aa r X i v : . [ m a t h . N T ] F e b A recurrence relation for elliptic divisibility sequences
Matteo Verzobio
Abstract
In literature, there are two different definitions of elliptic divisibility sequences. The firstone says that a sequence of integers { h n } n ∈ N is an elliptic divisibility sequence if it verifiesthe recurrence relation h m + n h m − n h r = h m + r h m − r h n − h n + r h n − r h m for every natural number m ≥ n ≥ r . The second definition says that a sequence of integers { β n } n ∈ N is an ellipticdivisibility sequence if it is the sequence of the square roots (chosen with an appropriate sign)of the denominators of the abscissas of the iterates of a point on a rational elliptic curve.It is well-known that the two sequences are not equivalent. Hence, given a sequence of thedenominators { β n } n ∈ N , in general does not hold β m + n β m − n β r = β m + r β m − r β n − β n + r β n − r β m for m ≥ n ≥ r . We will prove that the recurrence relation above holds for { β n } n ∈ N under someconditions on the indexes m , n , and r . The goal of this paper is to make a remark on the definition of elliptic divisibility sequences (alsocalled EDS). In literature, there are two different definitions of EDS. We want to show the linkbetween these two definitions.The first definition is due to Ward, in [7]. It is completely arithmetical.
Definition A.
A sequence of integers { h n } n ∈ N is an elliptic divisibility sequence if it verifiesthe following properties:• h = 0 ;• h = 1 ;• h divides h ;• for every m ≥ n ≥ r , h m + n h m − n h r = h m + r h m − r h n − h n + r h n − r h m . (1)In literature, there is an other definition of elliptic divisibility sequences. This definition is moregeometrical. Mathematics Subject Classification : Primary 11G05; Secondary 11B39.
Key words and phrases : Elliptic divisibility sequences, Recurrence sequences, Elliptic curves. efinition B. Let E be a rational elliptic curve defined by a Weierstrass equation with integercoefficients, and let P ∈ E ( Q ) . For every n ∈ N write x ( nP ) = A n ( E, P ) B n ( E, P ) with A n ( E, P ) and B n ( E, P ) two coprime integers and B n ( E, P ) ≥ . If nP = O , the identity ofthe curve, then we put B n ( E, P ) = 0 . Let ψ n be the n -th division polynomial of E , as defined inDefinition 2.1. Define β n ( E, P ) =
Sign ( ψ n ( x ( P ) , y ( P ))) · B n ( E, P ) B ( E, P ) . We say that the sequence { β n ( E, P ) } n ∈ N is an elliptic divisibility sequence . Remark 1.1.
One can easily show that B ( E, P ) divides B n ( E, P ) for every n and then thesequence of the β n ( E, P ) is a sequence of integers.The fact that the denominator of x ( nP ) is a square follows from the fact that the coefficientsof the Weierstrass equation are integers. Remark 1.2.
The sequences { β n ( E, P ) } n ∈ N and { B n ( E, P ) } n ∈ N are clearly strictly related. Insome papers, it is studied the sequence of the B n instead of the sequence of the β n . We will considerthe sequences of the β n ( E, P ) since it is easier to relate them with the sequences of Definition A.In the definition of β n , we divide by B in order to have the additional property β n ( E, P ) = 1 .Finally, the choice of the sign of β n is necessary in order to link the two definitions of EDS. Thereason for this choice will become clear during the paper.The study of the elliptic divisibility sequences is very interesting and has applications in a lotof fields, as for example cryptography or logic. These sequences are divisibility sequences. Recallthat a sequence of integers { a n } n ∈ N is a divisibility sequence if m | n = ⇒ a m | a n . Definition 1.3.
Given a sequence of integers { a n } n ∈ N , we say that the sequence is an EDSA if itis an elliptic divisibility sequence as in Definition A. We say that the sequence is an
EDSB if it isan elliptic divisibility sequence as in Definition B.There exist some sequences that are both EDSA and EDSB. Anyway, it is easy to show thatthe two definitions are not equivalent. The goal of this paper is to show the relation between thetwo definitions.First of all, we show an example of a sequence that is both an EDSA and an EDSB.
Example 1.4.
Consider the EDSA h = 1 , h = 2 , h = − , and h = − . Observe that onceone knows the value of h i for i ≤ , then every term can be computed using the recurrence relation(1). Take E the elliptic curve defined by the equation y = x + x + 1 and P = (0 , . Computingthe first terms, we obtain β ( E, P ) = 1 , β ( E, P ) = 2 , β ( E, P ) = − , and β ( E, P ) = − . Forexample, since P = (1 / , − / and ψ ( x, y ) = 2 y , we have β ( E, P ) =
Sign ( ψ ( x ( P ) , y ( P ))) B ( E, P ) B ( E, P ) =
Sign (2 y ( P )) √ √ . Hence, for i ≤ , we have β i ( E, P ) = h i . Using the work in the next pages and in particularTheorem 1.9, it is possible to show that in general h n = β n ( E, P ) .2nyway, in general it is not true that every EDSB is an EDSA. In the same way, it is not truethat every EDSA is an EDSB. We show two examples of this fact. Example 1.5.
