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Home , Local zeta function, Weil conjectures

THE WEIL ZETA FUNCTION AND THE IGUSA

MARGARET M. ROBINSON

Abstract. These are a set of notes and problem sets written for the Un- dergraduate Lecture Series section of ”Zeta Functions all the Way,” the 2006 topic of the Mentoring Program for Women in at the Institute for Advanced Study.

1. Introduction In the next four lectures I want to discuss two zeta functions the Weil zeta function and the Igusa local zeta function which, in comparison to the zeta functions we just studied, are more algebraic. These zeta functions are both associated with a polynomial f(x1, x2, ..., xn) in n variables with coefficients, for simplicity, in the integers. First I am going to introduce the two generating functions, often called Poincar´eSeries, which are associated with these zeta functions. Then I will define each zeta function and compute examples of them for different polynomials. I hope that this way we can compare the two zeta functions easily. I will give you some problems to work on in the afternoons which are supposed to make the zeta functions come to life for you. I will also discuss what is known about the functions and what is not known about then. Finally, I hope to give you a feeling for how these two functions are related. I also want to thank Diane Meuser from Boston University for reading these notes and making many very helpful suggestions,

e 1.1. Poincar´eSeries. Consider the unique finite field Fpe with p elements. To understand this Poincar´eseries and the associated Weil zeta function, one really needs to understand a lot about finite fields. So we will talk a lot more about finite fields. However, given a polynomial f(x1, x2, ..., xn) in n variables in the ring of polynomials Z[x1, x2, ..., xn] with coefficients in the integers, fix a prime p and consider f with coefficients reduced modulo p, so that it now has coefficients in the finite field Fp. To this polynomial, we associate the ∞ X e PW (T ) = |Ne| T e=1 (n) where Ne = {(x1, x2, ..., xn) ∈ Fpe |f(x1, x2, ..., xn) = 0 in Fpe } for e > 0 and | · | denotes set cardinality. The generating function P (T ) turns out to be a of T . The first part of the celebrated 1949 Weil (now theorems) was that the Weil zeta function should be a rational function. The rationality was proved by Bernie Dwork in 1960 using p-adic analysis and by showing that there is a nice relation between the |Ne|. The rationality of the Poincar´eseries follows right from the rationality of the zeta function. 1 2 MARGARET M. ROBINSON

For the same polynomial we can also associate another generating function to it as ∞ X e PI (T ) = |Ne| T e=1 e (n) e where Ne = {(x1, x2, ..., xn) ∈ (Z/p Z) |f(x1, x2, ..., xn) ≡ 0 mod p } for e > 0 and an integer. The generating function PI (T ) is also a rational function of T . Its rationality was conjectured in problem 9 of the exercises at the end of Chapter 1, section 5 in the textbook by Borevich and Shafarevich published in Russian in 1964 and in English in 1966. The rationality was settled by Igusa in 1974 when he showed that the associated Igusa local zeta function was rational. Igusa proved that the zeta function was rational using p-adic analysis and Hironaka’s resolution of singularities in characteristic 0. Hironaka won the ’s Medal in 1964 for this result. en Note: Both |Ne| and |Ne| are less than or equal to p . Sometimes you will see P∞ −en e the series (almost always, in the Igusa case) defined as PW (T ) = e=0 |Ne|p T P∞ −en e or PI (T ) = e=0 |Ne|p T with |N0| = |N0| = 1. Using this definition, each series begins with the constant 1 for e = 0 and converges for |T | < 1. However, we are not really interested in the convergence properties of the series because they are both rational functions of T and are thus meromorphic functions on the whole complex plane. Example 1. If we take the simplest non-constant polynomial f(x) = x. We find Ne = Ne = {0} so that |Ne| = |Ne| = 1 and the two Poincar´eseries are identical:

∞ X T P (T ) = P (T ) = T e = . W I 1 − T e=1

e Example 2. For f(x1, x2) = x1x2, we can show that |Ne| = {p roots where e e x1 = 0 and x2 is free to be any of the p elements in Fpe } + {p − 1 roots where e (2) x2 = 0 and x1 is free to be anything besides 0} = 2p − 1 total roots in Fpe . The Weil Poincar´eseries for this polynomial is:

∞ X e e PW (T ) = (2p − 1)T e=1 ∞ ∞ X X = 2(pT )e − T e e=1 e=1 2pT T = − 1 − pT 1 − T (2p − 1)T − pT 2 = . (1 − T )(1 − pT )

On the other hand, when we compute |Ne| things are a little more complicated: |N1| = 2p − 1, |N2| = {p(|N1| − 1) roots where x1 and x2 are not both divisible by 2 2 e−1 p} + {p roots where they are } = 3p − 2p, so that |Ne| = {p (|N1| − 1) roots 2 where x1 and x2 not both congruent to 0 modulo p}+{p (|Ne−2|) roots where they e e−1 are }. With careful counting we get that |Ne| = (e + 1)p − ep . THE WEIL ZETA FUNCTION AND THE IGUSA LOCAL ZETA FUNCTION 3

The Igusa Poincar´eseries for this polynomial is: ∞ X e e−1 e PI (T ) = [(e + 1)p − ep ]T e=1 ∞ X = (e + 1)(pT )e − ep−1(pT )e e=1 ∞ ∞ X X = e(1 − p−1)(pT )e + (pT )e e=1 e=1 (1 − p−1)(pT ) pT = + (1 − pT )2 (1 − pT ) (2p − 1)T − p2T 2 = (1 − pT )2 Note: We express the Igusa series in this way so that the comparison with the Weil series is more direct. The usual form of the Igusa Poincar´eseries, as mentioned before, would be to begin the series at e = 0 with |N0| = 1 and to replace T above with p−2T so that the series converges for |T | < 1. Thus, later we will write the Igusa Poincar´eseries for this example as 1 − p−2T P (T ) = . I (1 − p−1T )2 Questions: Now, of course, in these two examples we were able to find a formula for |Ne| and |Ne| without too much trouble. 1) However, it is not at all clear how formulas for |N1| and |N1| depend on p. We know that for any p we could compute these values but finding a formula in p is not easy. See papers Serre [12] , Koblitz [7], and Koblitz [8]. 2) The Local-Global Principle, first articulated by H. Hasse in his PhD thesis, asks whether information about solutions in Fp or really in Qp, the p-adic numbers, for all primes p can be put together in some way to say something about solutions in the rational numbers. Results like the proof that there are infinitely many primes by considering the Euler product expansion of the zeta function. 3) What is involved in finding a formula for |Ne| and |Ne| as e grows? What can we say about |Ne| − |Ne|? 4) The fact that the Poincar´eseries are rational means that |Ne| and |Ne| have nice recursion formulas. These nice formulas are tied to underlying geometric char- acteristics of the polynomial which are encapsulated in the .

