Gauss and Jacobi Sums and the Congruence Zeta Function

Home , Gauss sum, Legendre symbol

U.U.D.M. Project Report 2018:19

Einar Waara

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

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

Date: January 2018.

1 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 2πiat/p (t/p)e , where Fp is the finite field of order p and (·/p) is the quadratic reciprocity t∈Fp √ √ 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 AndréWeil (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 rational zeta functions Zf (u) and use this characterization to prove that Zf (u) is rational when m m m ∗ f(x0 . . . , xn) = a0x0 + a1x1 + ... + anxn 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. This is the main reference for the present thesis.

2 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  1 if a is a quadratic residue modulo p  (a/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. Pp−1 Corollary 2.3. (IR, 6.3 lemma 2) 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 Fp is a group homomorphism from the multiplica- ∗ tive group of Fp to the multiplicative group of the complex numbers: χ : Fp −→ (C \{0}, ×).

3 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)

2 2 ∗ Proof. χ(1) = χ(1 ) = χ(1) 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) = −1 −1 χ(1) = 1 hence χ(a) = χ(a ). 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 −1 complex multiplication we have χ(a) = χ(a). ( P 0 if χ 6= Proposition 2.6. Let χ ∈ Ωp, then χ(t) = t∈Fp 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 χ(t) = P χ(at) = P χ(t), hence P χ(t) = 0. t∈Fp t∈Fp t∈Fp t∈Fp

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 ∗ ∗ l g ∈ Fp be a generator. If a ∈ Fp then a = g for some l ∈ {0, 1, . . . , p−1}. Thus any character l l χ ∈ Ωp is completely determined by its action on g since χ(a) = χ(g ) = χ(g) . 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 k 2πi(k/(p−1)) follows that |Ωp| ≤ p − 1. We claim that a generator to Ωp is given by λ(g ) = e . Firstly λ is well deﬁned since λ(gk+p−1) = λ(gk)e2πi = λ(gk). Since λ(gk1 gk2 ) = 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. ∗ l Remark. Let a ∈ Fp \{1} then a = g for some l ∈ {1, 2, . . . , p − 2}, so p − 1 - l. Thus l 2πi(l/(p−1)) ∗ λ(a) = λ(g) = e 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 ∈ F ∗ \{1} then P χ(a) = 0. p χ∈Ωp

4 Proof. Let z be the complex number P χ(a). Since a 6= 1 there is (by the remark above) χ∈Ωp some character χ such that χ(a) 6= 1, thus X X λ(a)z = λ(a)χ(a) = λχ(a) = z.

χ∈Ωp χ∈Ωp

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.

∗ n Proposition 2.9. If a ∈ Fp , n|p − 1 and the equation x = 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 n l/n and by assumption x = a is not solvable, thus n - l (otherwise g 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 n p−1 χ = λ = since λ was determined to be a generator to Ωp. n n Let N(x = a) denote the number of solutions to the equation x = 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. n P Proposition 2.10. If n | p − 1, then N(x = a) = χn= χ(a) where the sum is taken over all characters χ such that χn = .

n Proof. If χ is a character such that χ = , then χ(g) ∈ C is a n’th root of unity since n n χ(g) = χ (g) = (g) = 1. Thus there are at most n such characters in Ωp. But in the previous proposition we deﬁned χ = λ(p−1)/n which meant that χ(g) = e2πi/n, thus , χ, χ2, . . . , χn−1 are 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 n n P n a solution, then χ(a) = χ(b ) = χ(b) = (b) = 1 and thus χn= χ(a) = n = N(x = a). Lastly, assume that a 6= 0 and that the equation xn = a is unsolvable. By proposition 2.9, n n there is a ρ ∈ Ωp such that ρ = and ρ(a) 6= 1. We have {χ : χ = } < Ωp, thus P P P P ρ(a) χn= χ(a) = χn= χρ(a) = χn= χ(a), hence ρ(a) − 1 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.

n Deﬁnition 4. A (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.

n+1 n (A (F ) \{(0, 0,..., 0)}) Deﬁnition 5. P (F ) = ∼ is called the projective n-space over F . The equivalence class of a ∈ An+1(F ) \{(0, 0,..., 0)} is denoted by [a].

5 n qn+1−1 The size of the aﬃne n-space is q , while the size of the projective n-space is q−1 = qn + qn−1 + ... + q + 1.