Consider the sequence a n = n . This is an EDSA, by direct computation. We caneasily show that this sequence is not an EDSB. Suppose, by absurd, that there exists E and P suchthat β n ( E, P ) = n for every n ∈ N . Since a n = 0 for n ≥ , we have that P is a non-torsion point.Using [6, Example IX.3.3], we have that lim n →∞ log | β n ( E, P ) | n = c > . The constant c depends on E and P . Anyway, it is always strictly positive, if P is a non-torsionpoint. Observe that lim n →∞ log log nn = 0 and then { n } n ∈ N cannot be an EDSB. Example 1.6.
Let E be the elliptic curve defined by the equation y = x + x +6 and take the point P = ( − , in E ( Q ) . Consider the EDSB { β n ( E, P ) } n ∈ N . One can compute that β ( E, P ) = 1 , β ( E, P ) = 1 , β ( E, P ) = − , β ( E, P ) = − , and β ( E, P ) = 1 . This is not an EDSA since doesnot verify (1). Indeed, if we put m = 3 , n = 2 , and r = 1 in (1) we should have β β = β β − β β . This is not true and then the sequence is not an EDSA.The problem of understanding when an EDSA is an EDSB has been studied in [7, Section IV].We will give some details on this problem at the beginning of Section 4.Instead, we will focus on the problem of understanding when an EDSB is an EDSA. This problemhas been studied in [5].
Theorem 1.7. [5, Theorem 5.1.1] Let E be an elliptic curve defined by a Weierstrass equation withinteger coefficients and let P ∈ E ( Q ) be a non-torsion point. There exists a multiple Q of P suchthat { β n ( E, Q ) } n ∈ N is an EDSA. The goal of this paper is to answer the following question.
Question 1.8.
Given an EDSB that it is not an EDSA, how far is the sequence from being anEDSA?
We know that β = 0 , that β = 1 , and that β divides β since it is a divisibility sequence. So,if the sequence is not an EDSA, then Equation (1) does not hold.We want to show that every EDSB verifies a subset of the equations in (1). Indeed, we willprove the following theorem. Theorem 1.9.
Let E be an elliptic curve defined by a Weierstrass equation with integer coefficientsand let P ∈ E ( Q ) be a non-torsion point. Consider the EDSB { β n } n ∈ N = { β n ( E, P ) } n ∈ N . Take n ≥ m ≥ r be three positive integers such that two of them are multiples of M ( P ) , a constant thatwe will define in Definition 3.4. Then, β n + m β n − m β r = β m + r β m − r β n − β n + r β n − r β m .
3s we will explain in Remark 4.4, this theorem is a generalization of Theorem 1.7.As we noticed before, if { h n } n ∈ N is an EDSA and we know h i for i ≤ , then we can compute h k for every k ∈ N , using the recurrence relation. Thanks to Theorem 1.9 we know that, if { β n ( E, P ) } n ∈ N is an EDSB and we know β i ( E, P ) for i ≤ M ( P ) , then we can compute β k ( E, P ) for every k ∈ N . Indeed, if k > M ( P ) , then we put r = M ( P ) , n = 2 M ( P ) , and m = k − M ( P ) and using the recurrence relation we can compute β k ( E, P ) using the induction. Remark 1.10.
The hypothesis that E is defined by a Weierstrass equation with integer coefficientsis necessary for two reasons. The most important is that the constant M ( P ) is not well-defined inthis case, since we cannot define the reduction of the curve modulo every prime. Moreover, if thecoefficients of the Weierstrass equation are not integers, then in general the sequence of the β n isnot a sequence of integers. Remark 1.11.
The hypothesis that P is a non-torsion point is necessary in order to prove theresults of Section 3. Indeed, we are not able to prove the results of that section without thishypothesis. In Corollary 4.6, we will prove that Theorem 1.9 holds in the case when P is torsionpoint, if M ( P ) = 1 . For some more considerations on the EDSB in the case when P is a torsionpoint, see [5, Section 5.5]. Observe that, in the case when P is a torsion point, the sequence β n ( E, P ) is quite simple. Indeed, in this case, the sequence is periodic with order small. Remark 1.12.
