DIRICHLET PRIME NUMBER THEOREM

Transcription

1 DIRICHLET PRIME NUMBER THEOREM JING MIAO Abstract. In number theory, the rime number theory describes the asymtotic distribution of rime numbers. We all know that there are infinitely many rimes,but how are they distributed? Dirichlet s theorem states that for any two ositive corime integers a and d, there are infinitely many rimes which are congruent to a modulo d. A stronger form of Dirichlet s theorem states that the sum of the recirocals of the rime numbers with the same modulo diverges, and different rogressions with the same modulus have aroximately the same roortions of rimes. We will walk through the roofs of Dirichlet s theorem, and introduce some related toics, such as the Riemann-zeta function and quadratic field. Contents. Introduction: the Euclidean method 2. Riemann zeta function 2 3. Dirichlet characters 3 4. Dirichlet L function 6 5. nonvanishing of L(χ, ) when χ is real-valued 8 Acknowledgements 9 Aendix A. Proofs of Theorems and Lemma 9 References 3. Introduction: the Euclidean method Many eole know that there are infinitely many rimes (Euclid s Theorem). The roof is easy. Suose that there are only finitely many rimes, say,..., r. Consider N =... r +. Then N is a rime that is not included in... r. Contradiction. Thus we rove Euclid s Theorem. What if we want to know how many rimes there are that are congruent to a modulo d (a,d are corime)? In some cases, we can generalize the Euclidean roof to rove there are infinitely many such rimes. We will show here for d = 3 and 4. Consider rimes of the form n + 2. Assume... r are rimes congruent to 2 modulo 3. Take N = 3... r. It is congruent to 2 modulo 3, so if it is not a rime, it must have a rime factor congruent to 2 modulo 3. Yet all such rimes do not divide N. Contradiction. Thus, there are infinitely many rimes of the form 3n + 2. Similar method can be alied to the roof of rimes of the form 4n + 3. Let N = 4... r + 3 where... r are rimes of the form 4n + 3. Then clearly N is congruent to 3 modulo 4. If N is not a rime, it must have a rime factor congruent Date: October, 203.

2 2 JING MIAO to 3 modulo 4. Contradiction. Therefore, there are infinitely many rimes of the form 4n + 3. The case of rimes of the form 4n + is a little bit tricky, because a non-rime number that is congruent to modulo 4 can have all rime factors not congruent to module 4. In this case, we let N = r +, and using the similar idea, we can rove by contradiction (For the roof, see aendix ). 2. Riemann zeta function However, not all cases can be shown in the Euclidean way. In this section, we will introduce Euler s roof of the infinitude of rimes. First, let s learn the Riemann-zeta function that will be frequently used in our later roofs. Definition 2.. The Riemann zeta function ζ(s), is a function of a comlex variable s that analytically continues the sum of the infinite series n s. The summation converges when the real art of s is greater than. What haens when s is very close to? Note that ζ() is unbounded because we know that does not have an uer bound. Also note that n n= ζ() = r n = + i n= i= + 2 i +... = i= n= r. i Since ζ() is unbounded, there are infinitely many rimes. Otherwise, the right hand side will be finite. We can then simify the Riemann zeta function into ζ(s) = n= = = n s = ( t n= t...t r N...t r) s ( + s + 2 s +...) s, where is indexed over all rime numbers. Then we want to see what function ζ(s) converges to. Since t s dt = s + t s+ = s,