A homogeneous polynomial f(x , x , . . . x ) = P a xi1 xi2 . . . xin in F [x , x , . . . , x ] 1 2 n (i1,i2,...,in) i1i2...in 1 2 n 1 2 n 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 n homogeneous) polynomial f ∈ F [x1, x2, . . . xn] we can view it as a function f : A (K) −→ K by evaluating f at points in An(K) in the usual way.

n Deﬁnition 6. Hf (K) = {a ∈ A (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. n Deﬁnition 7. Hg(K) = {[a] ∈ P (K): g(a) = 0} is called the projective hypersurface of g in P n(K).

∗ d The set Hg(K) is well defined since if γ ∈ F we have g(γx) = γ g(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]. n Definition 8. V = {a ∈ A (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 definitions 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.

Ps−1 qi Definition 9. The trace of α ∈ E from E to F is defined as trE/F (α) = i=0 α and the Qs−1 qi norm of α from E to F is defined as NE/F (α) = i=0 α . 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) s (VII) NE/F (aα) = a NE/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 ∈ Qs−1 qi (qs−1)/(q−1) ker(NE/F ) if and only if NE/F (α) = i=0 α = α = 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 indeed surjective.

6 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].

2πi/p 3.1. Gauss Sums. Let χ ∈ Ωp, a ∈ Fp and ζ = e . Deﬁnition 10. A sum of the form P χ(t)ζat is called a Gauss sum belonging to the t∈Fp character χ and will be denoted ga(χ), and g1(χ) in particular will be denoted g(χ). ( p if a ≡ 0 (p) Lemma 3.1. Pp−1 ζat = t=0 0 otherwise

a Pp−1 at Proof. If a ≡ 0 (p) then a/p is an integer so ζ = 1, hence t=0 ζ = p. Otherwise, if a Pp−1 at ζap−1 a 6≡ 0 (p), then ζ 6= 1 and t=0 ζ = 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. −1 (I) a 6= 0 and χ 6= =⇒ ga(χ) = χ(a )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)g (χ) = χ(a) P χ(t)ζat = a t∈Fp P χ(at)ζat = g (χ). Multiplying both sides by χ(a)−1 = χ(a−1) proves the ﬁrst statement. t∈Fp 1 Next we see that g () = P (t)ζat = P ζat = 0 by lemma 3.1, this proves the second a t∈Fp t∈Fp statement. Lastly, g (χ) = P χ(t), hence the third and fourth statement both follow from 0 t∈Fp proposition X. √ Proposition 3.3. If χ 6= , then |g(χ)| = p.

Proof. Our strategy will be to evaluate the sum X S = ga(χ)ga(χ)

a∈Fp in two diﬀerent ways. First, suppose that a 6= 0, then we have

−1 −1 ga(χ)ga(χ) = χ(a )g(χ)χ(a )g(χ) = χ(a−1)g(χ)χ(a)g(χ) = |g(χ)|2.

2 By proposition 3.2 we have g0(χ) = 0, thus S = (p − 1)|g(χ)| . Next we use the deﬁnition 10 directly to get X X ax−ay ga(χ)ga(χ) = χ(x)χ(y)ζ x y

7 thus, by lemma 3.1 we have X X X X X χ(x)χ(y)ζax−ay = χ(x)χ(y)δ(x, y)p = (p − 1)p a x y x y 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(χ , . . . , χ ) = P χ (t ) ··· χ (t ) is called a 1 k t1+...+tk=1 1 1 k k Jacobi sum.

2 2 2 2 We start by considering the familiar equation x + y = 1 where x, y ∈ Fp. Let N(x + y = 1) denote the number of solutions to this equation. First, notice that N(x2 + y2 = 1) = P N(x2 = a)N(y2 = b) = P (1 + (a/p))(1 + (b/p)) = p + P (a/p) + P (b/p) + Pa+b=1 P a+b=1 a b a+b=1(a/p)(b/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 (1) (a/p)(b/p). a+b=1

Let us pause this problem for a moment and consider N(x3 + y3 = 1) instead. Again, notice 3 3 P 3 3 that N(x + y = 1) = a+b=1 N(x = a)N(y = 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 2 χ ∈ Ωp \{} be a character of order 3. Since (2, 3) = 1, χ 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

2 2 X X X N(x3 + y3 = 1) = χi(a) χj(b) a+b=1 i=0 j=0 X X X (2) = χi(a)χj(b) i j a+b=1