Sometimes, in Definition A, Equation (1) is given only for r = 1 . Anyway, as isproved in [7, Lemma 30.1], the two definitions are equivalent. The aim of this section is to introduce the division polynomials. Even if our main theorem is forelliptic curves defined over Q , in the next two sections we will work with elliptic curves definedover a number field K , in order to give the most general results. We denote with O K the ring ofintegers of K . Define M K as the set of all finite places of K . Given ν ∈ M K , we define K ν as thecompletion of K with respect to ν .Let E be an elliptic curve defined by the equation y + a xy + a y = x + a x + a x + a (2)with coefficients in a number field K . Define the quantities b = 4 a + a , (3) b = 2 a + a a ,b = a + 4 a ,b = a a + 4 a a − a a a + a a − a . Given a point P ∈ E ( K ) and n ∈ N , we want to show how to effectively compute the coordinatesof the point nP . In order to do so, we need to define the so-called division polynomials . Definition 2.1.
Let ψ n ∈ Z [ x, y, a , a , a , a , a ] be the sequence of polynomials defined as follows:• ψ = 0 ; 4 ψ = 1 ;• ψ = 2 y + a x + a ;• ψ = 3 x + b x + 3 b x + 3 b x + b ;• ψ = ψ (2 x + b x + 5 b x + 10 b x + 10 b x + ( b b − b b ) x + ( b b − b )) ;• ψ n +1 = ψ n +2 ψ n − ψ n − ψ n +1 for n ≥ ;• ψ n ψ = ψ n ψ n +2 ψ n − − ψ n ψ n − ψ n +1 for n ≥ .Recall that the coefficients b i are defined in (3) and depend only on the coefficients a i . Thesepolynomials are the so-called division polynomials . For n ≥ , define also the polynomials φ n = xψ n − ψ n +1 ψ n − and ω n = ψ n +2 ψ n − − ψ n − ψ n +1 y . Observe that the points on the curve verify the equation (2 y + a x + a ) = 4 x + b x + b x + b . (4)This can be proved by substituting the coefficients b i with the coefficients a i using the definitionsgiven in (3) and obtaining the Weierstrass equation (2) that defines the curve.We will evaluate the polynomials of Definition 2.1 only on points of the curve. For such points,it holds (4) and so in the polynomials we can substitute (2 y + a x + a ) with x + b x + b x + b .For example, ψ = (2 y + a x + a ) = 4 x + b x + b x + b . Lemma 2.2.
Fix n ≥ . • Using the substitution (2 y + a x + a ) = 4 x + b x + b x + b , we can assume that the polynomial φ n does not depend on y . Therefore, the polynomial φ n isin Z [ x, a , a , a , a , a ] . • If n is odd, then the polynomial ψ n is in Z [ x, a , a , a , a , a ] . Instead, if n is even, then ψ n is a polynomial in Z [ x, a , a , a , a , a ] , multiplied by (2 y + a x + a ) . Therefore, using (2 y + a x + a ) = 4 x + b x + b x + b , we can assume that ψ n does not depend on y . So, ψ n ∈ Z [ x, a , a , a , a , a ] for every n .Proof. See [6, Exercise 3.7].
Lemma 2.3.
If the curve E is fixed, then the coefficients a i are fixed. Therefore, we say that φ n ( x ) and ψ n ( x ) depends only on x . The polynomial φ n ( x ) is monic and has degree n . • The polynomial ψ n ( x ) has degree n − and its leading coefficient is n . The zeros of thispolynomial are the x -coordinates of the non-trivial n -torsion points of E ( Q ) . • For every P ∈ E ( K ) that is not a n -torsion point, we have x ( nP ) = φ n ( x ( P )) ψ n ( x ( P )) and y ( nP ) = ω n ( x ( P ) , y ( P )) ψ n ( x ( P ) , y ( P )) . • For every n ≥ , the polynomials φ n ( x ) and ψ n ( x ) are coprime as polynomials in K [ x ] . If thecoefficients a i are in O K , then φ n ( x ) and ψ n ( x ) are in O K [ x ] .Proof. See [6, Exercise 3.7].
Example 2.4.
Let E be the elliptic curve defined by the equation y = x + x . Then, by definition, ψ = 1 , ψ = 2 y , and ψ = 3 x + 6 x − . Moreover, φ = xψ − ψ ψ = 4 xy − x − x + 1 . Using the substitution y = x + x , we obtain ψ = 4 x + 4 x and φ = x − x + 1 . Therefore, x (2 P ) = φ ( x ( P )) ψ ( x ( P )) = x ( P ) − x ( P ) + 14 x ( P ) + 4 x ( P ) . Remark 2.5.
