DIRICHLET S THEOREM ABOUT PRIMES IN ARITHMETIC PROGRESSIONS ANG LI Abstract. Dirichlet s theorem states that if q and l are two relatively rime ositive integers, there are infinitely many rimes of the form l+kq. Dirichlet s theorem is a generalized statement about rime numbers and the theory of Fourier series on the finite abelian grou (Z/qZ) lays an imortant role in the solution. Contents. Dirichlet s theorem on arithmetic rogressions 2. Fourier analysis, Direchlet characters, and reduction of the theorem 2 3. Dirichlett s L-functions 3 4. Logarithms 5 5. Non-vanishing of the L-function 8 5.. Comlex Dirichlet characters 8 5.2. Real Dirichlet characters 9 6. Conclusion 2 Acknowledgments 2 References 2. Dirichlet s theorem on arithmetic rogressions Dirichelt s theorem on arithmetic rogressions is a statement about the infinitude of rime numbers. Theorem.. If q and l are relatively rime ositive integers, then there are infinitely many rimes of the form l + kq with k Z This theorem was roved by Dirichlet in 837, and before that, there were several mathematicians whose work dealt closely with the achievements related to Dirichlet s theorem. We can easily rove by contradiction that there exist infinitely many rimes and by constructing a converging alternating series, we can also rove that there are infinitely many rimes in the form 4k +. Dirichlet roved this theorem by showing that the series (.2) lmodq
2 ANG LI diverges, where the sum is over all rimes congruent to l modulo q. The starting oint of Dirichlet s argument is Euler s roduct formula for the zeta function, and Legendre conjectured the theorem for his roof of the law of quadratic recirocity. Later, Riemann extended the zeta function to the comlex lane and indicated the non-vanishing of zeta function was essential in the understanding of the distribution of rime numbers. The main idea of this roof is to rove that l diverges as s +, and the roof starts with Fourier analysis, which reduces the theorem to a statement, which is easier to analyze and related to the Dirichlet L-function. We analyze the L-functions and introduce a general form of Euler s roduct formula, n= n s s, ( χ() s ), = where the roduct is over all rimes. We then rove the non-vanishing of L- functions and finish the roof. Moreover, the requirement of relatively rime integers q and l is indisensable for the result since we can easily construct a counterexamle by icking q = 2, l = 4 and there is only one rime, namely 2, contained in the arithmetic rogression of q, l. 2. Fourier analysis, Direchlet characters, and reduction of the theorem Definition 2.. Let Z (q) denote the abelian grou of all non-negative integers, which are smaller than q and relatively rime to q. Define the Euler hi-function by the order of Z (q). And the oeration on this grou is the normal multilication. Write φ(q) = Z (q) Set G = Z (q),consider the function δ l on G, the characteristic function of l; if n Z (q), then δ l (n) = { if n l mod q, 0, otherwise. Exand the function in a Fourier series as follows: δ l (n) = e Ĝ δ l(e)e(n) where, Ĝ denotes the grou of all characters of G, and ˆδ l (e) denotes the Fourier coefficient of δ l (n) with resect to e. ˆδ l (e) = G m G ˆδ l (m)e(m) = G e(l) Hence (2.2) δ l (n) = G e(l)e(n) Definition 2.3. The Dirichlet characters modulo q is a function defined on Z given by e Ĝ
