aa r X i v : . [ m a t h . N T ] N ov SMALL PRIME GAPS IN ABELIAN NUMBER FIELDS
ALEXANDRA MIHAELA MUSAT
Abstract.
In [5], Dan Goldston, Janos Pintz, and Cem Yildirim proved thatthere are infinitely many rational primes that are much closer together thanthe average gap between primes. Assuming the Elliott-Halberstam conjecture,they also proved the existence of infinitely many bounded differences betweenconsecutive primes.We will prove an analogue of their result for a number field. We let K be afinite extension of Q and consider ( q n ) the sequence of rational prime numbersthat split completely in K . For K an abelian extension of Q , we prove thatlim inf n →∞ q n +1 − q n log q n = 0 . Under the Elliott-Halberstam conjecture, we also prove that there are infinitelymany bounded gaps between consecutive members of the sequence q n .We also give another proof of the same result in the special case of a qua-dratic extension of class number 1, which relies on a generalization of theBombieri-Vinogradov theorem for quadratic number fields. Introduction
One of the most famous unresolved problems in number theory is the twin primeconjecture, which states that there are infinitely many prime numbers p for which p + 2 is also prime. In 2005, putting to rest a long-standing open problem, DanGoldston, Janos Pintz and Cem Yildirim proved that there are infinitely manyprimes that are very close together-much closer together than the average gapbetween primes. Under the Elliott-Halberstam conjecture, which concerns primesin arithmetic progressions, they also proved that there are infinitely many boundedgaps between consecutive prime numbers. More formally, they proved that(1.1) lim inf n →∞ ( p n +1 − p n ) ≤ , where p n denotes the n th prime number.If we let E = lim inf n →∞ p n +1 − p n log p n , then the main result proven by Goldston, Pintz and Yildirim in [5] is that(1.2) E = 0 , which is implied by the twin prime conjecture and by (1.1).Finding upper bounds on E has been a long studied problem. The famousprime number theorem states that the number π ( x ) of prime numbers up to x isroughly x ln x . This is equivalent to the statement that p n ∼ n ln n , which wouldimply that the average gap between consecutive prime numbers is ln n ∼ ln p n .This asymptotic expression implies that E ≤
1. Erd¨os [3] was the first to prove that
E <
1. Another important result was that
E <
12 , proven by Bombieri andDavenport in [2]. Maier [7] further improved the result to E ≤ . ... .In May 2005, Goldston, Pintz and Yildirim showed that E = 0. Their proofuses the Bombieri-Vinogradov theorem on primes in arithmetic progressions andshows a connection between the distribution of primes in arithmetic progressionsand small gaps between primes.In the present paper, we will prove a generalization of this result for the case ofan abelian number field. Following [5], we will prove both an unconditional resultand a conditional result, depending on the Elliott-Halberstam conjecture. We willprove the following unconditional result. Theorem 1.1 (Unconditional result) . Let K be an abelian extension of Q and let ( q n ) be the sequence of rational prime numbers that split completely in K . Then lim inf n →∞ q n +1 − q n log q n = 0 . Assuming the Eliott-Halberstam conjecture, we will also prove the followingtheorem.
Theorem 1.2 (Conditional Result) . Let K be an abelian extension of Q and let ( q n ) be the sequence of rational prime numbers that split completely in K . Then thereexist infinitely many bounded gaps between consecutive members of the sequence q n . Our method of proof of the two theorems above will follow the main ideas usedin [5]. Using results from class field theory, we know that the primes that split com-pletely in a number field are described, with finitely many exceptions, by congruenceconditions. In our proof, we will consider the analogue multiplicative functions asin [5] and we will impose some extra congruence conditions on them.The paper is organized as follows. In section 2, we give a quick overview ofthe method used by Goldston, Pintz and Yildirim in the original proof in [5]. Insection 3, we will find estimates for the main sums used in proving the existence ofconditionally bounded gaps and the unconditional result. In sections 4 and 5 weprove the conditional result (1.2) and the unconditional result (1.1), respectively.Section 6 of the paper provides a different approach for proving (1.1) and (1.2) inthe special case of a quadratic extension of class number 1, which could give someinsight into how we could prove analogous results for more general number fields.2.
