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Tobiaz Chavarria Finite expression for L(1; χ) Tob´ıasChavarr´ıaCastro Universidad de Costa Rica Abstract In 1839 the german mathematician Johann Peter Gustav Lejeune Dirichlet proof the prime number theorem in arithmetic progressions, this document provides a introduction to the theory of Dirichlet L-function, and a closed formula to compute the L-function for a particular case, which is important in the proof of the theorem. Introduction The theory of L-functions, plays a central role in number theory since Dirichlet's proof his famous theorem, this functions have the special property as the Riemann zeta-function, that their series representation they can also be described by an Euler product, i.e., a product taken over prime numbers. The famous Riemann zeta-function: 1 X 1 Y ζ(s) = = (1 − p−s)−1 ns n=1 p Theorem. (L.Dirichlet) Let q ≥ 1 and a ≥ 1 such that (a; q) = 1: Then there are infinity many primes p ≡ a(mod q) and more precisely: X 1 1 = log log(N) + O(1) p '(q) p≤N; p≡a(modq) An immediately consequence of this theorem is that in every arithmetic progression: a; a + q; a + 2q; ::: includes infinitely many primes. 1 A “first” proof of this result was only for the case when q is a prime number, that is the case for wich we are going to give a closed formula for L(1; χ). For this proof Dirichlet introduced the arithmetic functions called Dirichlet characters. Definition 1. Dirichlet characters mod q. A Dirichlet character modulo q is any function χ from the integers Z to the complex numbers C∗ such that χ has the following properties: • χ(n + q) = χ(n) for all n 2 Z • χ(nk) = χ(n)χ(k) for all n; k 2 Z • χ(n) 6= 0 if and only if (n; q) = 1 Notice that this is a completely multiplicative function. In practice the construction of χ can ∗ be done as follows: Let q a prime number, then the multiplicative group modulo q, ie (Z=qZ) is cyclic with generator, say g, called a primitive root modulo q, then we pick a complex number χ(g) with χ(g)q−1 = 1 . Then we define χ(gm) = χ(g)m.The simplest character is the trivial character χ0(n) = 1 if (n; q) = 1 and χ0(n) = 0 if (n; q) > 1. Definition 2. Dirichlet L-Functions Let χ a Dirichlet character modulo q. We define the Dirichlet L-function associated to χ by the formula: 1 X χ(n) L(s; χ) = ns n=1 An important observation of this definition is the convergence of this series, since jχ(n)j = 1 for all n 2 Z, we have jχ(n)n−sj <= n−s, this implies that the terms of L(s; χ) are dominated in absolute value by the corresponding terms in ζ(s). Thus L(s; χ) converges and is continuous for s > 1. Moreover since χ is completely multiplicative we have a product formula for L(s; χ) in exactly the same way as for ζ(s). Namely: Y L(s; χ) = (1 − χ(p)p−s)−1 p Dirichlet considered what can happen to log(L(s; χ)) as s ! 1. The character χ0 gives L(1; χ0) = (1 − q−s)ζ(s) and we know that this tends to 1 as s ! 1. The rest of the proof it will suffice to show that, for each character χ other than the trivial, log(L(s; χ)) is bounded as s ! 