The divisibility by 2 of rational points on elliptic curves
aa r X i v : . [ m a t h . N T ] F e b THE DIVISIBILITY BY OF RATIONAL POINTS ON ELLIPTICCURVES
BORIS M. BEKKER AND YURI G. ZARHIN
Abstract.
We give a simple proof of the well-known divisibility by 2 conditionfor rational points on elliptic curves with rational 2-torsion. As an applicationof the explicit division by 2 n formulas obtained in Sec.2, we construct versalfamilies of elliptic curves containing points of orders 4, 5, 6, and 8 from whichwe obtain an explicit description of elliptic curves over certain finite fields F q with a prescribed (small) group E ( F q ). In the last two sections we study 3-and 5-torsion. Introduction
Let E be an elliptic curve over a number field K . A famous Mordell-Weil theorem asserts that the (abelian) group E ( K ) of K -points on E is finitely generated [3, 18,21]. The first step in its proof (and actual finding a finite set that generates E ( K ))is a weak Mordell-Weil theorem that asserts that the quotient E ( K ) / E ( K ) is afinite (abelian) group. This step is called 2-descent and its basic ingredient is acriterion for when a K -point on E is twice another K -point (under an additionalassumption that all points of order 2 on E are defined over K ). In this paper wegive a new treatment of this criterion that seems to be less computational thanprevious ones ([10, Ch. 5, pp. 102–104], [4], [8, Th. 4.2 on pp. 85-87], [2, Lemma7.6 on p. 67] [1, pp. 331–332]). This approach allows us to describe explicitly 2-power torsion on elliptic curves. In addition we obtain explicitly families of ellipticcurves with various torsion subgroups over arbitrary fields of characteristic differentfrom 2 (the problem of constructing elliptic curves with given torsion goes back toB.Levi [14]).The paper is organized as follows. We work with elliptic curves E over an arbi-trary field K with char( K ) = 2. In Section 2 we discuss the criterion of divisibilityby 2 and explicit formulas for the “half-points” in E ( K ). Next we discuss a criterionof divisibility by any power of 2 in E ( K ) (Section 3). In Section 4 we collect usefulresults about elliptic curves and their torsion. In Sections 5,6, and 7 we will use ex-plicit formulas of Section 2 in order to construct versal families of elliptic curves E such that E ( K ) contains a subgroup isomorphic to Z / m Z ⊕ Z / Z with m = 2 , , versal family of elliptic curves E such that E ( K ) contains a subgroup isomorphic to Z / Z ⊕ Z / Z .) Such familiesare parameterized by K -points of rational curves that are closely related to certainmodular curves of genus zero (see [14, 9, 15, 16]); however, our approach remains The first named author (B.B.) is partially supported by RFFI grant N 14-01-00393.The second named author (Y.Z.) is partially supported by a grant from the Simons Foundation( quite elementary. In addition, in Sections 6 and 8 we construct versal families ofelliptic curves E such that E ( K ) contains a subgroup isomorphic to Z / Z ⊕ Z / Z and Z / Z ⊕ Z / Z , respectively. These two families are parameterized by K -pointsof curves that are closely related to certain modular curves of genus 1.As an unexpected application, we describe explicitly (and without computations)elliptic curves E over small finite fields F q such that E ( F q ) is isomorphic to a certainfinite group (of small order). Using deep highly nontrivial results of B. Mazur [12]and of S. Kamienny and M. Kenku–F. Momose [5, 7], we describe explicitly ellipticcurves E over the field Q of rational numbers and over quadratic fields K such thatthe torsion subgroup E ( Q ) t of E ( Q ) (resp. E ( K ) t of E ( K )) is isomorphic to acertain finite group. Acknowledgements . We are grateful to Robin Chapman for helpful comments.Our special thanks go to Tatiana Bandman for help with magma .2.
Division by 2
Let K be a field of characteristic different from 2. Let(1) E : y = ( x − α )( x − α )( x − α )be an elliptic curve over K , where α , α , α are distinct elements of K . This meansthat E ( K ) contains all three points of order 2, namely, the points(2) W = ( α , , W = ( α , , W = ( α , . The following statement is pretty well known ([3, pp. 269–270], [10, Ch. 5, pp.102–104], [4], [8, Th. 4.2 on pp. 85-87], [2, Lemma 7.6 on p. 67] [1, pp. 331–332],[21, pp. 212–214]; see also [22]).