A brief review of the Goldston, Pintz and Yildirim method
The general approach employed by Goldston, Pintz and Yildirim in [5] was thefollowing.Let k be a given positive integer which is at least 2 and take H = { h , . . . , h k } to be a set of nonnegative distinct integers. Define ̟ ( n ) = ( log p if n = p prime0 otherwise.Let P ( n, H ) = ( n + h ) · · · ( n + h k ) and suppose that for each prime p the number ν p of residue classes occupied by the elements of H is at most p. The Hardy-Littlewood
MALL PRIME GAPS IN ABELIAN NUMBER FIELDS conjecture implies that there exist infinitely many n such that the translate set n + H consists entirely of primes.The idea of Goldston, Pintz and Yildirim was to find a nonnegative function α ( n ) such that(2.1) X x ≤ n< x α ( n ) X h ∈ H ̟ ( n + h ) / X x ≤ n< x α ( n )is big for x → ∞ . This would be useful for the following reason. If we could choosea function α ( n ) such that for x ≫ X x ≤ n< x α ( n ) X h ∈ H ̟ ( n + h ) / X x ≤ n< x α ( n ) ≥ (1 + ǫ ) ln x, then it would follow that there would be at least two primes in the set n + H , forsome n ∈ [ x, x ), the gap being bounded by h k − h . Letting x → ∞ would provethe existence of bounded differences between consecutive primes. Motivated by theideas behind the Selberg sieve, they put α ( n, H ) = ( X d | P ( n,H ) λ d ) , with λ d given bythe formula λ d = ( k + l )! µ ( d )(log Rd ) k + l if d ≤ R l above is positive but o ( k ). This choice willbecome clearer later in the paper.In order to estimate the ratio (2.1), they invoked the Bombieri-Vinogradov theo-rem, which concerns the distribution of primes in arithmetic progressions. To statethe Bombieri-Vinogradov theorem, we must introduce some notation. Define theclassical von Mangoldt functionΛ( n ) = ( log p n = p a ψ ( x ; q, a ) = X n ≤ xn ≡ a (mod q ) Λ( n )for the arithmetic progression a mod q. Define the error quantity E ( x, q ) = max ( a,q )=1 max y ≤ x | ψ ( y ; q, a ) − yφ ( q ) | . Then the Bombieri-Vinogradov theorem asserts that the error term above is assmall on average over q as the Riemann hypothesis would predict. Theorem 2.1 (Bombieri-Vinogradov) . For a real number A , there exists a B suchthat X q< x / x ) B E ( x, q ) ≪ x (log x ) A . Let θ be the supremum of all θ ′ such that X q We call θ the level of distribution of primes in arithmetical progressions. Notethat the Bombieri-Vinogradov theorem implies that θ ≥ . The Elliot-Halberstamconjecture claims that θ = 1. In [5], it is shown that the truth of the Elliott-Halberstam conjecture implies that lim inf n →∞ ( p n +1 − p n ) ≤ Some estimates needed for an abelian extension In this section, we will prove the main lemma we will use in the proof of (1.1)and (1.2).3.1. Main ideas used in the proof. It is known that the primes that split com-pletely in an abelian extension are described, with finitely many exceptions, bycongruence conditions. Therefore, we need to prove that the sequence of rationalprime numbers ( q n ) with q n ≡ a (mod m ), for fixed a, m with ( a, m ) = 1 satisfies(1.1).As in [5], we take H to be a a set of k positive integers H = { h , h , . . . , h k } andwe impose the extra condition that H mod m = { } . Let(3.1) m = p a · · · p a r r be the prime factorization of m . Choose(3.2) α ( n, H ) = ( X d | P ( n,H ) λ d ) , with λ d given by the formula(3.3) λ d = ( k + l )! µ ( d )(log Rd ) k + l if d ≤ R R = R ( x ).We will estimate X x ≤ n< xn ≡ a (mod m ) α ( n ) and X x ≤ n< xn ≡ a (mod m ) α ( n ) X h ∈ H ̟ ( n + h ), and as-suming the Elliott-Halberstam conjecture, we will prove that X x ≤ n< xn ≡ a (mod m ) α ( n ) X h ∈ H ̟ ( n + h ) / X x ≤ n< xn ≡ a (mod m ) α ( n ) ≥ (1 + ǫ ) log x. Estimates. For a multiplicative function ρ on the squarefree integers, define S ( ρ ) = Y p prime (cid:18) − ρ ( p ) p (cid:19) (cid:18) − p (cid:19) − k . We impose the congruence condition H mod m = { } and we evaluate X x ≤ n< xn ≡ a (mod m ) α ( n )and X x ≤ n< xn ≡ a (mod m ) α ( n ) ̟ ( n + h ), for h not belonging to H and h ≡ m ),and for h ∈ H .The estimates are given by the following lemma. MALL PRIME GAPS IN ABELIAN NUMBER FIELDS Lemma 3.1. Let H be a set of k nonnegative integers such that H mod m = { } and max h ∈ H h ≪ log R. Let ǫ > and α ( n ) and λ d be as given by formula (3.3) . Then for R, x → ∞ such that log x ≪ log R , we have: (1) Provided R ≤ x / − ǫ , (3.4) X x ≤ n< xn ≡ a (mod m ) α ( n ) = xm (log R ) k +2 l ( k + 2 l )! (cid:18) ll (cid:19) ( S ( ρ ) + o (1)) , where the multiplicative function ρ ( p ) is defined by: ρ ( p ) = ( if p = p i , for i ∈ { , . . . , r } ν p if p prime = p i , for any i ∈ { , . . . , r } and extended multiplicatively to squarefree integers. Recall that p i are theprimes that appear in the factorization of m , as defined in (3.1) . (2) For an integer h / ∈ H such that h ≡ m ) , provided R ≤ x θ/ − ǫ :(3.5) X x ≤ n< xn ≡ a (mod m ) α ( n ) ̟ ( n + h ) = xφ ( m ) (log R ) k +2 l ( k + 2 l )! (cid:18) ll (cid:19) ( S ( ρ ) + o (1)) , where the multiplicative function ρ ( p ) is defined by ρ ( p ) = if p = p i , for i ∈ { , . . . , r } p ( ν p ( H ∪ { h } ) − φ ( p ) if p prime = p i , for any i ∈ { , . . . , r } . Recall that θ is the level of distribution of primes in arithmetic progressions,as defined in the previous section. (3) For an integer h ∈ H , provided R ≤ x θ/ − ǫ , (3.6) X x ≤ n< xn ≡ a (mod m ) α ( n ) ̟ ( n + h ) = xφ ( m ) (log R ) ( k +2 l +1) ( k + 2 l + 1)! (cid:18) l + 1) l + 1 (cid:19) ( S ( ρ ) + o (1)) , where the multiplicative function ρ ( p ) is defined by ρ ( p ) = if p = p i , for i ∈ { , . . . , r } p ( ν p ( H ) − φ ( p ) if p prime = p i , for any i ∈ { , . . . , r } . Proof. We will first prove that the main term of the sum (3.4) is of the form xm X d ,d ≤ R λ d λ d ρ ([ d , d ])[ d , d ] , and that the main terms of the sums (3.5) and (3.6) are of the form xφ ( m ) X d ,d ≤ R λ d λ d ρ ([ d , d ])[ d , d ] , ALEXANDRA MUSAT where ρ = ρ for the second sum and ρ = ρ for the third sum. We first evaluatethe following. X x ≤ n< xn ≡ a (mod m ) α ( n ) = X x ≤ n< xn ≡ a (mod m ) X d | P ( n,H )) λ d = X d ,d ≤ R λ d λ d X x ≤ n< xn ≡ a (mod m ) δ =[ d ,d ] | P ( n,H ) . If ( m, δ ) = 1, then there is some prime divisor p i of m ( i ∈ { , . . . , r } ) thatalso divides δ. If δ | P ( n, H ), then there would exist some element h ∈ H for which n + h ≡ p i ). Since H was chosen so that H mod m = { } , it follows that n ≡ p i ), which contradicts the condition n ≡ a (mod m ) , ( a, m ) = 1 . Hencewe have the following. X d ,d ≤ R λ d λ d X x ≤ n< xn ≡ a (mod m ) δ =[ d ,d ] | P ( n,H ) X d ,d ≤ R λ d λ d X x ≤ n< xn ≡ a (mod m ) δ | P ( n,H )( m,δ )=1 . Since δ is sqarefree, the condition δ | P ( n, H ) is equivalent to that for each primenumber p | δ, there is an element h ∈ H such that n + h ≡ p ). Since theelements of H occupy ν p congruence classes modulo p , it follows that the condition δ | P ( n, H ) is equivalent to saying that n belongs to one of ν p congruence classesmodulo p . Therefore, the conditions n ≡ a (mod m ) , δ | P ( n, H ) , ( m, δ ) = 1 areequivalent to that n belongs to one of ρ ( δ ) congruence classes modulo [ m, δ ] = mδ, where ρ is defined by: ρ ( p ) = ( p | mν p if p ∤ m. Then X x ≤ n< xn ≡ a (mod m ) δ | P ( n,H )( m,δ )=1 x ρ ( δ ) mδ + r δ , with | r δ | ≤ ρ ( δ ) . Hence we get X x ≤ n< xn ≡ a (mod m ) α ( n ) = X d ,d ≤ R λ d λ d X x ≤ n< xn ≡ a (mod m ) δ | P ( n,H )( m,δ )=1 xm X d ,d ≤ R λ d λ d ρ ( δ ) δ + X d ,d ≤ R λ d λ d r δ . We will show that the error term(3.7) X d ,d ≤ R λ d λ d r δ is o ( x ) for fixed k and l, provided that R ≤ x / − ǫ . MALL PRIME GAPS IN ABELIAN NUMBER FIELDS We have the inequality | r δ | ≤ ρ ( δ ) = Y p | δ ρ ( p ) ≤ Y p | δ k = k ω ( d ) , where ω ( d ) is the number of distinct prime divisors of d. Using the definition of λ d ,we get that λ d ≪ (log R ) k + l . Combining these two relations gives that the errorterm (3.7) is ≪ (log R ) k + l ) X δ ≤ R r δ X d ,d ≤ R [ d ,d ]= δ . Since d , d are squarefree, we have that X d ,d ≤ R [ d ,d ]= δ ω ( d ) , and hence the errorterm (3.7) is ≪ (log R ) k + l ) P δ ≤ R (3 k ) ω ( d ) . Using the estimate X n See [1]. (cid:3) Now we apply the lemma above for f ( x ) = 1 x , a ( n ) = (9 k ) ω ( n ) and y = 1, andwe get that(3.9) X n ≤ x (9 k ) ω ( n ) n = X n ≤ x (9 k ) ω ( n ) x + Z x X n ≤ t (9 k ) ω ( n ) t dt. Using the well-known asymptotics X n ≤ x k ω ( n ) ≪ x L , where L is defined as L = O ((log N ) O (1) ) , and (3.9), we get that P n ≤ x (9 k ) ω ( n ) n ≪ L , hence(3.10) X δ ≤ R (9 k ) ω ( δ ) mδ / ≪ L . Now we will use the asymptotics(3.11) X q The object of this section is to prove (1.2), which is conditional on the Elliott-Halberstam conjecture. We evaluate the ratio X x ≤ n< xn ≡ a (mod m ) α ( n ) X h ∈ H ̟ ( n + h ) / X x ≤ n< xn ≡ a (mod m ) α ( n ) , for the choice of α made.Using the lemma proved in the previous section, we get that for any δ > 0, theinequality X x ≤ n< xn ≡ a (mod m ) α ( n ) X h ∈ H ̟ ( n + h ) / X x ≤ n< xn ≡ a (mod m ) α ( n ) ≥ (cid:18) mφ ( m ) S ( ρ ) S ( ρ ) β − δ (cid:19) log R holds for sufficiently