DIRICHLET S THEOREM ABOUT PRIMES IN ARITHMETIC PROGRESSIONS 3 χ(m) = { e(m) gcd(m, q) = 0, otherwise. Clearly, the Dirichlet characters modulo q are multilicative on all of Z and for each character e of G, there is an associated Dirichlet character. Lemma 2.4. The Dirichlet characters are multilicative. Moreover, (2.5) δ l (m) = χ(l)χm, ϕ(q) where the sum is over all Dirichlet characters. This lemma shows that l = s χ δ l () = s φ(q) χ χ(l) χ(). s We divide the above sum in two arts deending on whether or not χ is trivial. Let χ 0 denotes the trivial Dirichlet character. So we have (2.6) l s = = χ 0 () ϕ(q) s + ϕ(q) ϕ(q) q s + ϕ(q) χ(l) χ χ 0 χ(l) χ χ 0 χ() s χ() s Since there are finitely many rimes dividing q, the first sum on the right-hand side diverges when s tends to. Therefore, Dirichlet theorem is a consequence of the following assertion. Theorem 2.7. If χ is a nontrivial Dirichlet character, then the sum remains bounded as s +. χ() s 3. Dirichlet s L-functions To rove the Dirichlet s theorem, we introduce Dirichlet s L-functions, which are general forms of the Euler zeta function. Definition 3.. The zeta function is defined by ζ(s) = n= = n s. Definition 3.2. Let χ be a Dirichlet character.then the L-function, L(s, χ) is defined for s > by (3.3) L(s, χ) = n s, where χ is a Dirichlet character. n= Proosition 3.4. Suose χ 0 is the trivial Dirichlet character, { if gcd(q, n) =, χ 0 (n) = 0, otherwise, and q = N n= an n is the rime factorization of q. Then N (3.5) L(s, χ 0 ) = ( ( s ))ζ(s). n= n
4 ANG LI Therefore L(s, χ 0 ) as s +. Proof. For the fact that = 0 if gcd(n, q), we have L(s, χ 0 ) = χ 0(n) n= n = χ 0(n) s i n n, s where the sum is over all ositive integers, which can be divided by i (i =, 2,..., N).Therefore L(s, χ 0 ) = i ( ) = ( N n= ( s n ))ζ(s). Moreover, ζ(s) as s, therefore L(s, χ 0 ). Proosition 3.6. If χ is a non-trivial Dirichlet character, then L(s, χ) converges for s > 0. Moreover: () The function L(s, χ) is continuously differentiable for 0 < s <. (2) There exists constants c, c > 0 so that L(s, χ) = + O(e cs ) as s, and L (s, χ) = O(e c s) as s. We first rove the cancellation roerty that non-trivial Dirichlet characters ossess. We will need the following lemma Lemma 3.7. If χ is a non-trivial Dirichlet character, then k (3.8) q, for any k. n= Proof. First rove that q n= = 0. Let S = q n= and select a Z (q), we have χ(a)s = χ(a) = χ(an) = = S. Since χ is nontrivial, χ(a) for some a. Therefore S = 0. We write k = aq + b with 0 b < q, then k n= = aq n= + qa+b n=aq+ = aq+b n=aq+ q We can now rove the roosition 3.6. Proof. Let s k = k n=, and s 0 = 0. Know that L(s, χ) is defined for s > as n= n, which converges absolutely and uniformly for s > δ >. We have s N χk N s k s k k s = k s (3.9) n= = = k= N k= N k= [ s k k s (k + ) s + s ] N N s f k (s) + s N N s,
DIRICHLET S THEOREM ABOUT PRIMES IN ARITHMETIC PROGRESSIONS 5 where f k (s) = s k [k s (k+) s ]. Letg(x) = x s. The Mean value theorem imlies g(k) g(k + ) = g (x) sx s. We also have s k q, so f k (s) qsk s. Therefore the series f k (s) converges uniformly and absolutely for s > δ > 0, and this roves that L(s, χ) is continuous. To rove L(s, χ) is also continuously differentiable, we differentiate L(s, χ) and get (log(n)) n s. Using a similar method, we can rove this series converges uniformly and absolutely for s > δ > 0 and thus L (s, χ) is continuously differentiable. To rove the second roosition, we notice that for s large enough (3.0) L(s, χ) = n= n s 2q n=2 n s 2 s a, where a is a constant. If we take a = log 2, then L(s, χ) = + O(e cs ) as s. For L (s, χ), we