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) g(χ)g(λ) (IV) χλ 6= =⇒ J(χ, λ) = g(χλ)

8 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 X X J(χ, χ−1) = χ(a)χ−1(b) = χ(1)χ−1(0) + χ(ab−1) a+b=1 a+b=1 ∧ b6=0 X = χa(1 − a)−1. a6=1 −1 −1 Let c = a(1 − a) , then if c 6= 1 we have a = c(1 + c) . Hence, when a varies over Fp \{1}, c varies over Fp \ {−1}. Thus, X X J(χ, χ−1) = χ(c) = χ(c) − χ(−1) = −χ(−1)

c6=−1 c∈Fp by proposition 2.6. To prove the last part, notice that ! ! X X g(χ)g(λ) = χ(x)ζx λ(y)ζy x y X = χ(x)λ(y)ζx+y x,y ! X X = χ(x)λ(y) ζt. t x+y=t

If t = 0 then X X χ(x)λ(−x) = λ(−1) χλ(x) = 0. x+y=0 x If t 6= 0, let x = tX and y = tY . Then we have X X X χ(x)λ(y) = χ(tX)λ(tY ) = χλ(t)χ(X)λ(Y ) = χλ(t)J(χ, λ). x+y=t X+Y =1 X+Y =1 Thus, X g(χ)g(λ) = J(χ, λ) χλ(t)ζt = J(χ, λ)g(χλ). t √ Corollary 3.5. χ, λ, χλ 6= =⇒ |J(χ, λ)| = p.

√ √ g(χ)g(λ) p p √

Proof. |J(χ, λ)| = = √ = p by proposition 3.3. g(χλ) p Using these identities, we can definitively evaluate (1) and make further progress with (2). P For (1) simply note that an application of part three yields 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. 2 2 Hence, the number of solutions to x + y = 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 ! X X X (3) χi(a)χj(b) = p − χ(−1) − χ2(−1) + J(χ, χ) + J(χ2, χ2). i j a+b=1

9 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 X X J(χ, χ) + J(χ2, χ2) = J(χ, χ) + J(χ, χ) = χ(ab) + χ(ab) = 2ReJ(χ, χ). a+b=1 a+b=1 Thus (3) simpliﬁes to (4) N(x3 + y3 = 1) = p − 2 + 2ReJ(χ, χ). Clearly, this is not as precise as the value of N(x2 + y2 = 1). However, we can now give a 3 3 estimate to the number of solutions to x + y = 1 in Fp. By corollary 3.5 we have √ (5) |N(x3 + y3 = 1) − p + 2| ≤ 2 p. 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 k X Y J0(χ1, . . . , χk) := χi(ti).

t1+...+tk=0 i=1 P This is the same as the Jacobi sum except that i ti = 0 instead of 1. Also, let ψ : Fp −→ C t P Q be the map t 7→ ζ , then ψ( i ti) = i ψ(ti). Proposition 3.6. k−1 (I) J0(, . . . , ) = J(, . . . , ) = p

(II) If some but not all χi = , then J0(χ1, . . . , χk) = J(χ1 . . . , χk) = 0

(III) Suppose χk 6= , then ( Qk 0 if i=1 χi 6= J0(χ1, . . . , χk) = (p − 1)χk(−1)J(χ1, . . . , χk−1) otherwise

k k Y Y (IV) χ1, . . . , χk, χ1 ··· χk 6= =⇒ g(χi) = J(χ1, . . . , χk)g( χi). i=1 i=1 P Proof. If k − 1 of the terms in i ti 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, . . . , χs 6= and χs+1, . . . , χk = . Then, for J0(χ1, . . . , χk) we have k s s X Y X Y k−s−1 Y X χi(ti) = χi(ti) = p χi(ti) = 0

t1+...+tk=0 i=1 t1,...,tk−1 i=1 i=1 ti∈Fp 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 k−1 X X Y J0(χ1, . . . , χk) = χi(ti) χk(s)(6)

s t1+...,tk−1=−s i=1 where s 6= 0 since this term gets deleted by χk(0) = 0. Thus we may deﬁne Ti by ti = −sTi. This gives k−1 k−1 k−1 k−1 X Y Y X Y Y (7) χi(ti) = χj(−s) χi(Ti) = χj(−s)J(χ1, . . . , χk−1),