large R, where β = 2 k (2 l + 1)( l + 1)( k + 2 l + 1) . Now by definition S ( ρ ) S ( ρ ) = Y p prime (cid:18) − ρ ( p ) p (cid:19) (cid:18) − p (cid:19) − k +1 Y p prime (cid:18) − ρ ( p ) p (cid:19) (cid:18) − p (cid:19) − k . For any prime p ∤ m we have (cid:18) − ρ ( p ) p (cid:19) (cid:18) − p (cid:19) − k +1 (cid:18) − ρ ( p ) p (cid:19) (cid:18) − p (cid:19) − k = (cid:18) − ν p ( H ) − p − (cid:19) (cid:18) − p (cid:19) − ν p ( H ) p = 1 . MALL PRIME GAPS IN ABELIAN NUMBER FIELDS On the other hand, for p | m we have ν p ( H ) = 1, so that (cid:18) − ρ ( p i ) p i (cid:19) (cid:18) − p i (cid:19) − k +1 (cid:18) − ρ ( p i ) p i (cid:19) (cid:18) − p i (cid:19) − k = 1 − p i . Hence S ( ρ ) S ( ρ ) = Y p prime (cid:18) − ρ ( p ) p (cid:19) (cid:18) − p (cid:19) − k +1 Y p prime (cid:18) − ρ ( p ) p (cid:19) (cid:18) − p (cid:19) − k = r Y i =1 (cid:18) − p i (cid:19) = φ ( m ) m . Thus, we get that X x ≤ n< xn ≡ a (mod m ) α ( n ) X h ∈ H ̟ ( n + h ) / X x ≤ n< xn ≡ a (mod m ) α ( n ) ≥ ( β − δ ) log R. If we assume that θ > and if we let k, l → ∞ with l = o ( k ), then β → . Thusfor a sufficiently small δ, there exists some small ǫ > β − δ ) log R ≥ (1 + ǫ ) log x, for R, x large enough.Combining the above two inequalities gives X x ≤ n< xn ≡ a (mod m ) α ( n ) X h ∈ H ̟ ( n + h ) / X x ≤ n< xn ≡ a (mod m ) α ( n ) ≥ (1 + ǫ ) log x, so any tuple with k elements each congruent to 0 mod m would have infinitely manytranslates containing two primes congruent to a mod m. Therefore,lim inf n →∞ ( q n +1 − q n ) < ∞ . The unconditional Result To obtain the unconditional result(5.1) lim inf n →∞ q n +1 − q n log q n = 0 , we fix δ > , and we define the function g ( n ) = X H ⊆ A | H | = k α ( n, H ) , where A = [1 , mδ log x ] ∩ m Z . We will prove that(5.2) X x ≤ n< xn ≡ m ) g ( n ) X h ≡ a (mod m ) h Lemma 5.1. Let B A ( k ) = X | H | = kH ⊆ A S ( H ) , where all sets H ⊆ A ⊆ [1 , N ] are counted with k ! multiplicity and | A | = h. Let S ∗ A ( k ) = B A ( k ) h k . If k < ǫ ( h ) h/ log N, then S ∗ A ( k + 1) ≥ S ∗ A ( k )(1 + O ( ǫ ( h )) + O ( 1log N )) . MALL PRIME GAPS IN ABELIAN NUMBER FIELDS We can apply the above lemma for A = [1 , mδ log x ] ∩ m Z , for which | A | = δ log x + O (1) . By the lemma, we have that B A ( k ) / | A | k is, apart from a factor of1 + o (1), non-decreasing as a function of k. We divide (5.4) by P | H | = kH ⊆ A S ( H ) . Thus,(5.4) is equivalent to proving that β log R + δ log X ≥ (1 + ǫ ) log x. Taking β → R → x θ/ = x / , notice that the above is true for any ǫ < δ. Hence we get that lim inf n →∞ q n +1 − q n log q n ≤ δ, and since δ can be chosen to be arbitrarly small, we get the desired result.6. An alternate approach for quadratic fields We will now give a second proof of (1.2) in the special case when K is a quadraticextension of Q of class number 1. The proof relies on a generalization of theBombieri-Vinogradov theorem for the case of a quadratic number field, proven byE. Fogels in [4].Although the result we prove in this section is a