have (3.) L (s, χ) = (log n) χ n s 2q (log n)n s 2 s a, where a is also a constant. Therefore, similarly L (s, χ) = O(e c s ) as s with c = c. Theorem 3.2. If s >, then (3.3) n s = n= where the roduct is over all rimes. n=2 ( χ() s ), 4. Logarithms To rove the theorem above, we need to construct two logarithms, which are different from the normal logarithms. Definition 4.. For the first logarithm, we define (4.2) log ( z ) = for z <. Proosition 4.3. The logarithm function log satisfies the following roerties: () If z <, then k= (4.4) ex(log ( z )) = z. (2) If z <,then (4.5) log ( z ) = z + E (z), where the error E satisfies E (z) z 2 if z < /2. z k k
6 ANG LI (4.6) (3) If z < /2, then log ( z ) 2 z. Proof. To establish the first roerty, let z = re iθ with 0 r <, and we rove that (4.7) ( z)e log( z ) = ( re iθ )e k= (reiθ ) k /k = We differentiate the left-hand side with resect to r, and get We also have [ e iθ + ( re iθ )( k= (reiθ ) k /k) ]e k= (reiθ ) k /k e iθ + ( re iθ )( k= (reiθ ) k /k) = e iθ + ( re iθ )e iθ re iθ = 0. Hence, the left-hand side of the equation (4.7) is constant. Let r = 0, we get the desired result. For the second roerty, we simly relace log ( z ) by the infinite series. E (z) = k=2 zk k z 2 k=0 z k z 2. The last inequality holds since z < /2. Therefore, log ( z ) = z + E (z) z + z 2 2 z for z < /2 for z < /2 and this roves the third roerty. Proosition 4.8. If a n converges, and a n for all n, then n= ( a n ) converges. Moreover, this roduct is non-zero. The roof of this roosition is omitted. We now can rove theorem 3.2, the Dirichlet roduct formula χn (4.9) n s = ( χ() s ). n Proof of Theorem 3.2. Let L denote the left-hand side of equation (4.9). Define S N = n N n s and N = N ( χ() s ). If we set a n = χ( n ) s n, where n is the n th rime, then a n converges when s. By the revious roosition, the infinite roduct = lim N N = ( χ() ) converges. Also, define s N,M = χ() N ( + +... + χ(m ) ) s Ms Now given ɛ > 0 and choose N large enough so that S N L < ɛ and N < ɛ.
DIRICHLET S THEOREM ABOUT PRIMES IN ARITHMETIC PROGRESSIONS 7 Given rime N, according to the fundamental theorem of arithmetic, there exists M Z such M doesn t divide any integer n N but there exists n 0 N such that M divides n 0. Together with the fact that the Dirichlet characters are multilicative, we can find M large enough so that S N N,M < ɛ and N,M N < ɛ. Therefore, we have (4.0) Then (4.) L L SN + n n s S N N,M + N,M N + < 4ɛ. N = ( χ() s ).. We can now define the logarithm for L(s, χ), and associate it with the logarithm defined for z. Definition 4.2. If χ is a nontrivial Dirichlet character and s >, we define (4.3) log 2 L(s, χ) = The following roosition links log and log 2. Proosition 4.4. If s > then (4.5) e log 2 L(s,χ) = L(s, χ). s L (t, χ) L(t, χ) dt. Moreover (4.6) log 2 L(s, χ) = log ( χ() s ). Proof. Differentiating e log 2 L(s,χ) L(s, χ) with resect to s, we get (4.7) L (s, χ) L(s, χ) e log 2 L(s,χ) L(s, χ) + e log 2 L(s,χ) L (s, χ) = 0 is immediate from the definition of log 2. This imlies that e log 2 L(s,χ) L(s, χ) should be a constant and take s, we have e log 2 L(s,χ) L(s, χ) =. To associate two logarithms, simly take the exonential of both sides. We have ( ) (4.8) ex log ( χ() s ) = ( χ() s ) = L(s, χ), This means for each s,there exists M(s) Z such that log 2 L(s, χ) log ( χ() ) = s 2πiM(s). The left side is continuous in s, but M(s) is integer-valued so M(s) should be constant. Moreover, let s, we have M(s) = 0.
