Bounded gaps between Gaussian primes

Transcription

1 Journal of Number Theory 171 (2017) Contents lists available at ScienceDirect Journal of Number Theory Bounded gaps between Gaussian primes Akshaa Vatwani Department of Mathematics, Queen s University, Kingston, Ontario K7L 3N6, Canada a r t i c l e i n f o a b s t r a c t Article history: Received 3 November 2015 Received in revised form 26 July 2016 Accepted 26 July 2016 Available online 6 September 2016 Communicated by R.C. Vaughan MSC: 11N36 11R44 Keywords: Primes of the form a 2 + b 2 Gaussian primes Bounded gaps Higher rank Selberg sieve We show that there are infinitely many distinct rational primes of the form p 1 = a 2 + b 2 and p 2 = a 2 +(b + h) 2, with a, b, h integers, such that h 246. We do this by viewing a Gaussian prime c + di as a lattice point (c, d) in R 2 and showing that there are infinitely many pairs of distinct Gaussian primes (c 1, d 1 )and (c 2, d 2 )such that the Euclidean distance between them is bounded by 246. Our method, motivated by the work of Maynard [9] and the Polymath project [13], is applicable to the wider setting of imaginary quadratic fields with class number 1 and yields better results than those previously obtained for gaps between primes in the corresponding number rings Elsevier Inc. All rights reserved. 1. Introduction The well-known twin prime conjecture states that there are infinitely many primes p such that p +2is also prime. The last decade has marked remarkable progress towards this conjecture. We now know that there are infinitely many bounded gaps between the primes and moreover this bound has been reduced to as low as 246. This progress began with the first such result by Zhang [19], building innovatively upon previous work of address: akshaa@mast.queensu.ca X/ 2016 Elsevier Inc. All rights reserved.

2 450 A. Vatwani / Journal of Number Theory 171 (2017) Goldston, Pintz and Yıldırım [3], followed by important contributions due to Maynard [9] and Tao, and eventually culminated in the collaboration of a number of mathematicians under the Polymath project [13]. One way to think about the twin prime conjecture is to view it as a special case of the more general prime k-tuples conjecture. A set H of distinct non-negative integers is said to be admissible in Z if for every prime p, there is some residue class which is not contained in H (mod p), that is, H (mod p) <pfor all primes p. Then, the prime k-tuples conjecture asserts that for any given admissible set H, there are infinitely many n such that n + h 1,..., n + h k are all prime. To view the twin prime conjecture as a special case of this, notice that any admissible set of size 2is given without loss of generality by {0, 2m}, where m Z +. Then, the prime k-tuples conjecture applied to this set with m = 1yields infinitely many twin primes while any m 2gives an infinitude of what are known as generalized twin prime pairs. We will formulate the analogue of the generalized twin prime conjecture in the setting of the Gaussian integers Z[i]. In order to do this, we first define the notions of primality and admissibility in Z[i]. We say that p Z[i] is a Gaussian prime if the ideal (p) Z[i] generated by it is a prime ideal. A set H = {h 1,..., h k } of distinct Gaussian integers is called admissible if H (mod p) < N(p) for every Gaussian prime p. Let us consider admissible sets of size 2. It is clear that admissibility must now be checked only for the primes 1 + i and 1 i having norm 2. As m 1 + m 2 i (mod 1 ± i) 0iff m 1 and m 2 have the same parity, without loss of generality, any admissible set of size 2is given by {0, m 1 + m 2 i}, where m 1 m 2 (mod 2). This gives us the following analogue of the generalized twin prime conjecture: Conjecture 1.1 (Generalized twin primes in Z[i]). Given any integers m 1 and m 2 having the same parity, there are infinitely many Gaussian primes p = a +bi such that a +m 1 + (b + m 2 )i is also prime. This conjecture has an interesting consequence in terms of the rational primes obtained by taking norms of the Gaussian primes above. Conjecture 1.2 (Pairs of rational primes of a special form). Given any integers m 1 and m 2 having the same parity, there are infinitely many pairs of rational primes (p 1, p 2 ) of the form p 1 = a 2 + b 2 and p 2 =(a + m 1 ) 2 +(b + m 2 ) 2. It is indeed intriguing that the problem of finding infinitely many pairs of primes of a special form in Z translates to the generalized twin prime conjecture when viewed in the ring Z[i]. This problem is still unsolved. A related recent result by Thorner [17] states that for any fixed 0 < ɛ < 1/2, there are infinitely primes p 1, p 2 of the form p = a 2 + b 2 with a < ɛ p, such that p 1 p 2 C(ɛ).