2. Weil zeta function For any polynomial with coefficients in the integers, fix a prime p and consider f(x1, x2, ..., xn) with coefficients reduced modulo p, so that it now has coefficients in Fp. We define the Weil zeta function by the following awkward looking expression:

∞ e X |Ne|T Z (T ) = exp( ). W e e=1 One of my friends says that he is always embarrassed when he has to explain why there is an exponential function in the formula for ZW (T ). But in addition 4 MARGARET M. ROBINSON to reasons of tradition, we will see that the zeta function takes on a particularly e simple form when this definition is used. For example, if |Ne| = α then X (αT )e Z (T ) = exp( ) W e e≥1 = exp(− log(1 − αT )) 1 = 1 − αT where we make use of the identity that ∞ X xe − log(1 − x) = . e e=1 The exponential function also allows the zeta function to have a nice multiplicative property with respect to the values of |Ne|. Suppose that |Ne| = Ae + Be (as happens when f(x) = g(x)h(x) where g and h have no common roots), then

∞ e ∞ e X AeT X BeT Z (T ) = exp( ) · exp( ). W e e e=1 e=1 e e e e e e So that if |Ne| = α1 +α2 +···+αm −β1 −β2 −· · ·−βn the properties of logarithms and exponenials mean that

(1 − β1T )(1 − β2T ) ··· (1 − βnT ) ZW (T ) = . (1 − α1T )(1 − α2T ) ··· (1 − αmT ) We will spend most of our time investigating the zeta function as defined above but before we continue with that I want to mention that there is always a product form of the Weil zeta function which is analogous to the Euler product of the . Although in general we just know that the Weil zeta function is rational, for a projectively non-singular polynomial (whatever that is) the Weil conjectures tell us a lot more about the form of the rational function. For in the projectively non-singular case, we also have a functional equation and a result analogous to the (this time a theorem) about the zeros of the Weil zeta function. The product form of the Weil zeta function comes about from looking at the algebraic set of points where f = 0 over the algebraic closure of the field Fp. Starting with any root α = (a1, a2, ..., an) of the polynomial f in this field, there is immediately a smallest finite field Fpe which contains each of the a1, a2, ..., an of the i i i i p p pi pi p p pi root α and all of the a1 , a2 , ..., an of its conjugate roots α = (a1 , a2 , ..., an ), where 0 ≤ i ≤ e − 1. We group together all the conjugate roots to form what is called a prime divisor P of the algebraic set. The norm of P is defined to be the e cardinality of the field Fpe , so that NP = p , and e is called the degree of the prime divisor, degP = e. With these definitions is is not hard (see Ireland and Rosen p. 157) to see that

Y degP −1 ZW (T ) = (1 − T ) P where P runs over all (usually infinitely many) prime divisors of the algebraic set. Now if we let T = p−s, we can use the norm of the prime divisors NP = pdegP to THE WEIL ZETA FUNCTION AND THE IGUSA LOCAL ZETA FUNCTION 5 put the zeta function in a product form that is directly analogous to that of the Riemann zeta function, as follows:

−s Y −s −1 ZW (p ) = (1 − (NP) ) . P

Finally, as a formal series, the zeta function ZW (T ) is related to its Poincar´e series PW (T ) as follows:

0  0 TZW (T ) PW (T ) = T log ZW (T ) = . ZW (T )

Example 1. When f(x) = x, |Ne| = 1 and the Weil zeta function is: X T e Z (T ) = exp( ) W e e≥1 = exp(− log(1 − T )) 1 = 1 − T 0 Note that PW (T ), as we calculated it above, is indeed equal to TZW (T )/ZW (T ). e Example 2.When f(x1, x2) = x1x2, |Ne| = 2p − 1 and we have that X (2pe − 1)T e Z (T ) = exp( ) W e e≥1 = exp(−2 log(1 − pT ) + log(1 − T )) 1 − T = (1 − pT )2 where we make use of the properties of the logarithm. Note again that this result agrees with our calculation of PW (T ) above. 2.1. Finite Fields. See Ireland and Rosen [4], chapter 7, and/or Lidl and Nieder- reiter [9] for proofs and more information. Theorem 1. If f(x) is an irreducible polynomial of degree d in Fp[x] then the d factor ring Fp[x]/(f(x)) is a finite field with p elements. Theorem 2. If F is a finite field with q elements in it then F∗ = F \{0} with multiplication is a . Theorem 3. If F is a finite field. Then F has q = pd elements, where p is the characteristic of the field and d is the degree of F over its prime subfield Fp. Theorem 4. If F is a finite field with q elements, then any a ∈ F satisfies aq = a. d Theorem 5. For any positive integer d there exists a finite field Fpd with q = p elements and this field is unique. (There always exist irreducible polynomials over d Fp for all positive integer degrees d. All finite fields with p elements are isomorphic q to the splitting field of the polynomial x − x over Fp.) ∗ Definition 1. A generator of the cyclic group Fq is called a primitive element of Fq. Theorem 6. Fpd ⊆ Fpe if and only if d divides e. ∗ Using the fact that Fq is cyclic we can prove the following. ∗ n n Proposition 1. If a ∈ (Fq ) then x = a has d solutions where d = gcd(n, q − 1). n d Corollary. For a ∈ Fq then |{x|x = a}| = |{x|x = a}| where d = gcd(n, q − 1). 6 MARGARET M. ROBINSON

For example, in F11 although the sets are not equal their cardinalities are equal, |{x|x6 = 3}| = |{3, 8}| = 2 and |{x|x2 = 3}| = |{5, 6}| = 2. 2 Note that if q is even, a ∈ Fq then |{x|x = a}| = |{x|x = a}| = 1 since gcd(2, q − 1) = 1.

2.2. Examples and Multiplicative Characters. . One variable case. In the one variable case when the polynomial f(x) is irreducible of degree d over Fp, |N1| = 0, |Ne| = 0 for e < d, |Nd| = d when e = d ( Fpd is the splitting field of f), and |Ne| = d when d divides e, otherwise |Ne| = 0 for d not dividing e. 2 For example√ when f(x) = x − x − 1 the two roots of f in the algebraic closure are (1 ± 5)/2. So by quadratic reciprocity we know that 5 is not a square modulo p exactly when p ≡ ±2 mod 5 and thus for those values of p f(x) is irreducible modulo p. Thus for p = 7, we have that |Ne| = 0 when e is odd and |Ne| = 2 for all even values of e. But for p ≡ ±1 mod 5, |Ne| = 2 for all e. If p = 5 then the 2 polynomial has a double root f(x) = (x − 3) so that |Ne| = 1 for all e. Two variable case. In the curve case f(x, y), we can still compute the values of |Ne| directly in some cases. From now on to distinguish between the different polynomials f(x, y), we will use |Ne(f(x, y))| to denote the number of solutions to (2) f(x, y) = 0 in Fpe . 2 2 1. For example if f(x, y) = x − y, the parabola, we can easily see that |Ne(x − e y)| = p since we are free to choose any value in Fpe for x but then y is determined. 2. When f(x, y) = xy − 1, the rotated hyperbola, it is easy to see that |Ne(xy − 1)| = pe − 1. Since there are pe − 1 non-zero choices for x and for each of those choices y is then determined. 3. When f(x, y) = x2 − y2 − 1, the hyperbola in standard position, and q = pe is odd, we are looking for points such that (x − y)(x + y) = 1. By making a change of variables so that u = x − y and v = x + y, we can argue that the map is one-to-one 2 2 e e and onto so that |Ne(x − y − 1)| = |Ne(uv − 1)| = p − 1 for odd q. When q = p 2 2 2 e is even (Fpe ) = Fpe so that |Ne(x − y − 1)| = |Ne(x − y − 1)| = p . 4. Suppose f(x, y) = x2 + y2 − 1, the unit circle and p is an odd prime. ( Note: If p is even, we have the same argument as in example 3 for the even case. ) We 2 2 (2) want to compute |Ne(x + y − 1)| the number of roots of f(x, y) in Fpe . Since y2 = 1 − x2, we need to know when 1 − x2 is a square. If 1 − x2 is a square(ie 2 ∗ 2 1 − x ∈ (Fpe ) )then there will be 2 possibilities for y with that value for x. If 1 − x2 = 0 then (x, y) = (x, 0) is a solution. And finally, if 1 − x2 is not a square ∗ in Fpe there are no solutions with that x value. To keep track of this situation, we use the quadratic character χ : Fpe → {−1, 0, 1} defined so that χ(ab) = χ(a)χ(b) and  ∗ 2  1 a ∈ (Fpe ) χ(a) = 0 a = 0 ∗ 2  −1 a 6∈ (Fpe ) and a 6= 0 2 Now we can write that for a fixed a ∈ Fpe , |Ne(y = a)| is 1 + χ(a) which just happens to equal 2, 1, 0 in the cases when a is a non-zero square, is zero, or is not THE WEIL ZETA FUNCTION AND THE IGUSA LOCAL ZETA FUNCTION 7 a square, respectively. Using this formulation we get that

