RATIONALITY OF ZETA FUNCTIONS OVER FINITE FIELDS

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1 RATIONALITY OF ZETA FUNCTIONS OVER FINITE FIELDS SUN WOO PARK Abstract. The zeta function of an affine variety over a finite field contains comprehensive information on the number of points of the variety for all the field extensions of the finite field. This expository paper follows Koblitz s treatment [2] of Dwork s proof [1] of the rationality of zeta functions of affine varieties over finite fields, also known as the rationality statement of the Weil Conjectures. Contents 1. Introduction 1 2. Power Series over C p Logarithms, Exponents, and Other Power Series Newton Polygons Overconvergent Power Series Rationality of Zeta Functions Zeta Functions over Finite Fields p-adic Meromorphic Rationality 20 Acknowledgments 24 References Introduction Let f(x 1,..., x n be a polynomial with n variables over a finite field F q where q = p r is a power of a prime p. The natural question then arises as to what are the roots of the polynomial, which motivates the definition of the affine hypersurface of the polynomial. Definition 1.1. Let F q be a finite field with q a power of a prime p. The n-dimensional affine space over the finite field F q is the set of ordered n-tuples (a 1,..., a n where each element a i is in the finite field F q. Denote the affine space as A n F q. Let f(x 1,..., x n be a polynomial with n variables over F q. Then the affine hypersurface H f of the polynomial f(x 1,..., x n is the set of all the roots of f(x 1,..., x n in the affine space A n F q. H f = {(a 1,..., a n A n F q f(a 1,..., a n = 0} Let s be a positive integer and let F q s be a field extension of F q. Observe that the coefficients of the polynomial f(x 1,..., x n are also in F q s for any s. This allows 1

2 2 SUN WOO PARK us to consider the set of all the roots of the polynomial f(x 1,..., x n in the affine space of F q s. Definition 1.2. Let F q s be a field extension of F q. Let H f be the affine hypersurface of a polynomial f(x 1,..., x n with n variables over F q. Then the set of F q s-points of H f, denoted as H f (F q s, is the set of all the roots of f(x 1,..., x n in the affine space of F q s. H f (F q s = {(a 1,..., a n A n F q s f(a 1,..., a n = 0} The order of the set H f (F q s can be understood as the number of roots of the polynomial f(x 1,..., x n over F q s. Using the definition above, we can construct a function over C p, the analogue of C in the field of p-adic numbers Q p. Before we construct the function, we state the definition of the field of p-adic numbers Q p. Definition 1.3. Let p be a prime number. The ring of p-adic integers Z p is the inverse limit of the inverse system ((Z/p n Z n N, (f nm n>m N where the transition morphism f nm : Z/p n Z Z/p m Z is given by reduction mod p m. The fraction field of Z p is the field of p-adic numbers Q p and the completion of the algebraic closure of Q p is the complex field of p-adic numbers C p. Notice C p is defined as above because the algebraic closure of Q p is not complete. The valuation over Q p and C p is defined as follows, the construction of which is described in Chapter 1 and 3 of Koblitz [1]. Definition 1.4. The valuation of an element a Q p, denoted as ord p a, is the largest integer power l of p such that p l divides a. The p-adic norm of a is defined as follows. { p ordpa if a 0 a p = 0 if a = 0 The norm defined above is a non-archimedean norm. Let b Q p have the irreducible polynomial f(x = x n + a 1 x n a n over Q p. Then the norm of b is defined as b p = a n 1/n p. The norm on Q p extends to C p by defining x p = lim n x n p where x C p and {x n } is a sequence in Q p. Definition 1.5. Let f(x 1,..., x n be a polynomial with n variables over F q for q = p r a power of p. Let H f be the affine hypersurface of f(x 1,..., x n. Denote N s = H f (F q s as the number of F q s-points of H f. Then the zeta function of H f is defined as follows, where Exp p is the exponential function in C p. ( N s T s Z(H f, T = Exp p s The zeta function contains all the information on the number of roots of a polynomial over each field extension of the finite field F q. Bernard Dwork proved the following property of the zeta function of H f in 1960, which is the primary focus of the paper. Theorem 1.6 (Dwork, Let H f be the affine hypersurface of a polynomial f with n variables over the finite field F q. Then the zeta function of H f is a quotient of two polynomials with coefficients in Q p. s=1

3 RATIONALITY OF ZETA FUNCTIONS OVER FINITE FIELDS 3 In fact, this theorem is one of the statements of the Weil conjectures proposed by André Weil in The Weil conjectures explain the properties of the zeta functions of algebraic varieties over finite fields. Weil claimed that the zeta functions are quotients of two polynomials over Q p and satisfy some forms of functional equations. He further claimed that the roots of the zeta functions appear in restricted places, an analogue of the Riemann hypothesis. As a particular example, the reciprocal roots of the zeta functions F (s = Z(H f, q s of projective 1-dimensional hypersurfaces H f are on the line Re(s = 1/2. Dwork s theorem corresponds to the rationality statement of the Weil conjectures. Alexander Grothendieck proved the statement on functional equations in 1965 and Pierre Deligne proved the analogue of the Riemann hypothesis in Dwork s proof consists of two steps. The first step, proved by induction on the number of variables, shows that the zeta function is a quotient of two power series over C p, both of which converge everywhere. Observe that N s is the sum of the number of roots of the polynomial where each coordinate is non-zero and the number of roots of the polynomial where at least one coordinate is zero. The latter case inductively relates N s with the number of roots of other polynomials with fewer variables. The former case can be expressed as a sum of a set of p-th roots of unity. By extending the coefficients of the polynomial from the finite field F q to the field of p-adic numbers Q p, Dwork constructs a power series over C p and uses the series to change the sum of p-th roots of unity to the sum of power series over C p. He then shows that the determinant of the power series under the basis of monomials converges everywhere on C p, which proves the first step. The second step uses the first step to prove the theorem. Using p-adic analysis, Dwork shows that there exists a polynomial and a power series over C p with certain radius of convergence such that the quotient of the polynomial and the power series is the zeta function. He then compares the coefficients of the quotient with those of the zeta function, which allows him to estimate the p-adic norm and the usual norm of the determinant of a matrix consisting of a finite set of coefficients of the zeta function. The comparison between the two norms shows that the determinant of the matrix is 0 which implies that the zeta function is a quotient of two polynomials over C p. This paper focuses on carefully following Koblitz s treatment [2] of the proof of Dwork s theorem [1]. The paper first provides important backgrounds on properties of some power series over C p. Using the properties, the paper then follows Dwork s proof on the rationality of zeta functions. The paper also states an alternate construction of the zeta functions given by Kapranov [5], extending the construction of the zeta functions over finite fields to zeta functions over perfect fields. 2. Power Series over C p Let F (x be a power series over C p of the form F (x = n=0 a nx n. Since C p is complete and the norm is non-archimedean, the series converges if and only if the norm of the term a n x n p 0 as n. In other words, for certain values of x C p we can evaluate F (x by the limit the infinite sum converges to. Observe that if all the coefficients a n are in Z p, then the series F (x converges for x p < 1 because a n x n p x n p 0 as n. This proves the following lemma. Lemma 2.1. If F (x is a power series in Z p, then F (x converges for x p < 1.

