3.2 we have g0(χ) = 0, thus S = (p − 1)|g(χ)|2. Next we use the definition 10
thus, by lemma 3.1 we have X a X x X y χ(x)χ(y)ζax−ay =X x X y χ(x)χ(y)δ(x, y)p = (p − 1)p
where δ(x, y) = 1 if x ≡ y (p) and δ(x, y) = 0 if x 6≡ y (p). Equating these two different ways
of expressing S proves the result.
3.2. Jacobi Sums and Applications. We define the Jacobi sum and show how it can be used to count the solutions to certain equations over Fp. Let χ1, . . . , χk∈ Ωp.
Definition 11. A sum of the form J (χ1, . . . , χk) = Pt1+...+tk=1χ1(t1) · · · χk(tk) is called a
Jacobi sum.
We start by considering the familiar equation x2+ y2 = 1 where x, y ∈ Fp. Let N (x2+ y2 =
1) denote the number of solutions to this equation. First, notice that N (x2 + y2 = 1) = P a+b=1N (x2 = a)N (y2 = b) = P a+b=1(1 + (a/p))(1 + (b/p)) = p + P a(a/p) + P b(b/p) + P a+b=1(a/p)(b/p) = p + P
a+b=1(a/p)(b/p) where the last equality follows from corollary 2.3.
We are left with evaluating the sum X
a+b=1
(a/p)(b/p). (1)
Let us pause this problem for a moment and consider N (x3+ y3 = 1) instead. Again, notice that N (x3+ y3 = 1) = P
a+b=1N (x3 = a)N (y3 = b). Now, if p ≡ 2 (3) this sum is easily
evaluated to be p by proposition 2.4. However if p ≡ 1 (3) then it is not as simple. Let χ ∈ Ωp\ {} be a character of order 3. Since (2, 3) = 1, χ2 is also a character of order 3. Since
Ωp is cyclic, the number of elements of order d | p − 1 is φ(d), where φ denotes Euler’s totient
function. Thus there are φ(3) = 2 characters of order 3, and since 3 is prime, a character must have order 3 (or 1) for its order to divide 3. Thus by proposition 2.10 we have
N (x3+ y3 = 1) = X a+b=1 2 X i=0 χi(a) 2 X j=0 χj(b) =X i X j X a+b=1 χi(a)χj(b) (2)
In summary, with both these examples we are eventually forced to deal with a Jacobi sum. This motivates the notion of such sums and we shall see how the properties of these sums will allow us to finally evaluate (at least partially) (1) and (2). The following theorem (which we soon shall generalize), provides us with the necessary tools for this purpose. Part four shows how the Jacobi and Gauss sums are related in an surprisingly simple way.
Theorem 3.4. Let χ, λ ∈ Ωp\ {}. Then
(I) J (, ) = p (II) J (, χ) = 0
(III) J (χ, χ−1) = −χ(−1)
Proof. The first part is trivial since the number of pairs a, b such that a + b = 1 is p and the second part follows from proposition 2.6. For part three, notice that
J (χ, χ−1) = X a+b=1 χ(a)χ−1(b) = χ(1)χ−1(0) + X a+b=1 ∧ b6=0 χ(ab−1) =X a6=1 χ a(1 − a)−1.
Let c = a(1 − a)−1, then if c 6= 1 we have a = c(1 + c)−1. Hence, when a varies over Fp\ {1},
c varies over Fp\ {−1}. Thus,
J (χ, χ−1) = X
c6=−1
χ(c) = X
c∈Fp
χ(c) − χ(−1) = −χ(−1) by proposition 2.6. To prove the last part, notice that
g(χ)g(λ) = X x χ(x)ζx ! X y λ(y)ζy ! =X x,y χ(x)λ(y)ζx+y =X t X x+y=t χ(x)λ(y) ! ζt. If t = 0 then X x+y=0 χ(x)λ(−x) = λ(−1)X x χλ(x) = 0. If t 6= 0, let x = tX and y = tY . Then we have
X x+y=t χ(x)λ(y) = X X+Y =1 χ(tX)λ(tY ) = X X+Y =1 χλ(t)χ(X)λ(Y ) = χλ(t)J (χ, λ). Thus, g(χ)g(λ) = J (χ, λ)X t χλ(t)ζt= J (χ, λ)g(χλ). Corollary 3.5. χ, λ, χλ 6= =⇒ |J (χ, λ)| =√p. Proof. |J (χ, λ)| = g(χ)g(λ) g(χλ) = √ p√p √ p = √ p by proposition 3.3.
Using these identities, we can definitively evaluate (1) and make further progress with (2). For (1) simply note that an application of part three yields P
a+b=1(a/p)(b/p) = −(−1/p) =
−(−1)(p−1)/2. The last equality follows from the first part of proposition 2.1 since
(−1)(p−1)/2 ≡ (−1/p) (p) ⇐⇒ p | (−1)(p−1)/2− (−1/p) ⇐⇒ (−1)(p−1)/2− (−1/p) = 0.