2 2 X 2 2 |Ne(x + y − 1)| = |Ne(x = a1)| · |Ne(y = a2)| a1,a2∈Fq ,a1+a2=1 X = (1 + χ(a1))(1 + χ(a2)) a1,a2∈Fq ,a1+a2=1 X = 1 + χ(a1) + χ(a2) + χ(a1)χ(a2) a1,a2∈Fq ,a1+a2=1 X = 1 + χ(a1) + χ(1 − a1) + χ(a1)χ(1 − a1) a1∈Fq X X X = q + χ(a1) + χ(1 − a1) + χ(a1)χ(1 − a1) a1∈Fq a1∈Fq a1∈Fq

Now, the two middle sums in the last equation above both evaluate to 0 because they are both equal to the sum of χ(c) as c runs through all elements in the finite field. We see easily that this sum must equal 0. For a non-zero b chosen so that χ(b) 6= 1, we have that X X χ(b) χ(c) = χ(b)χ(c)

c∈Fq c∈Fq X = χ(bc)

c∈Fq X = χ(c)

c∈Fq Bringing both sums to the same side of the equation, we have that X X χ(b) χ(c) − χ(c) = 0

c∈Fq c∈Fq or that X (2.1) (χ(b) − 1) χ(c) = 0.

c∈Fq Now since χ(b) 6= 1 we have that P χ(c) = 0. Thus from above c∈Fq

2 2 X (2.2) |Ne(x + y − 1)| = q + χ(a1)χ(1 − a1) a1∈Fq \{0,1}

2 2 We see that |Ne(x + y − 1)| ≈ q and consider the last sum as a kind of error term (also called a Jacobi sum) in our calculation. Note that q = |{(a1, a2)|a1 + a2 = 1}| −1 as a1 and a2 run through the finite field. By using the facts that χ = χ and −1 2 2 χ (a) = χ(1/a), we can rewrite our expression for |Ne(x + y − 1)| as:

2 2 X a1 |Ne(x + y − 1)| = q + χ( ). 1 − a1 a1∈Fq \{0,1} 8 MARGARET M. ROBINSON

a1 As a1 runs through all the elements in q \{0, 1}, c = runs though all the F 1−a1 c elements in Fq \{0, −1} because a1 = 1+c . As a result, 2 2 X |Ne(x + y − 1)| = q + χ(c) c∈Fq \{0,−1} X = q + χ(c) − χ(0) − χ(−1)

c∈Fq = q − χ(−1). e Thus for q = p odd, we have by Euler’s criterion in the finite field Fq that  e  q−1 1 q = p ≡ 1 mod 4 1 p ≡ 1 mod 4 χ(−1) = (−1) 2 = = −1 q = pe ≡ 3 mod 4 (−1)e p ≡ 3 mod 4 Finally for the unit circle when pe is odd, we have that  pe − 1 p ≡ 1 mod 4 |N (x2 + y2 − 1)| = e pe − (−1)e p ≡ 3 mod 4 and the “error term” we mentioned above is ±1. 2 2 Now to generalize this example, we might consider f(x1, x2, ..., xn) = x1 + x2 + 2 ··· + xn − 1. Without too much trouble we can show that 2 2 X 2 2 2 |Ne(x1 + ··· + xn − 1)| = Ne(x1 = a1) · Ne(x2 = a2) ··· Ne(xn = an) a1+···+an=1 n−1 X = q + χ(a1)χ(a2) ··· χ(an) a1,...,an∈Fq ,a1+···+an=1 Since all other terms turn out to be 0. n−1 So here again we see that the number of roots is approximately q = |{(a1, a2)|a1+ a2 + ··· + an = 1}| up to an error term which is a more general Jacobi sum. Note that this final Jacobi sum is clearly less than qn−1 for it is a sum of 0’s, 1’s, and -1’s n−1 over q points (a1, a2, ..., an) and when ai = 0 for any i (which it most definitely is for lots of points in the sum) that term in the Jacobi sum will be 0. Thus for this example |Ne| > 0 for all e. If z is a complex number such that |z| = 1, we call z a and write z = cos(θ) + i sin(θ) = exp(iθ) for some θ. If θ has the property that nθ = 2π then n z = 1 and z is an n-th root of unity. If we let ζn be the n-th root of 1 with the smallest θ then the other n-th roots of 1 are all powers of ζn. Thus if we let µn denote the set of n-th roots of 1 in the complex numbers then i µn = {ζn|0 ≤ i ≤ n − 1}. Definition. A multiplicative character of Fq is a function ∗ χ : Fq → µq−1 such that χ(ab) = χ(a)χ(b). Examples. 1. The, so called, “trivial character” is usually denoted by χ0 and is defined so ∗ that χ0(a) = 1 for all a ∈ Fq . In fact the definition of the trivial character is often extended to all of Fq by defining χ0(0) = 1. ∗ 2. Since Fq =< g > is cyclic of order q − 1 with a generator g, the choice of χ(g) determines the rest of the map. Thus, χ(g) can be any (q − 1)-st root of unity: 1, 2 q−2 ζq−1, ζq−1,...,ζq−1 . In this way, for any Fq there are q − 1 multiplicative characters: THE WEIL ZETA FUNCTION AND THE IGUSA LOCAL ZETA FUNCTION 9

χ0, χ1, ...., χq−2. The definition of a non-trivial character is extended to all of Fq by defining χ(0) = 0. ∗ 3. The definition χ0(0) = 1 is awkward because now when χ1 · χ2 = χ0 on Fq for χ1 and χ2 nontrivial, we have that χ1 · χ2(0) 6= χ1(0)χ2(0). ∗ 4. To use F7 =< 3 > (generated by powers of 3) as an example, we see√ that it has 6 characters associated with√ it: χ0(3) = 1, χ1(3) = exp(πi/3) = 1/2+i 3/2, χ2(3) = exp(2πi/√3) = −1/2 + i 3/2, χ3(3) = exp(3√πi/3) = −1, χ4(3) = exp(4πi/3) = −1/2 − i 3/2 χ5(3) = exp(5πi/3) = 1/2 − i 3/2. ∗ Looking more carefully at χ2, we see where each member of F7 is sent: √ χ (31) = χ (3) = exp(2πi/3) = −1/2 + i 3/2 2 2 √ 2 χ2(3 ) = χ2(2) = exp(4πi/3) = −1/2 − i 3/2 χ (33) = χ (6) = exp(2πi) = 1 2 2 √ χ (34) = χ (4) = exp(2πi/3) = −1/2 + i 3/2 2 2 √ 5 χ2(3 ) = χ2(5) = exp(4πi/3) = −1/2 − i 3/2 6 χ2(3 ) = χ2(1) = exp(2πi) = 1.