4 4 SUN WOO PARK Similar to power series over C, we can define the radius of convergence of a power series over C p. Definition 2.2. The radius of convergence r of a power series F (x = n=0 a nx n is the following limit. 1 r = lim sup n a n 1/n p As the term suggests, the series converges if x p < r and diverges if x p > r. This can be shown by observing whether the norm of the term a n x n approaches 0 as n. It is not true, however, that the series converges if x p = r. Definition 2.3. The closed and open disks of radius r R around a point a C p are defined respectively as follows. D a (r = {x C p x a p r} D a (r = {x C p x a p < r} We will abbreviate the disk around the origin D 0 (r as D(r. Using this notation, we can say that a power series F (x converges in D(r or D(r if it has r as the radius of convergence Logarithms, Exponents, and Other Power Series. Several power series defined over C can be analogously defined over C p. Definition 2.4. The logarithmic function Log p and the exponential function Exp p is defined as the following power series over C p. n+1 xn Log p (1 + x = ( 1 n n=1 x n Exp p (x = n! n=0 As in the case for complex numbers, the following proposition holds. Proposition 2.5. Let Log p and Exp p be the logarithmic and the exponential function in C p. Then the following holds. (1 Log p (1 + x converges in D(1. (2 Exp p (x converges in D(p 1 p 1. (3 Log p and Exp p are mutually inverse isomorphisms between the multiplicative group D 1 (p 1 p 1 and the additive group D(p 1 p 1. Proof. Observe that lim sup n 1/n 1/n p = lim n p ordpn n = 1. Since x n /n p = p ordpn 1 if x p = 1, Log p (1 + x converges in D(1. Notice that ord p n! = n Sn p 1 where S n is the sum of all p-adic digits of n!. Suppose that n has the p-adic expansion n = a k p k + a k 1 p k a 0. Observe that ord p n! = k n p as n i p contributes to a factor p, n p contributes to a square 2 factor of p, and so on. Then for every i we have n p = a i k p k i +a k 1 x k 1 i +...+a i, which implies the following. ord p n! = k n p i = k a i p i 1 p 1 = k i=0 (a ip i k i=0 a i = n S n p 1 p 1

5 RATIONALITY OF ZETA FUNCTIONS OVER FINITE FIELDS 5 This shows that the radius of convergence of Exp p is p 1 p 1 as from the equation below. Suppose x p ord p x n /n! = lim sup 1/n! 1/n p n = lim sup p n Sn 1 n(p 1 = p p 1 n = p 1 p 1. Choose n to be a power of p such that Sn = 1. Then pm p 1 pm 1 p 1 = 1 p 1, which shows xn /n! p = p 1 p 1. Hence Expp converges in D(p 1 p 1. From the definition of Log p, Log p (1 + x(1 + y = Log p (1 + x + Log p (1 + y, which shows that Log p is a group homomorphism from the multiplicative group D 1 (p 1 p 1 to the additive group D(p 1 p 1. Similarly, Expp is a group homomorphism from D(p 1 p 1 to D1 (p 1 p 1. By the definition of two functions, we also have Exp p (Log p (1 + x = 1 + x and Log p (Exp p (x = x if x D(p 1 p 1. Hence the two functions are mutually inverse isomorphisms. Definition 2.6. Let a C p. Then the binomial expansion (1 + x a is defined as the following power series. a(a 1...(a n + 1 B a (x = x n n! n=0 If a p > 1 then for any integer i, a i p = a p. Hence the norm of the nth term of the binomial power series is an x n n! p. This implies that the power series converges 1 in D( p p 1 a p. If a p 1 then for any integer i, a i p 1. This shows that the norm of the nth term of the binomial power series is a(a 1...(a n+1xn n! p xn n! p. This implies that the power series converges in D(p 1 p 1. We now define a new power series which will be used in Dwork s proof of the rationality of zeta functions. Definition 2.7. Let F (x, y be a power series with two variables over C p defined as follows. F (x, y = B x (y (y p i+1 This can be rewritten as follows. i=0 B x p i+1 x pi p i+1 F (x, y = (1 + y x (1 + y p xp x p...(1 + y pn xpn x pn 1 p n... Observe that the power series F (x, y has all its coefficients in Q p. The series is also well defined because for any term x n y m, the corresponding coefficient can be achieved by taking finitely many terms in Q p. We will analyze this power series further after proving the following lemma. Lemma 2.8 (Dwork s Lemma. Let F (x be a power series over Q p such that the F (x constant term is 1. Then F (x has all its coefficients in Z p if and only if p (F (x p has all its coefficients other than the constant term in pz p. Proof. Suppose F (x has all its coefficients in Z p with constant term 1. Notice that (a + b p a p + b p mod p and a p a mod p for any a, b Z p. Then there exists a

6 6 SUN WOO PARK power series G(x xz p [[x]] such that the following holds. (F (x p = F (x p + pg(x The above equation implies the following equation because (F (x p has all its coefficients in Z p and 1 + xz p [[x]] is a group under multiplication. F (x p (F (x p = 1 p G(x (F (x p 1 + pxz p[[x]] Suppose F (xp (F (x has all its coefficients other than the constant term in pz p p. Then we can find a power series G(x 1 + pxz p [[x]] such that F (x p = (F (x p G(x. Let F (x = i=0 a ix i and G(x = j=0 b jx j where b j pz p for all j. We want to show that a i Z p for all i. We prove the lemma by induction on the index i. The base case holds because a 0 = 1. Suppose a i Z p for i n 1. Consider the following equation. F (x p = (F (x p G(x The coefficients of x n on both sides of the equation must be the same. Notice that the coefficient of x n from the RHS of the equation above is determined by the product ( n i=0 a ix i p (1 + n j=1 b jx j. This shows that the coefficient from the RHS consists of terms which contain b j for some j and terms which only have a i as factors. The former terms are in pz p, which we will denote as pb. If p n, then the coefficient from the LHS is a n/p while the coefficient from the RHS is pb + pa n + a p n/p. For any a Z p we have a p a mod p. Hence a n is an element of Z p. If p n then the coefficient from the LHS is 0 while the coefficient from the RHS is pb + pa n. This also shows a n is an element of Z p. The generalization of Dwork s lemma is as follows, the proof of which is analogous to that of Lemma 2.8. Lemma 2.9 (Generalized Dwork s Lemma. Let F (x 1, x 2,..., x n be a power series with n variables over Q p such that the constant term is 1. Then F (x 1,..., x n has F (x all its coefficients in Z p if and only if p 1,...,xp n (F (x 1,...,x n has all its coefficients other than p the constant term in pz p. Using the generalized Dwork s lemma, we prove the following proposition. Proposition The power series F (x, y has all its coefficients in Z p. Proof. Observe that the following equation holds by Lemma 2.9. F (x p, y p (F (x, y p = B xp(y p i=0 B (x pi+2 x pi+1 /p (y i+1 pi+2 B px (y i=0 B (x pi+1 x pi /p (y i pi+1 B x p(y p i=0 = B (x pi+2 x pi+1 /p (y i+1 pi+2 B px (yb xp x(y p B (x pi+1 x pi /p (y i pi+1 = (1 + y p xp (1 + y px (1 + y xp x = (1 + yp x (1 + y px Hence it suffices to show that (1+yp x (1+y has all its coefficients other than the constant px term in pz p. Notice that 1 + y is an element of 1 + yz p [[y]]. By Lemma 2.8, there

