On the Rank of the Elliptic Curve y 2 = x(x p)(x 2)

Transcription

1 On the Rank of the Ellitic Curve y = x(x )(x ) Jeffrey Hatley Aril 9, 009 Abstract An ellitic curve E defined over Q is an algebraic variety which forms a finitely generated abelian grou, and the structure theorem then imlies that E = Z r Z tors for some r 0; this value r is called the rank of E. It is a classical roblem in the study of ellitic curves to classify curves by their rank. In this aer, the author uses the method of descent to calculate the rank of two families of ellitic curves. 1 Introduction An ellitic curve E defined over Q is the set of solutions (x, y) Q to an equation of the form y = x 3 + ax + bx + c, with a, b, c Q along with an additional oint at infinity, O. A classical theorem from Mordell shows that such an elliic curve forms a finitely generated abelian grou; the structure theorem then imlies that E = Z r Z tors,r 0. This nonnegative integer r is called the rank of the ellitic curve. It is a classical roblem to classify elliic curves, and the rank of a curve rovides a useful way to distinguish it from other curves, as well as to gain some insight into its algebraic structure. However, calculating the rank of a given elliic curve can be quite difficult in general. One method of doing so is the method of -descent. The machinery behind -descent is quite high-owered, but the idea is rather simle. The main idea behind -descent is a local-to-global method. One indirectly studies an ellitic curve E by examining whether certain related equations, called homogeneous saces, have rational oints over every local field Q for = or a rational rime. This information is then ieced together to yield information about E. 1

2 This aer uses two variations on the method of two-descent to gain results on the ranks of two different families of elliic curves. While studying comuter comutations of the ranks of ellitic curves of the form E : y + x(x )(x ), Jason Beers made the following conjecture. Conjecture. Let E be the ellitic curve defined by E : y = x(x )(x ) where and are twin rimes. Then 0 if 7 (mod 8), rank(e )= 1 if 3, 5 (mod 8), if 1 (mod 8). Curves similar to this form were considered in [1], and similar results for the curve E : y = x(x ( ))(x ) were recently made available on arxiv. In this aer, we rove the first case and art of the second; more secifically, we rove Theorem 1. Let and be twin rimes, and let E be the ellitic curve defined by E : y = x(x )(x ). (a) If 7 (mod 8), then rank(e )=0. In articular, we have E (Q) = Z/Z Z/Z. (b) If 7 (mod 8), then rank(e ) 1. In articular, we have E (Q) = Z/Z Z/Z. or E (Q) = Z Z/Z Z/Z. Before going through each roof, we state the method of two-descent to be used as resented in [1]. Proof of Theorem 1(a) In this section we rove the following theorem: Theorem 1(a). Let and be rime numbers in Z with 7 (mod 8). Then the ellitic curve E(Q) given by E : y = x(x )(x )

3 has rank 0. In articular, E(Q) = Z/Z Z/Z. To rove this theorem, we use the method of -descent resented as Proosition 1.4 of Chater X in [1], which we now state. Theorem. (Comlete -Descent, Version 1). Let E/Q be an ellitic curve given by an equation y =(x e 1 )(x e )(x e 3 ) with e 1,e,e 3 Q. Let S be a set of laces of Q including,, and all laces dividing the discriminant of E. Further, let Q(S, ) = {b Q /Q : ord ν (b) 0(mod ) for all ν/ S}. There is an injective homomorhism defined by E(Q)/E(Q) Q(S, ) Q(S, ) (x e 1,x e ) if x e q,e ((e 1 e 3 )/(e 1 e ),e 1 e ) if x = e 1 P =(x, y) (e e 1, (e e 3 )/(e e 1 )) if x = e (1, 1) if x = (i.e. if P = O). Let (b 1,b ) Q(S, ) Q(S, ) be a air which is not the image of one of the three oints O, (e 1, 0), (e, 0). Then (b 1,b ) is the image of a oint P =(x, y) E(Q)/E(Q) if and only if the equations b 1 z 1 b z = e e 1 b 1 z 1 b 1 b z 3 = e 3 e 1 have a solution (z 1,z,z 3 ) Q Q Q ; if such a solution exists, then one can take P =(x, y) =(b 1 z 1 + e 1,b 1 b z 1 z z 3 ). Thus, Theorem allows one to calculate E(Q)/E(Q) for elliic curves defined by a sufficiently nice equation. If E(Q) = Z r Z tors, then all of the odd-torsion (i.e. factors of E(Q) of the form Z/mZ where m is odd) is killed in E(Q)/E(Q). Furthermore, if is a rime of good reduction for E and gcd(, m)=1, then E(Q)[m] injects into Ẽ(F ), the F rational oints on the reduction of E modulo, so it is easy to calculate E tors. These facts together 3