Hence, the number of solutions to x2+ y2= 1 in Fp turns out to be
N (x2+ y2= 1) = p − (−1)(p−1)/2. We can use the first three parts of proposition 3.4 to simplify (2) as
Since χ is a character of order 3 we have χ(−1) = χ (−1)3 = χ3(−1) = (−1) = 1 and χ2 = χ−1 = χ. Hence J (χ, χ) + J (χ2, χ2) = J (χ, χ) + J (χ, χ) = X a+b=1 χ(ab) + X a+b=1 χ(ab) = 2ReJ (χ, χ). Thus (3) simplifies to N (x3+ y3= 1) = p − 2 + 2ReJ (χ, χ). (4)
Clearly, this is not as precise as the value of N (x2+ y2 = 1). However, we can now give a estimate to the number of solutions to x3+ y3 = 1 in F
p. By corollary 3.5 we have
|N (x3+ y3 = 1) − p + 2| ≤ 2√p. (5)
which tells us that, the number of solutions to x3+ y3= 1 is approximately p − 2 with an error
term 2√p, hence for large primes p there will always exist many solutions to this equation. To begin generalizing some of the results above, the following sum will be useful
J0(χ1, . . . , χk) := X t1+...+tk=0 k Y i=1 χi(ti).
This is the same as the Jacobi sum except that P
iti= 0 instead of 1. Also, let ψ : Fp−→ C
be the map t 7→ ζt, then ψ(P
iti) =Qiψ(ti).
Proposition 3.6.
(I) J0(, . . . , ) = J (, . . . , ) = pk−1
(II) If some but not all χi = , then J0(χ1, . . . , χk) = J (χ1. . . , χk) = 0
(III) Suppose χk6= , then
J0(χ1, . . . , χk) = ( 0 if Qk i=1χi6= (p − 1)χk(−1)J (χ1, . . . , χk−1) otherwise (IV) χ1, . . . , χk, χ1· · · χk6= =⇒ k Y i=1 g(χi) = J (χ1, . . . , χk)g( k Y i=1 χi).
Proof. If k − 1 of the terms in P
iti is chosen, the last term is uniquely determined by the
requirement that the whole sum equals 0 or 1. These k − 1 terms can be chosen in pk−1 ways, this proves part one. For the second part we can assume we have ordered the characters in such a way that χ1, . . . , χs6= and χs+1, . . . , χk= . Then, for J0(χ1, . . . , χk) we have
X t1+...+tk=0 k Y i=1 χi(ti) = X t1,...,tk−1 s Y i=1 χi(ti) = pk−s−1 s Y i=1 X ti∈Fp χi(ti) = 0
where the last equality follows from proposition 2.6. The proof for J (χ1, . . . , χk) = 0 is similar.
To prove part three we rewrite J0(χ1, . . . , χk) as
J0(χ1, . . . , χk) = X s X t1+...,tk−1=−s k−1 Y i=1 χi(ti) χk(s) (6)
where s 6= 0 since this term gets deleted by χk(0) = 0. Thus we may define Ti by ti = −sTi.
substituting the last expression in (7) into (6) yields J0(χ1, . . . , χk) = χ1· · · χk−1(−1)J (χ1, . . . , χk−1) X s6=0 χ1· · · χk(s). If Qk i=1χi 6= then P
s6=0χ1· · · χk(s) = 0, otherwise the sum equals p − 1 by proposition
(2.6). Also, χ1· · · χk−1(−1) = ±1 and χk(−1) = ±1. In the case when Qki=1χi 6= then
χ1· · · χk−1(−1)χk(−1) = (−1) = 1 hence we can replace χ1· · · χk−1 with χk.
To prove part four, notice that Y i g(χi) = Y i X ti χi(ti)ψ(ti) =X s X t1+...+tk=s Y i χi(ti) ψ(s). (8)
Since by assumption χ1, . . . , χk 6= , part three above implies that Pt1+...+tk=s
Q
iχi(ti) = 0
if s = 0, so we can suppose s 6= 0 and define ti = sTi as before. Then
X t1+...+tk=s Y i χi(ti) = X T1+...+Tk=1 χ1· · · χk(s) Y i χi(Ti) = χ1· · · χk(s)J (χ1, . . . , χk), (9)
substituting the last expression in (9) into (8) gives Y i g(χi) = J (χ1, . . . , χk) X s6=0 χ1· · · χk(s)ψ(s) = J (χ1, . . . , χk)g( Y i χi). Part four has a couple of corollaries of interest.
Corollary 3.7. Suppose that χ1, . . . , χk6= and χ1· · · , χk= then k
Y
i=1
g(χi) = pχk(−1)J (χ1, . . . , χk−1).