The sequence { ψ n ( x ( P ) , y ( P )) } n ∈ N is almost an EDSA. Indeed, it verifies everycondition of Definition A, except for the condition that the terms are integers. This follows from [6,Exercise III.7.g]. Lemma 2.6.
Let E be an elliptic curve defined over a number field K and consider the divisionpolynomials as defined in Definition 2.1. Let n, m ∈ N > . Then, (cid:16) φ m ( x ) ψ n ( x ) − φ n ( x ) ψ m ( x ) (cid:17) = ψ n + m ( x ) ψ | n − m | ( x ) . Proof.
Observe that both sides have degree n + m − . The leading term of both sides is ( n − m ) , thanks to Lemma 2.3. Then, we just need to check that the zeros of the two polynomialsare the same. Using the definition, φ m ( x ) ψ n ( x ) − φ n ( x ) ψ m ( x ) = ( x ( mP ) − x ( nP )) ψ n ( x ) ψ m ( x ) . If Q is a point of ( n + m ) -torsion, then nQ = − mQ and x ( nQ ) = x ( − mQ ) = x ( mQ ) . So, the left-hand side is annihilated in the x -coordinates of the ( n + m ) -torsion points. If Q is apoint of | n − m | -torsion, then x ( nQ ) = x ( mQ ) and hence the left-hand side is also annihilated in the x -coordinates of the | n − m | -torsion points. The non-trivial ( n + m ) -torsion points are ( n + m ) − and the non-trivial | n − m | -torsion points are ( n − m ) − . Therefore, the union of these two sets has n + m − elements, which is the degree of the polynomial. So, the roots of both polynomials arethe abscissas of the non-trivial ( n + m ) -torsion points and the non-trivial | n − m | -torsion points.6uppose now that E is defined over Q . Consider p a prime in Z and ν the place associated with p . Suppose that ν ( x ( P )) ≥ . This happens if P does not reduce to the identity modulo p . So, ν ( φ n ( x ( P ))) ≥ and ν ( ψ n ( x ( P ))) ≥ . Recall that, as we defined in the introduction, x ( nP ) = A n ( E, P ) B n ( E, P ) . Observe that ν ( B n ( E, P )) = max { , − ν ( x ( nP )) } = max { , ν ( ψ n ( x ( P ))) − ν ( φ n ( x ( P ))) } = ν ( ψ n ( x ( P ))) − min { ν ( ψ n ( x ( P ))) , ν ( φ n ( x ( P ))) } . For the previous equality, in order to study the sequence of the β n ( E, P ) = ± B n ( E, P ) /B ( E, P ) ,we need to study min { ν ( φ n ( x ( P ))) , ν ( ψ n ( x ( P ))) } . In the next section, we will study this quantity.
The goal of this section is to prove Proposition 3.12, that is necessary in order to prove Theorem1.9. Again, we will work assuming that the curve E is defined over a number field K , in order togive the most general results. Definition 3.1.
Let E be an elliptic curve defined by a Weierstrass equation with integer coefficientsin K and let P ∈ E ( K ) . Let ν ∈ M K and n ∈ N . If nP = O , then define g n,ν ( P ) := min { ν ( φ n ( x ( P ))) , ν ( ψ n ( x ( P ))) } . Moreover, if nP = O , we put g n,ν ( P ) = 0 . Lemma 3.2.
Let ν be a finite place of the number field K . Let E be an elliptic curve defined over K and let P ∈ E ( K ) be a non-torsion point. Assume ν ( x ( P )) ≥ . Therefore, ≤ g n,ν ( P ) ≤ n ( n − ν (∆) . Proof.
Let R n := Res( φ n ( x ) , ψ n ( x )) . With Res we denote the resultant of the two polynomials.Then, using [3, Exercise 3.1, page 78], there exist two polynomials P n and Q n with coefficients in O K such that P n ( x ) φ n ( x ) + Q n ( x ) ψ n ( x ) = R n . Evaluating the polynomials in x ( P ) , we have ν ( R n ) ≥ min { ν ( P n ( x ( P )) φ n ( x ( P ))) , ν ( Q n ( x ( P )) ψ n ( x ( P ))) }≥ min { ν ( φ n ( x ( P ))) , ν ( ψ n ( x ( P ))) } = g n,ν ( P ) .