4 with Theorem allow one to calculate the rank of certain ellitic curves. We now rove Theorem 1(a). Proof of Theorem 1(a). Our curve is E : y = x(x )(x ) = x 3 ( + )x +x which has discriminant = ( ), so our set S is and our set Q(S, ) is S = {,, } Q(S, ) = {±1, ±, ±, ±( ), ±, ±( ), ±( ), ±( )}. The two-torsion oints of E(Q) are those oints (x, y) E(Q) with y =0, hence the two-torsion of our curve is E(Q)[] = {O, (0, 0), (, 0), (, 0)}. Now, since 7, we have 3, and so we see that E tors injects into E(F 3 ). By hyothesis = 7 + 8k for some non-negative integer k. In fact, k 0 (mod 3), since if k 1 (mod 3) then 0 (mod 3) and so is not rime; likewise, if k (mod 3), then 0 (mod 3), so is not rime. Hence we have k 0 (mod 3), and E(F 3 )={O, (0, 0), (1, 0), (, 0)} where the fact that (1, 0) E follows from our congruence argument on k. Any odd m torsion must be an m grou, and since E[] E tors and both E[] and E(F 3 ) have cardinality 4, we see that E tors = E[]. Now we consider the ma φ : E(Q)/E(Q) Q(S, ) Q(S, ) given in Theorem with e 1 =0, e =, and e 3 =. There are 56 airs (b 1,b ) Q(S, ) Q(S, ), and for each air we must check to see whether it comes from an element of Q(S, ) Q(S, ). Using Theorem, we can comute the image φ(e[]) in Q(S, ) Q(S, ): O (1, 1) (0, 0) (, ) (, 0) (, ( )) (, 0) (, ). For the remaining airs (b 1,b ) we must determine whether the equations b 1 z 1 b z = (1) and b 1 z 1 b 1 b z 3 = () 4

5 have a simultaneous solution (z 1,z,z 3 ) Q 3. This is facilitated by a few facts. First, recall that Q Q q (the q-adic comletion of Q) for each rime q, so if an equation has no solutions over Q q then it has no solutions over Q, and the air (b 1,b ) in that case is not in Q(S, ) Q(S, ). Second, the ma φ is a homomorhism. Thus, if (b 1,b ) and (b 1,b ) are both in the image of φ, then so is (b 1 b 1,b b ); if (b 1,b ) is in the image and (b 1,b ) is not, then (b 1 b 1,b b ) is not. If the equations corresonding to a air (b 1,b ) have no solutions over some Q, we say that (b 1,b ) is Q -non-trivial. We follow Silverman s lead and list our results in a table whose entries list either the oint (x, y) E(Q) that gets maed to the air (b 1,b ) or the field over which the equations (1) and () have no solution. If (z 1,z,z 3 ) is a solution to equations (1) and (), then the re-image of (b 1,b ) is (b 1 z1 + e 1,b 1 b z 1 z z 3 ). Each table entry also has a suerscrit number (n) which refers to a note exlaining the entry. We do exclude half of the oints from the table, however: it is easy to see that if b 1 < 0 and b > 0 then equation (1) has no solutions in R, and if b 1 < 0 and b < 0 then equation () has no solution in R. Hence, we exclude the ortion of the table with b 1 < 0. b 1 \b 1 ( ) ( ) ( ) 1 O Q (3) Q (5) Q (7) Q (3) Q (5) Q (4) Q (1) Q (4) Q (16) Q (16) Q (4) Q (14) Q (9) Q (6) Q (4) ( ) Q (14) Q (3) Q (1) Q (9) (, 0) Q (6) Q (3) Q (10) Q (3) Q (9) Q (4) Q (10) Q (9) Q (4) Q (10) Q (1) Q (14) Q (16) ( ) Q (9) Q (10) Q (9) ( ) Q (4) Q (9) Q (4) Q (10) Q (9) Q (4) Q (3) Q (10) 1 Q (14) Q (1) Q (14) Q (15) Q (3) Q (4) Q (6) Q (4) (0, 0) Q (6) Q (4) Q (8) Q (9) ( ) Q (11) Q (3) Q (10) Q (11) Q (3) Q (10) Q (13) Q (16) Q (13) Q (4) Q (4) ( ) (, 0) Q (6) Q (9) Q (10) Q (9) Q (3) Q (4) Q (10) Q (8) Q (6) ( ) Q (9) Q (10) Q (9) Q (10) ( ) Q (4) Q (9) Q (4) Q (10) Q (9) Q (4) Q (10) Notes For Table 1. If b 1 < 0 and b > 0, equation (1) has no solutions in R.. If b 1 < 0 and b < 0, equation () has no sution in R. 5