special case of (1.2), the methodwe use could potentially be adapted to prove results about small prime gaps in moregeneral number fields.We first give a brief overview of Fogels’ result. In [4], Fogels gives a generalizationof the Bombieri-Vinogradov theorem in the special case of a quadratic number field,in which he considers rational primes that split completely in the number field.Let K be a fixed quadratic field. R stands for classes of ideals. For any naturalnumber q , we consider the group formed by the reduced classes of residues l mod q formed by the residues of the idealnorms N ( a ) with ( a , [ q ]) = 1 and a belonging tothe principal ideal class R . Denote by φ ( q ) the order of this group and by h thenumber of classes of the ideal class group . Then Fogels’ result is the following. Theorem 6.1 (Fogels) . Let a = a ( q, R ) be a normresidue (mod q ) with ( a, q ) = 1 for the class R of ideals in the quadratic number field K and let π ( x ; R , q, a ) denotethe number of primes p ≤ x such that p ≡ a (mod q ) and such that p = N ( p ) with p ∈ R . Then for any constant A > there is a corresponding constant B > suchthat X R X q ≤ z / / (log z ) − B max a ( q, R ) max x ≤ z | π ( x ; R , q, a ) − hφ ( q ) Li ( x ) | ≪ z (log z ) A , with the constant in the notation depending merely on A and the discriminant ofthe field. To prove (1.1) for the case of a quadratic number field, we proceed as follows. Proof. Let K be a fixed quadratic field of discriminant D with class number h = 1,and we take H = { h , . . . , h k } to be a set of nonnegative distinct integers such that H mod D = { } . We define the characteristic function f on the integers by f ( n ) = ( n is prime and splits completely in K Let α ( n ) be as defined in (3.2). We will estimate X x ≤ n< x α ( n ). We will also estimate X x ≤ n< x α ( n ) f ( n + h ), for h an integer not belonging to H such that h ≡ D ) and for h belonging to H , respectively.Using the estimate in the original paper we get that(6.1) X x ≤ n< x α ( n ) = x (log R ) k +2 l ( k + 2 l )! (cid:18) ll (cid:19) ( S ( H ) + o (1)) . Now let h / ∈ H be an integer such that h ≡ D ). We will denote by δ = [ d , d ]. We define π ∗ ( x ; q, a ) by π ∗ ( x ; q, a ) = π (2 x ; q, a ) − π ( x ; q, a ) , where π ( x ; q, a ) denotes the number of primes p ≤ x such that p ≡ a (mod q ) and suchthat p = N ( p ) for p an ideal in the number field (which is a principal ideal, since K has class number one). Then we have X x ≤ n< x α ( n ) f ( n + h ) = X x ≤ n< x f ( n + h ) X d | P ( n,H ) λ d = X x ≤ n< x f ( n + h ) X δ | P ( n,H ) λ d λ d = X d ,d ≤ R λ d λ d X x ≤ n< xδ | P ( n,H ) f ( n + h )= X d ,d ≤ R λ d λ d X m = n mod δδ | P ( m,H )( δ,m + h )=1 π ∗ ( X ; δ, m + h )= X d ,d ≤ R λ d λ d X m = n mod δδ | P ( m,H ) Li( x ) φ ( δ )+ X d ,d ≤ R λ d λ d X m = n mod δδ | P ( m,H ) π ∗ ( x ; δ, m + h ) − Li( x ) φ ( δ ) . Now we will evaluate the main term(6.2) X d ,d ≤ R λ d λ d X m = n mod δδ | P ( m,H ) Li( x ) φ ( δ )and the error term(6.3) X d ,d ≤ R λ d λ d