3 A. Vatwani / Journal of Number Theory 171 (2017) If we define the diameter of H as diam H = max N(h i h j ), (1) 1 i,j k then it is clear that the smallest possible diameter of an admissible set of size 2is attained when m 1, m 2 = ±1. This would be expected to yield the analogue of the twin prime conjecture in the Gaussian integers. We formulate this for the admissible set {0, 1 + i}. Conjecture 1.3 (Twin primes in Z[i]). There are infinitely many Gaussian primes p = a + bi such that p + (1 + i) is prime. Consequently, there are infinitely many pairs of rational primes (p 1, p 2 ) of the form p 1 = a 2 + b 2 and p 2 =(a + 1) 2 +(b + 1) 2, with a, b Z. It is worth noting that the case (m 1, m 2 ) = (0, 2) of Conjecture 1.2 would yield infinitely many primes p 1, p 2 of the form p 1 = a 2 + b 2 and p 2 = a 2 +(b +2) 2. As a step towards this conjecture for the case m 1 =0and m 2 even, we prove the following result in this paper. Theorem 1.4. There are infinitely many rational primes of the form p 1 = a 2 + b 2 and p 2 = a 2 +(b + h) 2, with a, b, h Z, such that 0 < h 246. Remark. If Lemma 10.2 in Section 10 is assumed to hold for any θ <1, that is, if we assume the Elliott Halberstam conjecture for Gaussian primes, the above result can be improved to 0 < h 12, using the methods of [13]. 2. Approach to the problem In view of the discussion of the previous section, it is not surprising that Conjecture 1.2 is addressed by considering the problem of bounded gaps between primes for the ring Z[i]. More precisely, we may view a Gaussian prime a + bi as a lattice point (a, b) in R 2. Then are there infinitely many pairs of distinct Gaussian primes (a, b) and (c, d) such that the Euclidean distance between them is bounded? We first discuss some work that has been done in this direction. One may view Gaussian primes p which are not rational, as arising from rational primes of the form p 1 (mod 4), that is, p = a + bi with the norm a 2 + b 2 equal to p. Then, from Theorem 1.2 of Thorner [16] and more recent work of Kaptan [8], bounds are known for gaps between rational primes in an arithmetic progression. In the context of the ring Z[i], this proves that there are infinitely many non-associate Gaussian primes a + bi and c + di such that the difference between their norms is bounded. However, the Euclidean distance between such primes (viewing them as lattice points in R 2 ) is at most the sum of the square roots of their norms and can satisfy better bounds. We establish a bound of 246 for this Euclidean distance.

4 452 A. Vatwani / Journal of Number Theory 171 (2017) Theorem 2.1. There are infinitely many distinct Gaussian primes p 1, p 2 N(p 1 p 2 ) < Moreover, one has such that σ(p 1 p 2 ) < 246 for every embedding σ of Q(i). In the proof of this theorem, the admissible set used allows us to conclude that p 2 is of the form p 1 + h for some h Z, thereby proving Theorem 1.4. At this point, let us remark that a version of bounded gaps between primes in number fields has been proved by Castillo, Hall, Oliver and Pollack [1]. We state their result below. Let K be a number field with ring of integers O K. A set H = {h 1,..., h k } of distinct elements of O K is called admissible if H (mod p) < O K /p for every prime ideal p O K. An element α O K is said to be prime if it generates a prime ideal. One can formulate the prime k-tuples conjecture in the setting of number rings as follows: Conjecture 2.2 (Prime k-tuples conjecture in K). Given any admissible set H = {h 1,..., h k } of elements of O K, there are infinitely many α O K such that α + h 1,..., α + h k are all prime. Then, the following is proved in [1]. Theorem 2.3 (cf. Theorem 1.1, [1]). For any m 2, there exists k (depending on K and m) so that for any admissible set {h 1,..., h k }, there are infinitely many α O K such that at least m many of α + h 1,..., α + h k are prime in O K. In section 12, we prove this result for the case K = Q(i) with m = 2and k = 54. Further refinement along the lines of [13] allows us to establish the bound in our main theorem above. As mentioned in the concluding remarks, this method can be generalized to any imaginary quadratic field with class number 1. Although it seems that some bound on gaps between Gaussian primes should also be attainable from the above result of [1], it is worth noting that the technique adopted in this paper is different from that of [1]. Indeed, our method allows us to exploit a greater level of distribution, thereby yielding better numerical bounds. The key conceptual idea of our approach that has not been executed previously in the context of bounded gaps is the generalization of an arithmetic progression of integers to a ray class of ideals. This is discussed in Section 5. Consequently, we run over integral ideals of bounded norm in the relevant sums: more precisely, the classical notion of a positive integer n x is generalized to an integral ideal of norm bounded by x. This natural setting allows us to use an equidistribution theorem of Bombieri Vinogradov type for the number of prime ideals in a ray class. Huxley [6] established such a result with level of distribution θ =1/2 (cf.lemma 10.1 below). On the other hand, the results of [1] rely on summing over elements of O K rather than over ideals and use an equidistribution

5 A. Vatwani / Journal of Number Theory 171 (2017) theorem proved by Hinz [5] which is compatible with their setting. Letting r 1, r 2 denote the number of real and non-real embeddings of K into C respectively, Hinz obtains a level of distribution θ =1/2 if K is totally real (i.e. r 2 =0) and θ = 1 (r 1 + r 2 1) + 5/2 if r 2 > 0. In the case K = Q(i), indeed for all imaginary quadratic fields, this gives θ =2/5 which is less than half. The method of [1] hence does not avail of a level of distribution θ =1/2 except in the totally real case. 3. Preliminaries We recall some facts about number fields that will be useful in our exposition. Given a number field K, we denote its ring of integers by O K. Let a, b, c denote elements of O K. Parenthesis around an element denotes the ideal generated by that element in O K. The divisibility condition c a is interpreted as the containment of the corresponding ideals: (a) (c). The congruence relation a b (mod c) is defined as a b (c), where (c) O K is the ideal generated by c. One may also express this congruence relation in terms of the ideal, that is, a b (mod (c)). Recall that any non-zero ideal a in O K has finite index in O K. The norm of the ideal a is N(a) = O K /a. For a principal ideal (c), the ideal norm coincides with the absolute value of the field norm NQ K (c). We also recall the φ function of an ideal a as the size of the multiplicative group (O K /a). More precisely, one has φ(a) =N(a) p a ( 1 1 ), (2) N(p) where the product goes over all prime ideals p of O K. We say that p O K is prime if (p) is a prime ideal. The Dedekind zeta function for the field K is defined for Re(s) > 1as the infinite series ζ K (s) = N(a) s, a O K 1 where the sum is over all non-zero ideals of O K. We can also write ζ K (s) = n=1 a n n s,