7e conclude using that R n = ∆ n n − , as is proved in [4].Thanks to the previous lemma, we know that g n,ν ( P ) = 0 only if ν (∆) > , assuming ν ( x ( P )) ≥ . Hence, we need to compute g n,ν ( P ) only in the case when ν (∆) > . Observe that ν (∆) > ifand only if E is singular modulo P , where P is the prime associated with ν .We will show that the terms of the sequence g n,ν ( P ) verifies a recurrence relation. Let E ( K ν ) = { M ∈ E ( K ν ) | M is not singular in E ( F P ) } where M is the reduction of the point M in the reduced curve E ( F P ) modulo P . With F P wedenote the field O K / PO K . This is a subgroup of E ( K ν ) and E ( K ν ) /E ( K ν ) is finite, thanksto [6, Corollary C.15.2.1]. Definition 3.3.
Let K be a number field, ν ∈ M K , and let E be an elliptic curve defined by aWeierstrass equation with integer coefficients in K ν . Let P ∈ E ( K ν ) . Denote with r ( P , P ) theorder of P in E ( K ν ) /E ( K ν ) .Recall that P is the prime associated with the finite valuation ν . Definition 3.4.
Let E be an elliptic curve defined by a Weierstrass equation with integer coefficientsin K and let P ∈ E ( K ) . Define M ( P ) := lcm P { r ( P , P ) } . If P does not divide ∆ , then E has good reduction and therefore E ( K ν ) = E ( K ν ) . So, r ( P , P ) = 1 only for finitely many P and then M ( P ) is a well-defined positive integer.The value of M ( P ) can be bounded using [6, Corollary C.15.2.1]. For example, if the j -invariantis integral, then M ( P ) divides . Observe that the point Q = M ( P ) P is non-singular moduloevery prime. Hence, M ( Q ) = 1 . Proposition 3.5.
Let E be an elliptic curve defined by a Weierstrass equation with integer coeffi-cients in K . Let P ∈ E ( K ) and assume ν ( x ( P )) ≥ . If r ( P , P ) = 1 , then g n,ν ( P ) = 0 for every n ∈ N .Proof. See [1, Theorem A].Now, we study g n,ν ( P ) in the case r ( P , P ) > . Theorem 3.6. [2, Theorem 4] Let E be an elliptic curve defined by a Weierstrass equation withcoefficients in O K . Let P be a non-torsion point of E ( K ) and assume ν ( x ( P )) ≥ . Let r = r ( P , P ) > and n > . Then, g n,ν ( P ) = µm if n = mr, µm ± ν ( ψ k ( x ( P ) ,y ( P )) ψ r − k ( x ( P ) ,y ( P )) ) + µ ) m +2 ν ( ψ k ( x ( P ) , y ( P ))) if n = 2 mr ± k with ≤ k < r, (5) where µ = g r,ν ( P ) . roposition 3.7. Let E be an elliptic curve defined by a Weierstrass equation with integer coeffi-cients in K and let P ∈ E ( K ) be a non-torsion point. Let ν be a finite place and P be the primeassociated with ν . Assume ν ( x ( P )) ≥ . If m is a multiple of r ( P , P ) , then g n + m,ν ( P ) + g | n − m | ,ν ( P ) = 2( g n,ν ( P ) + g m,ν ( P )) . If m is a multiple of M ( P ) , then the equation holds for every ν .Proof. If r ( P , P ) = 1 , then we conclude using Proposition 3.5. So, we assume r ( P , P ) > .If n = m , then m = n = kr , where r = r ( P , P ) > . So, using (5), g n + m,ν ( P ) + g | n − m | ,ν ( P ) = g kr,ν ( P ) = 4 µk = 2( g kr,ν ( P ) + g kr,ν ( P )) . In this case, the lemma holds.Now, we assume that n = m . Define n := max { n, m } and n := min { n, m } . So, n > n andsuppose that n is a multiple of r = r ( P , P ) . If n ≡ r , then n = m r and g n ,ν ( P ) = µm .In the same way, g n ,ν ( P ) = µm . Therefore, using Theorem 3.6, g n + n ,ν ( P ) = µ ( m + m ) , and g n − n ,ν ( P ) = µ ( m − m ) . Now, it is easy to conclude observing that ( m + m ) + ( m − m ) = 2 m + 2 m . So, we assume that n r . We put n = 2 rm ± k for < k < r and n = 2 rm or n = r (2 m + 1) since n ≡ r . Put n = n + n and n = n − n . We want to study L n ,n = g n + n ,ν ( P ) + g n − n ,ν ( P ) − g n ,ν ( P ) + g n ,ν ( P )) . We want to prove that L n ,n = 0 . We are going to use Theorem 3.6.1. If n = 2 m r + k and n ≡ r ) , then n = 2( m + m ) r + k and n = 2( m − m ) r + k .So, L n ,n =4 µ ( m + m ) + 2(2 ν ( ψ k ψ r − k ) + µ )( m + m ) + 2 ν ( ψ k )+ 4 µ ( m − m ) + 2(2 ν ( ψ k ψ r − k ) + µ )( m − m ) + 2 ν ( ψ k ) − µm + 2(2 ν ( ψ k ψ r − k ) + µ ) m + 2 ν ( ψ k )) − µm ) =0 .