Proof. Use part four of proposition 3.6 and multiply both sides by g(χk). Note that if k = 2
we have J (χ1) = 1 in the right hand side, by definition.
Corollary 3.8. Under the same assumptions as in corollary 3.7 we have J (χ1, . . . , χk) = −χk(−1)J (χ1, . . . , χk−1).
Proof. The case when k = 2 is the content of theorem 3.4 part three, hence suppose k > 2. Use equations (8) and (9) above combined with the assumptions to get
Y i g(χi) = J0(χ1, . . . , χk) + X s6=0 X t1+...+tk=s Y i χi(ti) ψ(s) = J0(χ1, . . . , χk) + X s6=0 (s)J (χ1, . . . , χk)ψ(s) = J0(χ1, . . . , χk) + J (χ1, . . . , χk) X s6=0 ψ(s) = J0(χ1, . . . , χk) − J (χ1, . . . , χk)
where the last equality follows from part two of proposition 3.2. By corollary 3.7 and part three of proposition 3.6 we have
pχk(−1)J (χ1, . . . , χk−1) = J0(χ1, . . . , χk) − J (χ1, . . . , χk)
= (p − 1)χk(−1)J (χ1, . . . , χk−1) − J (χ1, . . . , χk).
which concludes the proof.
(I) χ1· · · χk 6= =⇒ |J (χ1, . . . , χk)| = p(k−1)/2.
(II) χ1· · · χk = =⇒ |J0(χ1, . . . , χk)| = (p − 1)p(k/2)−1 and |J (χ1, . . . , χk)| = p(k/2)−1.
Proof. Since each χi 6= we have by proposition 3.3 |g(χi)| =
√ p for each i ∈ {1, 2, . . . , k}. Thus |J (χ1, . . . , χk)| = Q i|g(χi)| |g(χ1, . . . , χk)| = √ pk √ p = p (k−1)/2.
For the second part note that
J0(χ1, . . . , χk) = (p − 1)χk(−1)J (χ1, . . . , χk−1).
by proposition 3.6 part three. Since χ1· · · χk−1 6= we have by corollary 3.7 and 3.8
|J0(χ1, . . . , χk)| = (p − 1)Q i|g(χi)| p = (p − 1)√pk p = (p − 1)p (k/2)−1 and |J (χ1, . . . , χk)| = | − χk(−1)J (χ1, . . . , χk−1)| = Q i|g(χi)| p = √ pk p = p (k/2)−1. We now consider the most general case needed for our purposes, namely the number of solu-tions N to the equation a1xn11 + a2xn2 + . . . + akxnkk = b where a1, . . . , ak ∈ Fp∗ and b ∈ Fp.
Here is the main theorem of this section, and we will see later how it will be useful for our study of the congruence zeta function.
Theorem 3.10.
(I) b = 0 =⇒ N = pk−1+X
k
Y
i=1
χi(a−1i )J0(χ1, . . . , χk) where the sum is over all k-tuples
(χ1, . . . , χk) where χi 6= , χnii = for i = 1, 2, . . . , k and χ1· · · χk = . If M is the
number of such k-tuples, then |N − pr−1| ≤ M (p − 1)p(k/2)−1.
(II) b 6= 0 =⇒ N = pr−1+Xχ1· · · χk(b) k
Y
i=1
χi(a−1i )J (χ1, . . . , χk) where the sum is over
all k-tuples (χ1, . . . , χk) where χi 6= and χnii = for i = 1, 2, . . . , k. If M1 is the
number of such k-tuples with χ1· · · χk= and M2 is the number of such k-tuples with
χ1· · · χk 6= , then |N − pr−1| ≤ M1p(k/2)−1+ M2p(k−1)/2.