6 454 A. Vatwani / Journal of Number Theory 171 (2017) where a n denotes the number of integral ideals of norm n. By the work of Hecke [4], ζ K (s) extends analytically to the entire complex plane except for a simple pole at s = 1 with residue where ϱ K = 2r 1 (2π) r 2 h K R K ω, d K r 1 = number of real embeddings of K in C, 2r 2 = number of nonreal embeddings of K in C, h K = class number of K, R K = regulator of K, ω = number of roots of unity in K, d K = discriminant of K. Let N(x, K) denote the number of ideals in O K with norm x. By a result of Weber (see for example p. 144 of [11]), we have N(x, K) := n x a n = ϱ K x + O ( x 1 1 [K:Q] ). In the case K = Q(i), it is easy to see that ϱ K = π/4, so that N(x, Q(i)) = π 4 x + O( x). This result can also be derived by elementary means. The study of the error term is called the circle problem. It is conjectured that the error is O(x θ )for any θ >1/4. The last century has seen progress on this problem by a number of mathematicians. At the moment, it is known that we can take θ = 131/416 + ɛ for any ɛ > 0. This was proved by Huxley [7] in As Z[i] is a principal ideal domain and contains exactly four units, each ideal (n)gives four elements ±n, ±in of the same norm. Hence, 4. Notation #{n Z[i] :N(n) x} = πx + O( x). (3) We will denote elements of Z[i] by bold letters. In our study, we are led to consider sums ranging over tuples of Gaussian integers. It is convenient to set up some notation for such tuples. The k-tuple of Gaussian integers (d 1,..., d k ), with d j Z[i] for all j, is denoted by d. We use p to denote a prime element of Z[i]. This simply means that

7 A. Vatwani / Journal of Number Theory 171 (2017) (p) is a prime ideal in Z[i]. An element n Z[i] is said to be squarefree if p 2 n for any p Z[i]. This notion is well defined due to unique factorization in the ring Z[i]. A tuple is said to be square-free if the product of its components is square-free. The norm of a tuple is defined as the norm of the product of its components. Hence for example, for R R, the inequality N(d) R is to be interpreted as j N(d j) R. The notion of divisibility among tuples is defined component-wise, that is, d n d j n j for all 1 j k. It follows that the notion of congruence among tuples, modulo a tuple, is also defined component-wise. On the other hand, we say a Gaussian integer q divides the tuple d if q divides the product i d i. Similarly, d q j d j q However when we explicitly write the congruence relation d e (mod q), we mean that it holds in Z[i] for each component. We will also come across real valued functions acting on k-tuples in Z[i]. A vector function is said to be multiplicative if all its component functions are multiplicative. In this context, we define the function f(d) to mean the product of its component (multiplicative) functions acting on the corresponding components of the tuple, that is, f(d) = k f i (d i ). For example, the Möbius function μ can be defined on Z[i] in the usual manner as i=1 { ( 1) r if n = p 1...p r μ(n) = 0 if n is not square-free. Then on the tuple d, the Möbius function is given by k μ(d) = μ(d j ). j=1 We also use the notation [d, e] to denote the product of the component lcm s, that is, k [d j, e j ]. j=1

8 456 A. Vatwani / Journal of Number Theory 171 (2017) Ray classes in the Gaussian ring We would like to generalize the notion of an arithmetic progression in the rational integers to a general number ring. One way to do this is to consider ray class groups. Ray class groups were introduced by Weber in It was this ingenious discovery that paved the way for many developments in class field theory. Given an ideal q of O K, the set of all fractional ideals in K which are co-prime to q forms a group. We will define an equivalence relation on this set. We say that a b if there exist α, β O K satisfying (i) (α), (β) coprime to q (ii) α β q (iii) α/β 0, that is, σ(α/β) > 0for all real embeddings σ of K, such that (α)a =(β)b. This set of equivalence classes can be shown to be finite (cf. Theorem 1.7, [10]) and can be given the structure of an abelian group in the same way as done for the ideal class group. We call this the ray class group (mod q). We denote this group by C q and its cardinality by h(q). When q = 1, this is called the narrow ideal class group C. The cardinality h of C is called the narrow class number of the field. If r 1 =0, the narrow ideal class group and the usual ideal class group are identical. Thus in the case of imaginary quadratic fields, the two groups are the same. It can be shown (cf. Proposition 1.6, [10]) that every ray class can be represented by an integral ideal. Hence, without loss of generality, we may consider only integral ideal representatives. In order to see that ray classes are indeed a generalization of residue classes in the rational integers, let us consider the case K = Q. Here r 1 =1and r 2 =0. Fix q Z + and consider the ray class group modulo (q). Then two integral ideals (r) and (s), coprime to (q), are equivalent if there exist positive integers m, n, coprime to q, with q dividing m n, such that (m)(r) =(n)(s). This gives us the relation mr = ±ns. Without loss of generality, we may fix the generators r and s to be positive. Then we have mr = ns. Going modulo q gives r s (mod q). Conversely, if r s (mod q), then the trivial identity (s)(r) = (r)(s) implies (r) (s). Thus, we see that the ray class group (mod q) is isomorphic to the group of residue classes coprime to q: (Z/qZ). We are interested in the case K = Q(i). In this case, as there are no real embeddings of K, the condition (iii) above is vacuous. Fix an ideal (q) Z[i]. Then two ideals (r) and