Theorem 2.1.
Let P = ( x , y ) be a K -point on E . Then P is divisible by in E ( K ) if and only if all three elements x − α i are squares in K . While the proof of the claim that the divisibility implies the squareness isstraightforward, it seems that the known elementary proofs of the converse state-ment are more involved/computational. (Notice that there is another approach,which is based on Galois cohomology [17, Sect. X.1, pp. 313–315] and it works forhyperelliptic jacobians as well [13].)We start with an elementary proof of the divisibility that seems to be less com-putational. (In additional, it will give us immediately explicit formulas for thecoordinates of all four P .) Proof.
So, let us assume that all three elements x − α i are squares in K , and let Q = ( x , y ) be a point on E with 2 Q = P . Since P = ∞ , we have y = 0, andtherefore the equation of the tangent line L to E at Q may be written in the form L : y = lx + m. (Here x , y , l, m are elements of an overfield of K .) In particular, y = lx + m .By the definition of Q and L , the point − P = ( x , − y ) is the “third” commonpoint of L and E ; in particular, − y = lx + m , i.e., y = − ( lx + m ). Standardarguments (the restriction of the equation for E to L , see [18, pp. 25–27], [21, pp.12–14], [1, p. 331]) tell us that the monic cubic polynomial( x − α )( x − α )( x − α ) − ( lx + m ) coincides with ( x − x ) ( x − x ). This implies that − ( lα i + m ) = ( α i − x ) ( α i − x ) for all i = 1 , , . Since 2 Q = P = ∞ , none of x − α i vanishes. Recall that all x − α i are squares in K and they are obviously distinct. Consequently, the corresponding square roots[1, p. 331] r i := lα i + mx − α i = √ x − α i are distinct elements of K . In other words, the transformation z lz + m − z + x of the projective line sends the three distinct K -points α , α , α to the three dis-tinct K -points r , r , r , respectively. This implies that our transformation is not constant, i.e., is an honest linear fractional transformation and is defined over K .Since one of the “matrix entries”, −
1, is already a nonzero element of K , all othermatrix entries l, m, x also lie in K . Since y = lx + m , it also lies in K . So, Q = ( x , y ) is a K -point of E . (cid:3) Let us get explicit formulas for x , y , l, m in terms of r , r , r . We have α i = x − r i , lα i + m = r i ( x − α i ) , and therefore l ( x − r i ) + m = r i [ x − ( x − r i )] = r i + ( x − x ) r i , which is equivalent to r i + lr i + ( x − x ) r i − ( lx + m ) = 0, and this equality holdsfor all i = 1 , ,
3. This means that the monic cubic polynomial h ( t ) = t + lt + ( x − x ) t − ( lx + m )coincides with ( t − r )( t − r )( t − r ). Recall that − ( lx + m ) = y and get(3) r r r = − y . We also get l = − ( r + r + r ) , x − x = r r + r r + r r . This implies that(4) x = x + ( r r + r r + r r ) . Since y = lx + m and − y = lx + m , we obtain that m = − y − lx = − y + ( r + r + r ) x , and therefore y = − ( r + r + r )[ x + ( r r + r r + r r )] + [ − y + ( r + r + r ) x ] , i.e.,(5) y = − y − ( r + r + r )( r r + r r + r r ) . Notice that there are precisely four points Q ∈ E ( K ) with 2 Q = P ,(6) Q = ( x + ( r r + r r + r r ) , − y − ( r + r + r )( r r + r r + r r )) , Another way to see this is to assume the contrary. Then the determinant lx + m = 0, i.e., y = 0, and therefore P = 2 Q is the infinite point, which is not true. BORIS M. BEKKER AND YURI G. ZARHIN each of which corresponds to one of the four choices of the three square roots r i = √ x − α i ∈ K ( i = 1 , ,
3) with r r r = − y . Using the latter equality, wemay rewrite (5) as (7) y = − ( r + r )( r + r )( r + r ) . In addition,(8) x = α i + ( r i + r j )( r i + r k ) , where i, j, k is any permutation of 1 , ,
3. Indeed, x − α i = ( x − α i ) + r r + r r + r r = r i + r r + r r + r r = ( r i + r j )( r i + r k ) . The remaining four choices of the “signs” of r , r , r bring us to the same valuesof abscissas and the opposite values of ordinates and give the results of division by2 of the point − P .Conversely, if we know Q = ( x , y ), then we may recover the corresponding( r , r , r ). Namely, the equalities (8) and (7) imply that r j + r k = − y x − α i ,r i = − ( r j + r k ) + ( r i + r j ) + ( r i + r k )2= − y · (cid:18) − x − α i + 1 x − α j + 1 x − α k (cid:19) for any permutation i, j, k of 1 , , Example 2.2.
