POLYNOMIALS OF AN ELLIPTIC CURVE. non-torsion point M such that M mod p is non-zero and non-singular, and it is asymptotically

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1 EXPLICIT VALUATIONS OF DIVISION POLYNOMIALS OF AN ELLIPTIC CURVE y J. Cheon and z S. Hahn Abstract. In this paper, we estimate valuations of division polynomials and compute them explicitely at singular primes. We show that p( m(m)) is asymptotically equal to p(m) for a non-torsion point M such that M mod p is non-zero and non-singular, and it is asymptotically equal to c 1 m for some constant c 1 for a non-torsion point M such thatm mod p is either singular or zero. Furthermore, we show that the common factors of m(m) and m(m) have valuations at p asymptotically equal to c m for some constant c when M mod p is singular, which is a generalization of M. Ayad's result. Division polynomials were introduced to compute scalar multiplications of points of elliptic curves. They enable us to investigate properties of multiplications without direct multiplication. They are also used to compute n-torsion points and to derive their properties. However, computing division polynomials is not easier than computing direct multiplication of a point, because division polynomials are not usually relatively prime when evaluated at a point in an elliptic curve. In this paper, we estimate valuations of division polynomials. Especially, we can compute them explicitely at singular primes. Using these results, we can compute valuations of multiples of a point in an elliptic curve fromvaluations of associated division polynomials, and vice versa. Our results may be used in computing S-integral points of an elliptic curve of rank 1 as in [1] and in nding periods of elliptic series as in []. Moreover, we suspect that these might be used in measuring the dierence between the canonical height and the absolute height ofapoint in an elliptic curve. Let K be a number eld, R the ring of integers of K and the discrete valuation related to a prime ideal p of R with () = 1 for a uniformizer of p. Consider an elliptic curve E dened over K by a general Weierstrass equation : E : y + a 1 xy + a 3 y = x 3 + a x + a 4 x + a 6 a i R: Let M =(x y) be a non-zero point ofe(k), m an integer. The point mm is given by m (M) (1) mm = m(m)! m(m) 3 m(m) for division polynomials m, m,! m Z[a 1 a 6 x y]. Given a non-torsion point M such that M mod p is non-zero and non-singular, we show, in Theorem 1, that ( m (M)) is asymptotically equal to (m) as a function of m. Onthe other hand, given a non-torsion point M such thatm mod p is either singular or zero, we show, in Theorem 1, Theorem, and Theorem 3, that ( m (M)) is asymptotically equal to c 1 m for some constant c Mathematics Subject Classication. 11G05, 11G0, 14H5. Typeset by AMS-TEX 1

2 J. CHEON AND S. HAHN Furthermore, we show, in Theorem 4, that the common factors of m (M) and m(m) have valuations asymptotically equal to c m for some constant c when M mod p is singular, which is a generalization of M. Ayad's result. He showed in [1] that the common factors of m (M) and m(m) are those primes p of bad reduction such that M mod p is singular. The constants c 1, c depend only on M and are easily computable so that ( m (M)) for large m can be easily evaluated when M mod p is singular. We present some examples of evaluating valuations of m (M) when M mod p is singular. 1. Division Polynomials In this section, we introduce some properties of division polynomials to be used in proving theorems. For the denitions of division polynomials m m Z[a 1 a 6 x y], see [3,III, Exercises 3.7], and for more general situation see [1]. (a) m Z[a 1 a 6 ][x] is of degree m with leading coecient 1. For even m, m =(y+ a 1 x + a 3 ) Z[a 1 a 6 ][x] is of degree m ; with leading coecient m. For odd m, m Z[a 1 a 6 ][x] is of degree m ;1 with leading coecient m. (b) By denition, for all integers m () m = x m ; m;1 m+1 : (c) For all integers m, n, (3) mn(m) = n (M) m m(nm) mn (M) = n (M) m m (nm): (d) For all integers k, l, (4) k+l k;l = k+1 k;1 l ; l+1 l;1 k : (e) For any M E(K) with M 6 O mod p, wehave the following theorem [1] : (5) M mod p is singular if and only if ( m (M)) > 0and( m (M)) > 0 for any m:. Non-Singular Points From now on,we use the following notation : mm =(x m y m ), m = m (M) and m = m (M) for any integer m, unless otherwise specied. Moreover we dene two subsets E 0 (K) ande 1 (K) ofe(k) as follows : (6) E 0 (K) =fm E(K)jM mod p is nonsingular in E(R=p)g E 1 (K) =fm E(K)jM mod p = O in E(R=p)g: Note that E 0 (K)=E 1 (K) is isomorphic to the group of nonsingular points in E(R=p). In this section, we consider a non-torsion point M in E 0 (K). Lemma 1. Let M be a non-torsion point in E 0 (K). If(x) < 0, then (x m )=(x) ; (m)