9 A. Vatwani / Journal of Number Theory 171 (2017) (s) are equivalent (mod q) if there exist a, b Z[i], coprime to (q), with a b (mod q) so that (a)(r) =(b)(s). This gives ar = ubs for some unit u in Z[i]. Reducing modulo q gives r us (mod q), where u can be ±1, ±i. Thus ray classes in the Gaussian ring correspond to congruence conditions up to a unit. 6. Setting up the sieve We will formulate the problem of bounded gaps in the context of the Gaussian primes. Recall that the set of Gaussian integers H = {h 1,...h k }, is admissible if H is not the full set of coset representatives of Z[i]/p for any prime ideal p in Z[i]. This can be equivalently expressed as the inequality H (mod p) < Z[i]/(p) being satisfied for any Gaussian prime p. Here, (p) Z[i] is the ideal generated by the element p. One then has the following analogue of the classical prime k-tuples conjecture. Conjecture 6.1 (Prime k-tuples conjecture in Z[i]). Given an admissible set H of elements in Z[i], there are infinitely many Gaussian integers n such that all of n + h 1,..., n + h k are prime in Z[i]. We will emulate the classical approach to this conjecture. Throughout the rest of the paper, n, h 1,..., h k are understood to denote elements of Z[i]. We will also often refer to Gaussian primes p simply as primes and will explicitly say so if we mean primes in any other context. Henceforth we work with a fixed admissible set H in Z[i] Let χ P denote the characteristic function of the Gaussian primes: χ P (n) = { 1 if n Z[i] is prime 0 otherwise. For x R, the notation n x means x N(n) 2x. In practice we will let x. We use a technical modification known as the W -trick to eliminate the effect of primes with small norm. To do this, fix D 0 R and let W = N(p)<D 0 p.

10 458 A. Vatwani / Journal of Number Theory 171 (2017) Admissibility of the set H implies that H (mod p) misses some residue class modulo p for each p dividing W. In other words, there exists b p (mod p) so that none of b p +h 1,..., b p +h k are divisible by p. Solving the equations y b p (mod p) for all p W simultaneously, the Chinese remainder theorem yields a Gaussian integer b (mod W) such that (b + h j, W) = 1for all j. We choose D 0 = log log log x so that W log log x by an application of the prime number theorem. Consider the expressions S 1 = n x n b (mod W) ν(n) and S 2 = n x n b (mod W) k χ P (n + h j ) ν(n), j=1 where ν(n) are non-negative parameters. For ρ positive, we denote by S(x, ρ) the quantity k S 2 ρs 1 = χ P (n + h j ) ρ ν(n) (4) n x n b (mod W) The key idea is the following proposition. Proposition 6.2. Given a positive number ρ, if S(x, ρ) > 0 for all large x, then there are infinitely many n Z[i] such that at least ρ +1 many of n + h 1,..., n + h k are Gaussian primes. Proof. As ν(n) is non-negative, S(x, ρ) > 0 implies that the inequality j=1 k χ P (n + h j ) ρ>0, j=1 must hold for some n x. Since this can be done for all large x and χ P is integer valued, it is easy to see that the claim holds. The choice of the parameters ν(n) is motivated by Selberg [15] and recent progress on bounded gaps by Maynard and Tao [9,13]. Letting n denote the tuple (n+h 1,..., n+h k ), we put ( ) 2 ν(n) = λ d, d n

11 A. Vatwani / Journal of Number Theory 171 (2017) where λ d is a sequence of real numbers indexed by k-tuples in (Z[i]) k and the sum runs over only those tuples d n such that each component d j divides n + h j and is in a fixed quadrant (say Re(d j ), Im(d j ) > 0). As in the classical Selberg sieve, the λ d s are also subject to the constraint λ 1 =1and are supported only on square-free tuples. The choice of λ d s is the same as in [13], in terms of a fixed symmetric smooth test function F :[0, ) k R, supported on the simplex Δ k (1) := {(t 1,...,t k ) [0, ) k : t t k 1}. For fixed R R to be chosen later, we choose, λ d = μ(d)f which we abbreviate for the sake of notational brevity to ( log N(d1 ),...,log N(d ) k), (5) ( ) log N(d) μ(d)f. Observe that as F is a smooth function with compact support, it is bounded and so is λ d. With this approach in mind, we proceed to derive asymptotic formulas for the sums S 1 and S 2 as we did in [14] in the rational number field case. 7. The higher rank Selberg sieve for Gaussian integers We are interested in estimating sums of the form n x n b (mod W) w n( d n λ d ) 2, (6) where w n is a weight attached to the Gaussian integers n and λ d are parameters chosen as in (5). In [14], M. Ram Murty and the author give an axiomatic formulation of the higher rank sieve method. Following along the lines indicated in this work, we impose certain hypotheses on the above sum: H2. The function w n satisfies d n n b (mod W) w n = X f(d) + r d, for some multiplicative function f and some quantity X depending on x.

12 460 A. Vatwani / Journal of Number Theory 171 (2017) H3. On primes p Z[i], the components of f satisfy for some fixed α j N. f j (p) = N(p) α j + O(N(p) t ), with t<1 We denote the tuple of integers (α 1,..., α k )as α and the sum of the components k j=1 α j as α. H4. There exists θ >0and Y X such that for any A > 0. N([d,e])<Y θ r [d,e] Y (log Y ) A We state the main result of the higher rank sieve in this setting. Theorem 7.1. Fix K = Q(i). Let λ d s be as chosen in (5). Suppose hypotheses H2, H3 hold and H4 holds with Y = X. Set R = X θ/2 δ for small δ >0. Then, with n x n b (mod W) ( ) 2 (α) c(w) X w n λ d =(1+o(1))C(F, F) ϱ α d n K () α, c(w) := N(W)α φ(w) α, ϱ K = π 4, and C(F, F) (α) = 0 0 k t α j 1 j ( 2 F (t)) (α) dt (α j 1)! j=1 Proof. The proof is motivated by the work of Maynard [9] and the Polymath project [13], which built upon the early ideas of Goldston, Pintz and Yıldrım [3]. The proof is much along the lines of that of Theorem 3.6 and Lemma 3.4 in [14], with minor modifications to adapt it to the setting of Gaussian integers. Therefore, we sketch the main ideas of the proof, giving details where necessary. Expanding out the square, interchanging the order of summation gives us ( ) 2 ( w n λ d = λ d λ e w n ). n b (mod W) d n d,e [d,e] n n b (mod W)