7 RATIONALITY OF ZETA FUNCTIONS OVER FINITE FIELDS 7 exists a power series G(y Z p [[y]] such that the following holds. (1 + y p (1 + y p = 1 + pyg(y By the definition of binomial power series, the following equation holds. (1 + y p x (1 + y px = (1 + pyg(yx 1 + pxz p [[x, y]] + pyz p [[x, y]] The proposition shows that F (x, y can be expressed as follows in which for all non-negative integers n and m, a m,n Z p. F (x, y = (x n a m,n y m n=0 Notice we have m=n a m,ny m instead of m=0 a m,ny m because for each term in (y the power of x is less than or equal to the power of y, i.e. B x p i+1 x pi p i+1 (x pi+1 x pi p i Newton Polygons. m=n (x pi+1 x (x pi pi+1 x pi y npi p i p i+1 n + 1 n! Definition Let f(x = 1 + n a ix i be a polynomial of degree n over C p. The Newton polygon of f(x is the convex hull of the points (0, 0, (1, ord p a 1,..., (i, ord p a i,..., (n, ord p a n in the real coordinate plane. In other words, it is the highest convex polygonal line which starts at (0, 0, ends at (n, ord p a n, and joins or passes below all the points (i, ord p a i. Let F (x = 1 + a ix i be a power series over C p. Let f n (x be the n-th partial sum of F (x. Then the Newton polygon of F (x is the limit of the Newton polygons of f n (x. The way to construct the Newton polygon of a polynomial or a power series is as follows. Plot all the set of points (i, ord p i. Rotate the vertical line passing through (0, 0 counter-clockwise until it hits a point (i, ord p a i for some i. Then rotate the line with respect to the point (i, ord p a i and repeat the process. For convenience we will define the terms as follows. The set of vertices of the Newton polygon is the set of points where the slopes change. The length of a segment is the length of the projection of the segments onto the horizontal axis. The Newton polygon of a polynomial f(x over C p gives important properties of the roots of f(x. We omit the proof of the following lemma which is written in Chapter 4 of Koblitz [2]. Lemma Let f(x = n (1 x/a i be a polynomial over C p with roots {a i } n. Let ord pa i = m i. If m is a slope of the Newton polygon of f(x with length q, then precisely q of m i are equal to m. The Newton polygon of a power series can be classified into three categories. (1 There are infinitely many segments of finite length. Consider the power series F (x = 1 + n=1 pi2 x i. The Newton polygon is an infinite set of finite line segments which connects the lattices (n, n 2 on the graph y = x 2.

8 8 SUN WOO PARK (2 There are finitely many segment in which the last segment is infinitely long. The power series F (x = 1 + n=1 xn has an infinite horizontal line as its Newton polygon. (3 There may be a case such that it is not possible to draw the Newton polygon using the method above. Consider the power series F (x = 1 + n=1 pxn. Observe that there exists an integer n such that the point (n, 1 lies below a line segment which is obtained from rotating the horizontal line passing through the origin counterclockwise. In this case, we let the last segment of the Newton polygon have the slope to be the least upper bound of all slopes of which any point (i, ord p a i lies above or on the line. In the previous example, the Newton polygon is the infinite horizontal line passing through the origin. Newton polygons provide a different way of understanding the power series over C p. The slope of the Newton polygon of a power series gives additional information on the radius of convergence of the series as well as the valuations of the roots of the power series. The following theorem, which is an analogue of Lemma 2.12 for power series, will prove useful for understanding Dwork s proof in subsequent sections. Theorem 2.13 (p-adic Weierstrass Preparation Theorem. Let F (x be a power series over C p which converges on D(p m. Denote the terms of the power series as F (x = 1 + n=1 a nx n. Suppose the Newton polygon of F (x does not have the infinitely long last segment of slope m. Denote N as the total length of all finite segments having slope less than m. If the Newton polygon of F (x has the last segment of slope m, denote N as the greatest n such that (n, ord p a n lies on the last segment. Then there exists a unique polynomial h(x in C p with constant term 1 of degree N and a power series G(x = 1 + n=1 b nx n which converges and is nonzero in D(p m such that h(x = F (xg(x. Corollary Suppose F (x is a power series over C p with the constant term 1. If a segment of the Newton polygon of F (x has horizontal length N and slope m, then there are precisely N roots x of F (x counting multiplicity such that the valuation ord p x = m. We first prove the following lemmas which will help us prove Theorem Lemma Let F (x = 1 + n=1 a nx n be a power series over C p. Let m be the least upper bound of all the slopes of the Newton polygon of F (x. Here m may be infinite. Then the radius of convergence of F (x is p m. Proof. Suppose x C p such that ord p x = m > m. Then ord p (a n x n = ord p a n nm. Notice that (n, ord p a n lies above the Newton polygon while (n, nm lies above a line of slope m passing through the origin. By the definition of m, ord p a n nm as n, which implies that the series converges. Now assume ord p x = m < m. By the similar argument as above, for infinitely many n, the value ord p a n nm is negative, proving that the series does not converge. The above lemma does not explain whether the power series converges when ord p x is equal to the least upper bound of all slopes of the Newton polygon. Remark Let F (x = 1 + n=1 a nx n be a power series over C p. Suppose c C p such that ord p c = d and G(x = F (x/c = 1 + n=1 b nx n is also a power

