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KARGAPOLOvfM~AKov.Fundamenta~
of the Theory of Groops. 63 BOLLOBAS. Graph Theory. (comlflued after inde.x)
Michael Rosen
Number Theory in Function Fields
Springer
Michael Rosen Department of Mathematics Brown University Providence, RI 02912-1917
USA
[email protected] Editorial Board
S. Axler Mathematics Department San Francisco State University San Francisco, CA 94132 USA
University of Michigan Ann Arbor, 1v1I 48109
K.A. Ribet Mathematics Department University of California, Berkeley Berkeley, CA 94720-3840
USA
USA
F.W. Gehring Mathematics Department East Hall
I'vlalhematics Su!{ject Classification (2000): IIR29. IIR58. 14H05 Library of Congress Cataloging-in-Publication Data Rosen, Michael I. (Michael Ira), 1938Number theory in function llelds I Michael Rosen. p. cm. - (Graduate texts in mathematics ; 210) Includes bibliographical references and index. ISBN 0-387-95335-3 (alk. paper) 1. Number theory. 2. Finite fields (Algebra). l. Title. II. Serie~. QA241 .R6752001 512·.7-dc21 2001042962
Printed on acid-free paper. © 2002 Springer-Verlag New York, Inc. AI.1 rights reserved. This work may not bc trallSlate
Michael Rosen Brown University
Contents
Preface
vii
1
Polynomials over Finite Fields Exercises . . . . " . , . . . .
2
Primes, Arithmetic Functions, and the Zeta FUnction Exercises . , . . . . . , . " . . . . . . . . . . . . . .
3 The Reciprocity Law Exercises
1 7
11 19
23 30
33 43
4
Dirichlet Exercises
5
Algebraic FUnction Fields and Global Function Fields Exercises ' . . . . . . . . . . . . . . . . . . . . . . . .
45 59
6
Weil Differentials and the Canonical Class Exercises . . . . . . . . . . . , . . " . .
63 75
7
Extensions of Function Fields, Riemann-Hurwitz: and the ABC Theorem Exercises " , . " . . , . . , . . . . . . . . . .
77 98
.1.r'JU) =
~
t
':=1
i=l
II 1. In the classical case of the Riemann zeta function, «( s) = 2::::=1 n- s , it is easy to show the defining series converges for fR( s) > 1, but it is more difficult to provide an analytic continuation. Riemann showed that it can be analytically continued to a meromorphic function on the whole complex plane with the only pole being a simple pole of residue 1 at s = 1. Moreover, if r(s) is the classical gamma function and ~(s) = 1f-frr{~)«(s), Riemann showed the functional equation e(l s) = .(;"(s). What can be said about (A (S)? By Equation 1 above, we see clearly that (A(S), which is initially defined for !R(s) > 1, can be continued to a meromorphic function on the whole complex plane with a simple pole at s 1. A simple computation shows that the residue at s = 1 is lo:(q)' Now define ~A(S) = q-"(l q-s)-l(J\(S). It is easy to check that ';A (1-s) .(;"A (s) so that a functional equation holds in this situation as well. As opposed to case of the cia..'lsical zeta-function, the proofs are very easy for (A(S). Later we will consider generalizations of (A (s) in the context of function fields over finite fields. Similar statements will hold, but the proofs will be more difficult and will be based on the Riemann-Roch theorem for algebraic curves. Euler noted that the unique decomposition of integers into products of primes leads to the following identity for the Riemann zeta-function:
«(s) =
II
(1 _.
p prime
1
s
P
1'>0
This is valid for ?Res) > 1. The exact same reasoning (which we won't repeat here) leads to the following identity:
II P irreducible P monic
(1-
~~)-l. I I
(2)
This is also valid for all !R(s) > 1. One can immediately put E-quation 2 to use. Suppose there were only finitely many irreducible polynomials in k The right-hand side of the equation would then be defined at s = 1 and even ha.ve a non-zero value there. On the other hand, the left hand side has a pole at s = L This shows there are infinitely many irreducibles in A. One doesn't need the zeta-function to show this. Euclid's proof that there are infinitely many prime integers works equally well in polynomial rings. Suppose S is a finite set of irreducibles. Multiply the elements of S together and add one. The result is a polynomial of positive degree not divisible by any element of S. Thus, S cannot contain all irreducible polynomials. It follows, once more, that there are infinitely many irreducibles. Let x be areal number and 1l"(x) be the number of positive prime numbers less than or equal to x. The classical prime number theorem states that
2.