13 A. Vatwani / Journal of Number Theory 171 (2017) Recall that by construction of the sieve parameters in Section 6, the above sum runs over only those tuples d, e for which the real and imaginary parts of each component are both positive. Arguing as in the proof of Theorem 3.6 ([14]), applying hypothesis H2, and noting that λ d is absolutely bounded, we derive the following expression for the required sum X N(d),N(e)<R ( λ d λ e f([d, e]) + O N(d),N(e)<R ) r [d,e], (7) where the dash over the sum represents the additional constraint that [d, e] is square-free and co-prime to W. It is clear that the error term is taken care of by H4. By the choice (5) of the λ d s, the main term above is given by X d,e μ(d)μ(e) f([d, e]) F ( log N(d) ) F ( ) log N(e). ( ) Writing F as a Fourier transform as in Lemma 3.4 ([14]), we can write ( ) as where Z(u,v)= d,e X R k R k η F (u)η F (v)z(u,v)dudv, (8) μ(d)μ(e) f([d, e]) 1 1 N(d) (1+iu)/ N(e) (1+iv)/ and F(t) = R k η(u)exp( (1 + u) t)du. Note that here u, v are k-tuples in ordinary integers, 1 is the tuple (1,..., 1) and the notation d u represents the product k j=1 du j j. As Z[i] has unique factorization up to units, and the functions f and N( ) are multiplicative functions, one can write an Euler product for the series Z(u, v) in the same way as done in the case of the rational integers in [14]. Moreover, the Gaussian primes p appearing in the Euler product will be in the quadrant Re(p), Im(p) > 0and will also be co-prime to W due the dash over the sum. Hence, Z(u, v) is given by the product (p) Z[i] (p) (W) 1 k j=1 ( 1 f j (p) 1 N(p) 1+iu j + 1 N(p) 1+iv j ) 1. N(p) 1+iu j + 1+iv j

14 462 A. Vatwani / Journal of Number Theory 171 (2017) Applying H3 and performing some algebraic manipulations as done in Lemma 3.4 of [14], we obtain the following convenient expression for Z(u, v): Z(u,v) k j=1 (p):n(p)>d 0 ( )( ) 1 α j N(p) 1 1+iu j 1 α j N(p) 1 1+iv j 1 α j N(p) 1 1+iu j +1+iv j. (9) Arguing as in Lemma 3.4 of [14], we find that the main contribution to the sum (8) is from the region u, v < () ɛ. We write each term (p):n(p)>d 0 appearing in the product above as ( ( ζ K 1+ 1+is ) ( αj j 1 (p) (W) 1 α j N(p) 1+ 1+is j 1 N(p) 1+ 1+is j ) ) αj ( D j 1+ 1+is ) j, where ζ K (s) is the Dedekind zeta function of the field K = Q(i) and D j (s) = (p):n(p)>d 0 ( 1 1 ) αj ( N(p) s 1 α ) j N(p) s. The Euler product D j (s) above is supported only on prime ideals (p) with N(p) > D 0 and is absolutely convergent for Re(s) > 1/2. For Re(s) = 1, we have D j (s) =1+O ( ) 1 N(p) 2 =1+O (p):n(p)>d 0 1 as R (and hence D 0 ) goes to infinity. Using this along with the fact that ζ K has a simple pole at s = 1with residue ϱ K, we find that in the region s j < () ɛ, (p):n(p)>d 0 ( 1 α j N(p) 1+ 1+is j as R. Applying this to (9), we have ) =(1+o(1))ϱ α j K ϱ α K Z(u,v)=(1+o(1))c(W) () α k j=1 D 0 ( ) αj 1+isj N(W) α j φ(w) α, j (1 + iu j ) α j (1 + iv j ) α j (1 + iu j +1+iv j ) α j, where c(w) is as defined in the statement of the theorem. This is the analogue of the expression (3.10) obtained in [14]. From here on, the proof proceeds exactly as in

15 A. Vatwani / Journal of Number Theory 171 (2017) Lemma 3.4 of [14] to yield the required result, keeping in mind that ϱ K = π/4 for the field K = Q(i). 8. Norm of a fixed translate In our estimates below, we will often come across the problem of bounding the norm of a fixed translate of n, given a bound for N(n). Let h be a fixed Gaussian integer. Then, By the triangle inequality, we have N(n + h) = n + h 2 = n 2 +2nh + h 2. n 2 2nh + h 2 N(n + h) n 2 + 2nh + h 2. As h 2n+h cn(n) 1/2, for some constant c depending on N(h), we have obtained N(n + h) =N(n)+O N(h) (N(n) 1/2 ). (10) In practice, we will apply this when n x, in which case, the above relation gives us x + O( x) N(n + h) 2x + O( x). For convenience, we will represent this as n + h x + O( x). 9. The sum S 1 With choice of weights as in (5), the sum S 1 is given by S 1 (x) = n x n b (mod W) ( d j n+h j j λ d ) 2 Employing the notation of Theorem 7.1, we obtain the following asymptotic formula for S 1 (x). Theorem 9.1. Choosing θ <1, λ d s as in (5) in terms of the function F, and R = x θ/2 δ, we have, as x, where S 1 (x) =(1+o(1)) N(W)k 1 φ(w) k 4 k π k 1 x () k I(F),