9 RATIONALITY OF ZETA FUNCTIONS OVER FINITE FIELDS 9 series over C p. Notice that for each n, ord p b n = ord p (a n /c n = ord p a n dn. This shows that the Newton polygon of G(x is obtained by subtracting the line y = dx from the Newton polygon of F (x. Lemma Suppose m 1 is the first slope of the Newton polygon of a power series F (x = 1 + n=1 a nx n over C p. Let c C p such that ord p c = m m 1. Let G(x be the power series defined as G(x = (1 cxf (x. Suppose F (x converges on the closed disk D(p m. Then the Newton polygon of G(x is obtained by first joining (0, 0 to (1, m and then shifting the Newton polygon of F (x by 1 to the right and d upwards. Suppose further that the last slope of the Newton polygon of F (x is m f. Then the Newton polygon of G(x also has m f as the last slope. In addition, F (x converges in D(p m f if and only if G(x does. Proof. We will first prove the lemma when c = 1. Then G(x = (1 xf (x satisfies G(x = 1 + n=1 (a n a n 1 x n. Denote b n = a n a n 1. It follows that ord p b n min(ord p a n, ord p a n 1. Since both points (n, ord p a n, (n 1, ord p a n 1 lie above or on the Newton polygon of F (x, the points (n, ord p b n all lie above the Newton polygon of F (x translated by 1 to the right. Notice that the translation includes an extra segment connecting (0, 0 and (1, 0. If (n 1, ord p a n 1 is one of the vertices of the Newton polygon of F (x, then ord p b n = ord p a n 1 because ord p a n > ord p a n 1. This shows (n, ord p b n is also a corner of the translated Newton Polygon. Now we prove the second statement of the lemma. Notice that since ord p b n min(ord p a n, ord p a n 1, G(x converges whenever F (x converges. Suppose that the Newton polygon of G(x has a slope such that m g > m f. Then for some n, (n + 1, ord p a n lies below the Newton polygon of G(x. This shows that for all k n + 1, ord p b k > ord p a n. Since a n+1 = b n+1 + a n, ord p a n+1 = ord p a n. By the similar argument, ord p a k = ord p a n for all such k n + 1. This contradicts the assumption that the power series F (x converges on D(1. The converse holds analogously. Now we prove the lemma for arbitrary choice of c. Define the power series F 1 (x = F (x/c and G 1 (x = (1 xf 1 (x. Notice that the lemma holds for F 1 (x and G 1 (x because both power series satisfy the base case of the proof of the lemma. By Remark 2.16, we obtain the desired results for the Newton polygon of G(x. Lemma Let F (x = 1 + n=1 a nx n be a power series over C p. Suppose the Newton polygon of F (x has the first slope m 1. If F (x converges on the closed disk D(p m1 and the line through (0, 0 with slope m 1 passes through (n, ord p a n for some n, then there exists a root x of F (x such that ord p x = m 1. Proof. We first prove the lemma when m 1 = 0. This implies that for all n we have ord p a n 0 and ord p a n as n. Let M be the greatest integer n such that ord p a n = 0. Denote f n (x as the n-th partial sum of F (x. By Lemma 2.12, for n M, the polynomial f n (x has precisely M roots x n,1, x n,2,..., x n,m such that for every i, ord p x n,i = 0. Let S n be the set of roots of f n (x counting multiplicities. Consider the following sequence {x n } such that x M = x M,1 and x n+1 = a S n+1 such that ord p a = 0 and a x n is minimal. Notice that if a S n + 1 such that ord p a < 0, then 1 x n /a p = 1. Hence the following holds for n > M. The equation below shows that {x n } is a cauchy sequence because a n p as n.

10 10 SUN WOO PARK a n + 1 p = a n+1 x n+1 n p = f n+1 (x f n (x p = f n+1 (x n p = 1 x n a p = M 1 x n x n+1,i x n+1 x n p a S n+1 M = x n+1,i x n p p Suppose x is the limit of the sequence {x n }. Then the following holds. f n (x p = f n (x f n (x n p = x x n p n x x n p 0 as n k=1 a k x k x k n x x n Since F (x = lim n f n (x = 0, x is the root of the power series, proving the lemma. Notice that the general case follows directly from the base case. Let b C p such that ord p b = m 1. Define G(x = F (x/b. Then G(x satisfies the conditions for the base case of the proof. The analogous result holds for F (x as well. Lemma Let F (x = 1 + n=1 a nx n be a power series over C p which converges and vanishes to 0 at a C p. Let G(x = F (x 1 x/a = 1 + k=1 b kx k be a power series over C p. Then G(x converges on D( a p. Proof. Let f n (x be the n-th partial sum of F (x. Notice that G(x can be obtained by multiplying F (x with the power series m=0 xm /a m. This implies that for each n, b n = n a j j=0 a = fn(a n j a. Since f(a = 0, we have b n n a n p = f n (a p 0 as n. We now prove Theorem Theorem We prove the theorem by induction on N, the degree of the polynomial h(x. Without loss of generality, assume m = 0. The general case for any value of m is analogous to the proof of Lemma 2.17 and Lemma Suppose N = 0. It suffices to show that the inverse power series of F (x is convergent and nonzero in D(p m. Let G(x be the inverse of F (x. Assume for every n, ord p a n > 0. Observe that ord p a n as n because F (x converges in D(1. Since F (xg(x = 1, b n = n k=1 b n ka k, which shows that ord p b n > 0 for every n. Notice that ord p b n as n. For fixed C > 0, choose c such that for all n > c, ord p a n > C. Denote ɛ = min(ord p a 1, ord p a 2,..., ord p a c. We claim that for all n > ic, ord p b n > min(c, iɛ, which proves the theorem for N = 0. We prove the theorem by by induction on i. The case in which i = 0 is trivial. Suppose the claim holds for i 1. Then from the sum b n = n k=1 b n ka k, the terms b n k a k with k > c have ord p (b n k a k ord p a k > C while the terms b n k a k with k c have ord p (b n k a k ord p b n k + ɛ > min(c, (i 1ɛ + ɛ by the induction hypothesis. Hence ord p b n > min(c, iɛ. Now suppose N 1 and the theorem holds for N 1. Let m 1 m be the first slope of the Newton polygon of F (x. By Lemma 2.18, there exists a C p such that F (a = 0 and ord p a = m 1. Consider the following power series over C p. p