6 3. Suose there exists a solution (z 1,z,z 3 ). We have b 1 0 (mod ) and b 0(mod ). Then comaring adic valuations of the left-hand and right-hand sides of equation (1) easily imlies z 1,z Z. But then b 1 z 1 b 1 b z 3 0 (mod ), so equation () imlies 0 (mod ). Since is odd, this is a contradiction, so equations (1) and () have no solutions over Q 4. Adding the Q -non-trivial airs from (3) to the (airs corresonding to the) oints in E[] yields these Q -non-trivial airs. 5. If (b 1,b ) = (, 1) or (( ), 1), then valuation arguments again show that any solution (z 1,z,z 3 ) has z 1,z Z, hence equation (1) becomes z (mod ) which has no solutions since 7 (mod 8). 6. Adding the airs from (5) to the oints in E[] yields these Q -non-trivial oints. 7. If (b 1,b )=(, 1), then again z 1,z Z, so equation (1) imlies z (mod ( )), which has no solutions since 5 (mod 8). 8. Adding the airs from (7) yields these Q -non-trivial oints. 9. Suose b 0 (mod ) and b 1 0(mod ), and suose there exists a solution (z 1,z,z 3 ). Let Then equation () imlies that k = ν (z 1 ), j = ν (z, l = ν (z 3. 1=ν (b 1 z 1 b 1 b z 3) = min{k, 1+l} which imlies l =0and k>0. But equation (1) imlies 0=ν (b 1 z 1 b z ) = min{k, 1+j} which imlies k =0, which is a contradiction. Hence equations (1) and () have no solutions over Q in his case. 10. Adding airs from (9) to the oints in E[] yields these Q -non-trivial airs. 6

7 11. If (b 1,b ) = (1, ( )), then once again equation (1) imlies z 1,z Z. Subtracting from both sides of equation (1) yields so looking modulo ( ) we have z 1 +( )z =, z 1 (mod ( ))). But 5 (mod 8), so no such z 1 exists, hence equation (1) has no solutions over Q. 1. Adding airs from (11) to the oints in E[] yields these Q -non-trivial airs. 13. If (b 1,b ) = (, ( )) or (( ), ( ), then equation (??) again imlies z 1,z Z and reduces to z 1 (mod ) which has no solutions in Q since 3 (mod 4). 14. Adding the airs from (13) to the oints in E[] yields these Q non-trivial airs. 15. If (b 1,b ) = (( ), 1), then equation (1) imlies z 1,z Z and reduces to z (mod ) which has no solutions in Q since 5 (mod 8). 16. Adding the air from (15) to the oints in E[] yieldsthese Q -nontrivial airs. This table reveals that the only elements of Q(S, ) Q(S, ) in the image of the ma φ are those we got from E[], and Theorem then imlies E(Q)/E(Q) Q(S, ) Q(S, ) = (Z/Z). Since we showed before that E tors = E[], we know hence E(Q) = Z r (Z/Z), E(Q)/E(Q) (Z/Z) +r. Thus, we have shown that the rank of E is r =0, and we have E = (Z/Z). 7