3 DIRICHLET PRIME NUMBER THEOREM 3 we assume that ζ(s) = s + h(s), where the real art of s is greater than. The function h(s) is a holomorhic function that is bounded. To rove this, h(s) = ζ(s) s = n s n N = n N = n N n+ n 0 t s dt ( n s t s ) dt ( n s (n + t) s ) dt. Denote f n (t) = n s (n+t), f s n (0) = 0, f n(t) s = (n+t). When 0 < t <, s+ f n (t) su f n(t) s. ns+ Thus, h(s) n N s n, which converges when the real art of s is greater than s+ 0. Thus, log ζ(s) log s. On the other hand, we have log ζ(s) = log = s log( s ) = ( s + /2 2s + /3 3s +...). Note that when real art of s is greater than /2, /2 2s + /3 3s s s 2 2s. Thus, (/2 2s + /3 3s +...) converges. Therefore, s log s. 3. Dirichlet characters We showed in the revious section that rime ln s s. Now we want to show that for any (a, b) =, amodb s φ(b) ln s as s goes to, where φ(b) is the number of a < b, such that (a, b) =. The infinitude of rimes of the form a + bn follows from this. Thus, we want a nice way to write the characteristic function, such that (x) = { : n a mod b 0 : n a mod b. Then a mod b = () s. Such characteristic functions are called Dirichlet s characters χ. There are two basic roerties of Dirichlet characters: χ() = ; if a b mod N, then χ(a) = χ(b). Theorem 3.. If (a, N) =, then a mod N. This is called Euler s Thoerem. Thus, χ(a) is a -th root of unity.

4 4 JING MIAO Definition 3.2. A congruent class g mod N is called a rimitive root mod N if for any a Z and (a, N) =, a g k ( mod N) for some k. The result is that there exists a unique k in the interval 0 k, such that a g k (mod N). We label such k: k = ind g a and call it the index of a to base gmodn. So a g indga. Proosition 3.3. Let g be a rimitive root modulo N, and (a, N) = (b, N) =. Then ind g (ab) = ind g a + ind g b mod, ind g (a k ) = kind g a mod, If g is another rimitive root, then ind g a = ind g g ind g a mod. So for general natural numbern, do rimitive roots exist? The answer is that they do not always exist. Theorem 3.4. For only N =, 2, 4, α, 2 α, where is an odd rime and α N,there exists a rimitive root. Proof. It is easy to check that,2,4 have rimitive roots. Then we shall rove that all odd rimes have rimitive roots. Let n be the least universal exonent for, i.e. n is the smallest ositive integer such that x n mod, for all non-zero x 0,.... Notice that there is some element g, 2..., such that g n but g m mod for any m < n, i.e. the multilicative order of g is recisely n. Also, notice that by Euler s theorem, n.now, notice that the olynomial f(x) = x n has at most n roots over the field Z.(The roof is in the aendix.) f(x) 0 mod for all non-zero x mod. Thus, n. Hence, n = and g is of exact order. Therefore, g is a rimitive root. Next we shall show that numbers of the form of α have rimitive roots. First, we shall show if a is a rimitive root of, then a is a rimitive root of 2. Let n amod. Then n (a + k)mod for some k <. By Euler s theorem, we know (a + k) ( ) mod 2. By contradiction, if a is not a rimitive root, there exists some f such that (a + k) ( )/f mod 2. Since f, = tf + for some integer t.then (a + k) ( )/f = (a + k) )(t+/f) (i + k) t+/f = ( + k) t ( + k) /f ( + m) (b) mod 2, where b is any number but not that is congruent to. This is a contradiction. Thus, a is a rimitive root of 2. Then we will rove if a is a rimitive root of 2, a is also a rimitive root of α. Using similar method, we can rove that a is also a rimitive root for α. Next we will rove that if α has rimitive root, then 2 α also has rimitive root. If the rimitive root a is an odd number, since α and 2 α have the same order, a