Primes, Arithmetic FUnctions, and the Zeta FUnction
13
n(x) is asymptotic to xl log(x). Let d be a positive integer and x = qd. We will show that the number of monic irreducibles P such that JPI x is asymptotic to xl logq{x) which is clearly·in the spirit of the classical result. Define ad to be the number of monic irreducibles of degree d. Then, from Equation 2 we find 00
(A(S)
= IT (1 d=l
If we recall that (A(S) = 1/(1 - ql-s) and substitute u = q-S (note that lui < 1 if and only if !Jt(s) > 1) we obtain the identity
Taking the logarithmic derivative of both sides and multiplying the result by it yields
~ = f:dad'ud 1 - qu
d=l
1-
Finally, expand Doth sides into power series using the geometric series and compare coefficients of un. The result is the beautiful formula,
Proposition 2.1.
Ldad =qn. din
This formula is often attributed to Richard Dedekind. It is interesting to note that it appears, with essentially the above proof, in a manuscript of c.P. Gauss (unpublished in his lifetime), "Die Lehre von den Resten." Sec Gauss [1], pages 608··611.
Corollary an
1"
"-
= nL...,. j.I.(d)q""J .
(3)
din
Proof. This formula follows by applying the Mobius inversion formula. to the formula given in the proposition. The formula in the above proposition can also be proven by means of the algebraic theory of finite fields. In fact, most books on abstract algebra contain the formula and the purely algebraic proof. The zeta-function approach has the advantage that the same method can be used to prove many other things as we shall see in this and later cha.pters. The next task is to wdte an in a way which makes it easy to see how big it is. In Equation 3 the highest power of q that occurs is qn and the next highest power that may occur is q~ (this occurs if and only if 21n. All the other terms have the form where m ::; J. 'I'he total number of terms is
14
Michael Rosen
Ldln IfL( d) Il which is easily seen to be 2t , where t lS the number of distinct prime divisors of n. Let PbP2, .. . ,Pt be the distinct primes dividing n. Then, 2t ::; PtP2 ... p! :::; n. Thus, we have the following estimate:
Using the standard big 0 notation, we have proved the following theorem. Theorem 2.2. ('l'he prime number theorem for polynomials) Let D' n denote the number of monic irreducible polynomials in.4 = IF[TJ of degree n. Then, a,., = n +0
(-;:qt) .
Note that if we set x = qn the right-hand side of this equation is x/logq(x) + O( JX/logq(x) which looks like the eonjectured precise form of the classical prime number theorem. This is still not proven. It depends on the truth of the Riemann hypothesis (which will discussed later). We now show how to use the zeta function for other counting problems. What is the number of square-free monics of degree n? Let this number be bn . Consider the product
(4) As usual, the product is over all monic irreducibles P and the sum is over
all monics f. We will maintain this COllvention unless otherwise stated. The function r5(f) is 1 when f is square-free, and 0 otherwise. Tius is an easy consequence of unique factorization in A and the definition of q-S once again, the right-hand square-free. Making the substitution u side of Equation 4 becomes 2::::=0 bnu n . Consider the identity 1 + w 0= (1 w 2)!(1 - w). If we substitute w IPI-s and then take the product over all monic irreducibles P, we see that the left-hand side of Equa.tion 4 is equal to (A(S)/(A(2s) (1 ql-2s)!(1 ql-S). Putting everything in terms of u leads to the identity
Finally, expand the left-hand side in a geometric series and compare the coefficients of un on both sides. VVe have proven-
Proposition 2.3. Let bn be the n'umber of square-free monics in A of degree n. Then b1 = q and far n > I, bn qn(l _ q-l). It is amusing to compare this result with what is known to be true in Z. If Bn is the number of positive square-free integers less than or equal
2.