11 RATIONALITY OF ZETA FUNCTIONS OVER FINITE FIELDS 11 F 1 (x = F (x 1 x/a = 1 + a nx n By Lemma 2.19, F 1 (x converges on D(p m1. Let m 1 be the first slope of the Newton polygon of F 1 (x. Suppose m 1 < m 1. Then by Lemma 2.18, F 1 (x has a root a with ord p a = m 1. By the equation above, a is also a root of F (x. This is a contradiction to the base case N = 0. Hence m 1 m 1. By Lemma 2.17, the Newton Polygon of F 1 (x is the same as that of F (x with the segment from (0, 0 to (1, m 1 subtracted. In fact, Lemma 2.17 implies that F 1 (x also converges in D(p m if the Newton polygon of F (x has m as the last slope. Notice that F 1 (x satisfies the condition of the theorem with a total length of N 1. By the induction hypothesis, there exist a polynomial h 1 (x 1 + xc p [x] of degree N 1 and a power series G(x 1 + xc p [[x]] which is convergent and nonzero on D(p m such that h 1 (x = F 1 (xg(x. Multiply both sides of the equation by (1 x/a and set h(x = (1 x/ah 1 (x to derive h(x = F (xg(x. Notice that our choice of h(x is unique. Suppose there is another polynomial h (x and a power series G (x such that satisfy the conditions of the theorem. It suffices to show that h(x and h (x have the same zeroes with the same multiplicities. We prove the claim by induction on the degree N. The base case in which N = 1 is trivial. Suppose the claim holds for N 1. Without loss of generality, let a C p be a root of h(x such that ord p a = m. Since h(xg(x = h (xg (x, a is also the root of h (x. In other words, we can divide both sides of the equation by (1 x/a. By Lemma 2.19, the equation reduces to a case with degree N Overconvergent Power Series. We now consider power series over C p with n variables in the space of overconvergent power series. The results obtained from this subsection will be used for proving that the zeta function of a hypersurface is a quotient of two power series over C p, both of which converge everywhere. Definition Let R = C p [[x 1,..., x n ]] be the space of power series with n variables. Let W be the space of ordered n-tuples where each coordinate is a nonnegative integer. Define the norm of an element w = (w 1,..., w n W as the sum of all coordinates of w, i.e. w = n k=1 w k. The above definition allows us to write any power series G(x 1,..., x n R as G(x 1,..., x n = G(x = w W g wx w where for each w = (w 1,..., w n, x w = x w1 1 xw2 2...xwn n and g w C p. We also abbreviate the n-tuple (a 1,..., a n as a if a i = a N for every i. In this case the monomial x a is an abbreviation of x a = x a 1x a 2...x a n. Definition The space of overconvergent power series R 0 is defined as follows. R 0 = {G(x = w W n=1 g w x w R M > 0 such that w W, ord p g w M w } The definition implies that the power series in R 0 converges in some disk D(r that strictly contains D(1. It is clear that R 0, closed under multiplication, is a subring of R. Notice that we can consider the space R 0 as an infinite vector space of C p with the set of monomials {x w } w W as the basis. This allows us to view any linear operator of R 0 as an infinite matrix, which motivates the definition of trace and determinant as follows.

12 12 SUN WOO PARK Definition Let φ : R 0 R 0 be a linear operator on R 0. Suppose A = {a w,v } w,v W is the infinite matrix of φ under the basis of monomials. Then the trace of A is defined as follows if the infinite sum converges. T r(a = w W a w,w Let T be an independent variable. Then the matrix (1 AT is an infinite matrix with entries in C p [T ]. Analogous to finite matrices, the determinant of (1 AT is defined as det(1 AT = b k T k where b k is defined as follows. b k = ( 1 k w 1,w 2,...,w k W σ S k ( k=0 sgn(σ k a wi,σ(w i Before we construct an operator which we will use in the first step of Dwork s proof, we define the following operators on R 0. Definition Let q be a positive integer. Then the translation operator T q : R 0 R 0 is defined as follows where for any w W, qw = (qw 1, qw 2,..., qw n. ( T q g w x w = g qw x w w W w W Definition Let G(x be a power series in R 0. The G(x-multiplication operator G : R 0 R 0 is defined as multiplication by G(x. Definition Let q be a positive integer and let G(x be a power series in R 0. Then the power series G q (x is defined as G q (X = G(x q. We also define the operator ψ q,g = T q G where F G denotes the composition of power series F (x and G(x over C p. Theorem Let q be a positive number and G(x = w W g wx w be a power series in R 0 with n variables. Let ψ q,g = T q G and let A be the infinite matrix of ψ q,g under the basis of monomials {x w } w W. Then the following holds for every positive integer s. (1 The trace T r(ψ s q,g exists. (2 (q s 1 n T r(ψ s q,g = x C n p x qs 1 =1 s 1 j=0 G(xqj (3 The determinant det(1 AT is a well defined power series in C p with infinite radius of convergence. ( (4 det(1 AT = Exp p T r(ψ s q,g T s s=1 Proof. We first prove the theorem when s = 1. Observe that for any u W, the following holds. ( ψ q,g (x u = T q (x u G(x = T q g w x w+u = g qw u x w w W w W s

13 RATIONALITY OF ZETA FUNCTIONS OVER FINITE FIELDS 13 Here if qw u is not in W, define g qw u = 0. By the definition of the trace, T r(ψ q,g = w W g qw w = w W g (q 1w. Since ψ q,g R 0, the series converges. Observe that for each i = 1, 2,..., n and w i a non-negative integer, the following holds because each x i is the (q 1-th root of unity. x i C p x q 1 i =1 x wi i = { q 1 if q 1 w 0 if otherwise Hence the following equation holds for any w W. n x w = = x C n p x qs 1 =1 x q 1 i =1 x wi i { (q 1 n if q 1 w i for every w i 0 if otherwise This implies the following, which proves the theorem for the base case. G(x = g w x w = (q 1 n g (q 1w = (q 1 n T r(ψ q,g w W x q 1 =1 w W x C n p x qs 1 =1 Now suppose s 2. Notice that the following equation holds. G T q = T q G q For the LHS of the equation above, the following holds if q w. G T q (x w = G(xx w/q = v W g v x v + w/q = v W g v w/q x v = v W If not, then G T q (x w = 0. For the RHS the following holds. g qv x qv+w T q G q (x w = T q ( g v x qv+w = g qv x qv+w v W v W Hence the two operators G T q and T q G q are equal to each other. This implies the following where F G denotes the product of two power series F (x and G(x over C p. ψq,g s = T q G T q G ψ s 2 q,g = T q T q G q G ψ s 2 q,g = T q 2 (G G q ψ s 2 q,g = T q 2 (G G q T q G ψ s 3 q,g = T q 3 (G G q q G ψ s 3 q,g = T q 3 (G G q G q 2 ψ s 3 q,g =... = T q s (G G q... G q s 1 = ψ qs,g G q... G q s 1 Applying the theorem for the base case s = 1 proves (1 and (2. Notice from the proof of statement (1, each entry a i,j of A is of form g qw v for w, v W. From the definition of det(1 AT, we have the following. ( k ( k k ord p = ord p g qσ(wi w i M qσ(w i w i a wi,σ(w i ( k M qσ(w i k u i = M(q 1 k w i Notice 1 k ord pb k as k because 1 k k w i as k. Thus, the power series det(1 AT is well defined and has infinite radius of convergence.

