On the Ranks of the 2-Selmer Groups of Twists of a Given Elliptic Curve
aa r X i v : . [ m a t h . N T ] S e p On the Ranks of the 2-Selmer Groups of Twistsof a Given Elliptic Curve
Daniel M. KaneStanford University Department of MathematicsBuilding 380, Sloan HallStanford, CA 94305 [email protected]
October 16, 2018
Abstract
In [8], Swinnerton-Dyer considered the proportion of twists of an el-liptic curve with full 2-torsion that have 2-Selmer group of a particulardimension. Swinnerton-Dyer obtained asymptotic results on the numberof such twists using an unusual notion of asymptotic density. We build onthis work to obtain similar results on the density of twists with particularrank of 2-Selmer group using the natural notion of density.
Let c , c , c be distinct rational numbers. Let E be the elliptic curve definedby the equation y = ( x − c )( x − c )( x − c ) . We make the additional technical assumption that none of the ( c i − c j )( c i − c k )are squares. This is equivalent to saying that E is an elliptic curve over Q withcomplete 2-torsion and no cyclic subgroup of order 4 defined over Q . For b asquare-free number, let E b be the twist defined by the equation y = ( x − bc )( x − bc )( x − bc ) . Let S be a finite set of places of Q including 2 , ∞ and all of the places at which E has bad reduction. Let D be a positive integer divisible by 8 and by theprimes in S . Let S ( E b ) denote the 2-Selmer group of the curve E b . We willbe interested in how the rank varies with b and in particular in the asymptoticdensity of b ’s so that S ( E b ) has a given rank.The parity of dim( S ( E b )) depends only on the class of b as an element of Q ν ∈ S Q ∗ ν / ( Q ∗ ν ) . We claim that for exactly half of these values this dimension isodd and exactly half of the time it is even. In particular, we make the followingclaim which will be proved later. 1 emma 1. There exists a set S consisting of exactly half of the classes c in ( Z /D ) ∗ / (( Z /D ) ∗ ) , so that for any positive integer b relatively prime to D wehave that dim( S ( E b )) is even if and only if b represents a class in S . We put off the proof of this statement until Section 4.Let b = p p . . . p n where p i are distinct primes relatively prime to D . In[8] the rank of S ( E b ) is shown to depend only on the images of the p i in( Z /D ) ∗ / (( Z /D ) ∗ ) and upon which p i are quadratic residues modulo which p j .There are 2 n | S | + ( n ) possible sets of values for these. Let π d ( n ) be the fractionof this set of possibilities that cause S ( E b ) to have rank exactly d . Then themain theorem of [8] together with Lemma 1 implies that: Theorem 2. lim n →∞ π d ( n ) = α d . where α = α = 0 and α n +2 = n Q nj =1 (2 j − Q ∞ j =0 (1+2 − j ) . The actual theorem proved in [8] says that if, in addition, the class of b in Q ν ∈ S Q ∗ ν / ( Q ∗ ν ) is fixed, then the analogous π d ( n ) either converge to 2 α d for d even and 0 for d odd, or to 2 α d for d odd and 0 for d even.This tells us information about the asymptotic density of twists of E whose2-Selmer group has a particular rank. Unfortunately, this asymptotic densityis taken in a somewhat awkward way by letting the number of primes dividing b go to infinity. In this paper, we prove the following more natural version ofTheorem 2: Theorem 3.