3 and by extension χ2(0) = 0. We say that χ2 is a character of order 3 since χ2(a) = ∗ χ0(a) for each element a ∈ F7. n ∗ Definition. χ is a character of order n if χ = χ · χ ··· χ = χ0 on Fq and n is the smallest positive integer with this property. In other words, χn(a) = (χ(a))n = 1 ∗ 2 for all a ∈ Fq . (Note χ = χ · χ, is multiplication not function composition. This meaning is clear since χ is not defined on its image.) Before we do more calculations we need to know more about character sums. e Proposition 2. Given some Fq where q = p . (a) Let χ be a non-trivial multiplicative character extended so that χ(0) = 0, then X χ(a) = 0.

a∈Fq P (b) If χ = χ0 the trivial character then χ0(a) = q a∈Fq ∗ P (c) For a fixed a ∈ Fq , a 6= 1, the allχ χ(a) = 0 Proof. The proof of part (a) is give above in (2.1) for the quadratic character but is essentially exactly the same in general. Part (b) is clear. For part (c), P let S = allχ χ(a). Since a 6= 1, there is a character χ1 such that χ1(a) 6= 1. j i Note that for the generator g of the finite field, χ(g) = ζq−1 and χ1(g) = ζq−1 for i j ij 0 ≤ i, j ≤ q − 2, so that χ1(g)χ(g) = ζq−1ζq−1 = ζq−1 = χ1χ(g). Then X X χ1(a)S = χ1(a)χ(a) = χ1χ(a) = S, allχ allχ since χ1χ runs through all characters if χ does. Thus we have that (χ1(a)−1)S = 0 and since χ1(a) 6= 1 we have that S = 0. 2 In the calculations above we had the formula: |Ne(x = a)| = 1 + χ(a) for n the quadratic character χ. To generalize this expression, we know that |Ne(x = d a)| = |Ne(x = a)| if d = gcd(n, q − 1), so it is enough to look for a formula for n |Ne(x = a)| when n divides q − 1. 10 MARGARET M. ROBINSON

Proposition 3. For a ∈ Fq and n divides q − 1, n X X |Ne(x = a)| = χ(a) = 1 + χ(a). n n {allχ|χ =χ0} {allχ6=χ0|χ =χ0} n Proof. First note that there are n characters χ = χ0, called the characters of order dividing n. There are exactly n such characters because for each of them n ∗ χ(g) = 1 when g is the generator of Fq and so χ(g) must be equal to one of the 2 n−1 following n-th roots of 1: 1, ζn = exp(2πi/n), ζn,...,ζn . So there are n of them. 2 n−1 If χ(g) = ζn, then the characters of order dividing n are: χ0, χ, χ ,..., χ . n If a = 0, we know that Ne(x = 0) = 1. Since χ0(0) = 1 but for all other P characters χ(0) = 0, it is clear that n χ(0) = 1. So the formula holds {allχ|χ =χ0} in this case. n ∗ ∗ If a 6= 0 and x = a has a solution in Fq , then there exists a b ∈ Fq such that n n b = a. By Proposition 1 about finite fields above Ne(x = a) = n since n = gcd(n, q −1). Now if χ is any of the n characters in the sum of order n, then χ(a) = n n n P χ(b ) = (χ(b)) = χ (b) = χ (b) = 1. Thus it is clear that n χ(a) = n. 0 {allχ|χ =χ0} So the formula holds in this case. n ∗ n If a 6= 0 and x = a is not solvable in Fq , then Ne(x = a) = 0. On the other hand, if a is not an n-power then a = gi and n does not divide i. So that for the i i character for which χ1(g) = ζn = exp(2πi/n), we know that χ1(a) = χ1(g ) = ζn 6= 1. Thus the same argument we used above multiplying both sides of the sum S by χ1(a) will show that the sum must be 0. So the formula holds in all cases. Definition. Let χ1 and χ2 be characters of Fq and let X J(χ1, χ2) = χ1(a)χ2(b) a+b=1 then J(χ1, χ2) is called a Jacobi sum. Proposition 4. Let χ be a non trivial character on Fq. Then (a) J(χ0, χ0) = q (b) J(χ0, χ) = 0 (c) J(χ, χ−1) = −χ(−1). Proof. Part (a) is clear and part (b) follows from Proposition 2 above. Part (c) follows from essentially the same argument that we used above in (2.2) for the Jacobi sum of two quadratic characters. Now we will apply the information about characters to the equation xn +yn = 1. If if χ is a character of order n then we can write all other characters of order dividing n as powers of χ so that:

n n X n n Ne(x + y = 1) = Ne(x = a1) · Ne(y = a2) a1,a2∈Fq ,a1+a2=1 n−1 n−1 X X j X i = χ (a1) · χ (a2)

a1+a2=1 j=0 i=0 n−1 n−1 X X = J(χj, χi) j=0 i=0 Now we can estimate this sum using Proposition 4. 0 0 (a) If i = j = 0, we have that J(χ , χ ) = J(χ0, χ0) = q. THE WEIL ZETA FUNCTION AND THE IGUSA LOCAL ZETA FUNCTION 11

(b) When i = 0 and j 6= 0 then J(χj, χ0) = 0. When j = 0 and i 6= 0 then J(χ0, χi) = 0 (c) When i + j = n, χi = (χj)−1, so that J(χj, χi) = −χj(−1). With this information, we can go further with our calculation:

n−1 n−1 n n X X j i Ne(x + y = 1) = J(χ , χ ) j=0 i=0 n−1 X X = q + −χj(−1) + J(χj, χi) j=1 {(i,j)|1≤i,j≤n−1,i+j6=n}

j Now consider the middle term above. If −1 is an n-power in Fq then χ (−1) = 1 Pn−1 j for 1 ≤ j ≤ n − 1 and j=1 −χ (−1) = −(n − 1). If −1 is not an n-power then χj(−1) 6= 1 for 1 ≤ j ≤ n − 1 and by the trick used in Proposition 2 (c) Pn−1 j Pn−1 j Pn−1 j j=0 χ (−1) = 0 so that j=1 −χ (−1) = − j=0 χ (−1) + χ0(−1) = 1. In this way we have that

n−1 X  1 − n, when − 1 ∈ ( ∗)n  −χj(−1) = Fq = 1 − n δ (−1). 1, when − 1 6∈ ( ∗)n n j=1 Fq where δn(−1) = 1 or 0 depending upon whether -1 is an n-th power or not. Now we have that:

n n X j i Ne(x + y = 1) = q + 1 − n δn(−1) + J(χ , χ ) {(i,j)|1≤i,j≤n−1,i+j6=n} Note that when n = 2 the third term above is empty and we get our earlier result:  ∗ 2  2 2 q + 1 − 2 −1 ∈ (Fq ) q − 1 q ≡ 1 mod 4 Ne(x +y = 1) = q+1−2 δ2(−1) = ∗ 2 = q + 1 −1 6∈ (Fq ) q + 1 q ≡ 3 mod 4

n n 2 Consider the expression for Ne(x +y = 1) above. There are (n−1) −(n−1) = (n−1)(n−2) terms in the final sum. It is not too hard to show using an expression √ for J(χ1, χ2) that uses Gauss sums that |J(χ1, χ2)| = q when χ1, χ2, and χ1χ2 are all not equal to χ0. (See for example [4], pp. 93-4.) In this way we get the estimate that

n n X j i |Ne(x + y = 1) − (q + 1) + n δn(−1)| ≤ |J(χ , χ )| {(i,j)|1≤i,j≤n−1,i+j6=n} √ ≤ (n − 1)(n − 2) q