X m = n mod δδ | P ( m,H ) π ∗ ( x ; δ, m + h ) − Li( x ) φ ( δ ) . The conditions on n in the main term (6.2) above are equivalent to that n belongsto one of ν p ( H ∪ { h } ) − p , by a similar argument as MALL PRIME GAPS IN ABELIAN NUMBER FIELDS when evaluating (3.8). We define the multiplicative function ρ ′ by ρ ′ ( p ) = p ( ν p ( H ∪ { h } ) − φ ( p ) , for p primeand extend it multiplicatively to squarefree integers. Since for ( m, δ ) = 1 we havethat φ ( mδ ) = φ ( m ) φ ( δ ), the main term (6.2) is(6.4) Li( x ) X d ,d ≤ R λ d λ d ρ ′ ([ d , d ])[ d , d ] ∼ Li( x ) (log R ) k +2 l ( k + 2 l )! (cid:18) ll (cid:19) S ( ρ ′ ) , where the last asymptotics follows using the result proven in the original paper [5].We have the following identity S ( ρ ′ ) = Y p prime (cid:18) − ρ ′ ( p ) p (cid:19) (cid:18) − p (cid:19) − k (6.5) = Y p prime (cid:18) − ν p ( H ∪ { h } ) φ ( p ) (cid:19) (cid:18) − p (cid:19) − k . When p | D we get that ν p ( H ∪ { h } ) − h ≡ D ), and H waschosen such that H mod D = { } .When p ∤ D it follows by a well-known result that φ ( p ) = p − 1. Hence, (6.5)becomes S ( ρ ′ ) = Y p | D (cid:18) − p (cid:19) − k Y p ∤ D (cid:18) − ν p ( H ∪ { h } ) p − (cid:19) (cid:18) − p (cid:19) − k . When p | D , (cid:18) − p (cid:19) − k = (cid:18) − ν p ( H ∪ { h } ) p (cid:19) (cid:18) − p (cid:19) − ( k +1) . When p ∤ D we have that (cid:18) − ν p ( H ∪ { h } ) − p − (cid:19) (cid:18) − p (cid:19) − k = p − ν p ( H ∪ { h } ) p − · p − p (cid:18) − p (cid:19) − k − = (cid:18) − ν p ( H ∪ { h } ) p (cid:19) (cid:18) − p (cid:19) − ( k +1) . Combining the above two identities gives that S ( ρ ′ ) = S ( H ∪ { h } ). Hence themain term (6.4) is(6.6) ∼ Li( x ) (log R ) k +2 l ( k + 2 l )! (cid:18) ll (cid:19) S ( H ∪ { h } ) . Now we will prove that the error term (6.3) is o ( x ) . Let E ( x ; q, a ) = (cid:12)(cid:12)(cid:12)(cid:12) π ∗ ( x ; q, a ) − Li( x ) φ ( q ) (cid:12)(cid:12)(cid:12)(cid:12) . Following a similar argument as in evaluating the error of the term (3.5), we getthat (6.3) is ≪ (log R ) k + l ) X d ,d ≤ R X m = n mod δδ | P ( m,H ) E ( x ; δ, m + h ) ≪ (log R ) k + l ) X δ ≤ R (3 k ) ω ( δ ) max E ( x ; δ, m + h ) . The Cauchy-Schwarz inequality shows that the above is(6.7) ≪ (log R ) k + l ) X δ ≤ R (9 k ) ω ( δ ) δ / X δ ≤ R δ max E ( X ; δ, m + h ) / . Now, using (3.10) we get that X δ ≤ R (9 k ) ω ( δ ) δ / ≪ L , where we recall that L = O ((log N ) O (1) ) . Using the asymptotics π ( x ; q, a ) ≪ xq , for q ≪ x, and Li ( x ) φ ( q ) ≪ xq (which both follow by similar arguments as the ones used in evaluating the error ofthe term (3.5)), we get that(6.8) E ( x ; δ ) ≪ xδ , where E ( x, q ) = max a max z ≤ x E ( z ; q, a ) . Now we define θ to be the supremum of all θ ′ for which X q ≤ x θ ′ E ( x, q ) ≪ x/ (log x ) A , for fixed A . Notice that by Theorem 6, it follows that θ ≥ / . Hence, using theabove asymptotics (6.8), we get that (6.7) is equivalent to that the error term is(6.9) ≪ L (cid:18) x L x (log x ) A (cid:19) / ≪ x (log x ) A ′ , for R ≪ x θ − ǫ . Thus, the error term is o ( x ).Since Li ( x ) ∼ x log x , for h / ∈ H such that h ≡ D ), for any ǫ > 0, weget the estimate(6.10) X x ≤ n< x α ( n ) f ( n + h ) = Li( x ) (log R ) k +2 l ( k + 2 l )! (cid:18) ll (cid:19) ( S ( H ∪ { h } ) + o ( 1log x )) , provided R ≤ x θ/ − ǫ . We use the same method to evaluate X x 0, we havethe following asymptotics(6.11) X x ≤ n< x α ( n ) f ( n + h ) = Li ( x ) (log R ) k +2 l +1 ( k + 2 l + 1)! (cid:18) l + 2 l + 1 (cid:19) ( S ( H ) + o ( 1log x )) , provided R ≤ x θ/ − ǫ . MALL PRIME GAPS IN ABELIAN NUMBER FIELDS To get the desired result, we evaluate the ratio X x ≤ n< x α ( n ) X h ∈ H f ( n + h ) / X x ≤ n< x α ( n ) . Using the asymptotics (6.11) and (6.1), it follows that for any δ > 0, the inequality X x ≤ n< x α ( n ) X h ∈ H f ( n + h ) / X x ≤ n< x α ( n ) ≥ β Li ( x ) x log R − δ, holds for sufficiently large R , where β = 2 k (2 l + 1)( l + 1)( k + 2 l + 1) . Since Li ( x ) ∼ x log x , we have X x ≤ n< x α ( n ) X h ∈ H f ( n + h ) / X x ≤ n< x α ( n ) ≥ β log R log x − δ. Notice that if we let k, l → ∞ with l = o ( k ), then β → 4. If we assume an Elliott-Halberstam type conjecture for Fogels’ generalization, then we can take θ > , andhence we get that X x ≤ n< x α ( n ) X h ∈ H f ( n + h ) / X x ≤ n< x α ( n ) ≥ − δ. Letting δ → X x ≤ n< x α ( n ) X h ∈ H f ( n + h ) / X x ≤ n< x α ( n ) ≥ , for R, x large enough. Letting x → ∞ shows that for any x , there is some n ∈ [ x, x )such that n + h i and n + h j split completely in the number field K , for some i, j ∈ { , . . . , k } . Hence, for any x , there are two primes that split completelybetween x and 2 x , whose difference is bounded by h k − h . This proves the existenceof infinitely many primes that split completely, whose differences are bounded.The proof of the unconditional result (5.1) follows the same outline as the proofin section 2.4. (cid:3) Acknowledgements The research for this paper has been done during the author’s SURF (SummerUndergraduate Research Fellowship) at Caltech. The author would like to thankDinakar Ramakrishnan and Paul Nelson for their guidance of the project and theCaltech SURF office for the funding.Department of Mathematics, California Institute of Technology, Pasadena, Califor-nia 91125 Email address: [email protected] References [1] Tom M. Apostol. Introduction to analytic number theory . Springer-Verlag, New York, 1976.Undergraduate Texts in Mathematics.[2] E. Bombieri and H. Davenport. Small differences between prime numbers. Proc. Roy. Soc.Ser. A , 293:1–18, 1966.[3] P. Erd¨os. On the difference of consecutive primes. Bull. Amer. Math. Soc. , 54:885–889, 1948.[4] E. Fogels. A mean value theorem of Bombieri’s type. Acta Arith. , 21:137–151, 1972.[5] Daniel A. Goldston, J´anos Pintz, and Cem Y. Yıldırım. Primes in tuples. I. Ann. of Math.(2) , 170(2):819–862, 2009.[6] Daniel A. Goldston, J´anos Pintz, and Cem Yal¸cin Yıldırım. Primes in tuples. II. Acta Math. ,204(1):1–47, 2010.[7] Helmut Maier. Small differences between prime numbers.