Let us choose as P = ( x , y ) the point W = ( α ,
0) of order 2on E . Then r = 0, and we have two arbitrary independent choices of (nonzero) r = √ α − α and r = √ α − α . Thus Q = ( α + r r , − ( r + r ) r r ) = ( α + r r , − r ( α − α ) − r ( α − α ))is a point on E with 2 Q = P ; in particular, Q is a point of order 4. The same istrue for the (three remaining) points − Q = ( α + r r , r ( α − α ) + r ( α − α )),( α − r r , − r ( α − α ) + r ( α − α )), and ( α − r r , r ( α − α ) − r ( α − α )).Recall that, in formula (6) for the coordinates of the points P , one may ar-bitrarily choose the signs of r , r , r under condition (3). Let Q be one of P ’sthat corresponds to a certain choice of r , r , r . The remaining three halves of P correspond to ( r , − r , − r ), ( − r , r , − r ), ( − r , − r , r ). Let us denote thesehalves by Q , Q , Q , respectively. For each i = 1 , ,
3, the difference Q i − Q is apoint of order 2 on E . Which one? The following assertion answers this question. Theorem 2.3.
Let i, j, k be a permutation of , , . Then (i) If P = W i , then Q i = − Q . (ii) If P = W i , then all three points Q i , − Q, W i are distinct. (iii) The points Q i , − Q, W i lie on the line y = ( r j + r k )( x − α i ) . (iv) Q i − Q = W i . This was brought to our attention by Robin Chapman.
HE DIVISIBILITY BY 2 OF RATIONAL POINTS ON ELLIPTIC CURVES 5
Proof.
First, assume that P = W i . In this case, formulas (4) and (5) tell us that Q = ( α i + r j r k , − r j r k ( r j + r k )) , which implies that Q i = ( α i + r j r k , r j r k ( r j + r k )) = − Q and Q i − Q = − Q = − P = P = W i . This proves (i) and a special case of (iv) when P = W i . Now assume that P = W i and prove that the three points Q i , − Q, W i are distinct . Since none of Q i and − Q is of order 2, none of them is W i . On the other hand, if Q i = − Q , then2 Q = P = 2 Q i = − Q = − P, and so P has order 2, say P = W j . Applying (a) to j instead of i , we get Q j = − Q ;but Q i = Q j since i = j . Therefore Q i , − Q, W i are three distinct points. Thisproves (ii).Let us prove (iii). Since x − α i = ( r i + r j )( r i + r k ) , y = − ( r + r )( r + r )( r + r ) , we have y = ( r j + r k )( x − α i ). Further x ( −Q i ) − α i = ( r i − r j )( r i − r k ) ,y ( −Q i ) = ( r i − r j )( − r j − r k )( − r k + r i ) = ( r j + r k ) ( x ( −Q i ) − α i ) . Therefore Q i , − Q, W i lie on the line y = ( r j + r k )( x − α i ) . We have already proven (iv) when P = W i . So, let us assume that P = W i .Now (iv) follows from (iii) combined with (i). (cid:3) Division by n Using the formulas above that describe the division by 2 on E , one may easilydeduce the following necessary and sufficient condition of divisibility by any powerof 2. For an overfield L of K , we consider a sequence of points Q µ in E ( L ) such that Q = P and 2 Q µ +1 = Q µ for all µ = 0 , , , . . . . Let r ( µ )1 , r ( µ )2 , r ( µ )3 ( µ = 0 , , , . . . )be arbitrary sequences of elements of L that satisfy the relations( r ( µ ) i ) = x ( Q µ ) − α i . Then for each permutation i, j, k of 1 , , x ( Q µ +1 ) − α i = (cid:0) r ( µ ) i + r ( µ ) j (cid:1)(cid:0) r ( µ ) i + r ( µ ) k (cid:1) , which implies that ( r ( µ +1) i ) = ( r ( µ ) i + r ( µ ) j )( r ( µ ) i + r ( µ ) k ) . By changing the signs of r ( µ ) i , r ( µ ) j , r ( µ ) k in the product ( r ( µ ) i + r ( µ ) j )( r ( µ ) i + r ( µ ) k ), weobtain all possible values of the abscissas of Q ( µ +1) with 2 Q µ +1 = Q µ .Suppose that Q µ ∈ E ( K ). Then Q µ is divisible by 2 in E ( K ) if and only if onemay choose r ( µ ) i , r ( µ ) j , r ( µ ) k in such a way that ( r ( µ ) i + r ( µ ) j )( r ( µ ) i + r ( µ ) k ) are squaresin K for all i = 1 , ,