t1+...tk−1=−s i=1 j=1 T1+...+Tk−1=1 i=1 j=1

10 substituting the last expression in (7) into (6) yields X J0(χ1, . . . , χk) = χ1 ··· χk−1(−1)J(χ1, . . . , χk−1) χ1 ··· χk(s). s6=0 Qk P If i=1 χi 6= then s6=0 χ1 ··· χk(s) = 0, otherwise the sum equals p − 1 by proposition Qk (2.6). Also, χ1 ··· χk−1(−1) = ±1 and χk(−1) = ±1. In the case when i=1 χi 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 Y X X X Y (8) g(χi) = χi(ti)ψ(ti) = χi(ti) ψ(s).

i i ti s t1+...+tk=s i Since by assumption χ , . . . , χ 6= , part three above implies that P Q χ (t ) = 0 1 k t1+...+tk=s i i i if s = 0, so we can suppose s 6= 0 and deﬁne ti = sTi as before. Then X Y X Y (9) χi(ti) = χ1 ··· χk(s) χi(Ti) = χ1 ··· χk(s)J(χ1, . . . , χk),

t1+...+tk=s i T1+...+Tk=1 i substituting the last expression in (9) into (8) gives Y X Y g(χi) = J(χ1, . . . , χk) χ1 ··· χk(s)ψ(s) = J(χ1, . . . , χk)g( χi). i s6=0 i Part four has a couple of corollaries of interest.

Corollary 3.7. Suppose that χ1, . . . , χk 6= and χ1 ··· , χk = then k Y g(χi) = pχk(−1)J(χ1, . . . , χk−1). i=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 X X Y g(χi) = J0(χ1, . . . , χk) + χi(ti) ψ(s)

i s6=0 t1+...+tk=s i X = J0(χ1, . . . , χk) + (s)J(χ1, . . . , χk)ψ(s) s6=0 X = J0(χ1, . . . , χk) + J(χ1, . . . , χk) ψ(s) s6=0

= 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.

Theorem 3.9. If χ1, . . . , χk 6= , then

11 (k−1)/2 (I) χ1 ··· χk 6= =⇒ |J(χ1, . . . , χk)| = p . (k/2)−1 (k/2)−1 (II) χ1 ··· χk = =⇒ |J0(χ1, . . . , χk)| = (p − 1)p and |J(χ1, . . . , χk)| = p . √ Proof. Since each χi 6= we have by proposition 3.3 |g(χi)| = p for each i ∈ {1, 2, . . . , k}. Thus Q √ k i |g(χi)| p (k−1)/2 |J(χ1, . . . , χk)| = = √ = p . |g(χ1, . . . , χk)| p 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 √ (p − 1) Q |g(χ )| (p − 1) pk |J (χ , . . . , χ )| = i i = = (p − 1)p(k/2)−1 0 1 k p p and √ Q |g(χ )| pk |J(χ , . . . , χ )| = | − χ (−1)J(χ , . . . , χ )| = i i = = p(k/2)−1. 1 k k 1 k−1 p p We now consider the most general case needed for our purposes, namely the number of solu- n1 n2 nk ∗ tions N to the equation a1x1 + a2x + ... + akxk = 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. k k−1 X Y −1 (I) b = 0 =⇒ N = p + χi(ai )J0(χ1, . . . , χk) where the sum is over all k-tuples i=1 ni (χ1, . . . , χk) where χi 6= , χi = 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.

k r−1 X Y −1 (II) b 6= 0 =⇒ N = p + χ1 ··· χk(b) χi(ai )J(χ1, . . . , χk) where the sum is over i=1 ni all k-tuples (χ1, . . . , χk) where χi 6= and χi = 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 r−1 (k/2)−1 (k−1)/2 χ1 ··· χk 6= , then |N − p | ≤ M1p + M2p .