The number (n − 1)(n − 2) has geometric significance. The genus g of the curve xn + yn = 1 is exactly (n − 1)(n − 2)/2 and gives the number of holes in the complex corresponding to the curve. In general it is true that the numerator of the projective Weil zeta function for a curve is a polynomial in T of degree 2g. 12 MARGARET M. ROBINSON

2.3. Projective . Now while everything we have worked on is true and very interesting, the Weil Conjectures, the main set of theorems about this zeta function, do not hold for the values of |Ne(f)| counted as points in affine space over Fpe . The Weil conjectures hold for homogeneous polynomials f(x0, x1, ...., xn) with the values for |Ne(f)| counted as points in projective space over Fpe . (n) Definition. Affine n-space over Fpe is Fpe = {(a1, a2, ..., an) | ai ∈ Fpe } and is (n) e also denoted by A (Fq) when q = p . Definition. A polynomial f(x0, x1, ...., xn) is homogeneous of degree d if every term of the polynomial f has degree d. 2 n n n For example, x1x2 − x0 is homogeneous of degree 2, x1 + x2 − x0 is homogeneous of degree n. Definition. Given any polynomial f(x1, ...., xn) of degree d that is not homoge- neous we can homogenize it by introducing the new variable x0 and considering the d polynomial F (x0, x1, x2, ..., xn) = x0f(x1/x0, x2/x0, ..., xn/x0). The polynomial F (x0, x1, x2, ..., xn) is called the homogenization of f. Suppose that f(x) = f(x0, x1, ...., xn) is homogeneous of degree d. If f(x) = 0 has a solution a = (a0, a1, ..., an) then ca = (ca0, ca1, ..., can) is also a solution for ∗ all c ∈ Fq . Realizing this, we could define two points a = (a0, a1, ..., an) and b = (b0, b1, ..., bn) to be equivalent, a ∼ b, if b = (b0, b1, ..., bn) = c(a0, a1, ..., an) = ca ∗ for some c ∈ Fq . Thus, we have that for f(x) homogeneous, if a ∼ b then f(a) = 0 if and only if f(b) = 0. Rather than count solutions in affine space for a homogeneous polynomial, we can count inequivalent projective solutions. So that for a homogeneous f(x),

    # affine solutions to  # inequivalent or projective (n) = q − 1 · + 1 f = 0 in Fq solutions 6= (0, 0, ..., 0)

Definition. For a field F the set of points in projective n-space over F , P(n)(F ), are the set of points in A(n+1)(F )\{(0, 0, ...0)} where we identify equivalent points. Thus,

(n) P (F ) = {(a0, a1, ...an) | ai ∈ F and not all ai = 0}/equivalence.

Example 1. Given any line ax + by = c where a and b are not both 0, we found that there were q affine points on the line. In other words, |Ne(ax + by = c)| = q. We can consider its homogenization ax1 + bx2 − cx0 = 0 and count its solutions in (2) projective space P (Fq). (2) If x0 6= 0 then the number of inequivalent solutions [x0, x1, x2] in P (Fq) is equal to the number of solutions [1,X,Y ] where X = x1/x0 and Y = x2/x0. So that in the case where x0 6= 0, we just have solutions that correspond to the original q solutions of aX + bY = c. If x0 = 0, we need to count the number of inequivalent solutions of the form [0, x1, x2] such that ax1 + bx2 = 0. We know that either x1 or x2 is not equal to 0 since the point [0, 0, 0] is not included in projective space. The only inequivalent solution of the form [0, x1, x2] is [0, b, −a] which is equivalent to [0, −b, a], [0, 1, −a/b] (if b 6= 0), and [0, −b/a, 1] (if a 6= 0). This gives us one additional point on the THE WEIL ZETA FUNCTION AND THE IGUSA LOCAL ZETA FUNCTION 13 also called the point at infinity. So that  # Projective solutions  = q + 1 to ax1 + bx2 − cx0 = 0 2 2 Example 2. Given the circle x1 + x2 = 1 for which we found that  q − 1 q ≡ 1 mod 4 |N (x2 + y2 − 1)| = = q + 1 − 2δ (−1). e q + 1 q ≡ 3 mod 4 2 2 2 2 We can consider its homogenization x1 + x2 − x0 = 0 and count its solutions in (2) projective space P (Fq). (2) If x0 6= 0 then the number of inequivalent solutions [x0, x1, x2] in P (Fq) is equal to the number of solutions [1,X1,X2] where X1 = x1/x0 and X2 = x2/x0. So that in the case where x0 6= 0, we just need to count the number of solutions in 2 2 affine space to the original equation X1 +X2 = 1 which we have already computed. If x0 = 0, we need to count the number of inequivalent solutions of the form 2 2 [0, x1, x2] such that x1 + x2 = 0. We know that either x1 or x2 is not equal to 0 since the point [0, 0, 0] is not included in projective space. So assume that x1 6= 0, the number of inequivalent solutions of the form [0, x1, x2] is equal to solutions of 2 the form [0, 1,X2] where X2 = x2/x1 that solve 1+X2 = 0. The equation X2 = −1 ∗ 2 ∗ 2 has 2 solutions if −1 ∈ (Fq ) and 0 solutions if −1 6∈ (Fq ) . 2 2 2 Thus, the number of projective points of x1 + x2 − x0 = 0 is equal to   # Projective solutions 2 2 2 2 2 = |Ne(x + y − 1)| + 2δ2(−1) = q + 1 to x1 + x2 − x0 = 0 2 2 2 So that the number of projective points on x1 +x2 −x0 = 0 is exactly the number of points on a projective line. Example 3. Just as in the case of the circle, we showed that the number of affine points on the curve xn + yn = 1 was equal to n n |Ne(x + y = 1)| = (q + 1) − n δn(−1) + E where E is the sum of the (n−1)(n−2) Jacobi sums we discussed above and δn(−1) is 1 or 0 depending on -1 being an n-th power in Fq or not. The number of projective points at infinity (i.e. points where x0 = 0) on the n n n homogenization x1 + x2 − x0 = 0 is nδn(−1) so that   # Projective solutions n n n n n = |Ne(x + y − 1)| + nδn(−1) = q + 1 + E to x1 + x2 − x0 = 0 Thus for n > 2,  # Projective Solutions   # Points on the  E = n n n − . to x1 + x2 − x0 = 0 Projective Line 2 3 Example 4. Consider the affine curve x2 − x1 − 1 = 0. The homogenization of 2 3 3 this curve is f(x0, x1, x2) = x2x0 − x1 − x0 = 0. The projective points on this curve consist of the points equivalent to [1, x1, x2] which correspond to the affine points on the original curve and the points at infinity equivalent to [0, x1, x2] such 3 that f(0, x1, x2) = −x1 = 0. In this case since x1 = 0, the only point at infinity is [0, 0, x2] ∼ x2[0, 0, 1] ∼ [0, 0, 1]. 4 2 2 3 2 3 2 Example 5. Consider the quartic curve x1−x1x2+3x1+x1x2+2x2+2x2+8x1+3 = 4 2 2 3 2 0. The homogenization of this curve is f(x0, x1, x2) = x1 −x1x2 +3x1x0 +x1x2x0 + 3 2 2 3 4 2x2x0 + 2x2x0 + 8x1x0 + 3x0 = 0. In addition to the affine points on the curve, 14 MARGARET M. ROBINSON the projective curve includes 3 points at infinity of the form [0, x1, x2] such that 4 2 2 2 f(0, x1, x2) = x1 − x1x2 = x1(x1 − x2)(x1 + x2) = 0. Thus the points at infinity are the point where x1 = 0 so that [0, 0, x2] ∼ [0, 0, 1], the point where x1 = x2 so that [0, x2, x2] ∼ [0, 1, 1], and the point where x1 = −x2 so that [0, −x2, x2] ∼ [0, −1, 1].