8 ANG LI Recall that in order to rove theorem. is bounded for any nontrivial Dirichlet character χ. From the roosition above, we have log 2 L(s, χ) = log ( χ() s ) (4.9) = = χ() s χ() s χ() s + O( + O(). 2s ) The equation above imlies that if L(, χ) 0 for nontrivial Dirichlet character, then log 2 L(s, χ) remains bounded as s +. Thus remains bounded. χ() s 5. Non-vanishing of the L-function Therefore, our next goal is to rove that L(, χ) 0 for nontrivial Dirichlet character χ. Theorem 5.. If χ χ 0, then L(, χ) 0. Definition 5.2. A Dirichlet character is said to be real if it takes only real values and comlex otherwise. We rove L(, χ) 0 for real and comlex characters searately. 5.. Comlex Dirichlet characters. We rove by contradiction that for comlex Dirichlet characters, L(, χ) 0. Lemma 5.3. The roduct is real-valued and if s >, then (5.4) L(s, χ), where the roduct is taken over all Dirichlet characters. Proof. By the Proosition 4.6, we know that L(s, χ) = ex( log ( χ() χ χ ( ) χ( k ) = ex k ks (5.5) χ = ex χ ( k= k= χ k χ( k ) ks ) s )) Lemma 2.4 imlies that χ χ(k ) = χ χ()χ(k ) = ϕ(q)δ ( k ). Therefore ( (5.6) L(s, χ) = ex ϕ(q) ) δ ( k ) k ks, χ k= since every term in the exonential is non-negative. Lemma 5.7. The following three roerties hold: () If L(, χ) = 0, then L(, χ) = 0..
DIRICHLET S THEOREM ABOUT PRIMES IN ARITHMETIC PROGRESSIONS 9 (2) If χ is non-trivial and L(, χ) = 0, then (5.8) L(s, χ) C s, when s 2. (3) For the trivial Dirichlet character χ 0, we have (5.9) L(s, χ 0 ) C s, when < s 2. Proof. It s easy to show that L(, χ) = L(, χ), the first statement is obvious. To rove the second statement, we aly mean the value theorem to L(s, χ). We have (5.0) L(s, χ) L(, χ) = s L(a, χ), for some < a < s 2. Therefore, L(s, χ) C s. Finally, the last statement follows because (5.) L(s, χ 0 ) = ( s i )ζ(s) A( + dx x s ) = A( + s ) i q C s. The first equality holds according to roosition 3.4 and the first inequality holds because there are only finitely many i so the roduct is bounded by some A and ζ(s) is bounded by ( + ) dx x. s Now we can conclude the roof that L(, χ) 0 for a non-trivial comlex Dirichlet character χ. Proof. Suose we have L(, χ) = 0, then L(, χ) = 0 and χ χ. Therefore, there are at least two terms in χ L(s, χ), that vanish like s as s +. Moreover, by the last roosition, only the trivial character contributes to a increasing term as s +, and this growth is no better than O( s ). Thus, the roduct χ L(s, χ) vanishes as s +, which contradicts Lemma 5.3. 5.2. Real Dirichlet characters. As for the a real and non-trivial χ, we use the summation along hyerbolas to rove that L(s, χ) 0. Given a real and non-trivial Dirichlet character χ, let (5.2) F (m, n) = (nm) /2, and define (5.3) S N = F (m, n), where the sum is over all integers m, n that satisfy mn N. Before we start to rove that L(, χ) 0, we first analyze the sum n N n /2 Proosition 5.4. If N is a ositive integer, then (5.5) n = 2N /2 + c + O /2 n N ( N /2 ).