5 DIRICHLET PRIME NUMBER THEOREM 5 must also be a rimitive root for 2 α. If a is even, take a = a + α, a is odd and (a, 2 α ) =. Consider (a + α ) φ(2α )/f, it is not congruent to mod 2P α.thus, 2 α has rimitive roots. We also have to show that 2 n (n > 2) and any other comosite numbers excet the ower of rime numbers do not have rimitive. First, consider 2 n. For any odd number 2k +, (2k +) 2n 2 mod 2 n. So 2 n does not have rimitive root. Then, consider comosite number N = t...tr r. The order of N is the multilication of the orders of tj j, which are all even number. So there is no rimitive roots. Thus, we finish the roof of this theorem. Now let s learn how to construct the Dirichlet characters. Let g be a rimitive root of of α where is odd. For any (n, ) =, we write b(n) = ind g n. So g g b(n) mod α. For h = 0,..., φ( α ), define χ h (n) = { e 2πihb(n)/φ( α ) : (n, α ) = 0 : (n, α ). Examles 3.5. Take = 5, α =, g = 3. Thus, b() = 4, b(2) = 3, b(3) =, b(4) = 2. Notice that if (nm, α ) =, χ h (nm) = e 2πihb(nm) φ( α ) = e 2πih(b(n)+b(m)) φ( α = χ h (m)χ h (n). As there are φ( α ) h s, there exist φ( α ) distinct Dirichlet characters. This construction also works for 2,4. However, there is no rimitive root for modulo 2 n when n 3. Similarly, for N = α α2 2...αr r, there is no rimitive root. If χ i is a Dirichlet character modulo αi i, then χ = χ χ 2...χ r. Since φ( αi i αj j ) = φ(αi i )φ( αj j ), we know that = φ( α )φ(α2 2 )...φ(αr r ). It is easy to check that if χ, ψ are Dirichelet characters, then φ ψ is also a Dirichlet character. This roerty is very useful. Now consider χmodn χ(x). If we multily it with ψ(x) for some ψ(x), we get ψ(x) χ mod N χ(x) = χ mod N ψ(x)χ(x) = χ mod N (ψ χ)(x) = χ mod N χ(x), (ψ(x) ) χ mod N χ mod N χ(x) = 0, χ(x) = 0 if x mod N. If x mod N, then χmodn χ(x) =. Thus,we have (x) = χ { : χ mod N 0 : χ mod N.

6 6 JING MIAO Thus, we have for (a, b) =, a mod b s = = a mod b = φ(b) s ( φ(b) s where aa mod b χ mod b χ mod b(χ(a ) χ(a )) χ() s ). 4. Dirichlet L function In this section, we introduce the Dirichlet L function. We want to show that L(s, χ) is finite near s =, and also L(, χ) 0. Definition 4.. A Dirichlet L-series is a function of the form L(s, χ) = χ(n) n= n, s where χ is a Dirichlet character and s a comlex variable with real art greater than. Since we roved in the revious section that (x) = χ { : χ mod N 0 : χ mod N, we can write I(x) = χ χ(a )χ(x), where I(x) is the identity function, and a Z, such that xa mod N. Then let x = ay. If a =, then a =. Thus I a (y) = χ χ(y), when y mod N. Now, a mod N s = = = I a () s χ(a ) χ N s + χ() s χ nontrivial χ(a ) χ() s. The first term is the summation of the trivial character (χ() = ). This is the case when h = 0. And the second is the summation of non-trivial characters. We will show later that the contribution of the second term is finite, thus, imlying that different rogressions with the same modulo diverge, and different rogressions with the same modulus have aroximately the same roortions of rimes. We will show that for each χ, χ()/s is bounded as s goes to. Since as s goes to, is unbounded, while s 2 s, 3 s... are all bounded, we have

7 DIRICHLET PRIME NUMBER THEOREM 7 χ() s = = log = log χ() s + χ() 2 2s + 3 log( χ() s ) χ() s ( + χ() s + χ(2 ) 2d +...) χ() s = log χ(n) n s since χ(n) χ(m) = χ(nm). n Notice that what is inside the ln is L(s, χ). We will show that L(s, χ) is finite and non-zero for s > 0 when χ is non-trivial. Notice that x Z N χ(x) = 0 when χ is non-trivial; x Z N χ(x) = when χ is trivial. χ χ(x) = 0 when x mod N; χ χ(x) = when x mod N. We will first rove the finitude of the Dirichlet L-function when χ is non-trivial. = L(s, χ) = m=0 i= n= χ(n) n s = N m=0 i= i= χ(i) (i + mn) s N N χ(i) N ( (i + mn) s ( χ(i)) (mn) s ) = m=0 i= N χ(i)( (i + mn) s (mn) s ). The second ste is true because N χ(i) = 0. Note that (i+mn) s (mn) s = (+ i mn )s (i+mn) s. Let x = m. i mn N mn = m. Since s > 0, (i + mn)s m s = x s. Thus, we have mn )s (i + mn) s ( + i = (mn) s (i + mn) s N χ(i) N (i + mn) s ( χ(i)) m=0 i= i= ( + x)s x s (+sx) x s = s x s+, (mn) s m=0 c m s+, where c is some finite number. The right hand side of the above equation converges absolutely when Re(s) > 0. Thus, L(s, χ) has the least uer bound. Next, we want to show L(, χ) 0. First, suose that χ is not quadratic, χ(n) n s χ(n) n s when s is real. Thus, i.e. χ is not real-valued. Then n = n L(s, χ) = L(s, χ). In articular, if we assume L(, χ) = 0, then L(, χ) = 0. Then we have at least two secific characters,χ, χ, such that L(, χ) = L(, χ) = 0. By the equation amodn s = N s + χnontrivial χ(a ) χ() s,

