DIRICHLET S THEOREM ON PRIMES IN ARITHMETIC PROGRESSIONS. 1. Introduction

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1 DIRICHLET S THEOREM ON PRIMES IN ARITHMETIC PROGRESSIONS INNA ZAKHAREVICH. Introduction It is a well-known fact that there are infinitely many rimes. However, it is less clear how the rimes are distributed throughout the integers. A natural question to ask is whether there are infinitely many rimes in arithmetic rogressions: sequences of the form a, a + q,a + 2q,a + 3q,... Clearly, there will be none in some sequences. If a = 4 and q = 8 then all elements of the sequence will be divisible by 4, so there clearly can t be any rimes. In other rogressions it is less clear. If a = 5 and q = 7, then there are many rimes: the first 7 elements of the sequence are 5, 2, 9, 26, 33, 40, 47, which already contains 3 rimes. It turns out that if no number other than divides all elements of the sequence, there will be infinitely many rimes in the sequence. More formally, Theorem. If a,q are ositive integers with (a,q) =, then there are infinitely many rimes in the sequence {a + n} n N. We seek to rove this theorem. To do so, we first introduce two objects, the Dirichlet series A(s,f) and the Dirichlet character, and rove some useful results about them. We will then use these to show that lim s a (mod q) s = which will show that there must be infinitely many rimes in the sum on the left. We assume some basic knowledge of number theory (the roerties of the grou of units modulo n) and comlex analysis (some familiarity with analytic functions). 2. Dirichlet Series Definition 2. A Dirichlet series is a sum of the form f(n) () A(s,f) = n, s The Riemann-Zeta Function, ζ(s), is the Dirichlet series where f(n) = for all n. The question of convergence for Dirichlet series is a comlicated one. However, if we restrict our attention to absolute convergence, we can quickly obtain a useful result: n=

2 Theorem 3. For each Dirichlet series A(s,f) = n= f(n)/ns there exists a unique number σ f R {± } such that the series A(s,f) is absolutely convergent for Re s > σ f, and not for Res < σ f. Proof. We follow the roof in []. Let S be the set of all α such that f(n) n α <. n= Since n s = n Re s is a monotonically decreasing function of Res, we know from the comarison test that the series n= f(n)n s is absolutely convergent for Res > α; thus if α S then all β > α are in S. Let σ f = inf S. Then by the definition of σ f, we know that there does not exist s with Re s < σ f such that n= f(n)n s is absolutely convergent. So we have shown that if S is nonemty we are done. If S is emty, set σ f =, and the theorem will hold for σ f. Notice that if we aly the integral test to the series for ζ(s), we get that σ f =. If we make some extra assumtions on f we can obtain a useful factorization of the series, known as the Euler roduct. The existence of such a factorization hints at the connection between these series and theorems about rimes; it is often useful to take the logarithm of such a roduct to obtain facts about sums over rimes. Proosition 4. If A(s,f) = n= f(n)n s and f(nm) = f(n)f(m) for all n,m then we have (2) A(s,f) = ( f() ) s for all s such that Re s > σ f. Proof. Notice that for Res > σ f k=0 f( k ) sk n= f(n), n s so the series on the left converges, and thus equals ( f()/ s ). Also, notice that for the first N rimes,..., N, we have f(n) N ( f( ) m) f(n) = e f( e N N ) n s s n= m= m n s e s n= e,...,e N e N N f(n) n s n= N + which can be made arbitrarily small since A(s,f) exists. Thus the two formulas converge to the same value, and we are done. 2

3 The Riemann Zeta function is the most famous Dirichlet series. It is useful in roving theorems about the distributions of rimes. Here we rove a simle bound on ζ(s): Proosition 5. For real s >, uniformly. ζ(s) = s + O() Proof. Notice that (n + ) s n+ u s du n s for n N. Summing over n we see n that ζ(s) u s du ζ(s) for s >. Comuting the integral and rearranging the inequality, we find that for s >. Thus s ζ(s) s + ζ(s) = s + O(). 3. Dirichlet Characters Definition 6. A Dirichlet character is a function : Z C such that () if n m (mod q) then (m) = (n), (2) (mn) = (m) (n) for all m,n Z, and (3) (n) = 0 if and only if (n,q). Although the definition of Dirichlet characters does not seem restrictive, it turns out that it is very simle to classify all Dirichlet characters. First, in order to simlify our exosition, we introduce some notation: Definition 7. We denote the cardinality of of (Z/nZ) by ϕ(n). For any integer a such that (a,n) = we denote the smallest integer k such that a k (mod n) by ord n (a). Definition 8. We denote the comlex number ex(2π/a), a rimitive a-th root of unity, by ζ a. Proosition 9. There are exactly ϕ(q) distinct. If b,...,b m are the generators of (Z/qZ) then the are exactly the functions defined by (a) = ζ k e ord q(b ) ζkmem ord q(b m) where 0 k j < ord q (b j ) if a b e b em m (mod q) and (a) = 0 otherwise. We will denote the character where k j = 0 for all j =,...,m by ɛ q. 3