8 3 Proving Theorem 1(b) Proving Theorem 1(a) was somewhat long and tedious, but it was not very difficult. To establish the bound of the rank of our curve when 5 (mod 8), we will need a more general form of descent. First we restate the theorem to be roved. Theorem 1(b). Let and be rime numbers in Z with 5 (mod 8). Then the ellitic curve E(Q) given by E : y = x(x )(x ) has rank at most 1. In articular, it is either of the form E(Q) = Z/Z Z/Z. or E(Q) = Z Z/Z Z/Z. To rove this, we use the method of -descent described in chater X of [1] as Proosition 4.9, which we now state. Theorem 3. (Descent via Two-Isogeny.) Let E/Q and E /Q be ellitic curves given by equations and let E : y = x 3 + ax + bx and E : Y = X 3 ax +(a 4b)X; φ : E E φ(x, y) =(y /x,y(b x )/x ) be the isogeny of degree with kernel E[φ] ={O, (0, 0)}. Let S consist of and the laces dividing b(a 4b), an let Q(S, ) be defined as before. There is an exact sequence 0 E (Q)/φ(E(Q)) Q(S, ) WC(E/Q)[φ] O 1 (0, 0) a 4b d {C d /Q}, (X, Y ) X where C d /Q is the homogeneous sace for E/Q given by the equation The φ Selmer grou is then C d : dq = d adz +(a 4b)z 4. S (φ) (E/Q) = {d Q(S, ) : C d (Q ν ) for all ν S}. 8

9 This method of descent is obviously more comlicated than the first version we used. The theory behind this method is high-owered, and the interested reader is encouraged to see [1]. We will say only a few words about this theorem before using it to rove Theorem 1(b). First, we did not reviously define WC(E/Q); this is the Weil-Châtelet grou for E/Q, which is the set of equivalence classes of homogeneous saces for E/Q. A homogeneous sace for E/Q is a "twist" of the ellitic curve E, or some smooth curve on which E has a simly transitive algebraic grou action. What is imortant about these homogeneous saces is that they encode certain information about the ellitic curve E, and Theorem 3 says that we are able to calculate the Selmer grou S (φ) of our isogeny φ by looking at these homogeneous saces over local fields. What, then, is the Selmer grou? This, too, should be investigated in [1]. Essentially, the Selmer grou contains the homogeneous saces which have Q ν - rational oints for every ν. The imortant thing about S (φ) is that we have an exact sequence 0 E (Q)/φ(E(Q)) S (φ) (E/Q) X(E/Q)[φ] 0. The Shafarevich-Tate grou X is essentially the grou of homogeneous saces for E which have a Q ν -rational oint for every lace ν but no Q rational oints; again, see [1]. Together, the Selmer and Sha grous measure the failure of the Hasse rincile for these curves. Finally, comuting both S φ) and S ( ˆφ) and using the above exact sequence together with the exact sequence 0 E (Q)[ ˆφ] φ(e(q)[] E (Q) φ(e(q)) E(Q) E(Q) E(Q) ˆφ(E (Q)) 0, we can comute E(Q)/E(Q) as before and thus deduce the rank of E. Before roving Theorem 1(b), we first rove two lemmas which will greatly simlify the roof of the theorem. Lemma 1. Given the curves E : y = x 3 ( + )x +x and E : y = x 3 + ( + )x +( ) x and the isogeny ( y φ : E E φ(x, y) = x, y( x ) x ), the selmer grou S (φ) is Proof of Lemma 1. S (φ) = {1, 1}. We first note that Ker φ = {O, (0, 0)} and E[] = {O, (0, 0), (, 0), (, 0)} and E [] = {O, (0, 0)}. 9

10 Also, we have b(a 4b) = 4( ), so the set Q(S, ) is Q(S, ) = {±1, ±, ±, ±( ), ±, ±( ), ±( ), ±( )}. For each d Q(S, ), we must check whether the associated homogeneous sace C d : dw = d + ( + )dz +( ) z 4 has oints over each local field Q q for q =,,. From Theorem 3, we have O, (0, 0) 1 and so d =1 S (φ), and it remains to check whether the remaining d Q(S, ) are in the Selmer grou. 1. d = ± We resent the argument for C to show that d = is not in the Selmer grou; the argument for C is identical. Our homogeneous sace is C : w = + ( + )z +( )z 4. Now, we have that the adic valuation of the left-hand side, ν (LHS), is odd. Hence, the valuation of the right-hand side must also odd. But we have ν (RHS) = min{, 1+k, 4k} where k = ν (z). (We have equality in the above exression since no two of, 1+k, and 4 can ever be the same.) If k 0, then this minimum equals 4k, which is even. Similary, if k>0, then the minimum is. In neither case can the valuations of the left- and right-hand sides match, so C has no Q -rational oints. Hence ± / S (φ).. d = ± We resent the argument for C to show that d =is not in the Selmer grou; the argument for C is identiical. Our homogeneous sace is C :w = ( + )z +( ) z 4. The adic valuation of the left hand side is odd, while ν (RHS) min{, +k, 4k} where k = ν (z). Since ν (RHS) must be odd, we must have at least two of, +k, and 4k be equal and minimal; this is easily seen to be imossible, hence C has no Q -rational oints. Thus ± / S (φ). 10