Proof. Let L(u) = L(u1, . . . , uk) =Pki=1aiui and Ni(ui) = N (xnii = ui). Similarly as before,
notice that
N = X
L(u)=b
N1(u1)N2(u2) · · · Nk(uk)
(10)
where the sum is over all k-tuples u = (u1, . . . , uk) such that L(u) = b. By proposition 2.10
we know that
Ni(ui) =
X
χi
χi(ui)
where χi ranges over all character of order dividing ni. Inserting this into equation (10) gives
We consider the two cases b = 0 and b 6= 0 separately. If b = 0 we define ti= aiui, then X L(u)=b χ1(u1) · · · χk(uk) = X t1+...+tk=0 χ1(t1)χ1(a−11 ) · · · χk(tk)χk(a−1k ) = χ1(a−11 ) · · · χk(a−1k ) X t1+...+tk=0 χ1(t1) · · · χk(tk) = χ1(a−11 ) · · · χk(a−1k )J0(χ1, . . . , χk). (12) If b 6= 0 we define ti= b−1aiui, then X L(u)=b χ1(u1) · · · χk(uk) = X t1+...+tk=1 χ1(a−11 )χ1(b)χ(t1) · · · χk(a−1k )χk(b)χ(tk) = χ1· · · χk(b)χ1(a−11 ) · · · χk(a−1k )J (χ1, . . . , χk). (13)
Now, in the first case (11) becomes X
χ1,...,χk
χ1(a−11 ) · · · χk(a−1k )J0(χ1, . . . , χk)
(14)
If χi= for all i, then J0(χ1, . . . , χk) = pk−1. If some but not all χi = then J0(χ1, . . . , χk) = 0
and ifQk
i=1χi6= then J0(χ1, . . . , χk) = 0, all this follows directly from proposition 3.6. Given
these facts we are left with the expression for N as stated in the theorem. If b 6= 0 the proof is similar. Both estimates follow directly from the fact that (for any χ ∈ Ωp and a ∈ Fp∗)
|χ(a)| = 1 since χ(a) is a (p − 1)st root of unity by proposition 2.5. Using the tools outlined in the earlier sections, we can generalize the notions of Gauss and Jacobi sums in the following way. Let F be an arbitrary finite field of order q = pr. Let ψ : F −→ C be the map α 7→ ζptrF /Fp(α), where Fp = Z/pZ and ζp = e2πi/p. The multiplicative
group of any finite field is cyclic, thus the propositions of multiplicative characters (which hinges on this fact) naturally extend to fields of arbitrary order, simply replace p with q. Given the following definition, the same extension to arbitrary finite fields is also valid for Gauss sums, Jacobi sums and the interrelations we saw between them earlier in this section. Let χ be a character of F and α ∈ F∗.
Definition 12. A sum of the formP
t∈Fχ(t)ψ(αt) is called a Gauss sum on F belonging to
the character χ and will be denoted gα(χ).
Note that this definition reduces to the previous one if F = Fp. To generalize the definition
of Jacobi sums, simply let the characters be characters on F instead of Fp.
Next we prove a theorem which will be of great use later on. Consider the equation f (y0, . . . , yn) =
Pn
i=0aiymi = 0 where ai ∈ F∗. This is a homogeneous equation, so it defines a hypersurface
Hf(F ) ⊆ Pn(F ).
Theorem 3.11. Suppose F is a finite field of order q ≡ 1 (m). Then |Hf(F )| = n−1 X i=0 qi+ 1 q − 1 X χ0,...,χn J0(χ0, . . . , χn) n Y j=0 χj(a−1j )
where the sum is over all (n+1)-tuples (χ0, . . . , χn) such that χi 6= , χmi = and χ0· · · χn= .
Proof. The first part is a corollary of the first part of theorem 3.10, we have |Hf(F )| = qn+ X χ0,...,χn Y i χi(a−1i )J0(χ0, . . . , χn). (15)
where the sum is over all (n + 1)-tuples (χ0, . . . , χn) as given by the theorem. The number
|Hf(F )| is obtained from dividing (15) by q − 1. For the second part, recall that J0(χ0, . . . , χn) = χn(−1)(q − 1)J (χ0, . . . , χn−1)
(16)
where we have extended proposition 3.6 to fields of arbitrary order q as discussed. By the same proposition we have
J (χ0, . . . , χn−1) =
g(χ0) · · · g(χn−1)
g(χ0· · · χn−1)
. (17)
Multiply the numerator and denominator of the right hand side of (17) by g(χn) and apply
corollary 3.7 to get J0(χ0, . . . , χn−1) = g(χ0) · · · g(χn) g(χ0· · · χn−1)g(χn) = g(χ0) · · · g(χn) χn(−1)q , (18)
substituting (18) into (16) concludes the proof.
3.2.1. A Proof of the Law of Quadratic Reciprocity. Here we provide an elegant proof of the famous law of quadratic reciprocity, using what we know about Jacobi and Gauss sums. Theorem 3.12 (Law of Quadratic Reciprocity). Let p and q be two distinct odd prime num-bers, then
(p/q)(q/p) = (−1)p−12 q−1
2
Proof. Let χ denote the character on Fp of order two. Then χq+1= since q + 1 is even. By
corollary 3.7 we have g(χ)q+1= g(χ)2 q+1 2 = pq+12 (−1) p−1 2 q+1 2 = p(−1) p−1 2 J (χ, . . . , χ)
where there are q components in J (χ, . . . , χ). Thus, J (χ, . . . , χ) = X t1+...+tq=1 χ(t1) · · · χ(tq) = p q−1 2 (−1) p−1 2 q−1 2 . If t1 = . . . = tq, then ti = 1/q and χ(1/q) · · · χ(1/q) = χ(1/q)q = χ(q)−q = χ(q). If ti 6= tj
for at least two indexes, then there are q terms in J (χ, . . . , χ), all of equal value, found by permutating (t1, . . . , tq) cyclicly. Thus in modulo q all of these terms vanish, hence
pq−12 (−1) p−1 2 q−1 2 ≡ χ(q) (q). Thus, (−1)p−12 q−1 2 (p/q) ≡ (q/p) (q) (19)
4. The Zeta Function
E. Artin introduced the concept of the congruence zeta function in 1924. In 1949 A. Weil formulated a set of conjectures known as the Weil conjectures related to the zeta function. The first part of the Weil conjectures states that any algebraic set has a rational zeta function, this was proved in 1959 by B. Dwork.