Let E be an elliptic curve over Q with full 2-torsion defined over Q so that in addition for E we have that lim n →∞ π d ( n ) = α d . With α d as given in Theorem 2. Then lim N →∞ { b ≤ N : b square-free , ( b, D ) = 1 and dim( S ( E b )) = d } { b ≤ N : b square-free and ( b, D ) = 1 } = α d . Applying this to twists of E by divisors of D and noting that twists bysquares do not affect the Selmer rank we have that Corollary 4. lim N →∞ { b ≤ N : dim( S ( E b )) = d } N = α d . and Corollary 5. lim N →∞ {− N ≤ b ≤ N : dim( S ( E b )) = d } N = α d . E b can be expressed as theintersection of two Lagrangian subspaces, U and W , of a particular symplecticspace, V , over F . Although U, V and W all depend on b , once the numberof primes dividing b has been fixed along with its congruence class modulo D , these spaces can all be written conveniently in terms of the primes, p i ,dividing b , which we think of as formal variables. Using the formula | U ∩ W | = √ | V | P u ∈ U,w ∈ W ( − u · w , we reduce our problem to bounding the size of the“characters” ( − u · w when averaged over b . These “characters” turn out to beproducts of Dirichlet characters of the p i and Legendre symbols of pairs of the p i . The bulk of our analytic work is in proving these bounds. These boundswill allow us to discount the contribution from most of the terms in our sum (inparticular the ones in which Legendre symbols show up in a non-trivial way),and allow us to show that the average of the remaining terms is roughly whatshould be expected from Swinnerton-Dyer’s result.We should point out the connections between our work and that of Heath-Brown in [3] where he proves our main result for the particular curve y = x − x. We employ techniques similar to those of [3], but the algebra behind them isorganized significantly differently. Heath-Brown’s overall strategy is again tocompute the average sizes of moments of | S ( E b ) | and use these to get at theranks. He computes | S ( E b ) | using a different formula than ours. Essentiallywhat he does is use some tricks specific to his curve to deal with the conditionsrelating to primes dividing D , and instead of considering each prime individually,he groups them based on how they occur in u and w . He lets D i be the productof all primes dividing b that relate in a particular way (indexed by i ). He thengets a formula for | S ( E b ) | that’s a sum over ways of writing b as a product, b = Q D i , of some term again involving characters of the D i and Legendresymbols. Using techniques similar to ours he shows that terms in this sum wherethe Legendre symbols have a non-negligible contribution (are not all trivial dueto one of the D i being 1) can be ignored. He then uses some algebra to showthat the average of the remaining terms is the desired value. This step differsfrom our technique where we merely make use of Swinnerton-Dyer’s result tocompute our average. Essentially we show that the algebra and the analysisfor this problem can be done separately and use [8] to take care of the algebra.Finally, Heath-Brown uses some techniques from linear algebra to show that themoment bounds imply the correct densities of ranks, while we use techniquesfrom complex analysis.We also note the work of Yu in [9]. In this paper, Yu shows that for a widefamily of curves of full 2-torsion that the average size of the 2-Selmer groupof a twist is equal to 12. This work uses techniques along the lines of Heath-3rown’s, though has some added complication in order to deal with the greatergenerality.One advantage of our technique over these others is that we can, to somedegree, separate the algebra involved in analyzing the sizes of these Selmergroups from the analysis. When considering the distribution of ranks of Selmergroups of twists of an elliptic curve, there are two types of density estimatesthat have come up in the literature. The first is to use the natural notion ofdensity over some obvious ordering of twist parameter. The other is to use somenotion similar to that of Swinnerton-Dyer, which can be thought of as lettingthe number of primes dividing the twist parameter go to infinity. Although oneis usually