3 EXPLICIT VALUATIONS OF DIVISION POLYNOMIALS 3 for all positive integers m. Proof. Let K be the completion of K at. Let M = fx K j(x) > 0g, ^E the formal group associated to E and ^Ga (M) the additive group M with its usual addition. We have an isomorphism [3, IV, Theorem 6.4 (b)], (7) log ^E : ^E(M) ;! ^Ga (M) such that (z) =(log ^E z)forz M. Hence we identify ^E(M) with ^Ga (M). Moreover, letting E 1 (K )=fm E(K )jm O mod pg we have an isomorphism [3, VII, Proposition.], (8) z : E 1 (K ) ;! ^E(M) such that (x[m]) = ;(z(m)) for all M E 1 (K ) nfog, where x[m] denotes the x- coordinate of M. By (7) and (8), we get(x[mm]) = (x[m]) ; (m) for all non-torsion point M E 1 (K ) nfog because of the following (z(mm)) = (log ^E z(mm)) = (m log ^E z(m)) = (m)+(z(m)): This completes the proof. Lemma 1 says that (x m ) is asymptotically equal to ;(m)if(x) < 0. For M E 0 (K), the p-part of m is the p-part of the denominator of x m by (5). Sowemay expect that ( m ) is asymptotically equal to (m). Theorem 1. Let M be a non-torsion point in E 0 (K) and m 0 its order in E 0 (K)=E 1 (K). Then 8 0 if m 0 - m ( m )= >< >: ( m0 )+( m m 0 ) if m 0 jm m 0 6=1 1 (m ; 1)(x)+(m) if m 0 =1: Proof. If m 0 6= 1, then (x) 0, so ( m ) 0, ( m) 0 and theirfore (9) ( m )=maxf; 1 (x m) 0g by (1) and (5). If m 0 - m, mm 6 O mod p so that (x m ) 0. By (9), we get( m )=0, which proves the rst case. If m 0 j m and m 0 6=1,then(x m )=(x m0 ) ; ( m m 0 )by Lemma 1 because (x m0 ) < 0. Hence we get the second case by (9). Assume that m 0 = 1 i.e. (x) < 0. Since m is a monic polynomial of degree m of x and (x) < 0, ( m ) = m (x). Since x m = m = m by (1), we get ( m) = m (x) ; (x m ). Also we have (x m )=(x) ; (m) by Lemma 1 since (x) < 0. Hence ( m )=(m ; 1)(x)+(m), which proves the third case.

4 4 J. CHEON AND S. HAHN 3. Singular Points Recall that for any point M, there is a positive integer r such thatrm E 0 (K) since je(k)=e 0 (K)j < 1 [3, VII, Theorem 6.1]. Throughout this section, we suppose that M is a non-torsion point ine(k) n E 0 (K) and of order r in the nite group E(K)=E 0 (K): Lemma. For any integer m and k with r - k, we have where = min(( r ) ( r )). (i) ( k;1 k+1 ) > ( k ) (ii) ( mr;1 mr+1 )=m Proof. (i) If r - k, M km mod p so that x x k mod p. Hence we get ( k;1 k+1 )=( k)+(x ; x k ) > ( k ): (ii) Assume rm = E 1 (K). Since mrm E 1 (K), ( mr ) <( mr) so that (x mr ; mr )=( mr ). Hence we get, using () and (3), the following ( mr;1 mr+1 )=( mr ) =m ( r )+( m (rm)) =m ( r )+( m r = m ( r ): Assume that rm 6 O mod p. IfmrM O mod p, then (x mr ; mr )=( mr ). Hence we get, using () and (3), ( mr;1 mr+1 )=( mr ) m r =m ( r )+( m (rm)) =m ( r ) because p divides m (rm) and p does not divide both of m (rm) and m (rm). If mrm = E 1 (K), then (x ; x mr ) = 0. Hence because rm E 0 and mrm = E 1 (M). ( mr;1 mr+1 )=( mr)+(x ; x mr ) =( mr ) =m ( r )+( m (rm)) =m ( r ) Lemma 3. For any integer m and k with 1 k<r,wehave where = min(( r ) ( r )). Proof. Note that by (4) ( mr+k )+( mr;k) =m +( k ) mr+k mr;k = mr+1 mr;1 k ; k+1 k;1 mr: The rst term of the right-hand side has valuation m +( k ) atpby Lemma. The second term of the right-hand side has valuation greater than ( k )+( mr )by Lemma. However, since mr;1 mr+1 = mr(x ; x mr )and(x ; x mr ) 0, we have ( mr+1 mr;1) ( mr ). Hence the left-hand side has valuation m +( k ). Using Lemma and Lemma 3, we can estimate ( m ) for any integer m with r - m. )

