p-adic全纯曲线分享超平面的不计重数的唯一.doc

资源描述

《p-adic全纯曲线分享超平面的不计重数的唯一.doc》由会员分享，可在线阅读，更多相关《p-adic全纯曲线分享超平面的不计重数的唯一.doc（10页珍藏版）》请在三一文库上搜索。

1、精品论文p-adic全纯曲线分享超平面的不计重数的唯一性定理颜启明同济大学数学系，上海200092 摘要：本文证明了关于到Pn (Cp )的p-adic全纯曲线分享2n + 2个一般位置超平面的不计重数的唯一性定理，这改进了Ru在2001年证明的分享3n + 1个一般位置超平面的唯一性定理。关键词：多复变函数, p-adic全纯曲线, 超平面, 唯一性定理中图分类号： O174Uniqueness Theorem for p-adic Holomorphic Curves intersecting Hyperplanes without Counting MultiplicitiesYan

2、 QimingDepartment of Mathematics, Tongji University, Shanghai 200092Abstract: In this paper, a uniqueness theorem is proved for p-adic holomorphic curves into Pn (Cp ) sharing 2n + 2 hyperplanes located in general position without counting multiplicities, which gives an improvement of Rus result for

3、 3n + 1 hyperplanes located in general position . Key words: Several complex variables, p-adic holomorphic curve, hyperplane, uniqueness theorem0 IntroductionIt is well known that two nonconstant polynomials f, g over C are identical if there exist two distinct values a, b such that f (x) = a if and

4、 only if g(x) = a and f (x) = b if and only if g(x) = b.In 1926, R. Nevanlinna 1 extended this result to meromorphic functions. He showed that, for two distinct nonconstant meromorphic functions f and g on the complex plane C, they can not have the same inverse images for five distinct values.W. W.

5、Adams and E. G. Straus 2 showed that p-adic entire functions behave in manyways more like polynomials than like entire functions of a complex variable. They proved the基金项目： National Natural Science Foundation of China (No. 11171255 and No. 10901120)，Doctoral ProgramFoundation of the Ministry of Educ

6、ation of China (No. 20090072110053)作者简介： Yan Qiming（1979-），male，lecturer，ma jor research direction：several complex variables.- 10 -following theorem.Theorem A. Let f and g be two nonconstant p-adic entire functions. Let a and b be two distinct(finite) values. Assume that f (x) = a g(x) = a and f (x)

7、 = b g(x) = b. Then f g.For p-adic meromorphic functions, they obtained the following result.Theorem B. Let f and g be two nonconstant p-adic meromorphic functions. Let a1 , a2 , a3 anda4 be four distinct values. Assume that f (x) = ai g(x) = ai for i = 1, 2, 3, 4. Then f g.In 2001, Ru 3 extended Th

8、eorem B to p-adic holomorphic curves in projective space.A p-adic holomorphic curve f is a map f = f0 : : fn : Cp Pn (Cp ), where f0 , . . . , fn are p-adic entire functions without common zeros. (f0 , . . . , fn ) is called a reduced representation of f .A p-adic holomorphic curve f : Cp Pn (Cp ) i

9、s said to be linearly non-degenerate if f (Cp )is not contained in any proper subspace of Pn (Cp ).Hyperplanes H1 , . . . , Hq in Pn (Cp ) are said to be in general position if any n + 1 of them are linearly independent.In 3, Ru showed thatTheorem C. Let f, g : Cp Pn (Cp ) be two linearly non-degene

10、rate p-adic holomorphic curves. Let H1 , . . . , H3n+1 be hyperplanes in Pn (Cp ) located in general position. Assume that f 1 (Hj ) = g1 (Hj ) for 1 j 3n + 1 and f 1 (Hi ) f 1 (Hj ) = for i = j. If f (z) = g(z) forevery point z S3n+1 f 1 (H ), then f g.j=1jIn this paper, we will improve Theorem C a

11、s follows.Theorem 1.1. Let f, g : Cp Pn (Cp ) be two linearly non-degenerate p-adic holomorphic curves. Let H1 , . . . , H2n+2 be hyperplanes in Pn (Cp ) located in general position. Assume that f 1 (Hj ) = g1 (Hj ) for 1 j 2n + 2 and f 1 (Hi ) f 1 (Hj ) = for i = j. If f (z) = g(z) forevery point z

12、 S2n+2 f 1 (H ), then f g.j=1j1 PreliminariesLet p be a prime number, let | |p be the standard p-adic valuation on Q normalized with|p|p = p1 . Let Qp be the completion of Q with respect to this valuation, and let Cp be the completion of the algebraic closure of Qp . For simplicity, we denote the p-

13、adic norm | |p on Cp by | |. For more detail, we refer readers to 4,5.Recall that an infinite sum converges in a non-Archimedean norm if and only if its generalterm approaches zero. Thus a function of the formh(z) = X an zn , an Cpn=0is well-defined whenever|an zn | 0 as n .Functions of this type ar