2. If n = 2 m r − k and n ≡ r ) , then n = 2( m + m ) r − k and n = 2( m − m ) r − k .Repeating the proof as in the case 1, we can show L n ,n = 0 .3. If n = 2 m r + k and n = (2 m + 1) r , then n = 2( m + m + 1) r − ( r − k ) and n =2( − m + m ) r − ( r − k ) . Repeating the proof as in the case 1, we can show L n ,n = 0 .9. If n = 2 m r − k and n = (2 m + 1) r , then n = 2( m + m ) r + ( r − k ) and n =2( m − m − r + ( r − k ) . Repeating the proof as in the case 1, we can show L n ,n = 0 .In the case when n is the multiple of r , the proof is identical. This conclude the first part of theproof.Assume now that m is a multiple of M ( P ) . Then, it is a multiple of r ( P , P ) for every P andwe conclude with the first part of the proposition. Remark 3.8.
In the previous proposition, the hypothesis that m is a multiple of r ( P , P ) it isnecessary. For example, take E the elliptic curve defined by the equation y = x + x + 6 and P = ( − , ∈ E ( Q ) . Let ν be the place associated with and then ν ( x ) = ord ( x ) . By definition, φ ( x ) = x − x − x + 1 and ψ ( x ) = 4( x + x + 6) . So, by direct computation, g ,ν ( P ) = min { ord ( φ ( x ( P ))) , ord ( ψ ( x ( P ))) } = min { ord (48) , ord (16) } = 4 and, in the same way, g ,ν ( P ) = 12 . Putting m = 1 and n = 2 we have g n + m,ν ( P ) + g n − m,ν ( P ) − g m,ν ( P ) − g n,ν ( P ) = g ,ν ( P ) + g ,ν ( P ) − g ,ν ( P ) − g ,ν ( P )= g ,ν ( P ) − g ,ν ( P )=4 and then the equation of the previous proposition does not hold. Observe that d ( x + x + 6) dx (cid:12)(cid:12)(cid:12) x = − = 3 x + 1 (cid:12)(cid:12)(cid:12) x = − = 4 and so P is singular modulo since d ( x + x + 6) dx (cid:12)(cid:12)(cid:12) x = x ( P ) ≡ d ( y ) dy (cid:12)(cid:12)(cid:12) y = y ( P ) ≡ . In the same way P = (3 , − is singular. Instead, P = (2 , is not singular since d ( x + x + 6) dx (cid:12)(cid:12)(cid:12) x =2 = 3 x + 1 (cid:12)(cid:12)(cid:12) x =2 ≡ . So, in order to apply the previous lemma in this case we need to take m multiple of .From now on and until the end of the section, we will assume that K = Q . We show why thestudy of the g n,ν ( P ) is important for the study of the sequence β n ( E, P ) . Definition 3.9.
Let u and v be two integers. Define φ n ( u, v ) as the homogenization of φ n ( x ) evaluated in u and v , which is v n φ n ( u/v ) . In the same way, define ψ n ( u, v ) as the homogenizationof ψ n ( x ) . Remark 3.10.
The sequence ψ n ( u, v ) is the square of an EDSA. This follows from [6, ExerciseIII.7.g]. 10 xample 3.11. Let E be the elliptic curve defined by the equation y = x + x . Then, ψ = 4 x +4 x and φ = x − x + 1 . So, φ ( u, v ) = v φ (cid:16) uv (cid:17) = v h(cid:16) uv (cid:17) − (cid:16) uv (cid:17) + 1 i = u − u v + v and ψ ( u, v ) = 4 v h(cid:16) uv (cid:17) + uv i = 4 u + 4 uv . Let P be a non-torsion point in E ( Q ) . There exist two integers u and v so that x ( P ) = uv with ( u, v ) = 1 and v > and then x ( nP ) = φ n ( x ( P )) ψ n ( x ( P )) = v n φ n ( x ( P )) v n ψ n ( x ( P )) = φ n ( u, v ) vψ n ( u, v ) . (6)For n > , define g n ( P ) := gcd( φ n ( u, v ) , vψ n ( u, v )) > . (7)Moreover, put g ( P ) = 1 . Observe that, if ν ∈ M Q , then ν ( g n ( P )) = g n,ν ( P ) where g n,ν ( P ) is defined in Definition 3.1.Recall that B n ( E, P ) , as defined in the introduction, represents the denominator of x ( nP ) andthat β n ( E, P ) = B n ( E, P ) /B ( E, P ) . Observe that B ( E, P ) = v . So, using (6), β n ( E, P ) = B n ( E, P ) B ( E, P ) = vψ n ( u, v ) B ( E, P ) g n ( P ) = ψ n ( u, v ) g n ( P ) . (8) Proposition 3.12.