We define the congruence zeta function, show how it and the Riemann zeta function satisfy analogous relations, and prove that (under certain criteria) Zf(u) is rational. We follow [IR,
Ch. 11.1] and [IR, Ch. 11.3-4].
It can be shown that if F is a finite field of order q then there exists a field Fs⊇ F of order
qs where s ≥ 1 is an integer. Given a homogeneous polynomial f (x) ∈ F [x0, x1, . . . , xn] we
let Ns denote the number |Hf(Fs)|. We wish to study the numbers Ns by studying the power
seriesP∞
s=1 Nsus
s of a complex variable u. We can (and will) view this as a formal power series,
which enables the exclusion of all questions of convergence. However, this is not necessary. Notice that Ns≤ |Pn(Fs)| = qs(n+1)− 1 qs− 1 = n X k=0 (qs)k< (n + 1)qsn. If |u| < q−n and s ≥ 1, then
∞ X s=1 |Ns||us| < (n + 1) ∞ X s=1 xs (20)
where 0 ≤ x = qn|u| < 1. Thus (20) converges, i.e. P∞
s=1Nsus converges absolutely for all
|u| < q−n.
Definition 13. The zeta function of the hypersurface defined by f is the series given by Zf(u) = exp ∞ X s=1 Nsus s .
If the zeta function is expanded about the origin we see that the constant term is 1. We can therefore assume that Zf(u) = p(u)q(u) where p(0) = q(0) = 1. To see this, assume that is not
the case. Then Zf(0) = p(0)q(0) = 1 if and only if p(0) = q(0) = c. Now if p(u) =Pni=0aiui and
q(u) =Pm i=0biui then p(u) q(u) = cPn i=0c−1aiui cPm i=0c−1biui = Pn i=0c−1aiui Pm i=0c−1biui = p 0(u) q0(u)
where p0(u) and q0(u) have the desired property. Hence the zeta function can be expressed in the form Zf(u) = Q i(1 − αiu) Q j(1 − βju) (21)
where αi, βj ∈ C. We now characterize rational zeta functions, this will be useful later when
we consider Zf(u) for particular homogeneous polynomials f .
Proposition 4.1. Z(u) ∈ {p(u)q(u) : p(u), q(u) ∈ C[u]} if and only if there exist αi, βj ∈ C such
that Ns=
P
jβsj −
P
Proof. We start by assuming that there exist αi, βj ∈ C such that Ns =Pjβsj −
P
iαsi. By
inserting this expression for Ns we get ∞ X s=1 Nsus s = ∞ X s=1 P jβsj − P iαsius s ! = ∞ X s=1 P j(βju)s−Pi(αiu)s s ! =X j ∞ X s=1 (βju)s s −X i ∞ X s=1 (αiu)s s . Further simplify the last expression above by using the identity P∞
s=1z s s = −ln(1 − z) to get X i ln(1 − αiu) − X j ln(1 − βju) (22)
Finally, to find the zeta function, we exponentiate (22) to get Zf(u) = exp X i ln(1 − αiu) − X j ln(1 − βju) ! = exp P iln(1 − αiu) exp P jln(1 − βju) = Q iexp ln(1 − αiu) Q jexp ln(1 − βju) = Q i(1 − αiu) Q j(1 − βju)
which is a rational function of u. We now show the other direction. Suppose therefore that the zeta function is rational so that
Zf(u) = Q i(1 − αiu) Q j(1 − βju) (23)
where αi, βj ∈ C. Take the logarithmic derivative of both sides of (4.1) to get
Zf0(u) Zf(u) = d duln Q i(1 − αiu) Q j(1 − βju) ! = d du X i ln(1 − αiu) − X j ln(1 − βju) ! =X i −αi 1 − αiu −X j −βj 1 − βju .
Multiply the first and last expressions above by u and expand the denominators in the last expression in a geometric series to get
We now compare this last power series with an alternate way of computing uZ
0 f(u)
Zf(u). By
definition Zf(u) = exp
P∞
s=1 Nsu
s . By taking the logarithmic derivative and then multiplying
by u we get uZ 0 f(u) Zf(u) = u d du ∞ X s=1 Nsus s = ∞ X s=1 Nsus, hence we have ∞ X s=1 X j βjs−X i αsi ! us= ∞ X s=1 Nsus and thus Ns= X j βjs−X i αsi. Next we show that Ns is independent of the choice of Fs, so that Ns only depends on s. This
is a consequence of the following proposition.
