2 Asymptotic density and Dirichlet density

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1 8.785: Analytic Number Theory, MIT, sring 2007 (K.S. Kedlaya) Primes in arithmetic rogressions In this unit, we first rove Dirichlet s theorem on rimes in arithmetic rogressions. We then rove the rime number theorem in arithmetic rogressions, modulo some exercises. Dirichlet s theorem For short, I will say an arithmetic rogression is eligible if it has the form m, m + N, m + 2N,... where gcd(m, N) = ; it is equivalent to ask that any two consecutive terms are relatively rime. Theorem (Dirichlet). Any eligible arithmetic rogression of ositive integers contains infinitely many rimes. There are a few secial cases where one can rove this directly, but otherwise algebraic methods cannot touch this roblem. Dirichlet s idea was to rove, in some aroriate quantitative sense, that the rimes distribute themselves equally among the eligible arithmetic rogressions with a articular difference; this goes back to Euler s roof of the infinitude of rimes using the Riemann zeta function. 2 Asymtotic density and Dirichlet density In order to seak quantitatively about the distribution of certain tyes of rimes, or integers in general, we need some sort of measure theory on the set of rimes or the set of integers. Note that Lebesgue-tye measure theory is not an otion for countable sets: we can only hoe to make finitely additive measures. For S T two sets of ositive integers, with T infinite, the uer natural density and lower natural density of S in T are defined as #{n S : n N} #{n S : n N} lim su, lim inf. N #{n T : n N} N #{n T : n N} Of course the uer density is never less than the lower density. If they coincide, we call the common value the natural density (or asymtotic density) of S in T. Many interesting sets fail to have a natural density (e.g., see exercises). We get a less restrictive notion of density by using Dirichlet series. For S T two sets of ositive integers, with n T n divergent, the uer Dirichlet density and lower Dirichlet density of S in T are defined as s s n n lim su n S, lim inf n S. s + n T n s s + n T n s If they coincide, we call the common value the Dirichlet density of S in T.

2 Let us make this exlicit in two cases of interest. Recall that ζ(s) has a simle ole of residue at s =, so as s +, (s ) n s = + o(). Hence if T = N, then the Dirichlet density of S is given by lim (s ) n s s + log ζ(s) + log(s ) = log(s ) + ns = O(). ns Moreover, n=2 = O(), so s = log(s ) + O(). Hence if T is the set of rimes, then the Dirichlet density of S is given by n= if the limit exists. Taking logarithms, we see that as s +, n S s S lim s + log(s ) n= if the limit exists. Here are some easy facts about density. (If I don t secify natural vs. Dirichlet, I mean that the statement holds if you make a choice and use it consistently throughout the statement.) Any finite set has density 0 in any infinite set. Density is a finitely additive measure: if S,..., S m are disjoint subsets of T with densities ζ,..., ζ m in T, then their union has density ζ + + ζ m in T. Corollary: two subsets of T whose combined density exceeds must have infinite intersection. If S has density ζ in N, then for any ositive integer n, ns = {ns : s S} has density ζ/n. I can t hel mentioning a fun examle of the additivity of densities. Let α, β be ositive irrational numbers with /α + /β =. Put S α = { nα : n N} S β = { nβ : n N}. Then S α, S β have natural densities /α, /β. The fact that these add u to is exlained by the beautiful result (Beatty s theorem) that S α, S β are disjoint and their union is N! (If you ve never seen this before, I recommend this as an amusing exercise.) 2

3 Lemma 2. Let S T be subsets of N such that S has natural density ζ in T. Then S also has Dirichlet density ζ in T. Proof. See exercises. (The converse is false; also see exercises.) To rove Theorem, we will rove the following. Theorem 3. For any ositive integers m, N with gcd(m, N) =, the set of rimes congruent to m modulo N has Dirichlet density / in the set of all rimes (hence is infinite). 3 L-functions and discrete Fourier analysis For α a Dirichlet character of level N, we can write log L(s, α) = α( n ) ns n= for Re(s) > ; as s +, we have log L(s, α) = α() s + O(). For α nonrincial, we know that L(s, α) is holomorhic and nonvanishing at s =, so α() s = O(), whereas for α 0 the rincial conductor of level N, we saw above that α 0 () s = log(s ) + O(). (b) (Orthogonality of characters) If α, α 2 Ĝ, then { G α = α 2 α (g)α 2 (g) = 0 α = α 2. At this oint it may be clear how to roceed: form a certain linear combination of the log L(s, α) to isolate m (N) s, and comare the asymtotic contributions of log(s ). The fact that we can do this amounts to what is sometimes called discrete Fourier analysis; if you refer, it is the reresentation theory of the finite abelian grou (Z/NZ). Theorem 4 (Discrete abelian Fourier analysis). Let G be a finite abelian grou, and let Ĝ be the character grou (or dual grou) of G, i.e., the grou of homomorhisms G C. (a) The order of Ĝ is equal to the order of G. g G 3