Pk ni Proof. Let L(u) = L(u1, . . . , uk) = i=1 aiui and Ni(ui) = N(xi = ui). Similarly as before, notice that X N = N1(u1)N2(u2) ··· Nk(uk)(10) L(u)=b where the sum is over all k-tuples u = (u1, . . . , uk) such that L(u) = b. By proposition 2.10 we know that X Ni(ui) = χi(ui) χi where χi ranges over all character of order dividing ni. Inserting this into equation (10) gives X X X X X (11) N = χ1(u1) ··· χk(uk) = χ1(u1) ··· χk(uk). L(u)=b χ1 χk χ1,...,χk L(u)=b

12 We consider the two cases b = 0 and b 6= 0 separately. If b = 0 we deﬁne ti = aiui, then

X X −1 −1 χ1(u1) ··· χk(uk) = χ1(t1)χ1(a1 ) ··· χk(tk)χk(ak ) L(u)=b t1+...+tk=0 −1 −1 X = χ1(a1 ) ··· χk(ak ) χ1(t1) ··· χk(tk) t1+...+tk=0 −1 −1 (12) = χ1(a1 ) ··· χk(ak )J0(χ1, . . . , χk). −1 If b 6= 0 we deﬁne ti = b aiui, then

X X −1 −1 χ1(u1) ··· χk(uk) = χ1(a1 )χ1(b)χ(t1) ··· χk(ak )χk(b)χ(tk) L(u)=b t1+...+tk=1 −1 −1 (13) = χ1 ··· χk(b)χ1(a1 ) ··· χk(ak )J(χ1, . . . , χk). Now, in the ﬁrst case (11) becomes

X −1 −1 χ1(a1 ) ··· χk(ak )J0(χ1, . . . , χk)(14) χ1,...,χk

k−1 If χi = for all i, then J0(χ1, . . . , χk) = p . If some but not all χi = then J0(χ1, . . . , χk) = 0 Qk and if i=1 χi 6= 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 tr (α) F/Fp 2πi/p ψ : F −→ C be the map α 7→ ζp , where Fp = Z/pZ and ζp = e . 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 ∗. P Definition 12. A sum of the form t∈F χ(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 m ∗ i=0 aiyi = 0 where ai ∈ F . This is a homogeneous equation, so it defines a hypersurface n Hf (F ) ⊆ P (F ). Theorem 3.11. Suppose F is a finite field of order q ≡ 1 (m). Then

n−1 n X 1 X Y |H (F )| = qi + J (χ , . . . , χ ) χ (a−1) f q − 1 0 0 n j j i=0 χ0,...,χn j=0

m where the sum is over all (n+1)-tuples (χ0, . . . , χn) such that χi 6= , χi = and χ0 ··· χn = . Moreover, under the same conditions we also have n 1 1 Y J (χ , . . . , χ ) = g(χ ). q − 1 0 0 n q i i=0

13 Proof. The ﬁrst part is a corollary of the ﬁrst part of theorem 3.10, we have

n X Y −1 (15) |Hf (F )| = q + χi(ai )J0(χ0, . . . , χn). χ0,...,χn i 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

g(χ0) ··· g(χn−1) (17) J(χ0, . . . , χn−1) = . g(χ0 ··· χn−1)

Multiply the numerator and denominator of the right hand side of (17) by g(χn) and apply corollary 3.7 to get

g(χ0) ··· g(χn) g(χ0) ··· g(χn) (18) J0(χ0, . . . , χn−1) = = , g(χ0 ··· χn−1)g(χn) χn(−1)q 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 numbers, then

p−1 q−1 (p/q)(q/p) = (−1) 2 2

q+1 Proof. Let χ denote the character on Fp of order two. Then χ = since q + 1 is even. By corollary 3.7 we have

q+1 q+1 2 q+1 p−1 q+1 p−1 g(χ) = g(χ) 2 = p 2 (−1) 2 2 = p(−1) 2 J(χ, . . . , χ) where there are q components in J(χ, . . . , χ). Thus,

X q−1 p−1 q−1 J(χ, . . . , χ) = χ(t1) ··· χ(tq) = p 2 (−1) 2 2 .

t1+...+tq=1

q −q If t1 = ... = tq, then ti = 1/q and χ(1/q) ··· χ(1/q) = χ(1/q) = χ(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

q−1 p−1 q−1 p 2 (−1) 2 2 ≡ χ(q)(q).

Thus,

p−1 q−1 (−1) 2 2 (p/q) ≡ (q/p)(q)(19) by proposition 2.1. Since q divides the diﬀerence in (19), we must in fact have equality. Multiplying both sides by (p/q) concludes the proof.