5 EXPLICIT VALUATIONS OF DIVISION POLYNOMIALS 5 Theorem. Assume that M is a non-torsion point in E(K) n E 0 (K) and has order r in the nite group E(K)=E 0 (K): For any integer m and k with 1 k<r,wehave where = min(( r ) ( r )). ( mr+k )=m +(( k = r;k)+)m + ( k ) ( mr;k) =m ; (( k = r;k)+)m + ( k ) Proof. Let m and k be integers with 1 k<r. By Lemma 3, (10) (11) ( mr+k )+( mr;k) =m +( k ) ( mr;k)+( (m;)r+k) =(m ; 1) +( r;k): Subtracting (11) from (10), we get By replacing m with m, weget ( mr+k )=(m ; 1) + ( (m;)r+k)+( k = r;k): ( mr+k )=(4m ; 1) + ( (m;1)r+k)+( k = r;k): If we sum up the above equations from 1 to m, wehave ( mr+k )= mx i=1 Using Lemma 3, we also get which completes the proof. f(4i;1)+( k = r;k)g+( k )=(m +m)+( k = r;k)m+( k ): ( mr;k) =(m ; m) ; ( k = r;k)m + ( k ) Theorem says that ( m )isapproximately a quadratic function of m when m is larger than r. The next proposition says that ( m ) also increases like a quadratic function of m for 1 m r. Proposition. Assume that M is a non-torsion point in E(K) n E 0 (K) and has order r in the nite group E(K)=E 0 (K): For any integer m with 1 m r, ( m ) is strictly increasing. More precisely, we have ( m ) 1 (m ; 1)(m ; +( )): Proof. By (1) and (), we have x ; x m = m;1 m+1 m so that ( m+1 = m )=( m = m;1)+(x ; x m ):

6 6 J. CHEON AND S. HAHN If we sum up the above equations from to m ; 1, we obtain ( m = m;1) = m;1 X k= (x ; x k )+( ): Since (x ; x m ) 1 for 1 m<rand ( ) > 0, ( m ) is strictly increasing for m with 1 m<r.furthermore if we leta m = P m k= (x ; x m), then a m m ; 1 and so we have X m;1 ( m )=( 1 )+ (a k + ( = 1 )) m;1 X k=1 k=1 (k ; 1+( )) = 1 (m ; 1)(m ; +( )): Using Theorem 1, we can also estimate ( m ) for any integer m with rjm. Theorem 3. Assume that M is a non-torsion point in E(K)nE 0 (K) and of order r in the nite group E(K)=E 0 (K): Let m 0 be the order of rm mod p in E(R=pR). Then we have ( mr )= Proof. (3) gives that 8 >< >: m ( r ) if m 0 - m m ( r )+( m0 (rm)) + ( m m 0 ) if m 0 jm m 0 6=1 1 (m ; 1)( r )+( r )+(m) if m 0 =1: mr = m r m (rm): Since rm E 0,we can apply Theorem 1 for m (rm). Then the rst and second cases are clear. For the third case, by Theorem 1, we have ( m (rm)) = 1 (m ; 1)( r r )+(m). Since ( mr )=m ( r )+( m (rm)), we have which completes the proof. ( mr )= 1 (m ; 1)( r )+( r )+(m) At last, we estimate the common p-factor of m and m for M E(K) n E 0 (K). Theorem 4. Assume that M is a non-torsion point in E(K) n E 0 (K) and has order r in the nite group E(K)=E 0 (K): For any integer m and k with 1 k<r,wehave min(( n ) ( n )) = 8 >< >: where = min(( r ) ( r )). m if n = mr 4m +(( k = r;k)+)m +( k ) if n =mr + k 4m ; (( k = r;k)+)m +( k ) if n =mr ; k Proof. First, we prove the second and the third cases. For 1 k<r,(mr + k)m 6 O mod p and (mr ; k)m 6 O mod p so that (x mr+k ) 0and(x mr;k) 0. Hence