Let E be an elliptic curve defined by a Weierstrass equation with integer coeffi-cients. Let P ∈ E ( Q ) be a non-torsion point. Let m, n ∈ N . If m is a multiple of M ( P ) , as definedin Definition 3.4, then g n + m ( P ) g | n − m | ( P ) = g n ( P ) g m ( P ) . Proof. If p divides v , then φ k ( u, v ) ≡ u k p since φ k is monic from Lemma 2.3 and ( u, v ) = 1 . So, ord p ( g k ( P )) = 0 for every k ≥ . Supposenow that p does not divide v . If ν is the place associated with p , then ν ( x ( P )) ≥ and ord p ( g k ( P )) = g k,ν ( P ) . Therefore, we conclude using Proposition 3.7. 11
Proof of Theorem 1.9
Recall that a sequence of integers is an EDSA if it is a sequence as in Definition A and it is anEDSB if it is a sequence as in Definition B.Let E be a rational elliptic curve defined by a Weierstrass equation with integer coefficients andlet P ∈ E ( Q ) . Let x ( P ) = u/v with u and v coprime integers. Recall that v is a square and let v / be the positive square root of v . Define h n := v n − ψ n ( x ( P ) , y ( P )) , where ψ n is defined in Definition 2.1. This is a sequence of integers. As is shown in [6, Exercise 3.7.g],the sequence of the h n is an EDSA. In some sense, every non-singular EDSA is a sequence of the h n for some elliptic curves E and a point P . For a definition of non-singular EDSA, see [7, Section19]. Theorem 4.1. [7, Theorem 12.1] Let { h n } n ∈ N be a non-singular EDSA with h h = 0 . So, thereexists an elliptic curve E and a point P ∈ E ( Q ) such that h n = v n − ψ n ( x ( P ) , y ( P )) . Let h n be an EDSA as in the previous theorem and take E and P as in the theorem. Hence,for (8), h n = v n − ψ n ( x ( P )) = B n ( E, P ) g n ( P ) B ( E, P ) = β n ( E, P ) g n ( P ) . Observe that h n and β n ( E, P ) have the same sign and then h n = β n ( E, P ) p g n ( P ) . So, given a non-singular EDSA, every term of the sequence is equal to the term of the EDSB { β n ( E, P ) } n ∈ N , multiplied by p g n ( P ) .Now, we want to show that every EDSB β n ( E, P ) is similar to an EDSA. Recall that g n ( P ) isdefined in Equation (7) and that the sequence { β n ( E, P ) } n ∈ N is defined in Definition B. Observethat g n ( P ) is a square thanks to (8). Lemma 4.2.
Let E be a rational elliptic curve defined by a Weierstrass equation with integercoefficients. Let P ∈ E ( Q ) with x ( P ) = u/v for u and v two coprime integers and with v > . Let w = √ v > and define the sequence h n = w n − ψ n ( x ( P ) , y ( P )) . (9) Since g n ( P ) is a square, define p g n ( P ) as the positive square root of g n ( P ) . Hence, β n ( E, P ) = h n p g n ( P ) . (10) Proof.
Observe that h n = v n − ψ n ( x ( P )) = ψ n ( u, v ) ∈ Z . h n is a sequence of integers. Moreover, by definition, the sign of β n ( E, P ) agrees with the sign of h n . Using (8), h n g n ( P ) = ψ n ( u, v ) g n ( P ) = (cid:16) B n ( E, P ) B ( E, P ) (cid:17) = β n ( E, P ) . Taking the square root, we conclude.In general, the sequence of the β n is not an EDSA, as we showed in Example 1.6. Recall that,as we said in Definition A, given an EDSA { h n } n ∈ N , we have h m + n h n − m h r = h m + r h m − r h n − h n + r h n − r h m for every m ≥ n ≥ r . We will show an EDSB verifies a subset of these equations.Now, we are ready to prove Theorem 1.9. Recall that M ( P ) is defined in Definition 3.4. Proof of Theorem 1.9.