interested in natural densities, the Swinnerton-Dyer type results areoften easier to prove as they tend to be essentially algebraic in nature, whileresults about natural density will generally require some tricky analytic work.The techniques of this paper show how asymptotics of the Swinnerton-Dyertype can be upgraded to results for natural density. Although we have onlymanaged to carry out this procedure for the family of curves used in Theorem2, there is hope that this procedure might have greater applicability. For ex-ample, if someone were to obtain a Swinnerton-Dyer type result for twists ofan elliptic curve with full 2-torsion over Q that has a rational 4-isogeny, it isalmost certain that the techniques from this paper would allow one to obtain aresult for the same curve using the natural density. Additionally, in [4], Klags-brun, Mazur and Rubin consider the ranks of twists of an elliptic curve withGal( K ( E [2]) /K ) ≃ S , and obtain Swinnerton-Dyer type density results. It ispossible that ideas in this paper may be adapted to improve these results towork with a more natural notion of density as well. Unfortunately, working inthis extended context will likely complicate the analytic aspects of the argumentconsiderably. For example, while we make important use of the fact that therank of S ( E b ) depends only on congruence classes of primes dividing b andLegendre symbols between them, it is shown in [1] that for curves with cycliccubic field of 2-torsion that the Selmer rank can depend on more complicatedalgebraic objects (such as what they term the spin of a prime).In Section 2, we introduce some basic concepts that will be used throughout.In Section 3, we will prove the necessary character bounds. We use these boundsin Section 4 to establish the average moments of the size of the Selmer groups.Finally, in Section 5, we explain how these results can be used to prove our mainTheorem. Throughout the rest of this paper we will make extensive use of O , and similarasymptotic notation. In our notation, O ( X ) will denote a quantity that is atmost H · X for some absolute constant H . If we need asymptotic notation thatdepends on some parameters, we will use O a,b,c ( X ) to denote a quantity that is4t most H ( a, b, c ) · X , where H is some function depending only on a, b and c . In order to make use of Swinnerton-Dyer’s result, we will need to consider twistsof E by integers b ≤ N with a specific number of prime divisors. For an integer m , we let ω ( m ) be the number of prime divisors of m . In our analysis, we willneed to have estimates on the number of of such b with a particular number ofprime divisors. We defineΠ n ( N ) = { primes p ≤ N so that ω ( p ) = n } . In order to deal with this we use Lemma A of [2] which states:
Lemma 6.
There exist absolute constants C and K so that for any ν and x Π ν +1 ( x ) ≤ Kx log( x ) (log log x + C ) ν ν ! . By maximizing the above in terms of ν it is easy to see that Corollary 7. Π n ( N ) = O N p log log( N ) ! . It is also easy to see from the above that most integers of size roughly N have about log log( N ) prime factors. In particular: Corollary 8.
There is a constant c > so that for all N , the number of b ≤ N with | ω ( b ) − log log( N ) | > log log( N ) / is at most N exp (cid:16) − c p log log( N ) (cid:17) . In particular, the fraction of b ≤ N with | ω ( b ) − log log( N ) | < log log( N ) / goesto 1 as N goes to infinity. We will use Corollary 8 to restrict our attention only to twists by b with anappropriate number of prime divisors. Our main purpose in this section will be to prove the following Propositions:
Proposition 9.
Fix positive integers