16 464 A. Vatwani / Journal of Number Theory 171 (2017) I(F) := Δ k (F (1) (t)) 2 dt. Proof. The result is a direct consequence of Theorem 7.1 after checking that all the required hypotheses hold. The weight w n in this case is simply 1, so that in order to check H2, we need to consider n x n b (mod W) d j n+h j j By the Chinese remainder theorem, this can be written as n x n c (mod q) where q = W j d j and c Z[i] is one of the N(q) many residue classes modulo q. The above expression counts the number of Gaussian integers of bounded norm in a fixed arithmetic progression c (mod q). Writing n c (mod q) as n = mq + c, we see that the condition n x is equivalent to mq x + O( x)by (10). Using (3) to count the number of elements m satisfying we obtain for the above sum N(m) 1. 1, x N(q) + O( x/n(q)), π N(q) x + O( x/n(q)). Recalling our definition of q and interpreting this in the manner of H2, we have X = πx/n(w), f(d) = k N(d i ), j=1 r d = O ( ( ) ) 1/2 x. N(W[d, e]) As f j (p) = N(p), it is clear that H3 holds with α j =1for all j. Moreover, the sum α = j α j is k in this case. To verify H4, we pick Y = x and any θ<1 ɛ. Then recalling that W log log x, we have x 1/2 N([d,e])<x θ N(W[d, e]) 1/2 x 1/2 x 1+θ+ɛ 2. N(W[d,e])<x θ+ɛ N(W[d, e]) 1/2 Hence, H4 is satisfied.

17 A. Vatwani / Journal of Number Theory 171 (2017) A generalization of the Bombieri Vinogradov theorem One of the key ingredients in the proof of bounded gaps between primes in the rational case is the Bombieri Vinogradov theorem. Let π(x) denote the number of primes upto x. Put for (a, q) = 1, E P (x, q, a) = n x n a (mod q) χ P (n) π(x) φ(q). (11) Then for any A > 0and any θ <1/2, the Bombieri Vinogradov theorem states that max E P(x, q, a) x (a,q)=1 (log x) A. (12) q x θ We define the level of distribution of the primes as the supremum of all the values of θ for which (12) holds. There have been many generalizations of this celebrated theorem to the number field setting. One such generalization is the addition of a non-abelian splitting condition on rational primes. This was done by M. Ram Murty and V. Kumar Murty in [12]. In keeping with our perspective of a residue class as a very special case of a ray class, the variant of (12) that is more suitable to our setting is a result of Wilson [18] and Huxley [6] dating back to They established an equidistribution result on average for the number of prime ideals p O K lying in a ray class. It is interesting to note that they used different techniques. Wilson showed a level of distribution 1/(1 +[K : Q]), following a method of Gallagher. On the other hand, a method of Davenport and Halberstam was adopted by Huxley, enabling him to obtain a level of distribution 1/2. In order to state this result, we proceed by setting up some terminology. Let q be an ideal in O K. We denote the ray class group (mod q) by C q and a ray class in this group by C q. Let π(x, K) be the number of prime ideals in O K of norm x and χ P be the characteristic function of the primes. In other words, χ P acts on integral ideals a, taking the value 1 if a is a prime ideal and zero otherwise. We define E P (x, q, C q )= a O K :N(a) x a C q χ P (a) π(x, K), (13) h(q) where h(q) denotes the size of the ray class group C q. Denote the number of residue classes of integers relatively prime to q by φ(q) as in (2). Then, the following lemma can be derived from Theorem 1 of Huxley [6] using partial summation. Lemma 10.1 (Huxley). Using the notation established in (13), for any A > 0 and any θ<1/2, we have

18 466 A. Vatwani / Journal of Number Theory 171 (2017) h(q) φ(q) max max E P(y, q, C q ) x C q C q y x (log x) A. N(q) x θ In order to see that this reduces to (12) in the case K = Q, as well as gain some insight into what this result yields for a general number field, let us study the factor h(q)/φ(q) closely. The number of ray classes h(q) can be expressed in terms of the class number h of the field K. More precisely we have the following (cf. Theorem 1.7, Ch. 5 of [10] or (7) of [18]). where h(q) = 2r 1 φ(q)h [U : U q,1 ], (14) r 1 = number of real embeddings of K in C, U = O K, U q,1 = {α O K : α 1(modq),α 0}. Let σ 1,..., σ r1 denote all the real embeddings of K into C. Recall that α 0means that σ j (α) > 0for all 1 j r 1. As h and 2 r 1 are fixed for a given field K and do not depend on q in any way, they can be treated as constants as far as the expression of Lemma 10.1 is concerned. The index of the subgroup U q,1 in U can be estimated as follows. Consider the homomorphism ψ : U (O K /q) r 1 j=1 {±1}, defined as ψ(α) = (α (mod q), sgn(σ 1 (α)),..., sgn(σ r1 (α))). Let T (q) denote the number of residue classes (mod q) that contain a unit. Then, the kernel of this map is precisely U q,1 and its image has size at most 2 r 1 T (q). By the first isomorphism theorem, we have [U : U q,1 ] K T (q). Thus, upto a constant (depending on K), the factor h(q)/φ(q) can be thought of as 1/T (q). We now move back to our specific case K = Q(i), with q =(q). As there are no real embeddings of Q(i), the map ψ takes a unit u to u (mod q), giving that the index [U : U q,1 ]is precisely T (q). Let us assume that N(q) is sufficiently large (more precisely we need N(q) > 4), so that T (q) = U = 4. Indeed, if two units u 1 and u 2 are congruent modulo q, then N(q) must divide N(u 1 u 2 ), which is at most 4. This can be thought of as analogous to the W -trick described in Section 6. Moreover as the class number of Q(i) is 1, we derive that the number of ray classes (mod q) for the Gaussian integers is