11 3. d = ± Either of the two arguments above show that ± / S (φ). Notice that since S (φ) is a grou, we have ±( ), ±( ), ±( ) / S (φ). It remains to check whether C d has Q q -rational oints for d { 1, ±( )} and q {,, }. 4. d = 1 Our homogeneous sace is C 1 : w =1 ( + )z +( ) z 4. We show that C 1 has Q q -rational oints for each q {,, }, hence 1 S (φ). (a) q =We find a solution in Q as follows. Let Then we have f(w, z) =1 ( + )z +( ) z 4 + w. f(, 1) = 1 ( + ) + ( ) +4 =5 (7 + 8k) + (3 + 8k) k k (mod 3) since 5 (mod 8) 0 (mod 3). So ν (f(, 1)) 5, but f w =w =4, (w,z)=(,1) so ν ( f/ w) =4< 5. solution in Q. Hence by Hensel s lemma this lifts to a (b) q = We find a solution in Q as follows. With f(w, z) as before, we have hence f(w, z) =1 4z +4z 4 + w (mod ), f(w 0, 0) = 1 + w 0 (mod ) has a solution w 0 since 4 (mod 8). Since w 0 and is an odd rime, we have f w =w 0(mod ), (w,z)=(w0,0) so by Hensel s lemma this lifts to a solution in Q. 11

12 (c) q = We find a solution in Q as follows. With f(w, z) as before, we have f(w, z) 1 ( + )z + w (mod ( ) ) so f(w, 0) 1+w 0 (mod ( ) ) has a solution w 0 since ( ) 1 (mod 4). Once again, we have f w =w 0(mod ( )) (w,z)=(w0,0) so by Hensel s lemma this lifts to a solution in Q. Thus, we have 1 S (φ). We now show that C ( ) has no Q rational oints, hence ( ) / S (φ). Since S (φ) is a grou, this will imly ( ) / S (φ), comleting our calculation of S (φ). 5. d = ( ) Our homogeneous sace is C ( ) : ( )w =( ) ( + )( )z +( )z 4. We will show this sace has no oints over Q. Let ν (w) =k, ν (z) =j. Suose there is a solution (w, z). We have and There are two cases to consider. ν (LHS) =k ν (RHS) min{0, 1+j, 4j}. Suose j>0, which imlies k =0. Then w, z Z with w odd and z even, so f(w, z) ( ) +( )w 1+3w (mod 8) which has no solutions. Having no solutions modulo 8 imlies that there are no solutions in Q. Now suose j =0, which imlies k 0. Then w, z Z and z is odd. We examine f(w, z) modulo different owers of to gain information about w and z. Remember that 5 (mod 8). Looking modulo yields f(w, z) w (mod ) since z is odd, so this imlies w is even. 1

13 Looking modulo and 3 yields no new information, so we look modulo 4. Note that ( ) 9 and (+)( ) 5 modulo 4. Then, writing z = 1 + r, w =s, and = 5 + 8l, we have f(w, z) 9 10(1 + r) + 9(1 + r) 4 + (3 + 8l)(s) (mod 16) 8 + 1s (mod 16) and having this equal to 0 modulo 16 is equivalent to having +3s 0 (mod 4) which has no solutions. This imlies that there are no solutions to f(w, z) = 0 in Q. Now suose j<0, which imlies k =j. Changing the signs of j and k, write w = j w 0 and z = j z 0, where w 0 and z 0 are both odd. The equation defining our homogeneous sace is then ( ) 4j w 0 =( ) ( + )( ) j z 0 +( ) 4j z 4 0. Multilying through by 4j, this becomes ( )w 0 = 4j ( ) ( + )( ) j+1 z 0 +( ) z 4 0, and looking at the resulting function modulo 8 yields f(w, z) ( ) z 4 0 +( )w 0 (mod 8) z w 0 (mod 8). Since z 0,w 0 are odd, we have f(w, z) 4 0modulo 8. Since there are no solutions modulo 8, there are no solutions in Q. We conclude that ( ) / S (φ), and since 1 S (φ), we have ( ) / S (φ). Thus, S (φ) = {1, 1}. Lemma. Given the curves E : Y = X 3 + ( + )X +( )X and E : y = x 3 ( )x +x and the isogeny ( ) ˆφ : E E ˆφ(X, y x Y )=,y x x the selmer grou Sel ( ˆ (φ)) is S ( ˆφ) = {1,,, }. Proof of Lemma. 13