2.4. Singular Points. Given a curve f(x, y) = 0 and a point P = (a, b) on the curve, the tangent line to the curve at the point P is given by the equation ∂f ∂f (a, b)(x − a) + (a, b)(y − b) = 0. ∂x ∂y However, what happens if both the partial derivatives of f are 0 at the point P ? Definition. A point P on f(x, y) = 0 where both partial derivatives vanish is called a singular point of the curve. A curve with no singular points is called a non-singular or smooth curve. Note that (0, 0) is a singular point on the curve x3−y2 = 0 but that x3−y2+1 = 0 is non-singular. Definition. If f(x0, x1, x2) = 0 is homogeneous and considered as a curve in projective space then a point P = [a, b, c] ∈ P(2) is singular if and only if ∂f ∂f ∂f (a, b, c) = (a, b, c) = (a, b, c) = 0. ∂x0 ∂x1 ∂x2

Note that if c 6= 0 and [a, b, c] is a projective singular point on f(x0, x1, x2) = 0 then [a, b, c] is equivalent to the point [a/c, b/c, 1] and can be viewed as the affine singular point (x1, x2) = (a/c, b/c) on the affine curve f(1, x1, x2) = 0. If c = 0 then we could take a different piece of affine space to look at the singular point [a, b, c]. We know that in P(2) at least one of a, b, or c is not 0, so that if b 6= 0 we can look at the point [a, b, c] by considering the equivalent point [a/b, 1, c/b] as the affine singular point (x0, x2) = (a/b, c/b) on the affine curve f(x0, 1, x2) = 0. If P = [a, b, c] is a non-singular point of the projective curve f(x0, x1, x2) = 0 then the tangent line in projective space to the curve at the point P is ∂f ∂f ∂f (P )x0 + (P )x1 + (P )x2 = 0. ∂x0 ∂x1 ∂x2 n n n Example 1. If p does not divide n then the projective curve x1 + x2 − x0 = 0 has (2) no singular points in P (Fq). Since in this case (0, 0, 0) is the only point for which (2) all partial derivatives vanish and (0, 0, 0) 6∈ P (Fq). If p does divide n then every point on the curve is singular modulo p. 3 2 3 2 Example 2. The curve x1 − x2 = 0 (with its homogenization x1 − x2x0 = 0) has only one point at infinity, the point [x0, x1, x2] = [0, 0, 1] and this point is non- ∂f 2 singular since ([0, 0, 1]) = −x | = −1. However the point [x0, x1, x2] = ∂x0 2 ([0,0,1]) [1, 0, 0] is a singular point on the curve. √ 2.5. Weil Conjectures. We showed earlier (except we assumed |J(χ1, χ2)| = q when χ1, χ2, χ1χ2 6= χ0) that n n √ ||Ne(x + y = 1)| − (q + 1) + n δn(−1)| ≤ 2g q. n n n Now putting this in a projective setting and letting |Ne(x1 + x2 = x0 )| equal the (2) number of solutions to this equation in P (Fq) counted projectively, we have that n n n √ ||Ne(x1 + x2 = x0 )| − (q + 1)| ≤ 2g q. THE WEIL ZETA FUNCTION AND THE IGUSA LOCAL ZETA FUNCTION 15

This inequality and ones like it for other curves is known as the “Riemann Hypoth- esis for curves” because as we will see later it implies that the roots of the projective Weil zeta function with T = q−s all lie along the line in the complex plane where √ R(s) = 1/2 or equivalently the reciprocal roots have absolute value equal to q. (2) The Riemann Hypothesis is true for any non-singular curve in P (Fq). One consequence of the Riemann Hypothesis is that √ |Ne| ≥ (q + 1) − 2g q.

As q → ∞ this implies that |Ne| ≥ 1 so that if C is any projective curve f(x0, x1, x2) = 0 and we let p grow then f(x0, x1, x2) ≡ 0 mod p will have solutions for all p large e enough. OR fix p and let q = p then as e → ∞ f(x0, x1, x2) = 0 will have solutions (2) in P (Fq) for all large enough e. Now we have defined the Weil zeta function above for polynomials f ∈ Fp[x1, ..., xn]. Now we will extend this definition to include both the earlier affine Weil zeta function and what has actually been more studied the projective Weil zeta func- tion. So for a polynomial f ∈ Fp[x1, ..., xn] we can consider its homogenization ∗ f ∈ Fp[x0, x1, ..., xn] then we can either let (n) 1. |Ne| = the number of points of f = 0 in A (Fpe ) OR ∗ (n) 2. |Ne| = the number of points of f = 0 in P (Fpe ). With either of these two definitions we can then form the ∞ X |Ne| Z(T ) = exp T e e e=1 for either the 1) affine Weil zeta function or the 2) projective Weil zeta function. 2 2 Example 1. For the circle x1 + x2 = 1, the affine Weil zeta function is e  exp P∞ p −1 T e p ≡ 1 mod 4  1−T p ≡ 1 mod 4  e=1 e  1−pT ZAff,W (T ) = = e e  P∞ p −(−1) e  1+T p ≡ 3 mod 4 exp e=1 e T p ≡ 3 mod 4 1−pT However, the projective Weil zeta function for this curve is ∞ X pe + 1 1 Z (T ) = exp T e = . P roj,W e (1 − T )(1 − pT ) e=1 n n n Davenport-Hasse Theorem. For the curve x1 + x2 − x0 = 0 when n divides p − 1, we showed that X |N1| = p + 1 + J(χ1, χ2) (n−1)(n+2) Jacobi sums The Davenport-Hasse Theorem (see [4], page 163 and [9], page 197) from 1935 gives a relation between |N1| and |Ne| for e ≥ 1: Let

{β1, β2, ..., β(n−1)(n−2)} = {−J(χ1, χ2)} where the Jacobi sums on the left are the (n − 1)(n − 2) non-trivial sums from the calculation of |N1|, i.e. (n−1)(n−2) X |N1| = p + 1 − βi. i=1 16 MARGARET M. ROBINSON

Then (n−1)(n−2) e X e |Ne| = p + 1 − βi . i=1 Hence, from the Davenport-Hasse Theorem we know that the projective zeta func- n n n tion for the curve x1 + x2 − x0 = 0 has the form Q(n−1)(n−2)(1 − β T ) Z (T ) = i=1 i . W (1 − T )(1 − pT ) In particular we see that the degree of the numerator is 2g = (n − 1)(n − 2). 3 3 3 Example 2. For the curve x1 + x2 − x0 = 0 where 3 divides p − 1, we have that

|N1| = p + 1 + J(χ1, χ1) + J(χ2, χ2) √ where if ∗ =< g > then χ (g) = ζ = −1/2 + i 3/2 and χ (g) = ζ2 = −1/2 − √ Fp 1 3 2 3 i 3/2. Thus, is the set up above β1 = −J(χ1, χ1) and β2 = −J(χ2, χ2). So that now we have that

|N1| = p + 1 − β1 − β2 = p + 1 − β1 − β1 since clearly χ1(g) = χ2(g), J(χ1, χ1) = J(χ2, χ2) and β2 = β1. Now by Davenport-Hasse:

e e e |Ne| = p + 1 − β1 − (β1) and (1 − β T )(1 − β T ) 1 − 2Re(β )T + pT 2 Z (T ) = 1 1 = 1 . W (1 − T )(1 − pT ) (1 − T )(1 − pT )