14、e called p-adic analytic functions. If h is analytic on Cp , then h is called a p-adic entire function. Leth(z) = X an zn , an Cpn=0be a p-adic analytic function on |z| R. For 0 r R, define Mh (r) = max|z|=r |h(z)|. We have the following lemma.Lemma 2.1. 2 The following statements hold:n(1) We have

15、Mh (r) = maxn0 |an |r .(2) The maximum on the right of (1) is attained for a unique value of n except for a discrete sequence of values r in the open interval (0, R).(3) If r 6 r and |z| = r 0).(6) We have Mf g (r) = Mf (r)Mg (r) for any analytic functions f and g.For a nonzero p-adic entire functio

16、n h, we denote the divisor of h by h . For z0 Cp ,h (z0 ) := ordz0 (h).hDenote Mthe divisor of h with truncated multiplicity by a positive integer M . Thathmeans, for z0 Cp , M (z0 ) := minM, h (z0 )h,=kWe define 1 1multiplicity. Hence,be the divisor of all zeros of h with multiplicity k, without co

17、unting( 0if h (z0 ) = k,for z0 Cp .h,=k (z0 ) =1if h (z0 ) = k,2 Proof of Main ResultBy the assumption, f = f0 : : fn and g = g0 : : gn are linearly non-degenerate p-adic holomorphic curves. Let H1 , . . . , Hq be q hyperplanes, located in general position. We denote Hj = x0 : : xn Pn (Cp )|aj0 x0 +

18、 + ajn xn = 0, (f, Hj ) = aj0 f0 + + ajn fn and (g, Hj ) = aj0 g0 + + ajn gn , 1 j q. Obviously, (f, Hj ) 6 0 and (g, Hj ) 6 0 for 1 j q.Suppose that f 6 g. By changing indices if necessary, we may assume that(f, H1 )(f, H2 )(f, Hk1 )(f, Hk1 +1 )(f, Hk2 )(g, H1 ) (g, H2 ) (g, Hk ) 6 (g, Hk +1 ) (g,

19、Hk )| grouzp 1112s )| grouzp 2 (f, Hks1 +1 )(f, Hks ) ,6 6 (g, Hks1+1 ) (g, Hkwhere ks = q.| grouzp sSince f 6 g, the number of elements of every group is at most n. We define the map : 1, . . . , q 1, . . . , q by( i + n if i + n q,(i) = i + n qif i + n q.(g,Hi )It is easy to see that is bijective

20、and |(i) i| n(note that q 2n). Hence (f,Hi )and(f,H(i) )(g,H(i) ) belong to distinct groups, so that (f, Hi )(g, H(i) ) (f, H(i) )(g, Hi ) 6 0.We consider (f, Hi )(g, H(i) ) (f, H(i) )(g, Hi ), 1 i q. (n 2)Lemma 3.1. For each i 1, . . . , q, we haveqX 1 n1 1 1j=1,j=i,(i)(f,Hj ) + (f,Hi ) (n 1)(g,Hi

21、),=1 (n 2)(g,Hi ),=2 (g,Hi ),=n1+(r) (n 1)n(f,H(i) )1 (g,H(i) ),=11 (g,H(i) ),=21 (g,H(i) ),=n1 (f,Hi )(g,H(i) )(f,H(i) )(g,Hi ) . (1)Proof. For any j 1, . . . , q i, (i), since f = g on f 1 (Hj ) (= g1 (Hj ), we have that a zero of (f, Hj ) is also a zero point of (f, Hi )(g, H(i) ) (f, H(i) )(g, H

22、i ).For any z0 f 1 (Hi ) (= g1 (Hi ), z0 is a zero of (f, Hi )(g, H(i) ) (f, H(i) )(g, Hi ) with(f,Hi )(g,H(i) )(f,H(i) )(g,Hi ) (z0 ) min(f,Hi ) (z0 ), (g,Hi ) (z0 ).Note thatf 1 (Hi )= z| min(f,Hi ) (z), (g,Hi ) (z) = (f,Hi ) (z) z| min(f,Hi ) (z), (g,Hi ) (z) = (g,Hi ) (z).Case 1. If z0 z| min(f,

23、Hi ) (z), (g,Hi ) (z) = (f,Hi ) (z), then min(f,Hi ) (z0 ), (g,Hi ) (z0 ) = (f,Hi ) (z0 ) min(f,Hi ) (z0 ), n.Case 2. Consider z0 z| min(f,Hi ) (z), (g,Hi ) (z) = (g,Hi ) (z).For z0 z| min(f,Hi ) (z), (g,Hi ) (z) = (g,Hi ) (z) z|(g,Hi ) (z) n, we have min(f,Hi ) (z0 ), (g,Hi ) (z0 ) = (g,Hi ) (z0 )