As is shown in [6, Exercise 3.7.g], the sequence ψ n := ψ n ( x ( P ) , y ( P )) verifies ψ n + m ψ n − m ψ r = ψ m + r ψ m − r ψ n − ψ n + r ψ n − r ψ m for all m ≥ n ≥ r. Therefore, using the definition of h n in Lemma 4.2, h n + m h n − m h r = w ( n + m ) − ψ n + m w ( n − m ) − ψ n − m w r − ψ r = w n +2 m +2 r − ( ψ n + m ψ n − m ψ r )= w n +2 m +2 r − ( ψ m + r ψ m − r ψ n − ψ n + r ψ n − r ψ m )= w ( m + r ) − ψ m + r w ( m − r ) − ψ m − r w n − ψ n − w ( n + r ) − ψ n + r w ( n − r ) − ψ n − r w m − ψ m = h m + r h m − r h n − h n + r h n − r h m for every m ≥ n ≥ r . So, we have h n + m h n − m h r = h m + r h m − r h n − h n + r h n − r h m (11)and then, dividing both sides by g r ( P ) g n ( P ) g m ( P ) we obtain h n + m h n − m g n ( P ) g m ( P ) · h r g r ( P ) = h m + r h m − r g m ( P ) g r ( P ) · h n g n ( P ) − h n + r h n − r g n ( P ) g r ( P ) · h m g m ( P ) . Define, L m,n := g m + n ( P ) g n − m ( P ) g n ( P ) g m ( P ) . Using Lemma 4.2, we substitute h n with β n p g n ( P ) and we obtain β n + m β n − m p L m,n β r = β m + r β m − r β n p L m,r − β n + r β n − r β m p L r,n . Now, we conclude using Proposition 3.12 since L m,n = L m,r = L n,r = 1 .13 orollary 4.3. Let E be a rational elliptic curve defined by a Weierstrass equation with integercoefficients. Let P ∈ E ( Q ) be a non-torsion point that is non-singular modulo every prime. Thenthe EDSB { β n ( E, P ) } n ∈ N is an EDSA. In particular, given a non-torsion point P , there exists amultiple Q of P such that { β n ( E, Q ) } n ∈ N is an EDSA.Proof. If P is non-singular modulo every prime, then M ( P ) = 1 and we are done using Theorem1.9. Observe that the point Q = M ( P ) P is non-singular modulo every prime. Hence, the sequence { β n ( E, Q ) } n ∈ N is an EDSA, using the first part of the corollary. Remark 4.4.
The previous corollary is equivalent to Theorem 1.7. Anyway, the two proofs of thisresult are completely different. Our theorem is a generalization since it studies also the case when M ( P ) = 1 . Remark 4.5.
In general, we cannot replace the constant M ( P ) with a smaller positive integer inTheorem 1.9. We show an example of this fact.Let E be the elliptic curve defined by the equation y = x + x + 6 and take P = ( − , .Consider the EDSB { β n } n ∈ N = { β n ( E, P ) } n ∈ N . The point P is non-singular modulo every primeexcept modulo . As we showed in Remark 3.8, r ( P,
2) = 3 and then M ( P ) = 3 . Thanks toTheorem 1.9, we know that β n + m β n − m β r = β m + r β m − r β n − β n + r β n − r β m . (12)if at least two of the indexes are multiples of . In order to show that we cannot replace M ( P ) with a smaller constant, we just need to show that the equation does not hold for r = 2 , n = 4 , and m = 6 . Using the definition, we compute that β = 1 , β = − , β = 8 , β = − , and β = 463 .Hence, Equation (12) does not holds for these values and then we cannot replace M ( P ) with or in Theorem 1.9.Now, we briefly deal with the problem when P is a torsion point. Corollary 4.6.
Let E be a rational elliptic curve defined by a Weierstrass equation with integercoefficients and let P ∈ E ( Q ) be a torsion point. Assume that M ( P ) = 1 . Consider the EDSB { β n } n ∈ N = { β n ( E, P ) } n ∈ N . For every m ≥ n ≥ r , β n + m β n − m β r = β m + r β m − r β n − β n + r β n − r β m . Proof.
Using Proposition 3.5, we have that g n ( P ) = 1 for every n ∈ N . Hence, using (10), β n = h n . As we proved in (11), we have h n + m h n − m h r = h m + r h m − r h n − h n + r h n − r h m . for every m ≥ n ≥ r . Hence, for every m ≥ n ≥ r , β n + m β n − m β r = β m + r β m − r β n − β n + r β n − r β m . If one wants to deal with the problem when P is a torsion point and M ( P ) = 1 , then it isnecessary to obtain an analogue of Theorem 3.6 in the case when P a torsion point.14 eferences [1] M. Ayad , Points S -entiers des courbes elliptiques , Manuscripta Math., 76 (1992), pp. 305–324.[2] J. Cheon and S. Hahn , Explicit valuations of division polynomials of an elliptic curve ,Manuscripta Math., 97 (1998), pp. 319–328.[3]
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