D, n, N with | D , log log N > , and (log log N ) / < n < N , and let c > be a real number. Let d i,j , e i,j ∈ Z / for i, j = 1 , . . . , n with e i,j = e j,i , d i,j = d j,i , e i,i = d i,i = 0 for all i, j . Let χ i be a quadratic character with modulus dividing D for i = 1 , . . . , n . Let m bethe number of indices i so that at least one of the following hold: e i,j = 1 for some j or • χ i has modulus not dividing 4 or • χ i has modulus exactly 4 and d i,j = 0 for all j .Let ǫ ( p ) = ( p − / . Then if m > (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n ! X S N,n,D Y i χ i ( p i ) Y i Let n, N, D be positive integers with log log N > , and (log log N ) / < n < N . Let G = (( Z /D ) ∗ / (( Z /D ) ∗ ) ) n . Let f : G → C be a function with | f | ∞ ≤ . Then n ! X S N,n,D f ( p , . . . , p n ) = | G | X g ∈ G f ( g ) (cid:18) | S N,n,D | n ! (cid:19) + O D (cid:18) N (log log log N )log log N (cid:19) . (2) (Above f ( p , . . . , p n ) is really f applied to the vector of their reductions modulo D ). This Proposition says that the average of f over such S N,n,D is roughly equalto the average of f over G . This will allow us to show that the average valueof the remaining terms in our moment calculation equal what we would expectgiven Swinnerton-Dyer’s result.We begin with a Proposition that gives a more precise form of Proposition9 in the case when the e i,j are all 0. Proposition 11. Let D, n, N be integers with | D with log log N > . Let C > be a real number. Let d i,j ∈ Z / for i, j = 1 , . . . , n with d i,j = d j,i , d i,i = 0 .Let χ i be a quadratic character of modulus dividing D for i = 1 , . . . , n . Suppose hat no Dirichlet character of modulus dividing D has an associated Siegel zerolarger than − β − . Let B = max( e ( C +2) β log log N , e K ( C +2) (log D ) (log log( DN )) , n log C +2 ( N )) for K a sufficiently large absolute constant. Suppose that B n < √ N . Let m bethe number of indices i so that either: • χ i does not have modulus dividing 4 or • χ i has modulus exactly 4 and d i,j = 0 for all j .Then (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n ! X S N,n,D Y i χ i ( p i ) Y i 0, there is some p i so that no matter how weset the other p j , our character still depends on p i . We split into cases based onwhether p i > B . If p i > B , we fix the values of the other p j , and use bounds oncharacter sums. For p i ≤ B , we note that this happens for only about a log log Bn fraction of the terms in our sum, and for each possible value of p i inductivelybound the remaining sum. To deal with the first case we prove the followingLemma: Lemma 12. Let K be a sufficiently large constant. Take χ any non-trivialDirichlet character of modulus at most D and with no Siegel zero more than − β − , N, C > integers, and X any integer with X > max( e ( C +2) β log log N , e K ( C +2) (log D ) (log log( DN )) ) . Then (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X p ≤ X χ ( p ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ O ( X log − C − ( N )) . Where the sum above is over primes p less than or equal to X .Proof. Theorem 5.27 of [5] implies that for any Y that for some constant c > X n ≤ Y χ ( n )Λ( n ) = Y · O Y − β − + exp − c p log( Y )log D ! (log D ) ! . n a power of a prime is O ( √ Y ). Using Abel summation to reduce this to a sum over p of χ ( p ) ratherthan χ ( p ) log( p ), we find that X p ≤ X χ ( p ) ≤ X · O X − β − + exp − c p log( X )log D ! (log D ) ! + O ( √ X ) . The former term is sufficiently small since by assumption X > e ( C +2) β log log N .The latter term is small enough since X > e K ( C +2) (log D ) (log log( DN )) . The lastterm is small enough since clearly X > log C +4 ( N ).For positive integers n, N, D , and S a set of prime numbers, denote by Q ( n, N, D, k, S ) the maximum possible absolute value of a sum of the formgiven in Equation (3) with m ≥ k , with the added restriction that none of the p i lie in S . In particular a sum of the form1 n ! X S N,n,D ′ Y i χ i ( p i ) Y i Lemma 13. For integers, n, D, N, M, C and B with B > max( e ( C +2) β log log M , e K ( C +2) (log D ) (log log( DM )) , n log C +2 ( M )) , where − β − is the largest Siegel zero of a Dirichlet character of modulusdividing D and K a sufficiently large constant, ≤ k ≤ n and S a set of primes ≤ B , then Q ( n, N, D, k, S ) as described above is at most O ( N log( N ) log − C − ( M )) + 1 n X p Since k ≥ 1, there must be an i so that either χ i has modulus bigger than4 or has modulus exactly 4 and all of the d i,j are 0. Without loss of generality, n is such an index. We split our sum into cases depending on whether p n ≥ B .For p n ≥ B , we proceed by fixing all of the p j