0 ANG LI Proof. Let a n = n+ dx, we have n /2 n x /2 (5.6) 0 a n n /2 (n + ) /2 C n 3/2. Suose the series n= a n converges to a limit a. Moreover, we have C (5.7) a n n C dx 3/2 n = O( ). 3/2 N /2 Therefore (5.8) N n= N+ N n /2 n=n+ dx x /2 = a N n=n+ N+ + N dx x = a + O( ). /2 N /2 Proosition 5.9. The following statements are true for S N :: () S N c log N for some constant c > 0. (2) S N = 2N frm e L(, χ) + O(). Proof. For the first statement, we first rove the following lemma. Lemma 5.20. { 0 for all k n k, if k = l 2 for l Z, Proof. If k is a ower of a rime, k = a, then the divisor of k are only the owers of. We have = χ() + χ() +... + χ( a ) (5.2) n k = + χ() + χ() 2 +... + χ() a. Hence, n k 0 for any a, and n k for even a. Generally, if k = N n= an n, any divisor of k is of the form N b n a n. Therefore, N (5.22) = (χ() + χ( j ) + χ( 2 j) +... + χ( aj j )). n k j= Therefore, n k if k = l2 for some l Z. From lemma (5.20), we then have S N = (5.23) = nm=n (nm) /2 N /2 n N c log N, k/2 k=l 2,l N /2 n= bn n, where 0 for some constant c. To rove the second statement, we calculate the sum S N by taking the sum in a different way. Consider the sums over three searated regions,
DIRICHLET S THEOREM ABOUT PRIMES IN ARITHMETIC PROGRESSIONS S = F (m, n); S 2 = F (m, n); S 3 = F (m, n). where S, S 2, S 3 are sums over the regions { m < N /2, N /2 < n N/m}, { m N /2, n N /2 } and {N /2 < m N/n, n < N /2 } We write (5.24) S N = S + (S 2 + S 3 ), and we evaluate S by first fixing m, and S 2 + S 3 by first fixing n. To rove the statement, we also need the following lemma: Lemma 5.25. For all integers 0 < a < b we have () b n=a (2) b n=a = O(a /2 ), n /2 n = O(a ). Proof. We use summation by arts. Let s n = k n χ(k), and by Lemma 3.7, we have s n q for all n. Therefore (5.26) b n=a n /2 = = b n=a+ χ(a) (s n s n ) + n/2 a /2 b s n [n /2 (n + ) /2 ] + O(a /2 ) n=a = O( n 3/2 ) + O(a /2 ) n=a = O(a /2 ). Similarly we can rove the second statement. With this lemma, we can finish the roof of the roosition. Summing vertically for S, we have (5.27) S = = m /2 n /2 m O(N /4 ). /2 m<n /2 N /2 <n N/m m<n /2 Moreover, (5.28) O m /2 m<n /2 = O Therefore, S = O(). Finally, we sum horizontally for S 2 + S 3 : S 2 + S 3 = n /2 n N /2 (5.29) = ( N /2 m /2 m N /2 ) dm = O(N /4 ). m/2 n /2 {2(N/n)/2 + c + O((n/N) /2 )} n N /2 = 2N /2 n N /2 n + c n N /2 n /2 + O( ). N /2 n N /2
2 ANG LI Let A = 2N /2 n N /2 n, then (5.30) A = 2N /2 L(, χ) + 2N /2 ON /2 = 2N /2 L(, χ) + O(). n /2 Moreover, the Lemma 5.25 gives c n N = O() and we also have /2 O( N n N ) = O(). Thus S /2 /2 N = 2N /2 L(, χ) + O(), which is the second statement of Proosition 5.9. Suose L(, χ) = 0 for some non-trivial real Dirichlet character χ, by Proosition 5.9, we know that (5.3) S N = O() c log N, which is imossible. Therefore L(, χ) 0 for non-trivial χ. 6. Conclusion From the last section, we know that L(, χ) 0 for any non-trivial Dirichlet character χ. Then by equation 4.2, we have that χ() remains bounded as s s +, which by equation 2.6 imlies that l diverges as s +. s Thus there are infinitely many rimes such that l mod q and we have roved the Dirichlet s theorem. Acknowledgments. It is a leasure to thank my mentor, Wouter van Limbeek and Bena Tshishiku, for all their hel in icking this interesting theorem and writing this aer. References [] Elias M. Stein, Rami Shakarchi. Fourier Analysis : An Introduction. Princeton University Press. 2003. [2] M.A. Armstrong. Grous and Symmetry. Sringer-Verlag New York Inc. 988.