3. We proved the following statement.
BORIS M. BEKKER AND YURI G. ZARHIN
Theorem 3.1.
Let P = ( x , y ) ∈ E ( K ) . Let r ( µ )1 , r ( µ )2 , r ( µ )3 ( µ = 0 , , , . . . ) besequences of elements of L that satisfy the relations ( r i ) = r i = x − α i , ( r ( µ +1) i ) = ( r ( µ ) i + r ( µ ) j )( r ( µ ) i + r ( µ ) k ) for all permutations i, j, k of , , . Then P is divisible by n in E ( K ) if and onlyif all x − α i are squares in K , and, for each µ = 0 , , . . . n − , one may choosesquare roots r ( µ )1 , r ( µ )2 , r ( µ )3 in such a way that the products ( r ( µ ) i + r ( µ ) j )( r ( µ ) i + r ( µ ) k ) are squares in K ( and therefore all r ( µ ) i lie in K for µ = 0 , , . . . n − . The knowledge of sequences r ( µ )1 , r ( µ )2 , r ( µ )3 allows us step by step to find thepoints P, P, P etc. Example 3.2.
Let P = ( x , y ), let R be a point of E such that 4 R = P , and let Q = 2 R = ( x , y ). By formulas (4) and (7), x = x + ( r r + r r + r r ) , y = − ( r + r )( r + r )( r + r ) , where the square roots r i = √ x − α i , i = 1 , , , are chosen in such a way that r r r = − y . Further, let r (1) i = q ( r i + r j )( r i + r k )be square roots that are chosen in such a way that r (1)1 r (1)2 r (1)3 = − y = ( r + r )( r + r )( r + r ) . In light of (4) and (7), x ( R ) = x + r (1)1 r (1)2 + r (1)2 r (1)3 + r (1)3 r (1)1 ,y ( R ) = − ( r (1)1 + r (1)2 )( r (1)2 + r (1)3 )( r (1)3 + r (1)1 ) , which implies that(9) x ( R ) = x + ( r r + r r + r r ) + ( r (1)1 r (1)2 + r (1)2 r (1)3 + r (1)3 r (1)1 ) ,y ( R ) = − ( r (1)1 + r (1)2 )( r (1)2 + r (1)3 )( r (1)3 + r (1)1 ) . Torsion of elliptic curves
In the sequel, we will freely use the following well-known elementary observation.
Let κ be a nonzero element of K . Then there is a canonical isomorphism of theelliptic curves E : y = ( x − α )( x − α )( x − α ) and E ( κ ) : y ′ = (cid:16) x ′ − α κ (cid:17) (cid:16) x ′ − α κ (cid:17) (cid:16) x ′ − α κ (cid:17) that is given by the change of variables x ′ = xκ , y ′ = yκ and respects the group structure. Under this isomorphism the point ( α i , ∈ E ( K ) goes to ( α i /κ , ∈ E ( κ )( K ) for all i = 1 , , . In addition, if P = (0 , y ( P )) lies in E ( K ) , then it goes (under this isomorphism) to (0 , y ( P ) /κ ) ∈ E ( κ )( K ) . We will also use the following classical result of Hasse (Hasse bound) [21, Th.4.2 on p. 97].