4 Proof. Plugging n = m = into roerty 2 we see that ( ) = () 2 so () = 0 or. Since it can t equal 0 by roerty 3, () =. Suose that (a,q) =. Then a ϕ(q) (mod q), so (a) ϕ(q) = () =, so (a) must be a ϕ(q)-th root of. Notice that since (b ordq(b j) j ) =, we know that (b j ) = ζ k j ord q(b j ) for some 0 k j < ord q (b j ). Thus all must be of the desired form. Now fix k j for each b j, and define on the rest of (Z/qZ) by (b e b em m ) = (b ) k e (b m ) kmem. Define (a) = 0 if (a,q). Then by definition this function will satisfy the definition of a Dirichlet character. Thus all Dirichlet characters satisfy the given formulas. Notice that from the formula we see that there are m ord q(b j ) different Dirichlet characters. But m j= ord q(b j ) = #(Z/qZ) = ϕ(q) by definition. Theorem 0. For (a, q) =, (a) () = { ϕ(q) if a (mod q) 0 otherwise. Proof. Let b,...,b m be generators of (Z/qZ). If (,q) then the theorem is trivially true. So now suose that (,q) =. Let a = m j= be j j and let = m j= bf j j. Then ( m ) ( m ) (a) () = (b j ) e j (b j ) f j j= = m (b j ) f j e j j= j= If f j = e j (so a (mod q)) this is just = ϕ(q). Otherwise, suose that m is the largest index such that f m e m. Then if we let C = m j=m + ord q(b j ) then m (b j ) f j e j j= = C ord q(b ) k = k m = m j= ζ k j(f j e j ) ord q(b j ) = C = C = 0 ord q(b ) k = ord q(b ) k = ord q(b m ) k m = ord q(b m ) k m = m j= m j= ζ k j(f j e j ) ord q(b j ) k m = ζ k m (f m e m ) ζ k j(f j e j ζ (f m e m )ordq(b m ) ) ord q(b j ) ζ f m e m because ζ (f m e m )ordq(b m ) =. So we are done. 4

5 4. Proof of the Main Theorem Theorem 0 tells us that we can ick out one equivalence class modulo q in a sum of Dirichlet series by summing over A(s, ) with the correct coefficients. More secifically, we will take the logarithm of the Euler roduct decomositions of A(s, ) and take a linear sum of those. Since the sums in the logarithms will be over rimes, we will be able to make a sum that is over only rimes that are equivalent to a fixed a modulo q. All that will remain will be to show that those sums are infinite. Notice that by comarison with ζ, we see that σ f for A(s, ) is at most. We show the connection between log A(s, ) and an infinite sum over rimes: Theorem. Let be a Dirichlet character. Then for real s >, k () k ks = log A(s, ). Proof. We follow the roof in []. Take logarithms of both sides of (2). Then we find that log A(s, ) = log( () s ). Since () s <, we can use the ower series exansion for log( z) to get that log A(s, ) = k () k ks. (As in []) So we now have an infinite sum over rime owers. To reduce this to a sum over rimes, we note that for s k ks ks = 2s s ( ) k=2 k=2 which converges by comarison with /n 2. Thus from the above we see that () = log A(s,χ s q ) + O(). Then for integers a,q such that (a,q) = and s > (3) (a) log A(s, ) = ϕ(q) ϕ(q) () (a) s + O() = a (mod q) s + O(). So we have now written a sum over rimes equivalent to a modulo q. It remains to give an estimate of this sum. First, we give an asymtotic for A(s,ɛ q ), which has a ole at s = : 5

6 Proosition 2. Proof. Notice that for s > A(s,ɛ q ) = s + O(q). A(s,ɛ q ) = ζ(s) q ( s ). So if we could show that q ( s ) is O(q), we would be done. Indeed, ( s ) ( 2 s ) 2 q = O(q). q q q So we are done. Now we can estimate the Dirichlet series for the other characters. Proosition 3. If is a Dirichlet character not equal to ɛ q, then A(s, ) is a continuous function whenever Re s > 0. Proof. We follow [3]. Let S(x) = n x (n). Notice that if is a Dirichlet character not equal to ɛ q, we know that l+q (k) = k=l = = q (k) = ord q(b ) e = ord q(b ) e = ord q(b ) e = ord q(b m ) e m = ord q(b m ) e m = ord q(b m) e m= (b e b em m ) ord q(b m) (b e b e m m ) so S(x) is uniformly bounded by max k k<q j= χ q(j). Now observe that ( ) A(s, ) = S(n) n s (n + ) s = s n= S(n) n= n+ n e m= (b em m ) (b e b e m m ) (b m ) ordq(bm) (b m ) x s dx = s S(x)x s. Since S(x) is uniformly bounded the above integral is analytic and converges for all s with Re s > 0, so A(s, ) is an analytic function on the half-lane Res > 0. 6 = 0.

