Advanced Number Theory Note #8: Dirichlet's theorem on primes in arithmetic progressions 29 August 2012 at 19:01

Transcription

1 Advanced Number Theory Note #8: Dirichlet's theorem on primes in arithmetic progressions 29 August 2012 at 19:01 Public In this note, which is intended mainly as a technical memo for myself, I give a 'blow-by-blow' account of a proof of Dirichlet's theorem based on H. Shapiro, 1952, On primes in arithmetic progression II, Ann. of Math. 52: Dirichlet's famous theorem says that any arithmetic progression h, h+ k, h + 2k, h + 3k,..., h + nk,... contains infinitely many primes if gcd(h, k) = 1. The proof is very intricate and makes constant use of properties of Dirichlet characters (see Advanced Number Theory Note #7). In what follows, the positive integer k represents a fixed modulus and h is a fixed integer such that gcd(h, k) = 1. The φ(k) Dirichlet charaters are denoted by with the first denoting the principal character. For non-principal characters χ we write L(1, χ) and L'(1, χ) for the sums of the following series:

2 The convergence of each of these series was demonstrated in Advanced Number Theory Note #7, where it was also shown that L(1, χ) 0 if χ is realvalued. I have divided the demonstration into six steps, each with its own section below. The final sixth step ties all the previous steps together to show how they together constitute a coherent proof of Dirichlet's theorem Step 1 The first step is to prove the following rather complicated formula for the sum of logp/p, where the sum is extended over all those primes p x which are congruent to h modulo k. The formula says that for x > 1 we have To prove this, recall that in Advanced Number Theory Note #5, equation (8), it was shown on the basis of Shapiro's Tauberian theorem that where the sum is extended over all primes p x. To obtain the formula in (1) we need to 'extract' from (2) those terms corresponding to primes of the form

3 p h (mod k) This extraction can be achieved by using the orthogonality relations for Dirichlet characters obtained in Advanced Number Theory Note #7, equation (5), which said the following in relation to a reduced residue system modulo k: Note that this is valid for gcd(n, k) = 1. We now take m = p and n = h, where gcd(h, k) = 1, then multiply both sides by logp/p and sum over all p x to get In the sum on the left of (3) we take out the terms involving only the principal character χ 1 and rewrite (3) in the form

4 But χ 1 *(h) = 1 and χ 1 (p) = 0 unless gcd(p, k) = 1, in which case χ 1 (p) = 1. Therefore the first term on the right of (4) can be written as where the O(1) term arises from the fact that there are only a finite number of primes which divide k. Combining (5) with (4) we get

5 We can use (2) to replace the first term on the right by logx, then dividing through by φ(k) we get (1). QED Step 2 The next step is to prove the following formula for the sum of χ(p)logp/p which appears on the right hand side of (1) above. For x > 1 and nonprincipal χ we have To prove this, begin with the sum where Λ(n) is Mangoldt's function. Recall (see Advanced Number Theory Note #1) that Λ(n) = logp if n = pᵃ and Λ(n) = 0 whenever n is not a prime power. We therefore deduce that

6 We now separate out the terms corresponding to a = 1 and write But the second term on the right hand side of (7) is dominated by so (7) gives us

7 Now recall from Advanced Number Theory Note #1 that Λ(n) = d n μ(d)log(n/d) Therefore In the sum on the right hand side we write n = cd and use the multiplicative property of χ to get Since x/d 1, in the sum over c we can use formula (9) in Advanced Number Theory Note #7 to get Equation (9) above now becomes

8 The sum inside the big-oh term is

9 where the third equality follows from formula (9) in Advanced Number Theory Note #3. Therefore (10) above becomes

10 and substituting this into (8) gives (6). QED Step 3 The next step is to prove the following formula for L(1, χ) times the sum of μ(n)χ(n)/n (this sum appears on the right hand side of (6) above). For x > 1 and non-principal χ we have: To prove this, we use the generalised Möbius inversion formula obtained in the first section of Advanced Number Theory Note #3, which if α is a completely multiplicative arithmetical function says the following:

11 We take α(n) = χ(n) and F(x) = x to obtain where By equation (8) in Advanced Number Theory Note #7, we can write G(x) as G(x) = xl(1, χ) + O(1) Using this in (12) above we find

12 Dividing through by x gives (11). QED Step 4 The next step is to prove that if χ is non-principal and L(1, χ) = 0 then we must have To prove this we will again make use of the generalised Möbius inversion formula invoked in Step 3. This time we take α(n) = χ(n) and F(x) = xlogx to obtain where

13 Now we use formulas (8) and (9) in Advanced Number Theory Note #7 to get where the second equality follows from the fact that L(1, χ) = 0 by assumption here. Therefore (14) gives us

14 We have already seen (towards the end of Step 2) that the big oh term on the right is O(x). Therefore we have Upon dividing through by x we get (13). QED Step 5 In Advanced Number Theory Note #7 we showed that L(1, χ) 0 for real-valued non-principal χ. It is now necessary to show that L(1, χ) 0 for all non-principal χ, whether complex-valued or real-valued. To do this we let N(k) denote the number of non-principal complex-valued characters χ mod k such that L(1, χ) = 0. If L(1, χ) = 0 then L(1, χ*) = 0 and χ χ* since χ is not real. Therefore the complex-valued characters χ for which L(1, χ) = 0 occur in conjugate pairs, so N(k) is even.

15 The goal in this step is to prove that N(k) = 0 and this will be deduced from the following asymptotic formula: For x > 1 we have If N(k) 0 then N(k) 2 since N(k) is even, therefore the coefficient of logx in (15) is negative and the right hand side of (15) - as x. This is a contradiction since all the terms on the left hand side are positive. Therefore (15) implies that N(k) = 0. The proof of (15) will be based on (1) and (13) above. We first use (1) with h = 1 to get In the sum over p on the right hand side we use (6) in Step 2 above, which says that If L(1, χ r ) 0, (11) in Step 3 above shows that the right hand side is O(1) (because we can divide (11) by L(1, χ) to deduce that the sum is O(1), which is the sum appearing on the right hand side above). This situation corresponds to setting N(k) = 0 in (15).

16 However, if L(1, χ r ) = 0 then (13) implies and there are N(k) of these terms, so the sum on the right of (16) becomes and (16) itself becomes (15). This proves (15). QED Step 6 We are now in a position to pull everything together and prove Dirichlet's theorem, which can be formally stated as follows: If k > 0 and gcd(h, k) = 1 there are infinitely many primes in the arithmetic progression nk + h, n = 0, 1, 2,.... Dirichlet's theorem is a consequence of the following asymptotic formula: If k > 0 and gcd(h, k) = 1 we have, for all x > 1,

17 where the sum is extended over those primes p x which are congruent to h modulo k. Since logx as x, this formula implies that there are infinitely many primes p h (mod k), and thus infinitely many primes in the progression nk + h, n = 0, 1, 2,.... The proof of (17) consists of Steps 1 to 5 above, as follows. In Step 1 we proved equation (1), which clearly implies (17) if for all non-principal χ, because then the second term on the right hand side of (1) collapses to O(1) as follows:

18 But in equation (6), which we proved in Step 2, the sum in (18) is expressed in a form which is not extended over primes, and (6) implies (18) if it can be shown that But (19) is exactly what can be deduced from (11) in Step 3 above if L(1, χ) 0 because we can cancel L(1, χ) in (11) as long as it is non-zero. This is why Dirichlet's theorem depends ultimately on the non-vanishing of L(1, χ) for all non-principal characters. We proved in Advanced Number Theory Note #7 that L(1, χ) is non-vanishing for real-valued non-principal χ, and we proved that it is nonvanishing also for complex-valued non-principal χ in Steps 4 and 5 above.

19 Thus, we can obtain (19) from (11), then use (19) in (6) to obtain (18), and finally use (18) in (1) to obtain (17). This proves Dirichlet's theorem. QED