8 8 JING MIAO we have mod N s = + log L(s, χ) + log L(s, χ) + other characters. s We have roved that log s s. Since the following two L functions have ole at s =, log L(s, χ) = log L(s, χ) log(s ) when s aroaches to. Thus, the first two terms on the right hand side cancel, leaving the third term and the remaining other L functions from other characters. When s is very close to, say.000, the LHS > 0 while the RHS < 0. There is a contradiction. Thus, when χ is not real-valued, the Dirichlet L-function can never be nonvanishing of L(χ, ) when χ is real-valued In order to rove the nonvanishing of L(χ, ),we introduce the Dedekind zeta function: ζ N (s) = χ L(χ, s) near s =. We know that for each nontrivial χ, L(χ, s) is holomorhic at s =, whereas for trivial χ we get essentially the Riemann zeta function, which we have seen has a simle ole at s =. We have seen that (s )ζ(s) as s. It follows from basic function theory that ζ N (s) has at most a simle ole at s =, and it has a ole iff L(χ, ) 0 for all nontrivial χ. Thus, our goal is to show that the Dedekind zeta function ζ N (s) has a singularity at s =. The key is that the Dirichlet series ζ N (s) has a very articular form. To see this, we need just a little notation: for a rime not dividing N, let f() denote the order of in the unit grou U(N), and ut g() = f(p ), which is a ositive integer. Proosition 5.. a) We have ζ N (s) = N. ( f()s )g() b) Therefore, ζ N (s) is a Dirichlet series with non-negative integral coefficients, converging absolutely when the real art of s is greater than. Proof. Let µ f() be the grou of f()-th roots of unity as roots (with multilicity one). Then for all N we have the olynomial identity w µ f() ( wt ) = T f(). Indeed, both sides have the f()th roots of unity as roots (with multilicity one), so they differ at most by a multilicative constant; but both sides evaluate to when T = 0. Now by the Character Extension Lemma, for all w µ f() there are recisely g() elements χ X(N) such that χ() = w. This establishes art a), and art b) follows from the exlicit formula of art a). Since we know that ζ N (s) has non-negative real coefficients, therefore we can aly Landau s Theorem: if σ is the abscissca of convergence of the Dirichlet series, then the function ζ n (s) has a singularity at σ. Clearly σ, so, if ζ N (s) does not have a singularity at s =, then not only does ζ N (s) extend analytically to some larger halflane where the real art of s is greater than ɛ, but it extends until it meets a singularity on the real line. But we have already seen that each Dirichlet L-function is holomorhic when the real art of s is greater than 0 and less than, so Landau s theorem tells us that σ 0. However, in our cases, it is unlikely that