14 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 s q 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 s P∞ Nsu series s=1 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 n qs(n+1) − 1 X N ≤ |P n(F )| = = (qs)k < (n + 1)qsn. s s qs − 1 k=0 If |u| < q−n and s ≥ 1, then

∞ ∞ X s X s (20) |Ns||u | < (n + 1) x s=1 s=1

n P∞ s where 0 ≤ x = q |u| < 1. Thus (20) converges, i.e. s=1 Nsu converges absolutely for all |u| < q−n.

Deﬁnition 13. The zeta function of the hypersurface deﬁned by f is the series given by

∞ s X Nsu Z (u) = exp . f s s=1

If the zeta function is expanded about the origin we see that the constant term is 1. We can p(u) therefore assume that Zf (u) = q(u) where p(0) = q(0) = 1. To see this, assume that is not p(0) Pn i the case. Then Zf (0) = q(0) = 1 if and only if p(0) = q(0) = c. Now if p(u) = i=0 aiu and Pm i q(u) = i=0 biu then Pn −1 i Pn −1 i 0 p(u) c i=0 c aiu i=0 c aiu p (u) = Pm −1 i = Pm −1 i = 0 q(u) c i=0 c biu i=0 c biu q (u) where p0(u) and q0(u) have the desired property. Hence the zeta function can be expressed in the form Q i(1 − αiu) (21) Zf (u) = Q j(1 − βju) 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.

p(u) Proposition 4.1. Z(u) ∈ { q(u) : p(u), q(u) ∈ C[u]} if and only if there exist αi, βj ∈ C such P s P s that Ns = j βj − i αi

15 P s P s Proof. We start by assuming that there exist αi, βj ∈ C such that Ns = j βj − i αi . By inserting this expression for Ns we get ∞ s ∞ P s P s s ! X Nsu X j βj − i αi u = s s s=1 s=1 ∞ P s P s ! X j(βju) − i(αiu) = s s=1 ∞ s ∞ s X X (βju) X X (αiu) = − . s s j s=1 i s=1 P∞ zs Further simplify the last expression above by using the identity s=1 s = −ln(1 − z) to get X X ln(1 − αiu) − ln(1 − βju)(22) i j Finally, to ﬁnd the zeta function, we exponentiate (22) to get ! P X X exp i ln(1 − αiu) Z (u) = exp ln(1 − αiu) − ln(1 − βju) = f exp P ln(1 − β u) i j j j Q i exp ln(1 − αiu) = Q j exp 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 Q i(1 − αiu) (23) Zf (u) = Q j(1 − βju) where αi, βj ∈ C. Take the logarithmic derivative of both sides of (4.1) to get Z0 (u) Q ! f d i(1 − αiu) = ln Q Zf (u) du j(1 − βju) ! d X X = ln(1 − α u) − ln(1 − β u) du i j i j X −αi X −βj = − . 1 − α u 1 − β u i i j j Multiply the ﬁrst and last expressions above by u and expand the denominators in the last expression in a geometric series to get 0 Zf (u) X −αiu X −βju u = − Z (u) 1 − α u 1 − β u f i i j j ∞ ∞ X X s X X s = − αiu (αiu) − − βju (βju) i s=0 j s=0 ∞ ∞ X X s X X s = (βju) − (αiu) j s=1 i s=1 ∞ X X s X s s = βj − αi u . s=1 j i

16 Z0 (u) We now compare this last power series with an alternate way of computing u f . By Zf (u) P∞ Nsu deﬁnition Zf (u) = exp s=1 s . By taking the logarithmic derivative and then multiplying by u we get 0 ∞ s ∞ Zf (u) d X Nsu X u = u = N us, Z (u) du s s f s=1 s=1 hence we have ∞ ! ∞ X X s X s s X s βj − αi u = Nsu s=1 j i s=1 and thus X s X s Ns = βj − αi . j i

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 ∼ 0 is an isomorphism σ : E −→ E such that σ F = id. We can use σ to induce a map from the respective projective n-spaces in a natural way by n n 0 letting σ : P (E) −→ P (E ) 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 = n 0 −1 −1 [α0, . . . , αn] ∈ P (E ) be arbitrary, then σ maps [σ (α0), . . . , σ (αn)] 7→ α, hence σ is surjective. Also, both P n(E) and P n(E0) are finite sets, hence σ is injective. It is moreover true 0 that σ is bijection to Hf (E ). We already know that the restriction is injective. Let Hf (E) 0 −1 −1 α = [α0, . . . , αn] ∈ Hf (E ), then f(α0, . . . , αn) = 0 and σ [σ (α0), . . . , σ (αn)] = α. We −1 −1 need to show that [σ (α0), . . . , σ (αn)] ∈ Hf (E). Now, σ acts as the identity on F , hence −1 −1 −1 −1 −1 −1 f σ (α0), . . . , σ (αn = σ f([α0, . . . , αn]) = σ (0) = 0, hence [σ (α0), . . . , σ (αn)] ∈ 0 Hf (E). We conclude that |Hf (E)| = |Hf (E )| and hence Ns indeed only depends on s as claimed, since this result holds also for the trivial field extensions F ⊆ F and F ⊆ F 0 =∼ F .