4 (c) If g, g 2 G, then { G g = g 2 α(g )α(g 2 ) = 0 g = g 2. χ Ĝ Proof. (a) If G = G G 2, then clearly Ĝ = Ĝ Ĝ 2. Since every finite abelian grou G is a roduct of cyclic grous, we may reduce to the case where G is cyclic, and then the result is clear. (For G = (Z/NZ), we can make this more exlicit: we can use the Chinese remainder theorem to slit N into distinct rime-ower factors, then use the fact that (Z/ n Z) is cyclic unless = 2, in which case it is {±} times a cyclic grou.) (b) We saw this argument once before, but here it goes again: the left side is invariant under multilication by α (h)α 2 (h) for any h G, because there is no difference between summing over g or over gh. If α = α 2, then we can make that multilier different from by choosing suitable h. So the sum vanishes if α = α 2. If α = α 2, each summand is equal to because characters of finite grous takes values which are roots of unity. (c) See exercises. So now it is clear what to do: given a choice of m corime to N, taking sums of α over all Dirichlet characters of level N, we obtain α(m) log L(s, α) = α()α(m) s + O() χ χ = s + O() m (N) as s +. On the other hand, α(m) log L(s, α) = log L(s, α 0 ) + α(m) log L(s, α) χ χ =χ 0 = log(s ) + O(). This yields Theorem 3. 4 The rime number theorem in arithmetic rogressions The roof of Dirichlet s theorem only uses information about the behavior of the L(s, α) near s =. Using the fact that L(s, α) = 0 on the entire line Re(s) =, we can rove a much stronger result. 4

5 Theorem 5 (Prime number theorem in arithmetic rogressions). For m, N ositive integers with gcd(m, N), the set of rimes congruent to m modulo N has natural density /. x In other words, the number of rimes x with m (mod N) is asymtotic to β( N) log x as x. Proof. Given what we now know, this is a straightforward adatation of our roof of the rime number theorem. For α a Dirichlet character of level N, define β χ (x) = α() log. Given a choice of m corime to N, ut β m (x) = α(m)β χ (x) χ = log. As in the roof of the rime number theorem, if we rove that the imroer integral β m (x) x dx x 2 converges, we may then deduce that β m (x) β( x as desired. N) It suffices in turn to check that for α rincial, β χ (x) x dx x 2 x x: m (N) converges, and for α nonrincial, β χ (x) 2 dx x converges. The former is an immediate consequence of the corresonding fact for β (which we roved in the unit on the rime number theorem), since β and β χ differ in only finitely many terms. For the latter, see exercises. As with the roof of the rime number theorem, we get very little information about the error term, i.e., the difference between the actual number of rimes x with m x (mod N) and the asymotic count β(. That becomes a roblem if, for instance, we N) log x want to know how long it takes to find one rime in an arithmetic rogression. To address this, we must first get better results on zero-free regions for the L(s, α), then make a better analytic argument to take advantage of the imroved analytic information. We turn to this in the next few lectures. 5

6 Exercises. Prove Lemma 2. (Hint: use artial summation.) 2. Let S be the set of ositive integers which have first digit when written in base 0. (a) Comute the uer and lower natural density of S, and verify that S does not have a natural density. (b) Prove that S has a natural Dirichlet density, and comute it. Otional (not to be turned in): generalize to an arbitrary base b 3. Even more otional: rove the analogous result for the set of rimes with first digit in base b. 3. Prove that there exists a constant c such that = log log x + c + o(). x (Hint: you established asymtotics for artial summation.) 4. (a result of Mertens; tricky, otional) In the revious exercise, rove that ( c = ϑ + log( ) + ), 5. Deduce oint (c) of Theorem 4 from oints (a) and (b). (Hint: form the matrix A with rows indexed by g G, columns indexed by α Ĝ, and entries α(g). Then comare AA with A A, where denotes conjugate transose. Or if you refer, rove that the dual of Ĝ is canonically isomorhic to G.) α 6. Prove that χ (t) dt converges for α nonrincial, by alying the Tauberian theorem t 2 from the unit on the rime number theorem. (Hint: use the fact that L(s, α) = 0 for Re(s) to argue that L (s, α)/l(s, α) is holomorhic in a neighborhood of Re(s). There will be an extra term to deal with, just as there was a term I neglected in the original notes from the rime number theorem unit; see the corrected notes online.) x where ϑ is Euler s constant. Then deduce that log χ ( ) e. log x x on a revious homework. Aly 6