9 DIRICHLET PRIME NUMBER THEOREM 9 a Dirichlet series with non-negative integral coefficients has absissca of convergence σ 0. To see it more exlicitly, ( = ( + ) g() + f()s f()s 2f()s +...)g() f()s 2f()s Ignoring all the crossterms gives a crude uer bound: this quantity is at least s 2s Multilying this over all, it follows that ζ N (s) n (n,n)= n s. When we evaluate at s =, we get (n,n)= n. Since the set of integers rime to N has ositive density, it is substantial. More concretely, since every n of the form Nk + is corime to N, this last sum is at least as large as k= Nk+ =. Thus, the ζ N (s) has a singularity at s =. Therefore, L(χ, ) 0 for all nontrivial χ. In conclusion, if we define the density of the rimes of the form a mod b to be ρ(a mod b) = a mod b, then we get the result of Dirichlet s Theorem: different rogressions (different a) with the same modulus have the same density. This is true because amodn s = N s + χnontrivial χ(a ) χ() s, and the second term on the right hand side is finite. Thus, we comlete the roof of Dirichlet s Theorem on the density of rimes. Acknowledgements I would like to thank my mentor, Henry Chan, for roviding me with knowledge about number theory and advice on this aer. Aendix A. Proofs of Theorems and Lemma Theorem A.. In number theory, Euler s theorem states that if n and a are corime ositive integers, then a φ(n) mod n where φ(n) is the number of all integers less than n that are corimes to n. Proof. This theorem can be roven using concets from the theory of grous: The residue classes (mod n) that are corime to n form a grou under multilication. The order of any subgrou of a finite grou divides the order of the entire grou, in this case φ(n). If a is any number corime to n, then a is in one of these residue

10 0 JING MIAO classes, and its owers a, a 2,..., a k mod n are a subgrou. Then k must divide φ(n), i.e. there is an integer M such that km = φ(n). Then a φ(n) = a km = (a k ) M M = mod n. Theorem A.2. Let F be a field and let (x) be a non-zero olynoial in F[x] of degree n 0. Then (x) has at most n roots in F (counted with multilicity). Proof. We roceed by induction. The case n = 0 is trivial since (x) is a non-zero constant, thus (x) cannot have any roots. Suose that any olynomial in F[x] of degree n has at most n roots and let (x) F[x] be a olynomial of degree n +. If (x) has no roots then the result is trivial, so let us assume that (x) has at least one root a F. Then,there exists a olynomial q(x) such that (x) = (x a) q(x). Hence,q(x) F[x] is a olynomial of degree n. By the induction hyothesis, the olynomial q(x) has at most n roots. It is clear that any root of q(x) is a root of (x) and if b a is a root of (x) then b is also a root of q(x). Thus, (x) has at most n + roots, which concludes the roof of the theorem. Lemma A.3. (Character Extension Lemma) Let H be a subgrou of a finite commutative grou G. For any character ψ : H C, there are [G : H] characters ψ : G C such that ψ H = ψ. Proof. The result is clear if H = G, so we may assume that there exists g G\H. Let G g =< g, H > be the subgrou generated by H and g. Now we may or may not have G g = G, but suose that we can establish the result for the grou G g and its subgrou H. Then the general case follows by induction, since for any H G, choose g,..., g n such that G =< H, g,..., g n >. Then we can define G 0 = H and for i n, G i =< G i, g i >. Alying the Lemma in turn to G i as a subgrou of G i gives that in all the number of ways to extend the character ψ of H = G 0 is [G : H][G 2 : G ]...[G n : G n ] = [G : G 0 ] = [G : H]. So let us now rove that the number of ways to extend ψ from H to G g =< H, g > is [G g : H]. For this, let d be the order of g in G, and consider Ḡ := H < g >. The number of ways to extend a character ψ of H to a character of Ḡ is equal to # < g >= d: such a homomorhism is uniquely secified by the image of (, g) in µ d C, and all d such choices give rise to homomorhisms. Moreover, there is a surjective homomorhism ψ : H < g > to G g : we just take (h, g i ) hg i. The kernel of ψ is the set of all airs (h, g i ) such that g i = h. In other words, it is recisely the intersection H < g >, which has cardinality, say e, some divisor of d. It follows that so #H g = #H < g > #H < g > = d e #H, [H g : H] = d e. But a homomorhism f : H < g > C descends to a homomorhism on the quotient H g iff it is trivial on the kernel of the quotient ma, i.e., is trivial on