14 14 SUN WOO PARK We now prove statement (4 of the theorem. Let A : R 0 R 0 be a matrix with entries {a w,v } w,v W such that det(1 AT and T r(a s are well defined for all positive integers s. Let {A (m } be a sequence of matrices with finite supports. For each m and for every w, v W, the set of entries {a (m w,v } w,v W of the matrix A (m is defined as a (m w,v = a w,v or a (m w,v = 0. We first show that statement (4 holds for all A (m. It suffices to show that the statement holds for any finite dimensional matrix V = {v ij } i,j=1,2,...,r. Notice we can assume V is an upper triangular matrix because there exists a change of basis such that V is upper triangular, such as the Jordan canonical forms. This implies that det(1 V T = r (1 v iit. Since T r(v s = r vs ii, we have the following. r r Exp p ( viit s s /s = Exp p ( (v ii T s /s s=1 = = s=1 r Exp p (Log p (1 v ii T r (1 a ii T = det(1 V T Now we return to the sequence of matrices {A (m }. By the aforementioned claim, for all m, A (m satisfies the theorem. To prove that the theorem holds for A, it suffices to show the following. (1 lim m det(1 A (m T = det(1 AT (2 lim m T r((a (m s = T r(a s for every positive integer s. For each m, let det(1 A (m T = k=1 b(m k T k and det(1 AT = k=1 b kt k. From the definition of the determinant, the following holds. ( k b (m k = ( 1 k sgn(σ w 1,w 2,...,w k W σ S k a (m w i,σ(w i Notice that the construction of A (m implies that every product in the sum is 0 or equal to the term sgn(σ k a w i,σ(w i in b k. For sufficiently large m, the term sgn(σ k a w i,σ(w i appears for any choice of σ S k and w 1, w 2,..., w k. Hence the first statement holds. The argument for the second statement is analogous. 3. Rationality of Zeta Functions Using the results obtained from the previous section on power series over C p, we will follow Dwork s proof of the rationality of zeta functions. In the first subsection we recall the definition of the zeta function and explore its properties. Next we prove that the zeta function is a quotient of two p-adic entire power series by constructing a suitable overconvergent power series in C p. We will then use these properties to show that the zeta function is rational Zeta Functions over Finite Fields. We recall the definition of the zeta function of a hypersurface over a finite field. The following is a restatement of Definition 1.5.

15 RATIONALITY OF ZETA FUNCTIONS OVER FINITE FIELDS 15 Definition 3.1. Let f(x 1,..., x n be a polynomial with n variables over F q for q = p r a power of p. Let H f be the affine hypersurface of f(x 1,..., x n. Then the zeta function of H f is defined as follows, where N s = H f (F q s denotes the number of F q s-points of H f. ( N s T s Z(H f, T = Exp p s Example 3.2. We first calculate the zeta functions for some affine hypersurfaces. (1 Let A n F q be the affine space over the finite field F q. For each s, N s = q ns. Hence the zeta function of the affine space is as follows. s=1 ( Z(A n q ns T s F q, T = Exp p = Exp p ( Log p (1 q n T = s s=1 1 1 q n T (2 Let P n F q be the projective space over the finite field F q. Notice that the projective space P n F q is the union n k=1 Ak F q. Hence we have N s = n k=1 qks for each s. The zeta function of the projective space is a product of those of affine spaces, which is Z(P n F q, T = n k=1 1 1 q k T. (3 The equation x 1 x 2 + x 3 x 4 1 = 0 has q 3s q s solutions over a finite field F q s. Hence the zeta function of the hypersurface of the polynomial is Z(H f, T = 1 qt 1 q 3 T. Using the examples, we can derive some properties of the zeta function. Proposition 3.3. Let H f be an affine hypersurface over a finite field and let Z(H f, T be the zeta function of the hypersurface. Then the following holds. (1 The coefficients of T s in Z(H f, T are at most q ns. (2 Z(H f, T has coefficients in Z. Proof. From Example 3.2, the maximum value of N s of a hypersurface is q ns = A n F q. Since the coefficients of Z(H f, T are less or equal to those of Z(A n F q, T, the statement holds. Order the F q s-points P = (x 1, x 2,..., x n of H f according to the least s such that all x i F q s. For j = 1, 2,..., s, let P j = (x 1,j,..., x n,j be the conjugates of P where P 1 = P and each x i,j is a conjugate of x i over F q. Notice that all the points P j are distinct. Suppose all x i are fixed by σ Gal(F q s/f q. Then the point P should be in a fixed field of σ, which is a field smaller than F q s. This is a contradiction. We will now observe how each point P 1, P 2,..., P s contributes to the terms of the zeta function Z(H f, T. Notice that for each j and for some r, P j is an F q r-point of H f if F q s F q r, or q s q r. Hence the point P j contributes to N ks for any positive integer k. The contributions the points P 1,..., P s give to the zeta function are as follows. Exp p ( k=1 st ks = Exp p ( Log p (1 T s = ks 1 1 T s = Extending the analogous argument for all possible s, we obtain that the zeta function is a product of the above series, which implies that the zeta function has integer coefficients. We recall the statement of the main theorem. k=0 T ks