Primes, Arithmetic F\mctions, and the Zeta Function
15
to n, then limn~oo Bn/n = 6/1f 2 . In less precise language, the probability that a positive integer is square-free is 6/1T 2 . The probablity that a monic polynomial of degree n is square-free is btJqn, and this equals (1 _ q-l) for n > 1. Thus the probabilty that a monic polynomial in A is squareq-l). Now, 6/1T 2 = 1/((2), so it is interesting to note that free is (1 (1 q-I) l/(A (2), This of course, no accident and one can give good heuristic reasons why this should occur. The interested reader may want to find these reasons and to investig'a.te the probablity that a polynomial be cube-free, fourth-power-free, etc. Our next goal is to introduce analogues of some well-known numbertheoretic functions and to discuss their properties. We have already introduced (flU). Let fLU) be 0 if I is not square-free, and (_l)t if f is a constant times a product of t distinct monic irreducibles. This is the polynomial version of the Mobius functlon. Let d(J) be the number of monic divisors of f and 0-(1) .EDit Ig! where the sum is over all monic divisors
of I. These functions I like their classical counterparts, have the property of being multiplicative. More precisely, a complex valued function A on A-{O} is called multiplicative if AUg) = >.(1),\(g) whenever f and 9 are relatively prime. \Ve assume A is 1 on IF"'. Let
f
aP{l p~2 ... Pt'
be the prime decomposition of I. If A is multiplicative)
A(1)
A(Pfl »,(P;2) ... A(Pte ,).
Thus) a multiplicative function is completely determined by its values on prime powers. Using llluitiplicativity, one can derive the following formulas for these functions. Proposition 2.4. Let the prime decomposition of
I
be given as above.
Then, if?(J)
III II (l_IPI~l),
dU)
(el
CJ(f)
[PIleI+! - 1 IPll-1
PI!
+ 1)(ez + 1) ... (et + 1). IHdez+i IPzl-1
1
IPtle t +l ~ 1
"'IPJ=-1
Proof. The formula for iP{n) has already been given in Proposition 1.7. If P is a monic irreducible, the only monic divisors of pe are 1, P, p2 I • . . , pe so d( pe) e + 1 and the second formula follows. By the above paragraph, CJ(PC) = 1 + IPI + IP l2 + ... [PI" (IPle+! - l/([PI 1), and the formula for CJ(J) also follows.
As a final topic in this chapter we shall introduce the notion of the average values in the context of polynomials. Suppose he:!:) is a complexvalued function on N, the set of positive integers. Suppose the following
16
Michael Rosen
limit exists lim
~~ h(n) = L
11.--+00 12
0;.
k=l
We then define 0; to be the average value of the function h. For example, suppose h(n) = 1 if n is square-free and 0 otherwise. Then, as noted above, the average value of h is known to be B/7r 2. The sum 'E~""1 h(k) sometimes grows too fa.. 0 we define 1 Aven(h) h(f). . =qn f
monic
deg(f)=n
This is clearly the average value of h on the set of monic polynomials of degree n. We define the average value of h to be lim n --+ oo Aven(h) provided this limit exists. This is the natural way in which average values arise in the context of polynomials. It is an exercise to show that if the average value exists in the sense just given, then it is also equal to the following limit: 1 lim h(f). .. '+qn n-HlO 1 + q + f monic
deg(f)::;n
As we pointed out above, this limit does not always exist. However, even . when it doesn't exist, one can speak of the average rate of growth of h(f). Define H(n) to equal the 811m of h(f) over all monic polynomials of degree n. As we will see, the function H(n) sometimes behaves in a quite regular manner even though the values h(J) va.ry erratically. Instead of approaching these problems directly we use the method of Carlitz which uses Dirichlet series. Given a function h as above, we define the associated Dirichlet series to be
2:
h(f)
[monlc
If Is
(5)
In what follows, we will work in a formal manner with these series. If one wants to worry about convergence, it is useful to remark that if jh(J)l
2.
Primes, Arithmetic Functions, and the Zeta Function
17
O(jflf:l), then Dh(S) converges for !R(s) > 1 + p. The proof just uses the comparison test and the fact that (A (s) converges for !R( s) > 1, 'I'he right-hand side of 5 is simply 2.:::';'0 H(n)ut! l so the Dirichlet series in s becomes a, power series in 'U whose coefficients are the averages H(n). 'lb see how this is useful, recall the function d(f) which is the number of monic divisors of f. Let D(n) be the sum of d(f) over all monies of degree n (hopefully, this notation will not cause to