Note that |N1| = p + 1 − β1 − β1 = p + 1 − (β1 + β1) = p + 1 − 2Re(β1). Now we have two methods for computing the zeta function:

• If we know |N1| from just counting solutions modulo p, we can compute Re(β1) and with that we have ZW (T ). • We can always compute β1 and hence ZW (T ) by actually computing the Jacobi sum. Thus, by the first method when p = 7, we find that

N1 = {(x0, x1, x2)} = {(0, 1, 3), (0, 1, 5), (0, 1, 6), (1, 0, 1), (1, 0, 2), (1, 0, 4), (1, 1, 0), (1, 2, 0), (1, 4, 0)}

3 3 3 2 are the roots of x1 + x2 − x0 = 0 counted projectively in P (F7). So we know that |N1| = 9. Now since 9 = 7 + 1 − 2Re(β1), we have that 2Re(β1) = −1 and 1 + T + 7T 2 Z (T ) = . W (1 − T )(1 − 7T )

2 Factoring 1 + T +√ 7T = (T − γ1)(T − γ2) = (1 − β1T )(1 −√β2T ) = 0 we get that γ1, γ2 = −√1/14 ± 3 3i/14 and hence β1 = 1/γ2 = −1/2 + 3 3i/2 and β2 = 1/γ1 = −1/2 − 3 3i/2. So that √ √ e e e |Ne| = p + 1 − (−1/2 + 3 3i/2) − (−1/2 − 3 3i/2) for all e. Note that for n = 3 the value of |N1| determines all other |Ne|. THE WEIL ZETA FUNCTION AND THE IGUSA LOCAL ZETA FUNCTION 17

By the second method, we need to compute β1 = −J(χ1, χ1) and we need the values of χ1 (which we unfortunately called χ2 in the discussion above where we ∗ listed all characters of F7). X J(χ1, χ1) = χ1(c1)χ1(c2) c1+c2=1, c1,c2∈F7 = 2χ1(0)χ1(1) + 2χ1(2)χ1(6) + 2χ1(3)χ1(5) + χ1(4)χ1(4) = 0 + 2(exp(4πi/3))(1) + 2(exp(2πi/3))(exp(4πi/3)) + (exp(2πi/3))2 2 = 2 + 3(exp(4πi/3)) = 2 + 3ζ3 √ √ 2 Thus, β1 = −(2 + 3ζ3 ) = −1/2 + 3 3i/2 and β2 = −(2 + 3ζ3) = −1/2 − 3 3i/2. Weil Conjectures for non-singular curves (NOW THEOREMS). For f(x0, x1, x2) = (2) 0 non-singular in P (Fp).

• Rationality: The zeta function ZW (T ) is always a rational function of the form Q2g (1 − β T ) Z (T ) = i=1 i W (1 − T )(1 − pT ) where g is the genus of the curve. This rationality implies that always e P2g e |Ne| = p + 1 − i=1 βi . • Integer Coefficients: the numerator and denominator of ZW (T ) have integer coefficients. p • Functional Equation: The map βi → permutes the {βi}. √ βi • Riemann Hypothesis. |βi| = p. (Hardest part to prove.) Weil Conjectures for non-singular projective hypersurfaces (NOW THE- (n) OREMS). For f(x0, x1, x2, ..., xn) = 0 non-singular of degree d in P (Fp).

• Rationality: The zeta function ZW (T ) is always a rational function of the form n P (T )(−1) Z (T ) = W (1 − T )(1 − pT ) ··· (1 − pn−1T ) Qm −1 n+1 n+1 where P (T ) = i=1(1 − βiT ) and m = d [(d − 1) + (−1) (d − 1)] is called the of the hypersurface and is related to the topology (n−1)e of the hypersurface. This rationality implies that always |Ne| = p + (n−2)e e n+1 Pm e p + ··· + p + 1 + (−1) i=1 βi . • Integer Coefficients: P (T ) has integer coefficients. pn−1 • Functional Equation: The map βi → permutes the {βi}. βi (n−1)/2 • Riemann Hypothesis. |βi| = p 2 4 The Riemann Hypothesis was shown by Gauss for y = x − 1 over Fp. E. Artin showed it was true for cases associated with quadratic number fields, and Hasse proved it for all elliptic curves (genus 1 curves). A. Weil proved the Riemann Hypothesis for curves of arbitrary genus in 1940-41. In 1949, A. Weil posed these conjectures in more generality than stated above (for projectively non-singular algebraic varieties over Fq). The rationality of Z(T ) was proved in both the affine and the projectively non-singular case by B. Dwork in 1960, items 2 and 3 above were proved in the mid-1960’s, and item 4, the Riemann Hypothesis and the hardest part of the conjectures, was proved by in 1973. 18 MARGARET M. ROBINSON

Item 4 is called the Riemann Hypothesis because if you let T = p−s then in the curve case item 4 is equivalent to all the zeros of the numerator of Z(p−s) being on the line Re(s) = 1/2 What is left to understand?

• General facts about the values of βi in the numerator. What can you say if you consider all curves of a given genus g? • What if we drop the non-singular assumption? We know from Dwork that Z(T ) is still rational, but what can one say about the numerator and de- nominator and hence the |Ne|’s. • What happens if you fix a curve and consider varying the prime p? What happens to Zp(T )? The usual way to study variation in p when for example your curve is C : 3 3 3 x1 + x2 − x0 = 0 is to take 1 − (β + β )T + pT 2 Z (T ) = 1 1 p (1 − T )(1 − pT ) and let ap = β1 + β1. Then consider Y 2 −1 (1 − apT + pT ) p or with T = p−s Y −s −2s −1 L(C, s) = (1 − app + pp ) . p Hasse first conjectured that this L-function for a curve should have an analytic continuation to the whole complex plane and a functional equation. The for a non-singular is known as the Hasse-Weil conjecture. THE WEIL ZETA FUNCTION AND THE IGUSA LOCAL ZETA FUNCTION 19

3. Exercises Exercises taken from N. Koblitz, Ireland and Rosen, R. Lidl and H. Niederreiter. ASK ME IF YOU NEED MORE EXERCISES CONCERNING FINITE FIELDS. Exercise 1. Compute PI (T ), PW (T ) and ZW (T ) for the polynomial in n variables 0 that is identically 0. Check the identity PW (T ) = TZW (T )/ZW (T ).

Exercise 2. Compute PI (T ) (for first 6 examples), PW (T ) and ZW (T ) (later in the week you can also compute ZP roj,W (T ) for homogenized curves and ZI (T )) for the following examples • f(x, y) = ax + by − c for any line where a, b, c ∈ Z • f(x, y) = xy − 1, a hyperbola • f(x, y) = x2 − y2 − 1, a hyperbola in standard form • f(x, y) = x2 − y, a parabola. How do the results for the parabola compare with the results for any polynomial f(x, y) = g(x) − y where g(x) is any polynomial? • f(x1, x2, x3, x4) = x1x2 + x3x4 • f(x1, x2, x3, x4) = x1x2 + x3x4 − 1 • f(x1, x2) = x1x2(x1 + x2 + 1) 2 3 • f(x1, x2) = x1 − x2 3 2 2 • f(x1, x2) = x1 + x1 − x2 0 Check the identity PW (T ) = TZW (T )/ZW (T ). ∗ Exercise 3. Consider Fq with q odd and let Fq =< g > . Let χ be the character 2πi/(q−1) (q−1)/2 of order q − 1 satisfying χ(g) = ζq−1 = e . Show that χ (x) is the quadratic character (the unique character of order 2 which is 1 when x is a square ∗ in Fq and −1 otherwise) for • the fields F5 and F7 • in general for Fq.