16 16 SUN WOO PARK Theorem 3.4. Let H f be an affine hypersurface over a finite field and let Z(H f, T be the zeta function of the hypersurface. Then Z(H f, T is rational, i.e. a quotient of two polynomials with coefficients in Q. Before we prove the theorem in subsequent sections, we first show that the theorem also holds for any affine variety over a finite field. Corollary 3.5. The zeta function of an affine variety over a finite field is rational. Proof. Let f 1 (x and f 2 (x be polynomials with n variables over F q. It suffices to show that if Dwork s theorem holds for hypersurfaces H f1 and H f2, then it holds for the affine variety H f1,f 2 = {x A n F q f 1 (x = f 2 (x = 0}. Notice that the following holds where H f1f 2 is a hypersurface for the polynomial f 1 f 2 (x. H f1,f 2 (F q s = H f1 (F q s + H f2 (F q s H f1f 2 (F q s The above equation implies the following. Z(H f1,f 2, T = Z(H f 1, T Z(H f2, T Z(H f1f 2, T Since all the power series on the RHS is rational, the power series in the LHS is rational. The analogous proof shows that Dwork s theorem also holds for a union of hypersurfaces over a finite field. Corollary 3.6. The zeta function of a union of hypersurfaces over a finite field is rational p-adic Meromorphic. We first show that the zeta function of an affine hypersurface Z(H f, T is p-adic meromorphic, meaning that it is a quotient of two power series with infinite radius of convergence. Definition 3.7. Let a F q such that q = p r for prime p. Then the Teichmüller representative t of a is an element of C p which satisfies the following. (1 t is a root of the equation x q x = 0. (2 t a mod p The existence and uniqueness of Teichmüller representatives follow from the following lemma whose proof is in Chapter 1 of Koblitz [1]. Theorem 3.8 (Hensel s Lemma. Let f(x be a polynomial over Z p and let f (x be the derivative of f(x. Let a 0 be an element in Z p such that f(a 0 0 mod p and f (a 0 0 mod p. Then there exists a unique element a Z p such that the following holds. (1 f(a = 0 (2 a a 0 mod p The Teichmüller representatives allow us to extend any element in the finite field to an element in p-adic complex numbers. In fact, by the definition the representative is the uniformizer of the unramified field extension K of Q p. This implies two observations. One is that the p-adic ordinal of the Teichmüller representative is either 0 or 1. The other is that the trace of the element t is given by T r K (t = r 1 i=0 tpi. Notice that we have ζp T r(a = ζ T r K(t p because the trace of t satisfies T r K (t r 1 i=0 api = T r(a mod p.

17 RATIONALITY OF ZETA FUNCTIONS OVER FINITE FIELDS 17 Definition 3.9. Let θ(t be a power series over C p with the following construction. ( θ(t = F (T, ζ p 1 = a m,n (ζ p 1 m T n = a n T n n=0 m=n In the equation above, F (x, y is the power series from Definition 2.7 and ζ p is a primitive p-th root of unity. The coefficient a n is the sum m=n a m,n(ζ p 1 m for each n. Lemma The power series θ(t converges in the disk D(p 1 p 1. Before we prove the lemma, we state the analogue of Eisenstein s criterion for a polynomial over Z p, the proof of which is similar to that of Eisenstein s criterion for a polynomial over Z. Theorem 3.11 (Eisenstein s Criterion. Let f(x = a n x n a 1 x + a 0 be a polynomial over Z p. Suppose the following conditions hold. (1 a i 0 mod p for i = 0, 1,..., n 1. (2 a n 0 mod p. (3 a 0 0 mod p 2. Then the polynomial f(x is irreducible over Q p. Lemma We first show that ord p (ζ p 1 = 1 p 1. Observe that ζ p 1 is a root of the polynomial f(x = x p 1 + p 1 ( p n=1 n x p 1 n because (1 + ζ p 1 p = 1. By Theorem 3.11, f(x is irreducible. This shows that the field extension Q p (ζ p = Q p (ζ p 1. Let G = Gal(Q p (ζ p /Q p. Notice that the Galois conjugates of ζ p 1 are ζp i 1 for i = 1, 2,..., p 1. Since the p-adic norm is invariant under the action of the Galois group G, we have ζ p 1 p = ζp i 1 p. From cyclotomic polynomials, we have p 1 n=0 xn = p 1 (x ζi p. Substituting x = 1 gives p 1 (1 ζi p = p, 1 which implies ζp i p 1 1 p = p p = p 1 p 1. From Proposition 2.10, the power series F (x, y has all its coefficients in Z p. This implies that a m,n p 1 for all m n. The above claim and the proposition shows that for all the coefficients a n of θ(t, ord p a n. Hence θ(t converges in the disk D(p 1 p 1. The lemma above shows that the power series θ(t converges in a disk strictly bigger than D(1. Recall that the Teichmüller representative has p-adic ordinal of 1 or 0. This allows us to evaluate the power series θ(t at T = t pi for any i = 0, 1,..., r 1, where t is the Teichmüller representative of an element a F q. Using the Teichmüller representative, we prove the following lemma. The lemma gives an insight to understanding the sum of a set of p-th roots of unity, which appears in the first step of the proof of Dwork s theorem. Lemma Let a F q with q = p r and let t C p be the Teichmüller representative of a. Let ζ p be a primitive p-th root of unity. Then the following holds. r 1 ζp T r(a = θ(t pk k=0 n p 1 n=0

18 18 SUN WOO PARK Proof. Observe that the following holds for any independent variable Y because t pr = t and the telescoping sum in the power of (1 + Y pi is 0. r 1 F (t pi, Y = (1 + Y t+tp +...+t pr 1 i=0 = (1 + Y t+tp +...+t pr 1 = (1 + Y t+tp +...+t pr 1 (1 + Y pi t Substituting Y = ζ p 1 gives the desired equation. p i t pi 1 +t pi+1 t pi +...+t pr+i t pr+i 1 p i (1 + Y pi t p t+t p 2 t p +...+t pr t pr 1 p i Definition Let f(x 1, x 2,..., x n = M i=0 b ix ui be a polynomial over the field F q s where u i is an n-tuple of non-negative integers and q = p r for prime p. Let x 0 be an independent variable in F q s. Let F (x 0, x 1,..., x n = M i=0 a ix wi be a polynomial over C p where each a i is the Teichmüller representative of b i and w i = (1, u i is an (n + 1-tuple of non-negative integers. In other words, F (x 0, x 1,..., x n is a polynomial induced from x 0 f(x 1,..., x n by changing each coefficient to its teichmüller representative. Denote G(x as a power series over C p which is defined as follows. G(x = M r i=0 k=0 θ(a pk i x pk w i Lemma G(x is an overconvergent power series over C p Proof. Since the space of overconvergent power series R 0 is a subring of the space of power series in C p, it suffices to show that each factor θ(a pk i x pk w i is an overconvergent power series. From the definition of θ(t = n=0 b nt n, θ(a pk i x pk w i = n=0 b na npk i x npk w i for each i and k. Since each a i is the Teichmüller representative, we have the following. b n a npk i p = b n p p n p 1 p N nw i 1 where N = min i=0,1,...,m (p 1 w i. Hence the power series θ(apk i x pk w i R 0. Using the derivation of the power series G(x and the remark from the introduction, we prove that the zeta function is p-adic meromorphic. Theorem Let p be a prime number and f(x 1,..., x n = M i=0 b ix ui be a polynomial over F q with n variables where q = p r and u i is an n-tuple of non-negative integers for each i. Then the zeta function Z(H f, T of an affine hypersurface H f over F q is p-adic meromorphic. Proof. We prove the theorem by induction on the number of variables n. The base case where n = 0 is trivial. Suppose the theorem holds for n 1 variables. Let N s be the number of F q s-points (x 1, x 2,..., x n of H f such that all the coordinates are non-zero. This is same as the number of F q s-points ( of H f coordinates satisfy x qs 1 i = 1. Define Z (H f, T = Exp p s=1 N s T s s such that all the. Then the