24、n = min(f,Hi ) (z0 ), n.For z0 z| min(f,Hi ) (z), (g,Hi ) (z) z|(g,Hi ) (z) = k, k = 1, . . . , n 1, we haveiiii(g,Hi )min(f,H ) (z0 ), (g,H ) (z0 ) = (g,H ) (z0 ) = k min(f,H ) (z0 ), n (n k) 1(z0 ).For any z0 f 1 (H(i) ) (= g1 (H(i) ), z0 is a zero of (f, Hi )(g, H(i) ) (f, H(i) )(g, Hi )with(f,Hi

25、 )(g,H(i) )(f,H(i) )(g,Hi ) (z0 ) min(f,H(i) ) (z0 ), (g,H(i) ) (z0 ).By the same argument, if z0 z| min(f,H(i) ) (z), (g,H(i) ) (z) = (f,H(i) ) (z), then min(f,H(i) ) (z0 ), (g,H(i) ) (z0 ) = (f,H(i) ) (z0 ) min(f,H(i) ) (z0 ), n.If z0 z| min(f,H(i) ) (z), (g,H(i) ) (z) = (g,H(i) ) (z) z|(g,H(i) )

26、(z) n, we have min(f,H(i) ) (z0 ), (g,H(i) ) (z0 ) = (g,H(i) ) (z0 ) n = min(f,H(i) ) (z0 ), n.1If z0 z| min(f,H(i) ) (z), (g,H(i) ) (z) z|(g,H(i) ) (z) = k, k = 1, . . . , n 1, we have min(f,H(i) ) (z0 ), (g,H(i) ) (z0 ) = (g,H(i) ) (z0 ) = k min(f,H(i) ) (z0 ), n (n k)(g,H(i) ) (z0 ). Note that f

27、1 (Hi ) f 1 (Hj ) = for all 1 i j q. We haveqX 1 n1 1 1j=1,j=i,(i)(f,Hj ) + (f,Hi ) (n 1)(g,Hi ),=1 (n 2)(g,Hi ),=2 (g,Hi ),=n1+(r) (n 1)n(f,H(i) )1 (g,H(i) ),=11 (g,H(i) ),=21 (g,H(i) ),=n1 (f,Hi )(g,H(i) )(f,H(i) )(g,Hi ) . (n 2)On the other hand, for each j, 1 j q,(n 1)+ (n 2)1 (g,Hi ),=1By (1) a

28、nd (2),+ + 1 (g,Hi ),=2= n1 (g,Hi ),=n1 1(g,Hj )n(g,Hj ). (2)qX 1 nn1 n n 1j=1,j=i,(i)(f,Hj ) + (f,Hi ) + (g,Hi ) n(g,Hi ) + (f,H(i) ) (r) + (g,H(i) ) n(g,H(i) ) (f,Hi )(g,H(i) )(f,H(i) )(g,Hi ) . (3)Take summation of (3) over 1 i q, we haveq(q 2) X 1(f,Hj )j=1q+(X n(f,Hi )+ i=1n+ (g,Hi )q) +(r) + X

29、 n(f,H(i) )i=1n(g,H(i) ) )qn (X 1(g,Hi )1(g,H(i) ) )qXi=1i=1(f,Hi )(g,H(i) )(f,H(i) )(g,Hi ) .Since is bijective, this givesq(q 2) X 1(f,Hj )j=1q+ 2(+ X n(f,Hi )i=1n(g,Hi )q) 2n X 1(g,Hi )i=1qXi=1(f,Hi )(g,H(i) )(f,H(i) )(g,Hi ) .Similarly, we haveq(q 2) X 1(g,Hj )j=1q+ 2(+ X n(f,Hi )i=1n(g,Hi )q) 2

30、n X 1(f,Hi )i=1qXi=1(f,Hi )(g,H(i) )(f,H(i) )(g,Hi ) .For q = 2n + 2, we have2n+22(X n(f,Hj )j=1n+ (g,Hj )2n+2i) X (f,H )(g,Hi=1(i) )(f,H(i)(g,Hi ) . (4)Denote by W (f0 , . . . , fn )(W (g0 , . . . , gn ) the Wronskian of f0 , . . . , fn (g0 , . . . , gn ). Since fand g are linearly non-degenerate,

31、we have W (f0 , . . . , fn )(W (g0 , . . . , gn ) 6 0.Lemma 3.2. Let H1 , . . . , H2n+2 be the hyperplanes in Pn (Cp ), located in general position.Then2n+22n+2X (f,H ) W (f ,.,f ) X njj=10nj=1(f,Hj ) . (5)Proof. Since f 1 (Hi ) f 1 (Hj ) = for all 1 i 0.|gjk (zk )| C2 max1ln+1|(g, Hl )(zk )| = C2 |

展开阅读全文