1, the characters are separated into the ones that take complex values (an they come in pairs, with their conjugates), and the real character with χ(g) = −1. Dirichlet could easily consider the complex characters, and see that L(1; χ) 6= 0 and note that that the real one, is much harder to treat. Now we are going to prove that L(1; χ) 6= 0 when χ is a real Dirichlet character modulo q, ie, χ(N) = f1; −1; 0g, in this case the Dirichlet character coincides with the Legendre symbol: 8 1 if a is quadratic residue mod q a <> ( ) = −1 if a is a quadratic nonresidue mod q q :>0 if qja: To do this, first we need to define a Gauss sum, and some results on this. 2πi Definition 3. Gauss Sum. Let ζ = e q the Gauss sum of a Dirichlet Character χ is: q−1 X g(χ) = χ(k)ζk k=1 Lemma 1. The value of the quadratic Gauss sum g(χ) is given by: (p q if q ≡ 1(4) g(χ) = p i q if q ≡ 3(4) 2pii Lemma 2. Let g(χ) the the classical Gauss sum, χ the Legendre symbol, and ζ = e p , q prime. Q N Q R − Define Q = N (1 − ζ ) R(1 − ζ ) 1 Where N runs over all non-quadratic residues modulo p and R runs over all quadratic residues modulo p. Then: Q = exp(L(1; χ)g(χ))) Pp−1 a Proof. Let g(χ) be the classical Gauss Sum a=1 χ(a)ζ , where χ is the Legendre symbol and 2πi ζ = e p . Define: Y Y Q = (1 − ζN ) (1 − ζR)−1 N R Where N runs over all non-quadratic residues modulo p and R runs over all quadratic residues modulo p. P1 xk Remember that: − log(1 − x) = k=1 k Y Y log(Q) = log( (1 − ζN ) (1 − ζR)−1) = N R X X = log(1 − ζN ) + − log(1 − ζR) N R 1 1 X X ζNk X X ζRk = − + k k N k=1 R k=1 1 X X ζNk X ζRk = ( χ(N) + χ(R) ) k k k=1 N R 1 ! X χ(k) X X = χ(k)χ(N)ζNk + χ(k)χ(R)ζRk k k=1 N R 1 ! X χ(k) X X = χ(kN)ζNk + χ(kR)ζRk k k=1 N R 1 p−1 X χ(k) X = χ(a)ζa k k=1 a=1 P1 χ(k) Pp−1 a Since L(1; χ) = k=1 k and g(χ) = a=1 χ(a)ζ , we have: Q = exp(L(1; χ)g(χ))) Since Q 6= 0 and g(χ) 6= 0, this prove that L(1; χ) is bounded and to prove that L(1; χ) 6= 0 is equivalent to proving that Q 6= 1. For this, Dirichlet had recourse to a result that had been proved by Gauss in his work on cyclotomy. This is that, for an ,indeterminate x, Y 1 p (x − ζR)= [Y (x) − qW (x)] 2 R and Y 1 p (x − ζN )= [Y (x) + qW (x)] 2 N Where Y (x);W (x) 2 Z[x]. Assuming this we have the identity: q−1 1 Y [Y 2(x) − qW 2(x)] = (x − ζa) = Φ (x) 4 q a=1 Where Φq(x) denotes the the q-th cyclotomic polynomia, and since q is a prime number we have: q−1 q−2 Φq(x) = x + x + ::: + x + 1 Now putting x = 1, we obtain integers, Y (1) = Y and W (1) = W , which satisfies the Diophantine equation: Y 2 − qW 2 = 4q We note that W 6= 0 since 4q is not a perfect square. Therefore expressing Q in terms of this polynomials we have: p Y Y Y + W q Q = (1 − ζN ) (1 − ζR)−1 = p Y − W q N R Since W 6= 0, we have the desired conclusion that Q 6= 1: We will show by an example the usefulness of this formula. Example. Consider the non-trivial Dirichlet character modulo 3. First note, A = ( Z )∗ = f1; 2g, since A ∼= A , where A is the set of Dirichlet characters. 3Z A = fχ0; χg is cyclic group generated by χ, ie, A =< χ >, with: −! ∗ −! ∗ Z C χ (k): Z C χ(k): k ≡ 2(3) −! −1 0 k −! 1 k ≡ 1(3) −! 1 Then χ match with the Legendre symbol,and: 1 X χ(n) L(1; χ) = = n n=1 1 X 1 1 = ( − ) 3k + 1 3k + 2 k=0 1 X 1 = (3k + 1)(3k + 2) k=0 This series looks easy to compute, but is not. p 2πi −1 3 3 Now applying the lemma, with ζ = e = 2 + 2 i, and χ the Legendre symbol modulo 3 we have: Y Y (1 − ζN ) (1 − ζR)−1) = N R = (1 − ζ2)(1 − ζ)−1 πi = (1 + ζ) = e 3 p And g(χ) = i 3 , since 3 ≡ 3(4) Therefore: πi p e 3 = exp(i 3)L(1; χ) and π L(1; χ) = p 3 3 The importance of this result is because in the proof of the Dirichlet Theorem he needs to prove that this value is not zero, and this is a proof in a special case, moreover is a closed formula to compute this value. In the book Multiplicative Number Theory by Harold Davenport, we found a more specific formula for this value, he split the formula in the case when p ≡ 1(4) and p ≡ 3(4) and gets: 8 π <− 3 if p ≡ 3(4) p 2 L(1; χ) = log(Q) p : p if p ≡ 1(4) Where: Q sin( πN ) Q = N p Q πR R sin( p ) ..
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