19 RATIONALITY OF ZETA FUNCTIONS OVER FINITE FIELDS 19 following equation holds. ( Z (N s N (H f, T Exp st s p = Z(H f, T s s=1 ( (N Observe that the expression Exp s N s T s p s=1 s corresponds to the zeta function of a union of n affine varieties H i where each H i is defined by two equations: f(x 1,..., x n = 0 and x i = 0. Since each H i is an affine hypersurface of a polynomial with at most n 1 variables, the induction hypothesis and Collorary 3.6 imply that the expression is p-adic meromorphic. Hence it suffices to show that Z (H f, T is p-adic meromorphic. We first derive an expression for N s for each s. Let a be an element in F q s and t be the Teichmüller representative of a. Then Lemma 3.12 implies the following equation. rs 1 ζp T r(a = θ(t pk Since ζ p is a primitive p-th root of unity, the following holds for u F q s. { 1 if u (F q s ζ T r(x0u p = x 0 (F q s k=0 q s 1 if u = 0 Substitute u = f(x 1, x 2,..., x n and sum the expression over all the variables x 1, x 2,..., x n (F q s. ζ T r(x0f(x1,x2,...,xn p x 0,x 1,...,x n (F q s = N s(q s 1 + ((q s 1 n N s( 1 = N sq s (q s 1 n Let F (x 0, x 1,..., x n = M i=0 a ix wi be a polynomial over C p such that each a i is the Teichmüller representative of the corresponding coefficient b i of the polynomial x 0 f(x 1,..., x n. For each i, w i = (1, u i where w i is an (n + 1-tuple of non-negative integers. This gives an expression for N s for all s by Lemma q s N s = (q s 1 n + = (q s 1 n + ζ T r(x0f(x1,x2,...,xn p x 0,x 1,...,x n (F q s M rs 1 x 0,x 1,...,x n C p k=0 x qs 1 i =1 for all i θ(a pk i x pk w i As in Definition 3.13, define the power series G(x over C p as follows. G(x = M r i=0 k=0 θ(a pk i x pk w i By definition, the following expression for N s also holds. q s N s = (q s 1 n + x 0,x 1,...,x n C p j=0 x qs 1 i =1 for all i s 1 G(x qj

20 20 SUN WOO PARK By Lemma 3.14, G(x is an overconvergent power series over C p. Consider the operator ψ q,g = T q G where T q is the translation operator and G is the multiplication operator by G(x. By Theorem 2.26, the operator ψ q,g satisfies the following equation. q s N s = (q s 1 n + (q s 1 n+1 T r(ψ s q,g This gives the following expression for the zeta function Z (H f, T. ( ( Z (q s 1 n T s (q s 1 n+1 T r(ψq,g s (H f, T = Exp p sq s Exp T s p sq s s=1 The first factor on the RHS reduces as follows, which is clearly p-adic meromorphic. ( ( (q s 1 n T s n ( n Exp p sq s = Exp p ( 1 i q s(n 1 i T s /s i s=1 s=1 i=0 ( n ( n = Exp p ( 1 i q s(n 1 i T s /s i = = i=0 n i=0 s=1 s=1 Exp p ( Log p (1 q n i 1 T ( 1i ( n i n (1 q n i 1 T ( 1i ( n i i=0 The second factor on the RHS reduces as follows. In the equation below, A is an infinite matrix of the operator ψ q,g = T q G under the basis of monomials. ( (q s 1 n+1 T r(ψq,g s Exp T s p sq s s=1 ( n+1 ( n + 1 = Exp p ( 1 i q s(n i T r(ψ i q,gt s s /s s=1 i=0 ( n+1 ( n + 1 = Exp p ( 1 i T r(q s(n i ψ s i q,gt s /s = i=0 n+1 i=0 s=1 det ( 1 A(q n i T ( 1 i+1 ( n+1 i By Theorem 2.26, for each i = 0, 1,..., n + 1 the term det ( 1 A(q n i T is a power series over C p with infinite radius of convergence. This shows that the second factor on the RHS is also p-adic meromorphic. Hence Z (H f, T is an alternating product of p-adic power series with infinite radius of convergence. Thus Z (H f, T is p-adic meromorphic, which implies that the zeta function Z(H f, T is p-adic meromorphic Rationality. We first prove the following lemma which we will use to show the rationality of the zeta function.

21 RATIONALITY OF ZETA FUNCTIONS OVER FINITE FIELDS 21 Lemma Let F (x = n=0 a nx n be a power series over C p. Let A s,m = {a s+i+j } 0 i,j m be the matrix as follows. a s a s+1 a s+m a s+1 a s+2 a s+m a s+m a s+m+1 a s+2m Then F (x is a quotient of two polynomials over C p if and only if there exist integers m 0 and S such that det(a s,m = 0 whenever s S. Proof. Suppose F (x is a quotient of two polynomials F (x = P (x/q(x such that P (x = M k=0 p kx k and Q(x = N k=0 q kx k. Equating the coefficients of x i for i > max(m, N gives the following. N a i N+l c N l = 0 l=0 Let S = max(m N + 1, 1 and m = N. Then if s S, the following equation holds, which shows that det(a s,n = 0 for every s S. a s a s+1 a s+n c N a s+1 a s+2 a s+n+1 c N = 0 a s+n a s+n+1 a s+2n c 0 Now suppose there exist S and m such that for s S, det(a s,m = 0. Choose the smallest m such that the property det(a s,m = 0 holds for all sufficiently large s. We first show that det(a s,m 1 0 for all s S. Assume not. Then some linear combination of the rows r 0, r 1,..., r m 1 of the matrix A s,m vanish, except possibly for the last column. Let r k be the first row which has non-zero coefficient in the linear combination. In other words, r k = a k+1 r k+1 + a k+2 r k a m 1 r m 1 except possibly at the last column. Through row operations we can replace the k-th row r k by r k a k+1 r k+1 +a k+2 r k a m 1 r m 1. If k > 0 then the matrix is of the following form. a s a s+1 a s+m 1 a s+m a s+1 a s+2 a s+m a s+m b a s+m a s+m+1 a s+2m 1 a s+2m Then the matrix A s+1,m 1 with the first row from a s+1 to a s+n has determinant 0. The case for k = 0 analogously follows. This implies that det(a s,m 1 = 0 for all s s by induction on s, a contradiction to our choice of m as the minimal integer satisfying the condition. Now we have for all s S, det(a s,m = 0 while det(a s,m 1 0. This shows that there exists a linear combination of rows of the matrix A S,m such that the coefficient of the last row of the matrix is non-zero. In other words, the last row of