7 EXPLICIT VALUATIONS OF DIVISION POLYNOMIALS 7 ( mr+k ) ( mr+k ) and ( mr;k) ( mr;k). Both cases then follow immediately from Theorem. To prove the rst case, we need the following equality from (3) : mr = m r m (rm) mr = r m m (rm): If rm 6 O mod p, then ( m (rm)) 0, ( m (rm)) 0 and theirfore one of them is 0 by (5). Since ( r ) ( r), we get If rm O mod p, min(( mr ) ( mr )) = m ( r )=m : ( mr )=m ( r )+m ( r )=m ( r ) since (x r ) < 0and m is of degree m.to calculate ( mr ), we apply Theorem 3. Since m 0 =1, Since ( r ) <( r), we alsoget ( mr )= 1 (m ; 1)( r )+( r )+(m): r min(( mr ) ( mr )) = m which completes the proof of the rst case. 4. Examples In this section, we introduce two examples of evaluating ( m (M)) using Theorem and Theorem 3, when M mod p is singular. We use the same notations as in Section 3. Example 1. Let E : ; y = x x and M =( 49) E(Q). E = ; Then M = (3 0) mod 7, 4M O mod 7 so that M mod 7 is singular, M mod 7 is non-singular and 4M mod 7 is zero in E(Z=7Z). Hence we getr =,m 0 =. We have ( )=, =4and( (M)) = 1. By applying Theorem and Theorem 3, we get the following for all integers m : ( 4m;1) =8m ; 4m ( 4m )=8m + (m)+1 ( 4m+1 )=8m +4m ( 4m+ )=8m +8m +: We can see that they agree with the following calculation, which is obtained directly from the inductive denition of division polynomials. 1(M) =1 (M) = 7 3(M) =; (M) =; (M) =; (M) =

8 8 J. CHEON AND S. HAHN 7(M) = (M) = (M) =; (M) =; (M) =; (M) = (M) = Example. Let E : y = x 3 ; x + 1 and M =(1 1) E(Q). E = ; 4 3. Then M =(;1 1) and 3M =(0 1) so that M mod and M mod are singular but 3M mod is non-singular in E(Z=Z) and6m O mod. Then we have r =3,m 0 =,( 3 )=3, =6( =, 3 =8and 3 = 0) and ( (3M)) = () = 1. By applying Theorem and Theorem 3, we get the following for all integers m : ( 6m )=1m +1+(m) ( 6m;) =1m ; 8m +1 ( 6m;1) =1m ; 4m ( 6m+1 )=1m +4m ( 6m+ )=1m +8m +1 ( 6m+3 )=1m +1m +3: We can see that they agree with the following calculation : 1(M) =1 (M) = 3 (M) = 3 4 (M) = 5 5 (M) =; 8 6(M) = ; 13 7 (M) = ; (M) = ; (M) = ; 7 10(M) = (M) = (M) = (M) = (M) =; (M) =; (M) =; (M) = ; (M) = ; (M) = (M) = (M) = (M) = (M) = ; (M) =; (M) =; Acknowledgement The second author was partially supported by Korea Research Foundation References 1. M. Ayad, Points S-Entiers des Courbes Elliptiques, Manuscripta Math. 76 (199), 305{34.. M. Ayad, Periodicite(modq) Des Suites Elliptiques et Points S-Entiers sur les Courbes Elliptiques, Ann. Inst. Fourier, Gronoble 43 (1993), J. Silverman, The Arithmetic of Elliptic Curves, Springer-Verlag, y Electronics and Telecommunications Research Institute Taejon , Republic of Korea cheon@dingo.etri.re.kr z Department of Mathematics Korea Advanced Institute of Science and Technology Taejon , Republic of Korea sghahn@math.kaist.ac.kr