19 A. Vatwani / Journal of Number Theory 171 (2017) h(q) = φ(q) T (q), by (14). This agrees with the discussion of Section 5. If we assume N(q) to be large enough, h(q) is simply φ(q)/4. With this assumption in place, we proceed to rewrite Lemma 10.1 in a manner which is amenable to our setting of Gaussian primes in arithmetic progression. Using the dictionary between ray classes and arithmetic progressions established in Section 5, consider (n) Z[i]:N(n) x (n) C q (a) χ P (n), where C q (a) is a ray class (mod q) containing the ideal (a). The condition (n) C q (a) means that n ua (mod q) for some unit u. Furthermore, as a is coprime to q and N(q) is large, this unit must be unique for each fixed generator n of (n). Suppose the unit in question for the generator n is denoted u 0. Pick any other generator u n of (n). We have u n congruent to a (mod q) up to a unique unit, but that unit must be u u 0 since n u 0 a (mod q). Hence, no matter which generator of (n) we pick, the above sum simply counts {n Z[i] :(n) is prime,n(n) x, n u 0 a (mod q)}. As u 0 is fixed, this is simply the number of Gaussian primes of norm bounded by x, in the arithmetic progression a (mod q). We denote this quantity by π(x, q, a). As a result of our discussion we obtain the following form of Lemma 10.1 applied to K = Q(i), which will be useful in our study of the sum S 2. Lemma Letting and we set π(y, Q(i)) = #{p Z[i] :N(p) y} π(y, q, a) =#{p Z[i] :N(p) y, p a (mod q)}, E P (y, q, a) =π(y, q, a) Then for any A > 0 and any θ <1/2, we have max max (a,q)=1 4<N(q) x θ 4π(y, Q(i)). φ(q) y x E P(y, q, a) x (log x) A. (15)

20 468 A. Vatwani / Journal of Number Theory 171 (2017) We define the level of distribution of the Gaussian primes as the supremum of all the values of θ for which (15) holds. 11. The sum S 2 For 1 m k, we define the sums S (m) 2 (x) = n x n b (mod W) ( χ P (n + h m ) d j n+h j j λ d ) 2, so that S 2 (x) = k m=1 S(m) 2 (x). We obtain the following asymptotic formula for S (m) 2 (x). Theorem With θ chosen so that Lemma 10.2 holds, λ d s chosen as in (5) in terms of F, and R = x θ/2 δ, we have as x, S (m) 2 (x) =(1+o(1)) N(W)k 1 φ(w) k 4 k π(2x, Q(i)) π(x, Q(i)) π k 1 () k 1 J m (F), with J m (F) given by the integral ( 2 F m (1) (t 1,...,t m 1,t m+1,...,t k )) dt1...dt m 1 dt m+1...dt k. Δ k 1 (1) Here F m is the function F restricted to tuples with mth component zero, more precisely, F m (t 1,...,t m 1,t m+1,...,t k )=F(t 1,...,t m 1, 0,t m+1,...,t k ). (16) Proof. To obtain H2, we consider the sum n x, d j n+h j j n b (mod W) χ P (n + h m ) This sum is non-zero if and only if (d m ) = (1) and hence this additional condition must be imposed. Let us denote χ(n) as the function taking the value 1if n generates the ideal (1) and zero otherwise. Writing n + h m as n and using (10), the above sum can be written as χ(d m ) χ P (n ) n x+o( x), n h m h j (mod d j ) j m n b+h m (mod W) Note that if a prime p divides h m h j, for j m, then p divides W. Hence (d j, W) = 1 implies that h m h j must be co-prime to d j for j m. It is also clear that b + h m

21 A. Vatwani / Journal of Number Theory 171 (2017) is co-prime to W. Then, using the Chinese remainder theorem, the above sum can be written as the sum over a single residue class a modulo q = W j m d j. Thus, we have that the required sum equals χ(d m ) n x+o( x), n a (mod q ) χ P (n ). Implementing the notation of Lemma 10.2, the inner sum is simply and can be written as π(2x + O( x), q, a) π(x + O( x), q, a), 4(π(2x, Q(i)) π(x, Q(i))) φ(q ) + E P (x, q, a)+o( x/φ(q )). The constraint d m = 1 implies that the sieve has now collapsed to a (k 1)-rank sieve. Hence fixing d m =1in the sum that appears in H2, this hypothesis holds with X = (π(2x) π(x))/φ(w) and f j (d j ) = φ(d j )for all j m. Furthermore, the error term is given by r d = E P (x, q, a)+o( x/φ(q )). This shows that H3 holds with α j =1for each j m. Keeping the additional constraint on the mth component in mind, to verify H4, we consider N[d,e]<x θ (d m ),(e m )=(1) ( E P (x, q, a ) + O N[d,e]<x θ (d m ),(e m )=(1) ) x/φ(q ), where q is now W j m [d j, e j ]. As θ is chosen such that Lemma 10.2 holds, there exists some ɛ > 0such that this lemma still holds for θ = θ + ɛ. Then, since N(W) log log x, the condition N[d, e] <x θ implies the bound N(q ) <x θ. Moreover, we must have N(q ) > 4for large x because N(W ) as x. Hence, the above sum is of the order of E P (x, q, a ) + x/φ(q ). 4<N(q )<x θ N(q )<x θ By Lemma 10.2, the first term is of the order of x/(log x) A for any A > 0. In order to estimate the second term, we note that for any Gaussian integer n, the inequality φ(n) N(n) holds. This gives us the bound x/n(q ) x log x N(q )<x θ