Proposition 4.2. Let E and E0 be field extensions over F of the same order qs. Then there is an isomorphism σ : E−∼→ E0 such that σ
F = id.
We can use σ to induce a map from the respective projective n-spaces in a natural way by letting σ : Pn(E) −→ Pn(E0) map [α0, . . . , αn] 7→ [σ(α0), . . . , σ(αn)]. This map is
well-defined since if [αo, . . . , αn] = [β0, . . . , βn], then by definition there is some γ ∈ E∗ such
that αi = γβi for i = 0, 1, . . . , n, thus [σ(α0), . . . , σ(α(n)] = [σ(γ)σ(β0), . . . , σ(γ)σ(βn)] =
[β0, . . . , βn]. By virtue of the isomorphic property of σ, σ is a bijection. Indeed, let α =
[α0, . . . , αn] ∈ Pn(E0) be arbitrary, then σ maps [σ−1(α0), . . . , σ−1(αn)] 7→ α, hence σ is
sur-jective. Also, both Pn(E) and Pn(E0) are finite sets, hence σ is injective. It is moreover true that σ
Hf(E) is bijection to Hf(E
0). We already know that the restriction is injective. Let
α = [α0, . . . , αn] ∈ Hf(E0), then f (α0, . . . , αn) = 0 and σ [σ−1(α0), . . . , σ−1(αn)] = α. We
need to show that [σ−1(α0), . . . , σ−1(αn)] ∈ Hf(E). Now, σ acts as the identity on F , hence
f σ−1(α0), . . . , σ−1(αn = σ−1 f ([α0, . . . , αn]) = σ−1(0) = 0, hence [σ−1(α0), . . . , σ−1(αn)] ∈
Hf(E). We conclude that |Hf(E)| = |Hf(E0)| and hence Ns indeed only depends on s as
claimed, since this result holds also for the trivial field extensions F ⊆ F and F ⊆ F0∼= F . 4.1. The Zeta Analogy. The congruence zeta function and the Riemann zeta function sat-isfy a highly analogous relation. In the case of the Riemann zeta function we have
ζ(s) = ∞ X n=1 n−s=Y p 1 (1 −p1s) , s > 1 where the product is over all primes p > 0.
So far we have dealt with the congruence zeta function defined by one polynomial f . Let us now consider it over a set of polynomials {f1, . . . , fm} ⊆ F [x1, . . . , xn]. Let F be a finite field
of order q and let V = {a ∈ An(F ) : fj(a) = 0 for j = 1, 2, . . . , m} be an algebraic set in
An(F ).
Definition 14. The function
ZV(u) = exp ∞ X s=1 Nsus s !
where Ns is the number of points in An(Fs) satisfying the equations defined in V , is called
Let K be the algebraic closure of F . One can show that K is the countable union of isomorphic copies to Fs of order qs for all positive integers s. It is natural to consider a field containing
all fields related to the numbers Ns and K will serve as that field. We can then extend V so
that V ⊆ An(K) and with Ns points whose coordinates all lie in Fs.
Suppose a = (a1, . . . , an) ∈ V and let L be the smallest field containing F ∪ {a1, . . . , an}.
Suppose |L| = qd, we then call a to be a point of degree d.
Lemma 4.3. The points a, aq, . . . , aqd−1 all lie in V and are pairwise distinct.
Proof. Suppose that a = aqj for some j ∈ {1, 2 . . . , d − 1}, and let j be the minimal element with this property. Notice that this implies that aqkj = a for all k ∈ Z+. Additionally, we
have that aqd= a since ai ∈ L for i = 1, 2, . . . , n. Thus, by the choice of j we have that j | d.
By basic facts about finite fields, there exists a subfield L0⊆ L of order qj and since a = aqj
is equivalent to ai ∈ L0 or i = 1, 2 . . . , n we have a contradiction by definition of L. This proves
that the points a, aq, . . . , aqd−1 are pairwise distinct.
To show that the points all lie in V , we need to verify that they satisfy the polynomial equations defining V . Let ψ : L −∼→ L be the automorphism which maps x 7→ xq, so that
ψ
F = id. Let a = (a1, . . . , an) ∈ V and let j ∈ {1, . . . , m} correspond to a polynomial
defining V . Then, by definition, fj(a1, . . . , an) = 0. But then ψk(0) = ψk fj(a1, . . . , an) =
fj ψk(ai), . . . , ψk(an) = fj aq
k
i , . . . , a qk
i = 0. Since this holds for any j ∈ {1, 2, . . . , m} and
k ∈ {1, 2, . . . , d − 1}, we are done.