Exercise 4a. Consider Fq with q ≡ 1 mod 3. What are all the multiplicative characters of order 3? First consider F7.

Exercise 4b. Consider Fq with q ≡ 1 mod 4. What are all the multiplicative characters of order 4? First consider F5.

Exercise 5. Consider Fq with q odd. If χ is a multiplicative character of degree d (thus d divides q − 1), show that   1 if d odd χ(−1) = 1 if d even and (q − 1)/d even  −1 if d even and (q − 1)/d odd

Exercise 6. For each of the affine curves C below find a projective curve whose affine part is C and then find all points at infinity on the projective curve. • 3x + 2y + 6 = 0 • x2 + xy − 2y2 + x − 5y + 7 = 0 • x3 + x2y − 3xy2 − 3y3 + x2 + 7y + 4 = 0

Exercise 7. For each of the following curves C and points P , either find the tangent line to C at P or else verify that P is singular. 20 MARGARET M. ROBINSON

• y2 = x3 − x and P = (1, 0) 2 2 2 • x1 + x2 = x0 and P = [3, 4, 5] • x2 + y4 + 2xy + 2x + 2y + 1 = 0 and P = (−1, 0) 3 3 • x1 + x2 = x1x2x0 and P = [1, −1, 0]

2 2 2 2 Exercise 8a. Show that the curve x1 + x2 + x1x2 = 0 has two points at infinity in (2) P (Fp), both of which are singular. Exercise 8b. Show that the curve y2 = x3 + ax + b has no singular points in (2) 3 2 P (Fp) if 4a + 27b 6= 0. 3 3 3 Exercise 9. For the Fermat curve x1 + x2 − x0 = 0 over Fp when p ≡ 1 mod 3 there are two ways to compute the projective Weil zeta function.

• The first way is to compute the Jacobi sums over Fp and to use the Davenport-Hasse Theorem to get |Ne| explicitly for all e. With these val- ues ZW (T ) is a straightforward computation. Calculate ZW (T ) this way for p = 7 and p = 13. • The second way is based on the observation that since |N1| = p + 1 − (β1 + β1), and (1 − (β + β )T + pT 2) Z (T ) = 1 1 . W (1 − T )(1 − pT )

Check your answer above by counting the number of solutions in |N1| di- rectly (by checking solutions modulo p without using the Jacobi sums) and

determining the coefficient (β1 + β1) in the zeta function.

Exercise 10. Determine the projective Weil zeta function ZW (T ) for the x3 +y2 = 1 when p = 7 and when p = 13 by the two methods outlined above.

4 4 4 Exercise 11. For the curve x1 + x2 − x0 = 0, determine ZW (T ) for p = 5 and p = 13 by using explicit calculations of the Jacobi sums and the Davenport-Hasse Theorem. Note that for this example |N1| alone does not determine the numerator of the zeta function (ie |N1| does not determine |Ne| for all e. How many values of |Ne|, e = 1, 2, ..., are needed to determine the numerator of the zeta function? How n n n about for the more general curve x1 + x2 − x0 = 0? Exercise 12. Prove that the rationality of the Weil zeta function is equivalent to: There exist algebraic complex numbers α1, ...., αn, β1, ..., βm, such that all conju- gates of an α is an α, all conjugates of a β is a β, and n m X e X e Ne = αi − βj i=1 j=1 for e = 1, 2, 3....

3 3 3 Exercise 13. For the curve x1 + x2 − x0 = 0 when 3 divides p − 1, we showed that (1 − β T )(1 − β T ) 1 − 2Re(β )T + pT Z (T ) = 1 1 = 1 . W (1 − T )(1 − pT ) (1 − T )(1 − pT )

Show that the third item in the Weil Conjectures implies that ZW (1/(pT )) = −s −s ZW (T ). Setting T = p , and thinking of ZW (T ) = ZW (p ) = ZW (s), show that ZW (1 − s) = ZW (s), a functional equation for the zeta function. Assuming THE WEIL ZETA FUNCTION AND THE IGUSA LOCAL ZETA FUNCTION 21 the Weil conjectures, can you find the functional equation for the zeta function of n n n x1 + x2 − x0 = 0 when n divides p − 1. n n n (2) Exercise 14. Consider solutions to x1 + x2 − x0 = 0 in P (Fq). Call a solution (a0, a1, a2) trivial if one of the ai = 0. Using the Riemann Hypothesis for curves, determine explicitly how large p must be in order for the equation to have non- trivial solutions when n = 3, when n = 4, and when n = 5. For a general n give an n n n expression C(n) such that x1 + x2 − x0 = 0 has non-trivial solutions for all primes p > C(n). Exercise 15. Generalize the methods we used to get the number of solutions to m m (2) m m m (2) x1 + x2 = 1 in A (Fq) and x1 + x2 − x0 = 0 in P (Fq) to get formulas m m m (n) for the number of solutions to a1x1 + a2x2 + ··· + anxn = a0 in A (Fq) and m m m m (n) a1x1 +a2x2 +···+anxn −a0x0 = 0 in P (Fq), finding formulas that are explicit m m as possible. Use these formulas to find the Weil zeta function for a1x1 + a2x2 + m m ··· + anxn − a0x0 = 0 as explicitly as possible. Exercise 16. See very good problems at the end of Chapter V of [6] on pages 106-7 in edition 1 or pages 114-115 in edition 2. 22 MARGARET M. ROBINSON

References [1] E. Artin. Quadratische K¨orper in Gebiete der h¨oheren Kongruenzen I and II, Math. Zeit. 19 (1924), 153-246. [2] B. Dwork.On the rationality of the zeta function, Am. J. Math., 82 (1960), 631-648. [3] P. Deligne. La Conjecture de Weil I IHES 43 (1974), 273-307. [4] K. Ireland and M. Rosen. A Classical Introduction to Modern Number Theory. Springer Verlag. 1972. [5] N. Katz. An overview of Deligne’s proof of the Riemann Hypothesis for varieties over finite fields. [6] N. Koblitz. p-adic Numbers, p-adic Analysis, and Zeta functions, Springer Verlag 1977. [7] N. Koblitz. The p-adic approach to solutions of equations over finite fields, The American Mathematical Monthly, Vol. 87, 1980, 115-118. [8] N. Koblitz. Why study equations over finite fields?. Mathematics Magazine, vol. 55, no. 3, May 1982. [9] R. Lidl and H. Niederreiter. Finite Fields. Vol 20 of Encyclopedia of Mathematics and its applications. See also other version which is selections from the above: Introduction to Finite fields and their Applications. Cambridge Univ. Press, 1986. [10] M. Rosen. Number Theory in Function Fields. Springer-Verlag, New York, 2002. [11] P. Roquette, The Riemann Hypothesis in characteristic p: Its origin and development. Mitt. Math. Ges. Hamburg. 1-79. [12] J-P. Serre. On a theorem of Jordan, Bulletin of the AMS, Vol 40, Number 4, 2003, 429-440. [13] F.K. Schmidt. Analytische Zahlentheorie in K¨orpern der Characteristik p. Math. Zeit. 33 (1931), 1-32. [14] I. Shparlinski. Finite Fields: theory and Computation Kluwer Acdemic Publishers, 1999. [15] C. Small. Arithmetic of Finite Fields. Dekker. New York, 1991. [16] A. Weil. Number of solutions of equations in finite fields, Bull. Amer. Math. Soc., 55 (1949), 497-508.

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