22 470 A. Vatwani / Journal of Number Theory 171 (2017) for the second error term. This shows that H4 holds. The result is now an application of Theorem Numerical bounds We summarize our results below. As F was chosen to be a symmetric function, it is clear from Theorem 11.1 that S 2 := k m=1 S (m) 2 = ks (k) 2. Choosing some θ for which the bound of Lemma 10.2 holds, so that θ is admissible in the derivation of the asymptotic formula for S 1 as well as S 2, one obtains the following after using the prime ideal theorem. Lemma Let J m (F) and I(F) be as in the statements of Theorems 11.1 and 9.1 respectively. Then, with θ <1/2, λ d s chosen as in (5) in terms of F, and R = x θ/2 δ, we have as x, S(x, ρ) :=S 2 ρs 1 =(1+o(1)) 4k N(W) k 1 π k 1 φ(w) k (( ) ) x θ () k 2 δ kj k (F) ρi(f). Combining this asymptotic formula for S(x, ρ) with Proposition 6.2, we need the inequality ( ) θ ρ< 2 δ kjk (F) I(F). (17) To maximize the right hand size we will consider the quantity M k =sup F kj k (F) I(F), with supremum taken over all symmetric smooth functions F supported on the simplex Δ k (1). In order to establish concrete bounds for gaps between primes, it is necessary to recall the notion of diameter of a set in Z[i] defined in (1). Let us define H(k) as the minimal diameter of an admissible set H Z[i] of size k, that is, H(k) = min diam H. H Z[i] An admissible set that attains this diameter is called the narrowest admissible set of size k in Z[i]. Then, the following general bound can be obtained from (17).

23 A. Vatwani / Journal of Number Theory 171 (2017) Lemma Suppose (17) holds for some k and ρ > 0. Let m be the least integer greater than ρ, that is, m = ρ +1. Then, there are infinitely many distinct Gaussian primes p 1,..., p m such that max N(p j p l ) H(k). 1 j,l k Moreover, if H = {h 1,..., h k } is the narrowest admissible set of size k in Z[i], then for each embedding σ of Q[i], we have max σ(p j p l ) max σ(h j h l ). 1 j,l k 1 j,l k Proof. As (17) holds for some ρ > 0, this means that S(x, ρ) > 0for all large x. Applying Proposition 6.2, we see that for any admissible set of size k, there are infinitely many n Z[i] such that at least m many elements of the n-th translate of the admissible set are Gaussian primes. In other words, there are infinitely many distinct Gaussian primes p 1,..., p m such that N(p j p l )is bounded by the diameter of the admissible set for all 1 j, l k. Choosing an admissible set H of Gaussian integers which has the minimal diameter, the inequalities follow. With this general result in hand, we consider M k for specific values of k. Picking θ = 1 2 ɛ for some ɛ > 0, one sees that for (17) to hold with ρ > 1, M k must be greater than 4. A thorough numerical analysis of M k for both large and small k has been carried out by the Polymath 8 project [13]. The smallest value of k for which M k > 4is 54. Moreover, as done in Theorem 26 of [13], it is also possible to extend the definition of M k in such a way that the supremum is taken over a larger class of functions F, namely those supported in the enlarged simplex Δ k (1 + ɛ), provided 1 + ɛ < 1/θ. Then, as done in [13], k = 50 suffices for (17) to hold with ρ > 1. We can now apply Lemma 12.2 with m = 2and k = 50. This leads us to the problem of finding the diameter H(50) of the narrowest admissible set of Gaussian integers of size 50. As a corollary of Lemma 3.1 of [1], any admissible set of integers is also admissible in Z[i]. In particular, the narrowest admissible set H of integers of size 50 has diameter 246. This proves Theorem There are infinitely many distinct Gaussian primes p 1, p 2 N(p 1 p 2 ) Moreover, one has such that σ(p 1 p 2 ) 246 for every embedding σ of Q(i). In fact, since H is a set of integers {h 1,..., h k }, we see that p 1, p 2 have the form n+h i and n + h j for some Gaussian integer n and some 1 i < j k, with h i h j 246. This proves our main result: Theorem 1.4.

24 472 A. Vatwani / Journal of Number Theory 171 (2017) Remark. If one had the analogue of the Elliott Halberstam conjecture for the Gaussian ring, that is, if one could establish Lemma 10.2 for any θ <1, this would yield infinitely many Gaussian primes p 1 and p 2 satisfying σ(p 1 p 2 ) 12 for every embedding σ of Q(i). 13. Concluding remarks It is easy to see that the method of this paper is applicable to all imaginary quadratic fields with class number 1. Since such rings have only finitely many units, by Huxley s equidistribution result, we have a level of distribution θ =1/2 in each case. Thus, the bound of Theorem 12.3 as well as the remark following it hold for gaps between primes in every such ring. For example, applying this to the imaginary quadratic field Q( 163) with ring of integers Z( ), we obtain infinitely many primes p 1, p 2 of the form p 1 = a 2 + b 163 2, p 2 = a 2 + h + b 163, a,b,h Z 2 with 0 < h 246. As the norms of p 1 and p 2 are rational primes, we have Theorem There are infinitely many rational primes p 1, p 2 of the form with 0 < h 246. p 1 = a b 2, p 2 = a b 2 + ah + h 2 (a, b, h Z) 4 4 As before, one can improve this bound to 12 under the Elliott Halberstam conjecture for primes in the ring of integers of Q( 163). One can also expect the following special case of Conjecture 1.2 for imaginary quadratic fields K having class number 1, due to the admissibility of the set {0, 2}. Conjecture There are infinitely many primes p O K such that p +2 is also prime. Another related but independent problem that suggests itself in the discussion following Lemma 12.2 is the problem of finding admissible sets in number rings other than Z. For example, if one can find an admissible set of Gaussian integers of size 50 with diameter below 246, then the bound for Gaussian prime gaps can be accordingly improved. In general, one would expect this bound to depend on the ambient ring. The problem of finding the diameter of the narrowest admissible set of rational integers of size k has been solved by Clark and Jarvis [2]. Analogous results in the setting of other number rings would be expected to give better numerical results for gaps between primes in that ring.

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