for j = n and summing over p n .Letting P = Q n − i =1 p i , we have N/B X P =1 n ! X P = p ...p n − p i distinct p i S ( D,P )=1 a X B ≤ p n ≤ N/Pp n = p j χ ( p n )8here a is some constant of norm 1 depending on p . . . p n − , and χ is a non-trivial character of modulus dividing D , perhaps also depending on p , . . . , p n − .The condition that p n = p j alters the value of the inner sum by at most n . Withthis condition removed, we may bound the inner sum by applying Lemma 12(taking the difference of the terms with X = N/P and X = B ). Hence thevalue of the inner sum is at most O ( N/P log − C − ( M ) + n ). Since N/P ≥ B ≥ n log C +2 ( M ), this is just O ( N/P log − C − ( M )). Note that for each P , there areat most ( n − n − /n factor, the sumabove is at most N/B X P =1 O ( N/P log − C − ( M )) = O ( N log( N ) log − C − ( M )) . For p n < B , we fix p n and consider the sum over the remaining p i . We notethat for p a prime not in S and relatively prime to D , this sum is plus or minusone over n times a sum of the type bounded by Q ( n − , N/p, D, k − , S ∪{ p } ). Inparticular, we note that since by assumption the value of m for our original sumwas at least k , that upon fixing this value of p n , the value of m for the resultingsum is at least k − Q ( n − , N/p, D, k − , S ∪ { p } ).This completes our proof.We are now prepared to Prove Proposition 11 Proof. We prove by induction on k that for n, N, D, C, M, β, B as above with B > max( e ( C +2) β log log M , e K ( C +2) (log D ) (log log( DM )) , n log C +2 ( M )) , and S a set of primes ≤ B , and c a sufficiently large constant that Q ( n, N, D, k, S ) ≤ c N p log log( N/B n ) ! (cid:18) c log log Bn (cid:19) k (4)+ cN log( N ) log − C − ( M ) k − X a =0 (cid:18) c log log Bn (cid:19) a . Plugging in M = N , k = m , S = ∅ , and B = max( e ( C +2) β log log N , e K ( C +2) (log D ) (log log( DN )) , n log C +2 ( N )) , yields the necessary result.We prove Equation 4 by induction on k . For k = 0, the sum is at most thesum over b = p . . . p n with appropriate conditions of n ! . Since each such b canbe written as such a product in at most n ! ways, this is at most Π n ( N ), whichby Corollary 7 is at most c (cid:16) N √ log log N (cid:17) for some constant c , as desired.9or larger values of k , we use the inductive hypothesis and Lemma 13 tobound Q ( n, N, D, k, S ) by cN log( N ) log − C − ( M ) + 1 n X p
First note that we can assume that 4 | D . This is because if that is notthe case, we can split our sum up into two cases, one where none of the p i are2, and one where one of the p i is 2. In either case we get a sum of the sameform but now can assume that D is divisible by 4. We assume this so that wecan use Proposition 11.It is clear that the difference between the left hand side of Equation 2 andthe main term on the right hand side is1 | G | X χ ∈ b G \{ } n ! X S N,n,D χ ( p , . . . , p n ) X g ∈ G f ( g ) χ ( g ) . | G | p | G || f | X χ ∈ b G \{ } (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n ! X S N,n,D χ ( p , . . . , p n ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) / . We note that | f | ≤ p | G | and hence that | G | p | G || f | ≤ 1. Bounding thecharacter sum using Proposition 11 (using the minimal possible value of B ), weget O (cid:16) N log log N (cid:17) times X χ ∈ b G \{ } O D (cid:18) log log log N log log N (cid:19) s . Where above s is the number of components on which χ (thought of as a productof characters of ( Z /D Z ) ∗ ) is non-trivial. Since each component of χ can either betrivial or have one of finitely many non-trivial values (each of which contributes O D ((log log log N ) / (log log N ) )) and this can be chosen independently for eachcomponent, the inner sum is O D (cid:18) log log log N log log N (cid:19) ! n − (cid:18) O D (cid:18) (log log log N ) log log N (cid:19)(cid:19) − O D (cid:18) (log log log N ) log log N (cid:19) . Hence the total error is at most1 | G | p | G | p | G | O D (cid:18) N log log log ( N )log log ( N ) (cid:19) / ! = O D (cid:18) N log log log( N )log log( N ) (cid:19) . The proof of Proposition 9 is along the same lines as the proof of Proposition11. Again we induct