tailieunhanh - Analytic Number Theory A Tribute to Gauss and Dirichlet Part 12

Tham khảo tài liệu 'analytic number theory a tribute to gauss and dirichlet part 12', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | 212 PER SALBERGER j 4 that 2 d 2 f -4 C1C2C3Ci 5-3 8Vd and d8-g d C2C3C4 4 C1C2C3C4 5-g d 4. Hence if we sum over all dyadic intervals Cj 2Cj 1 j 4 for 2-powers Cj as above and argue as in op. cit. we will get a set of xd B15 3 2 d 16 5 4 rational lines on X such that there are Odj B15g d 16 5 4 rational points of height B on the union of all geometrically integral hyperplane sections n n X with H n 5B 1 4 which do not lie on these lines. We now combine this with and . Then we conclude that there are xd B15g d 16 5 4 B5 2 points in S X B outside these lines if X is not a cone of a Steiner surface and - B15g d 16 5 4 B11 4 such points if X is a cone of a Steiner surface. To finish the proof note that max 15g d 16 5 4 5 2 max 45 16 vd 5 4 5 2 in the first case and that max 15g d 16 5 4 11 4 1205 448 in the second case. 4. The points on the lines We shall in this section estimate the contribution to N X B from the lines in Theorem . Lemma . X c P4 be a geometrically integral projective threefold over Q of degree d 2. Let M be a set of Od B5 2 rational lines A on X each contained in some hyperplane n c P4 of height 5B 1 4 and N Uagma B be the number of points in S X B n Uagm a Q . Then the following holds. a N Ua ma B Od e B11 4 e B5 2 3 2d . b N Ua mA B Od e B5 2 3 2d e if X is not a cone over a curve. c N Uagma B Odj B5 2 B9 4 3 2d if there are only finitely many planes on X . Proof. We shall for each A G M choose a hyperplane n A c P4 of minimal height containing A . Then H n A kH A 1 3 for some absolute constant k by . The contribution to N UagMa B from all A G M where n A n X is not geometrically integral is Odj B11 4 in general and Odj B5 2 if X is not a cone over a curve see . The contribution from the lines A G M with N A B 1 is Od B5 2 . We may and shall therefore in the sequel assume that n A n X is geometrically integral and N A B 2 for all A G M. From N A B 2 we deduce that H A 2B2 N A B c B2 H A and N Ua ma B N A B B2 H A -1 AeM .