11 DIRICHLET PRIME NUMBER THEOREM H < g >. In other words, the extensions of ψ to a character of H g corresond recisely to the number of ways to ma the order d element g into C such that g d e gets maed to. Thus we must ma g to a d e th root of unity, and conversely all such maings induce extensions of ψ. Thus the number of extensions is d e = [H g : H]. Theorem A.4. Let σ k (x) be the number of integers n x which are the roduct of just k rime factors so that n = 2... k, and let π k (x) be the number of such n for which all the i are different. The behavior of π k (x) and σ k (x) as x is given by π k (x) σ k (x) x(loglogx)k (k )!logx. Proof. A.Selberg found an elementary roof of x log x, which is equivalent to the Prime Number Theorem. We use an elementary deduction of the Landau s Theorem for k 2 from the above finding and the well-known elementary result x loglogx. We wrtie c n for the number of ways of exressing n in the form of n = 2... k, order being relevant. Clearly c n = 0, unless n is a roduct of just k rime factors; in this case, c n = k! or c n < k! according as the k rimes are or are not, all different. We write (x) = c n =, n x and so have (A.5) k k!π k (x) k 2... k x (x) k!σ k (x). Again, there are just σ k (x) π k (x) values of n x, each of which is reresentable in the form n = 2... k with two at least of the i eqaul. We may take k = k and so (A.6) σ k (x) π k (x) = (x)... k We write Ω 0 (x) = and,for k, so that kϑ k+ (x) =... k+ x Ω k (x) = n x 2... k 2 2 k c n n =... k x... k, k log( k+ ) + log( 3... k+ ) log( 2... k ) = (k + ) x ϑ( x ). Hence, if we have φ k (x) = ϑ k (x) kxω k x), kφ k+ (x) = (k + ) φ k ( x ), (k ). x

12 2 JING MIAO If, for some fixed k, (A.7) φ k (x) = o(loglogx) k, it follows that where, for any ɛ > 0, Hence φ k+ (x) x(loglogx) k x f(x ), 0 < f(x) A(x ), f(x) < ɛ(x x 0 = x 0 (ɛ)). x. for x x x 0, and so f(x ) ɛ + A x/x 0 x/x 0 leqx ɛloglog( x logx ) + Alog( ) + O() x 0 logx logx 0 2ɛloglogx φ k+ (x) = o[x(loglogx) k ], which is (A.7) with k+ for k. But, for k =, (A.7) is equivalent to x log x. Hence (A.7) is true for all k. Next we have ( )k Ω k (x) ( )k x and so, by x loglogx, Ωk(x) (loglogx) k. Hence, by (A.7), Trivially, (A.8) ϑ k (x) kx(loglogx) k. ϑ k (x) = c n logn (x)logx n x k and, if X = x logx, ϑ(x) X<n x But logx logx and,for k 2, by (A.8). Hence c n logn Π k (x) Π k (X)logX. Π k (X) = O(X) = O( x logx ) = o(ϑ k(x) logx ) = o(π K(x)) Π k (x) ϑ k(x) logx kx(loglogx)k. logx and so, by (A.5) and (A.6), This comletes our roof. π k (x) σ k (x) x(loglogx)k, (k 2). (k )!logx

13 DIRICHLET PRIME NUMBER THEOREM 3 Lemma A.9. There are infinitely many rimes of the form 4n +. Proof. For every k, all rime divisors of k 2 + are mod4. This is because any n 2 + fulfills n 2 mod and therefore ( ) =, which is, since must be odd, equivalent to mod4. Assume that there are only t rimes,..., t of the form 4m +, where m is a rime. Then we can derive a contradiction from considering the rime factors of (2... t ) 2 +. References [] E. Landau, Ueber Eine Verallgemeinerung Des Picardschen Satzes. Sitzungsber. Preuss. Akad. Wiss., 904. [2] S. Stoilov, The theory of functions of a comlex variable. Transl. Math. Monogr., 962. [3] L. Xiao, Course Notes on Number Theory [4] Tom M. Aostol Introduction to Analytic Number Theory. Sringer, [5] Serge Lang, Comlex Analysis. Sringer, New York, 4th Edition, 999.