4.1. The Zeta Analogy. The congruence zeta function and the Riemann zeta function satisfy a highly analogous relation. In the case of the Riemann zeta function we have ∞ X Y 1 ζ(s) = n−s = , s > 1 (1 − 1 ) n=1 p ps 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 n of order q and let V = {a ∈ A (F ): fj(a) = 0 for j = 1, 2, . . . , m} be an algebraic set in An(F ). Definition 14. The function ∞ s ! X Nsu Z (u) = exp V s s=1 n where Ns is the number of points in A (Fs) satisfying the equations defined in V , is called the zeta function of V over F .

17 Let K be the algebraic closure of F . One can show that K is the countable union of isomorphic s copies to Fs of order q 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 n that V ⊆ A (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 qkj with this property. Notice that this implies that a = a for all k ∈ Z+. Additionally, we qd have that a = 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 0 equivalent to ai ∈ L 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 k k defining V . Then, by definition, fj(a1, . . . , an) = 0. But then ψ (0) = ψ fj(a1, . . . , an) = k k qk qk fj ψ (ai), . . . , ψ (an) = fj ai , . . . , ai = 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. P Lemma 4.5. Ns = d|s dnd 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 Fs could 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. Y 1 Theorem 4.6. Z (u) = . V 1 − uδ(P) P

Proof. By merging the factors corresponding to prime divisors of the same degree, the product can be rewritten as ∞ n Y 1 k (24) . 1 − uk k=1 By taking the logarithmic derivative of (24) we get ∞ n ! ∞ ! d Y 1 k d X ln = −n ln(1 − uk) du 1 − uk du k k=1 k=1 ∞ 1 X kuk = n . u k 1 − uk k=1

18 Expand the denominator into a geometric series and compute the coeﬃcient of um to get ∞ ∞ 1 X X X (25) dn um = N um−1 u d m m=1 d|m m=1 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 m m m ∗ Zf (u) of f(x0, . . . , xn) = a0x0 + a1x1 + ... + anxn , 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 n−1 n X 1 X Y (s) (s) N = qsi + χ (a−1)g(χ )(26) s qs i i i i=0 (s) (s) j=0 χ0 ,...,χn (s) (s)m (s) (s) (s) where χi are characters of Fs such that χi = , χi 6= and χ0 ··· χn = . Let 0 χ be a character of F , compose it with NFs/F to get χ = χ ◦ NFs/F : Fs −→ C. This 0 mapping is in fact a character on Fs since χ (ab) = χ NFs/F (ab) = χ NFs/F (a)NFs/F (b) = 0 0 χ NFs/F (a) χ NFs/F (b) = χ (a)χ (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 m m ∗ χ ◦ NFs/F (a) = χ NFs/F (a) = NFs/F (a) = 1 for all a ∈ Fs , hence it is the trivial 0 s s character. For the third part, simply note that χ (a) = χ NFs/F (a · 1) = χ(a ) = χ(a) , by proposition 2.11. This lemma shows that by letting χ vary over all characters of F of order diving m, the same 0 happens for the corresponding characters χ of Fs. Thus equation (26) can now be rewritten as n−1 n X 1 X Y N = qsi + χ (a−1)sg(χ0 )(27) s qs i i i i=0 χ0,...,χn j=0 m where χi are characters of F such that χi = , 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)

0 Remark. We remind that s is an integer which depends on χ , 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, n n−1 n Ch. 11.4]). Given a monic polynomial f(x) = x − a1x + ... + (−1) an ∈ F [x], we define a mapping λ by λ(f) = ψ(a1)χ(an) and let λ(1) = 1. Lemma 4.9. λ(fg) = λ(f)λ(g) for all monic f, g ∈ F [x].