7 In articular, this means that A(s, ) tends to a finite limit as s. If we could show that A(, ) 0 then log A(, ) would be finite, and by considering (3) we would know that = s ϕ(q) log + O(log q). s a (mod q) This tends to infinity as s, so there must be infinitely many rimes in the left-hand sum. This is Dirichlet s theorem: Theorem 4. If a,q are ositive integers such that (a,q) = then there are infinitely many rimes in the sequence {a + kq} k N. We finish the roof by showing the following: Theorem 5. If is a Dirichlet character not equal to ɛ q, then A(, ) 0. Proof. We follow [3]. First we will rove a lemma: Lemma 5.. For real s > 0, Proof. Recall that log A(s, ) = A(s, ). k ( k ) ks. Summing over and using theorem 0 we see that log A(s, ) = ( k ) ks = ϕ(q) k χq For real s the right-hand side is nonnegative. Thus A(s, ). k (mod m) k ks. Now suose that is a comlex character (so there exists some n Z such that (n) / R). From the definition of A(s, ) we see that A(s, ) = A(s, ). Thus if A(, ) = 0 then A(, ) = 0. In roosition 2 we saw that A(s,ɛ q ) had a simle ole at s = ; A(s,χ) is analytic in the right half-lane for χ ɛ q. Thus if A(, ) has a zero at s = so does A(, ), and so it will have a double zero but only a single ole at s =, so the roduct will also be zero. However, from lemma 5. we saw that the roduct is at least : contradiction. So for all comlex characters, A(, ) 0. 7

8 So now we simly need to rove the statement for real, meaning (n) = ± for all n relatively rime to q. To facilitate this, we follow the exosition of a roof due to de la Vallée Poussin in [3] and rove the following lemma: Lemma 5.2. Suose f is a nonnegative function on N such that if (m,n) = then f(mn) = f(m)f(n). Suose also that there is a constant c such that f( k ) < c for all rime owers k. Then n= f(n)n s converges for all real s >, and n= f(n) n s = ( + f( k ) ks Proof. Fix s >, and let a() = f(k ) ks. Then a() < c = c s ks < s 2c s. Thus, using the fact that + x < e x for x > 0 we have + a()) < N( e a() = ex a(). N N But we know that N a() < 2c s = M. But from the multilicativity of f we see that N ( + a()) < ex M n= f(n)n s < N for all N. Since f is nonnegative, we see that n= f(n)n s converges. The last art follows from alying the same reasoning as in roosition 4 to this function. Now suose that is a real character and that A(, ) = 0. Consider ψ(s) = A(s,)A(s,ɛ q ). A(2s,ɛ q ) The zero of A(s, ) cancels out the ole of A(s,ɛ q ), so the numerator is analytic for Re s > 0. The denominator is analytic for Re s > /2, so ψ is analytic for Re s > /2. The denominator has a ole at s = /2, so as s /2, ψ(s) 0. Suose that s is real and s > /2. Then ψ(s) = ( () s ) ( ɛ q () s ) ( ɛ q () 2s ) = q ). ( 2s ) ( s )( () s ). Notice, however, that if () = then this is equal to. So we can reduce this to ψ(s) = + s = ( + s ) ks = ( ) + 2 ks. s ()= ()= 8 k=0 ()=

9 Alying lemma 5.2, we see that ψ(s) = n= a nn s where a n 0, the series converges for s >, and a =. Now exand ψ as a ower series in s about s = 2. We know that it is analytic for Re s > /2, so the radius of convergence is at least 3/2. The n-th coefficient will be equal to ψ (n) (2)/n!, and ψ (n) (2) = a m ( log m) n m 2 = ( ) n c n, m= with c n 0 and c 0 = ψ(2) = n= a nn 2 a =. Thus we can write ψ(s) = c n (2 s) n c 0 = /2 < s < 2. n= However, this contradicts the fact that ψ(s) 0 as s /2. So A(, ) Conclusion We have now roven theorem. It is ossible to rove something stronger: that, on average, there is the same number of rimes in the sequences {a + kq} and {a 2 + kq}, assuming that (a,q) = (a 2,q) =. This means that the rimes are equidistributed modulo q. For an exosition of this fact, as well as a more detailed and algebraic exosition of the roof of this theorem, see [2]. References [] Niven, I., Zuckerman, H. S., Montgomery, H. L. An Introduction to the Theory of Numbers. New York: John Wiley & Sons, Inc., 5th ed. 99. [2] Fröhlich, A., Taylor, M. J. Algebraic Number Theory. Cambridge: Cambridge University Press, 99. [3] Ireland, Kenneth. Rosen, Michael. A Classical Introduction to Modern Number Theory. New York: Sringer-Verlag, 99. 2nd ed. 9