on m . This time, we use Lemma 13 as our base case (whenall of the e i,j are 0). If some e i,j is non-zero, we break into cases based onwhether or not p i and p j are larger than some integer A (which will be somepower of log( N )). If both, p i and p j are large, then fixing the remaining primesand summing over p i and p j gives a relatively small result. Otherwise, fixingone of these primes at a small value, we are left with a sum of a similar formover the other primes. Unfortunately, doing this will increase our D by a factorof p i , and may introduce characters with bad Siegel zeroes. To counteract this,we will begin by throwing away all terms in our sum where D Q i p i is divisibleby the modulus of the worst Siegel zero in some range, and use standard resultsto bound the badness of other Siegel zeroes.We begin with some Lemmas that will allow us to bound sums of Legendresymbols of p i and p j as they vary over primes.11 emma 14. Let Q and N be positive integers with Q ≥ N . Let a be a function { , , . . . , N } → C , supported on square-free numbers. Then we have that X χ quadratic characterof modulus p or p, ≤ Q (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) N X n =1 a n χ ( n ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = O (cid:16) Q √ N || a || (cid:17) where the sum is over quadratic characters whose modulus is either a prime orfour times a prime and is less than or equal to Q , and where || a || = P Nn =1 | a n | is the squared L norm. Note the similarity between this and Lemma 4 of [3]. Proof. Let M be the largest positive integer so that Q ≤ N M ≤ Q . Let b : { , , . . . , M } → C be the function b n = M and b = 0 on non-squares.Let c = a ∗ b be the multiplicative convolution of a and b . Note that since a is supported on square-free numbers and b supported on squares that || c || = || a || || b || = || a || /M . Applying the multiplicative large sieve inequality (see [5]Theorem 7.13) to c we have that X q ≤ Q qφ ( q ) ∗ X χ mod q (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)X n c n χ ( n ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ ( Q + N M − || c || . (5)The right hand side is easily seen to be O ( Q ) || a || /M = O ( Q || a || / ( p Q /N )) = O ( Q √ N || a || ) . For the left hand side, we may note that it only becomes smaller if we removethe qφ ( q ) or ignore the characters that are not quadratic or do not have modulieither a prime or four times a prime. For such characters χ note that X n c n χ ( n ) = X n a n χ ( n ) ! X n b n χ ( n ) ! = Ω( X n a n χ ( n )) . Where the last equality above follows from the fact that χ is 1 on squares notdividing its modulus, and noting that since its modulus divides four times aprime, the latter case only happens at even numbers of multiples of p . Hencethe left hand side of Equation 5 is at least a constant multiple of X χ quadratic characterof modulus p or 4 p, ≤ Q (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) N X n =1 a n χ ( n ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . This completes our proof. 12 emma 15. Let A ≤ X be positive numbers, and let a, b : Z → C be functionsso that | a ( n ) | , | b ( n ) | ≤ for all n . We have that (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X A ≤ p ,p p p ≤ X a ( p ) b ( p ) (cid:18) p p (cid:19)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = O ( X log( X ) A − / ) . Where the above sum is over pairs of primes p i bigger than A with p p ≤ X ,and where (cid:16) p p (cid:17) is the Legendre symbol.Proof. We first bound the sum of the terms for which p ≤ √ X .We begin by partitioning [ A, √ X ] into O ( A / log( X )) intervals of the form[ Y, Y (1 + A − / )). We break up our sum based on which of these intervals p lies in. Once such an interval is fixed, we throw away the terms for which p ≥ X/ ( Y (1 + A − / )). We note that for such terms p p ≥ X (1 + A − / ) − .Therefore the number of such terms in our original sum is at most O ( XA − / ),and thus throwing these away introduces an error of at most O ( XA − / ).The sum of the remaining terms is at most X A ≤ p ≤ X/ ( Y (1+ A − / )) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X Y ≤ p ≤ Y (1+ A − / ) a ( p ) (cid:18) p p (cid:19)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . By Cauchy-Schwarz, this is at most p X/Y X A ≤ p ≤ X/ ( Y (1+ A − / )) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X Y ≤ p ≤ Y (1+ A − / ) a ( p ) (cid:18) p p (cid:19)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) / . In the evaluation of the above, we may restrict the support of a to primesbetween Y and Y (1 + A − / ). Therefore, by Lemma 14, the above is at most p X/Y · O (cid:18)q ( X/Y ) Y / ( Y A − / ) (cid:19) = O (cid:16) XY − / A − / (cid:17) = O ( XA − / ) . Hence summing over the O ( A / log( X )) such intervals, we get a total contri-bution of O ( X log( X ) A − / ) . We get a similar bound on the sum of terms for which p ≤ √ X . Finally weneed to subtract off the sum of terms where both p and p are at most √ X .This is X A ≤ p ≤√ X X A ≤ p ≤√ X a ( p ) b ( p ) (cid:18) p p (cid:19) . This is at most X A ≤ p ≤√ X (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X A ≤ p ≤√ X a ( p ) (cid:18) p p (cid:19)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . 13y Cauchy-Schwarz and Lemma 14, this is at most p X / O (cid:16)p X / X / X / (cid:17) = O ( X / ) = O ( XA − / ) . Hence all of our relevant factors are O ( X log( X ) A − / ), thus proving ourbound.As mentioned above, in proving Proposition 9, we are going to want to dealseparately with the terms in which D Q i p i is divisible by a particular bad Siegelzero. In particular, for X ≤ Y , let q ( X, Y ) be the modulus of the Dirichletcharacter with the worst (closest to 1) Siegel zero of any Dirichlet characterwith modulus between X and Y . In analogy with the Q defined in the proofof Proposition 11, for integers n, N, D, k, X, Y and a set S of primes, we define Q ( n, N, D, k, X, Y, S ) to be the largest possible value of (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n ! X S ′ N,n,D Y i χ i ( p i ) Y i Let n, N, D, k, X, Y be as above. Let β be a real number so that theworst Siegel zero of a Dirichlet series of modulus at most D other than q ( X, Y ) is at most − β − . Let M, A, B, C be integers so that B > max( e ( C +2) β log log M , e K ( C +2) (log D ) (log log( DM )) , n log C +2 ( M ) , A ) for a sufficiently large constant K . Then for S a set of primes ≤ A , we havethat Q ( n, N, D, k, X, Y, S ) is at most the maximum of N O (cid:18) log log Bn (cid:19) k + O (log( N ) log − C − ( M )) k − X a =0 O (cid:18) log log Bn (cid:19) a ! and O ( N log ( N ) A − / ) + 2 n X p
DP p n − p n is only relevant if DP is missing only one ortwo primes of q ( X, Y ). In the former case, it is equivalent to making one morevalue illegal for the p i . In the latter case it eliminates at most two terms. Thecondition that the p i are distinct removes at most p N/P terms from our sum.Therefore, perhaps after setting a and b to 0 on some set of primes, the aboveis ± n ! O ( p N/P ) + X A ≤ p n − ,p n p n − p n ≤ N/P a ( p n − ) b ( p n ) (cid:18) p n − p n (cid:19) . By Lemma 15, this is at most1 n ! O ( N/P log( N ) A − / ) . Now for each P ≤ N , it can be written in at most ( n − p n − , p n ≥ A is at most N X P =1 O ( N/P log( N ) A − / ) = O ( N log ( N ) A − / ) . Next, we consider the case where p n < A . We deal with this case by setting p n to each possible value of size at most A individually. It is easy to check thatafter setting p n to such a value p , the sum over the remaining p i is 1 /n timesa sum of the form bounded by Q ( n − , N, Dp, k − , X, Y, S ∪ { p } ). Hence the15um over all terms with p n < A is at most1 n X p
Let n, N, D, k, X, Y, S, M, A, B, C, β be as above. Assume further-more that Y ≥ DA n , B > max( e ( C +2) β log log M , e K ( C +2) (log Y ) (log log( Y M )) , n log C +2 ( M ) , A ) , and that S contains only elements of size at most A . Let L = n − k , then Q ( n, N, D, k, X, Y, S ) is at most N O (cid:18) log log BL (cid:19) k + O (cid:16) log ( N ) A − / + log( N ) log − C − ( M ) (cid:17) k − X a =0 O (cid:18) log log BL (cid:19) a ! . Note that we will wish to apply this Lemma with n about log log N , D aconstant, A polylog N , X polylog N , M = N , Y = DA n , and B its minimumpossible value. Proof.