PUMaC 2015 · 加试 · 第 5 题

Problem 5.4 (Integrality, Integrality!; 8, 12 ) . a) We define a recursive sequence { c } by c = c = c = c = c = 1 and for n ≥ 5 , c c = c c + n 0 1 2 3 4 n n − 5 n − 4 n − 1 c c . Prove that this sequence is integral for n ≥ 0 . n − 2 n − 3 b) We define a recursive sequence { d } by d = 1 , d = 2 , d = 1 , and d = − 3 and for n ≥ 4 , n 0 1 2 3  2 d d − d  n − 1 n − 3 n − 2   if n ≡ 0 , 1 (mod 3) d n − 4 d = n 2 d d − 3 d  n − 1 n − 3 n − 2   if n ≡ 2 (mod 3) . d n − 4 21 PUMaC 2015 Power Round Section 5 page 22 Prove that the sequence { d } is an integral sequence for n ≥ 0 . n Proof. a) The proof of this is very similar to the proof of the Somos-4 sequence. The main difference is that we ′ 2 ′ ′ use s := c + c c which gives c s = c s , which by corollary gives s := − 2 c c if n n − 2 n +2 n − 3 n +1 n n − 1 n +1 n n n n − 2 is even and s = 3 c c if n is odd, and the rest of the proof is nearly identical. A full proof is given n n − 1 n +1 at http://www.maths.ed.ac.uk/ mwemyss/Somos5proof.pdf with all due credit to its authors. b) The flavor of proof is identical to that of the problem above and the outline of the Somos-4 sequence. A full proof is attached in Appendix A 22 PUMaC 2015 Power Round Section 6 page 23 6 Interlude As the reader may have noticed by now, this Power Round is a rather eclectic collection of math topics. The following rephrases our previous work into a form we may use later. 2 We introduce projective space, specifically projective 2-space, denoted P ( R ) (this means that coordinates 2 are elements of R ; we can also work instead in P ( Q ), but that is not necessary). The motivation of such a system of numbers is hard to flesh out fully here. (For the interested reader, consider this system as an attempt to fix the “problem” that two parallel lines do not intersect by adding points at infinity. For the artists out there, this is a formalization of the concept of perspective drawings in which parallel lines do in fact converge. Unfortunately, the implementation we present here may not make it clear why these things are true.) 2 3 Elements of P ( R ) represent the lines in R , real 3-space, that pass through the origin. Examine such a line ` that passes through the origin (0 , 0 , 0). We represent ` by a triplet of coordinates ( a : b : c ) where passes through points (0 , 0 , 0) and ( a, b, c ). This clearly doesn’t give a unique representation of . Under this representation, for all real numbers s , ( sa : sb : sc ) and ( a : b : c ) will always represent the same line. For 2 example, if ` is a line that passes through (0 , 0 , 0) and (2 , 4 , 3), then we can denote this line in P ( R ) in many 3 π ∼ ∼ ways: (2 : 4 : 3) π : 2 π : ), and so forth. When possible, it is convention to standardize the = (4 : 8 : 6) = ( 2 a b way we represent these vectors by making the last coordinate 1; thus if c 6 = 0, then ( a : b : c ) = ( : : 1), c c and the latter is the preferred form. 2 2 Definition 23. Let P ( R ) represent projective 2-space. Then elements α ∈ P ( R ) are represented as α = ( a : b : c ) where if c 6 = 0 , we may assume c = 1 . The reason we introduce this space is because it is in some sense the “correct” medium in which to examine elliptic curves. 2 3 2 Definition 24. Let E : y + a xy + a y = x + a x + a x + a be an elliptic curve. Denote another curve 1 3 2 4 6 2 2 3 2 2 3 in three variables (adding z ) as F : y z + a xyz + a yz = x + a x z + a xz + a z . We say that F has 1 3 2 4 6 been homogenized because each term has the same degree (degree is the sum of the powers of all variables). 2 F is an elliptic curve in P ( R ) . By the three variables given by homogenization, we can start to look at F as a curve lying in projective 2-space. It’s clear that if ( a, b ) is a solution to E , then ( a : b : 1) is a solution to F . We may say more. Proposition 25. There is a bijective correspondence between an elliptic curve E and a homogenized F with a further bijective correspondence between points on E and F . We check this by proof by example! (Note this is not actually a proof. Never actually do this, but 2 this example should illustrate clearly why this proposition is true.) Examine the elliptic curve E : y = 3 2 3 2 3 x − 20 x − 15 again. Then the homogenization is F : y z = x − 20 xz − 15 z . Since ( − 4 , 1) is a solution to E , we see ( − 4 : 1 : 1) is clearly a solution to F . Conversely note (1284 : − 5601 : 64) is a solution to F (you ( ) ( ) 321 − 5601 321 − 5601 may way want to check this). Then it is clear : : 1 is also a solution to F , and so , is 16 64 16 64 a solution to E . Note the importance of homogenization in this work. For example, what would have failed if F were made by simply multiplying every term by exactly one factor z ? And finally, here is why we needed projective space: how do we look at elliptic curves over a finite field? An example of a finite field is F (you know enough to verify that is indeed a field). Fields are explored more p thoroughly in section 7. 23 PUMaC 2015 Power Round Section 6 page 24 Definition 26. Let K be a ring. Furthermore let K also have the additional property that for all non-zero − 1 − 1 − 1 elements a , ∃ a ∈ K such that a · a = a · a = 1 . Then K is called a field. Perhaps it is unclear here why we would want to look at elliptic curves over F , but you‘ll see why soon p enough. So, is there a notion of an elliptic curve over F where p is a prime? In case you haven’t noticed by p 2 3 2 now how these rhetorical questions go... Yes! There is. Suppose E : y + a xy + a y = x + a x + a x + a 1 3 2 4 6 is an elliptic curve with integer coefficients. We can look at points in E/ F (read E over F meaning that we p p take E as an equation and points on E all inside F ) in two ways: by solving E in F from the start, or by p p 2 3 2 reducing points from E/ Q . The first is easy to do: simply solve E : y + a xy + a y ≡ x + a x + a x + a 1 3 2 4 6 (mod p ). 2 3