19 Lemma 4.10. g(χ0) = P δ(f)λ(f)s/δ(f) where the sum is over all irreducible monic polynomials of degree s in F [x].

Given this deﬁnition of λ we have the identity X Y 1 λ(f)tδ(f) = 1 − λ(f)tδ(f) f 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 X = (λ(f)tδ(f))k. 1 − λ(f)tδ(f) k=0 Using the fact that λ is multiplicative by lemma 4.9 we get ∞ Y 1 Y X = λ(f k)tkδ(f) 1 − λ(f)tδ(f) f f k=0 Y = 1 + λ(f)tδ(f) + ... + λ(f k)tkδ(f) + ... . 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 Qn ai Qn ai sum. Inductively applying lemma 4.9 gives i=1 λ(fi ) = λ( i=1 fi ). This together with Qn ai Pn the basic fact that δ( i=1 fi ) = i=1 aiδ(fi) shows that the identity holds.

4.2.1. Proof of the Hasse-Davenport Relation. By collecting terms of equal degree, we have ∞ X X X (28) λ(f)tδ(f) = λ(f) ts. f s=0 δ(f)=s P where the left hand side is as in (28). We proceed by analyzing the sums δ(f)=s λ(f). If s = 1, then X X X λ(f) = λ(x − a) = χ(a)ψ(a) = g(χ)(29) δ(f)=s a∈F a∈F If s > 1 it turns out that the sums vanish, since X X s s−1 s (30) λ(f) = λ(x − a1x + ... + (−1) as).

δ(f)=s ai∈F

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 ! ! s−2 X s−2 X X q χ(as)ψ(a1) = q χ(as) ψ(a1) = 0 a1, an as a1 by proposition 2.6 extended to characters on arbitrary finite fields. Thus we have X Y 1 λ(f)tδ(f) = 1 + g(χ)t = . 1 − λ(f)tδ(f) f f By taking logarithmic derivatives, we get d g(χ) ln(1 + g(χ)t) = dt 1 + g(χ)t

20 and ! d Y 1 d X ln = − ln1 − λ(f)tδ(f) dt 1 − λ(f)tδ(f) dt f f X λ(f)δ(f)tδ(f)−1 = . 1 − λ(f)tδ(f) f By multiplying these expressions by t, we have g(χ)t X λ(f)δ(f)tδ(f) = . 1 + g(χ)t 1 − λ(f)tδ(f) f Expand the denominators in geometric series to get ∞ ∞ X X X (−1)s−1g(χ)sts = λ(f)kδ(f)tkδ(f) . s=1 f k=1 By comparing coeﬃcients for ts we get X (−1)s−1g(χ)s = δ(f)λ(f)s/δ(f) = g(χ0) δ(f)|s 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 m m m Ns associated with Zf (u), where f(x0, . . . , xn) = a0x0 + a1x1 + ... + anxn ∈ F [x0, . . . , xn]. Substituting g(χ0) = (−1)s+1g(χ)s into equation (27) we get n−1 n !s X X (−1)n+1 Y (31) N = qsi + (−1)n+1 χ (a−1)g(χ ) , s q i i i i=0 χ0,...,χn j=0 m where χi are characters of F such that χi = (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. ∗ m Theorem 4.11. Let a0, . . . , an ∈ F where |F | = q ≡ 1 (m) and f(x0, . . . , xn) = a0x0 + m m a1x1 + ... + anxn . Then the congruence zeta function Zf (u) is a rational function of the form P (u)(−1)n Z (u) = , f (1 − u)(1 − qu) ··· (1 − qn−1u) where Y 1 P (u) = 1 − (−1)n+1 χ (a−1) ··· χ (a−1)g(χ ) ··· g(χ )u q 0 0 n n 0 n χ0,...,χn m and the product is over all (n+1)-tuples (χ0, . . . , χn) such that χi = , 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 Pn−1 s P s j n+1 1 −1 −1 Ns = j=0 βj − i αi , where βj = q and αi = (−1) q χ0(a0 ) ··· χn(an )g(χ0) ··· g(χn) for some (n + 1)-tuple (χ0, . . . , χn) subject to the conditions above. If n is odd, then (31) can P P n be written as j βj − i αi where βj = Ns and αi = 0. Hence the exponent (−1) in the Q i(1 − αiu) numerator of Zf (u), since the zeta function is of the form Q . j(1 − βju)