Definition 15. Let a ∈ V be a point of degree d. A set of the form P = {a, aq, . . . , aqd−1} is called a prime divisor on V . The degree of P is denoted δ(P) and is equal to d.
Proposition 4.4. Let F be a finite field such that [F : Z/pZ] = n. Then the subfields of F are in one-to-one correspondence with the divisors of n.
Lemma 4.5. Ns=
P
d|sdnd where nd is the number of prime divisors on V of degree d.
Proof. The prime divisors partition V . Let α ∈ V be a point such that all coordinates lie in Fs. This Fscould be a larger field than necessary i.e. there is some d such that Fd⊆ Fs is the
smallest field such that α ∈ Fd. We know from proposition 4.4 that d | s, so that α defines a
(unique) prime divisor of degree d which is a divisor of s. This proves the lemma. Now we are in a position to show how the congruence zeta function and the Riemann zeta function bear considerable resemblance.
Theorem 4.6. ZV(u) =
Y
P
1 1 − uδ(P).
Proof. By merging the factors corresponding to prime divisors of the same degree, the product can be rewritten as ∞ Y k=1 1 1 − uk nk . (24)
Expand the denominator into a geometric series and compute the coefficient of um to get 1 u ∞ X m=1 X d|m dnd um = ∞ X m=1 Nmum−1 (25)
by lemma 4.5. Integrating the right hand side of (25) and then taking the exponential concludes
the proof.
4.2. The Rationality of the Zeta Function. In this section we prove that the zeta function Zf(u) of f (x0, . . . , xn) = a0xm0 + a1xm1 + . . . + anxmn, where a0, . . . , an∈ F∗ and F is a field of
order q ≡ 1 (m), is rational. In order to do this, we first outline a proof of the Hasse-Davenport relation, which is an interesting result in its own right.
Clearly f is a homogeneous polynomial, it therefore defines a projective hypersurface Hf(Fs)
where F ⊆ Fs is a field extension of degree s. Theorem 3.11 gives
Ns= n−1 X i=0 qsi+ 1 qs X χ(s)0 ,...,χ (s) n n Y j=0 χ(s)i (a−1i )g(χ(s)i ) (26)
where χ(s)i are characters of Fs such that χ(s)mi = , χ (s) i 6= and χ (s) 0 · · · χ (s) n = . Let
χ be a character of F , compose it with NFs/F to get χ
0 = χ ◦ N
Fs/F : Fs −→ C. This
mapping is in fact a character on Fs since χ0(ab) = χ NFs/F(ab) = χ NFs/F(a)NFs/F(b) =
χ NFs/F(a)χ NFs/F(b) = χ0(a)χ0(b) by proposition 2.11.
Lemma 4.7.
(I) χ 6= ρ =⇒ χ0 6= ρ0.
(II) χm = =⇒ χ0m= .
(III) χ0(a) = χ(a)s for all a ∈ F.
Proof. The first follows from the last part of proposition 2.11. For the second part, note that χ ◦ NFs/F(a)
m
= χm NFs/F(a) = NFs/F(a) = 1 for all a ∈ F
∗
s, hence it is the trivial
character. For the third part, simply note that χ0(a) = χ NFs/F(a · 1) = χ(as) = χ(a)s, by
proposition 2.11.
This lemma shows that by letting χ vary over all characters of F of order diving m, the same happens for the corresponding characters χ0 of Fs. Thus equation (26) can now be rewritten
as Ns= n−1 X i=0 qsi+ 1 qs X χ0,...,χn n Y j=0 χi(a−1i )sg(χ0i) (27)
where χi are characters of F such that χmi = , χi 6= and χ0· · · χn = . It turns out that
g(χ0) and g(χ) satisfy a simple relationship. Theorem 4.8 (Hasse-Davenport Relation).
− g(χ)s
= −g(χ0)
Remark. We remind that s is an integer which depends on χ0, which is a character on Fs⊇ F
defined above.
In order to prove this, we first need two lemmas, the proofs of which will be omitted (see [IR, Ch. 11.4]). Given a monic polynomial f (x) = xn− a
1xn−1+ . . . + (−1)nan ∈ F [x], we define
a mapping λ by λ(f ) = ψ(a1)χ(an) and let λ(1) = 1.
Lemma 4.10. g(χ0) = P δ(f )λ(f )s/δ(f ) where the sum is over all irreducible monic
polyno-mials of degree s in F [x].
Given this definition of λ we have the identity X f λ(f )tδ(f ) =Y f 1 1 − λ(f )tδ(f )
where the sum is over all monic polynomials, and the product is over all monic irreducible polynomials in F [x]. Let us spend some time outlining the legitimacy of this identity. We express the factors in the right hand side as a geometric series
1 1 − λ(f )tδ(f ) = ∞ X k=0 (λ(f )tδ(f ))k. Using the fact that λ is multiplicative by lemma 4.9 we get
Y f 1 1 − λ(f )tδ(f ) = Y f ∞ X k=0 λ(fk)tkδ(f ) =Y f 1 + λ(f )tδ(f )+ . . . + λ(fk)tkδ(f )+ . . . .