14 We first note that Ker ˆφ = {O, (0, 0)} and E[] = {O, (0, 0), (, 0), (, 0)} and E [] = {O, (0, 0)}. Our set Q(S, ) is once again Q(S, ) = {±1, ±, ±, ±( ), ±, ±( ), ±( ), ±( )}. For each d Q(S, ), we must check whether the associated homogeneous sace C d : dw = d 4( + )dz +z 4 has oints over each local field Q q for q =,,. From Theorem 3, we have O 1 (0, 0) (, 0) (, 0) and so {1,,, } S ˆφ; it remains to check whether the remaining d Q(S, ) are in the Selmer grou. The set of d which remain to be checked is { 1,,, ±( ), ±, ±( ), ±( ), ±( )}. We roceed as in the last lemma, only this time less work is necessary. Consider d = ±( ); we show that C has no solutions over Q, and the argument for d = ( ) is identical. Our homogeneous sace is C : ±( )w =( ) ± 4( + )( )z +z 4 Letting k = ν (w) =k and ν (z) =j, we have ν (LHS)= 1 + k and ν (RHS) min{, 1 + j, 4j}, which imlies j = k =0, hence w, z Z, and in articular, ( ) does not divide w, z. Taking as our function f(w, z) = ( ) ± 4( + )( )z +z 4 ( )w, then looking at the function modulo yields f(w, z) 4z 4 (mod ( )) which has no solutions (w, z) with z, hence f(w, z) has no solutions in Q ±( ). We conclude that ±( ) / S ( ˆφ). But by the grou structure of S ( ˆφ), this eliminates all of the other ossible d Q(S, ). Hence we have S ( ˆφ) = {1,,, }. We now rove the theorem. Proof of Theorem 1(b). First we show that E tors = E[] = (Z/Z). Suose >5. As in the roof of Theorem 1(a), 3 is a rime of good reduction. Since 5 (mod 8), we have + k (mod 3) for some 0 kleq. But if k =, then 0 (mod 3) and 14

15 is not rime (since >5). Similarly, if k =0, then 0 (mod 3) and so is not rime (since >5 imlies > 3). Then one easily checks that E(F 3 )={O, (0, 0), (1, 0), (, 0)}, and since E tors E(F 3 ) and E[] E tors, we have If =5, one easily checks that E tors = E[] = (Z/Z). E(F 7 )={O, (0, 0), (1, ), (1, 5), (, 0), (3, 1), (3, 6), (5, 0)} = (Z/4Z) (Z/Z) and a similar argument gives the desired result. Now we comute the rank. To aly Theorem 3, we need to comute recisely the Selmer grous from Lemmas 1 and. Having comuted both Selmer grous, we may now use the exact sequences mentioned earlier to comute the rank of E. Recall that we have the following exact sequences: Now, we have comuted 0 E φ(e) Sφ (E) X(E)[φ] 0 (3) 0 E ˆφ(E) S ˆφ(E) X(E )[ ˆφ] 0 (4) 0 E [ ˆφ] φ(e[]) E φ(e) E E S (φ) = Z/Z and S ( ˆφ) = (Z/Z), E ˆφ(E ) 0 (5) and we actually showed that every oint in S ( ˆφ) came from a oint on our curve, hence X(E )[ ˆφ] = 0. Sequence (4) then imlies E/ˆφ(E) = (Z/Z). It is also easy to check that φ(e[]) = E [ ˆφ], hence E [ ˆφ]/φ(E[]) = 0. Thus, sequence (5) becomes 0 E φ(e) E E (Z/Z) 0. (6) We do not know if X(E)[φ] =0; thus, from sequence (3) we have 0 E Z/Z X(E)[φ] 0, φ(e) E so we have either φ(e) = Z/Z or 0; comaring the orders of the grous in (6) shows that rank(e) =1in the first case and 0 in the second, roving the theorem. 15

16 4 Acknowledgements The author would like to thank Dr. Thomas Hagedorn for all of his hel and guidance with this aer, and Jason Beers for making the conjectures that feature as this aer s theorems. References [1] Joseh H. Silverman. The Arithmetic of Ellitic Curves. Sringer-Verlag, New York, NY,