Now, the ring F [x] is a UFD so in particular each monic polynomial can be expressed as a product of irreducible (monic) polynomials in a unique way. Now, since each power of each monic irreducible polynomial is found in the sum above, we see, by multiplying the factors in the last expression, that each monic polynomial occurs in the argument of λ in the resulting sum. Inductively applying lemma 4.9 gives Qn
i=1λ(f ai i ) = λ( Qn i=1f ai
i ). This together with
the basic fact that δ(Qn
i=1f ai
i ) =
Pn
i=1aiδ(fi) shows that the identity holds.
4.2.1. Proof of the Hasse-Davenport Relation. By collecting terms of equal degree, we have X f λ(f )tδ(f )= ∞ X s=0 X δ(f )=s λ(f ) ts. (28)
where the left hand side is as in (28). We proceed by analyzing the sums P
δ(f )=sλ(f ). If s = 1, then X δ(f )=s λ(f ) = X a∈F λ(x − a) =X a∈F χ(a)ψ(a) = g(χ) (29)
If s > 1 it turns out that the sums vanish, since X δ(f )=s λ(f ) = X ai∈F λ(xs− a1xs−1+ . . . + (−1)sas). (30)
Now, λ only depends on the coefficients a1 and an. By fixing those and letting the others vary
over F , we see that (30) is equal to qs−2 X a1, an χ(as)ψ(a1) = qs−2 X as χ(as) ! X a1 ψ(a1) ! = 0 by proposition 2.6 extended to characters on arbitrary finite fields. Thus we have
X f λ(f )tδ(f ) = 1 + g(χ)t =Y f 1 1 − λ(f )tδ(f ).
and d dt ln Y f 1 1 − λ(f )tδ(f ) ! = d dt −X f ln 1 − λ(f )tδ(f ) =X f λ(f )δ(f )tδ(f )−1 1 − λ(f )tδ(f ) .
By multiplying these expressions by t, we have g(χ)t 1 + g(χ)t = X f λ(f )δ(f )tδ(f ) 1 − λ(f )tδ(f ).
Expand the denominators in geometric series to get
∞ X s=1 (−1)s−1g(χ)sts=X f ∞ X k=1 λ(f )kδ(f )tkδ(f ) . By comparing coefficients for ts we get
(−1)s−1g(χ)s = X
δ(f )|s
δ(f )λ(f )s/δ(f )= g(χ0)
where the last equality is the content of lemma 4.10. This concludes the proof of theorem 4.8. We now make use of the Hasse-Davenport relation to conclude our analysis of the numbers Ns associated with Zf(u), where f (x0, . . . , xn) = a0xm0 + a1xm1 + . . . + anxmn ∈ F [x0, . . . , xn].
Substituting g(χ0) = (−1)s+1g(χ)s into equation (27) we get Ns= n−1 X i=0 qsi+ (−1)n+1 X χ0,...,χn (−1)n+1 q n Y j=0 χi(a−1i )g(χi) !s , (31)
where χi are characters of F such that χmi = (where m is as in the polynomial f ), χi 6=
and χ0· · · χn= .
Finally, we are in a position to state the main theorem of this section, which establishes the rationality of Zf(u) under certain conditions.
Theorem 4.11. Let a0, . . . , an ∈ F∗ where |F | = q ≡ 1 (m) and f (x0, . . . , xn) = a0xm0 +
a1xm1 + . . . + anxmn. Then the congruence zeta function Zf(u) is a rational function of the
form Zf(u) = P (u)(−1)n (1 − u)(1 − qu) · · · (1 − qn−1u), where P (u) = Y χ0,...,χn 1 − (−1)n+11 qχ0(a −1 0 ) · · · χn(a−1n )g(χ0) · · · g(χn)u
and the product is over all (n+1)-tuples (χ0, . . . , χn) such that χmi = , χi 6= and χ0· · · χn= .
Proof. This is a direct result of our characterization of rational zeta functions i.e. proposition 4.1 applied to the current situation. Note that if n is even, then (31) can be written as Ns =
Pn−1 j=0 βjs−
P
iαsi, where βj = qj and αi = (−1)n+1 1qχ0(a−10 ) · · · χn(a−1n )g(χ0) · · · g(χn)
for some (n + 1)-tuple (χ0, . . . , χn) subject to the conditions above. If n is odd, then (31) can
be written as P
jβj −Piαi where βj = Ns and αi = 0. Hence the exponent (−1)n in the
numerator of Zf(u), since the zeta function is of the form
Q
i(1 − αiu)
Q
j(1 − βju)