HE DIVISIBILITY BY 2 OF RATIONAL POINTS ON ELLIPTIC CURVES 7
Theorem 4.1. If q is a prime power, F q a q -element finite field and E is an ellipticcurve over F q , then E ( F q ) is a finite abelian group whose cardinality | E ( F q ) | satisfiesthe inequalities (10) q − √ q + 1 ≤ | E ( F q ) | ≤ q + 2 √ q + 1 . Another result that we are going to use is the following immediate corollary ofa celebrated theorem of B. Mazur ([12], [11, Th. 2.5.2 and p. 187]).
Theorem 4.2. If E is an elliptic curve over Q and the torsion subgroup E ( Q ) t of E ( Q ) is not cyclic, then E ( Q ) t is isomorphic to Z / m Z ⊕ Z / Z with m = 1 , , or . In particular, if m = 3 or and E ( Q ) contains a subgroup isomorphic to Z / m Z ⊕ Z / Z , then E ( Q ) t is isomorphic to Z / m Z ⊕ Z / Z . The next assertion follows readily from the list of possible torsion subgroups ofelliptic curves over quadratic fields obtained by S. Kamienny [5] and M.A. Kenku- F. Momose [7] (see also [6, Th. 1]).
Theorem 4.3.
Let E be an elliptic curve over a quadratic field K . Assume thatall points of order 2 on E are defined over K . Let E ( K ) t be the torsion subgroupof E ( K ) . Then E ( K ) t is isomorphic either to Z / Z ⊕ Z / Z or to Z / m Z ⊕ Z / Z with ≤ m ≤ .In particular, E ( K ) t enjoys the following properties. (1) If m = 5 or and E ( K ) contains a subgroup isomorphic to Z / m Z ⊕ Z / Z ,then E ( K ) t is isomorphic to Z / m Z ⊕ Z / Z . (2) If E ( K ) contains a subgroup isomorphic to Z / Z ⊕ Z / Z , then E ( K ) t isisomorphic to Z / Z ⊕ Z / Z . Rational points of order 4
We are going to describe explicitly elliptic curves (1) that contain a K -point oforder 4. In order to do that, we consider the elliptic curve E ,λ : y = ( x + λ )( x + 1) x over K . Here λ is an element of K \ { , ± } . In this case, we have α = − λ , α = − , α = 0 . Notice that E ,λ = E , − λ . All three differences α − α = λ , α − α = 1 , α − α = 0 are squares in K . Dividing the order 2 point W = (0 , ∈ E ,λ ( K ) by 2, we get r = 0 and the four choices r = ± λ, r = ± . Now Example 2.2 gives us four points Q with 2 Q = W , namely,( λ, ∓ ( λ + 1) λ ) , ( − λ, ± ( λ − λ ) . This implies that the group E ,λ ( K ) contains the subgroup generated by any Q and W , which is Z / Z ⊕ Z / Z . BORIS M. BEKKER AND YURI G. ZARHIN
Remark 5.1.
Our computations show that if Q is a K -point on E ,λ , then2 Q = W if and only if x ( Q ) = ± λ. Both cases (signs) do occur.
Remark 5.2.
There is another family of elliptic curves ([9, Table 3 on p. 217] (seealso [15, Part 2], [11, Appendix E])) E ,t : y + xy − (cid:18) t − (cid:19) y = x − (cid:18) t − (cid:19) x whose group of K -points contains a subgroup isomorphic to Z / Z ⊕ Z / Z . If weput y := y + x − ( t − )2 , then the equation may be rewritten as y = x − (cid:18) t − (cid:19) x + (cid:20) x − ( t − )2 (cid:21) = (cid:18) x − t + 116 (cid:19) (cid:18) x + t (cid:19) (cid:18) x − t (cid:19) . If we put x := x − t + 1 /
16, then the equation becomes y = x x + (cid:18) t + 14 (cid:19) ! x + (cid:18) t − (cid:19) ! , which defines the elliptic curve E ,λ (1 /κ ) with λ = t − t + , κ = t + 14 . In particular, E ,t is isomorphic to E ,λ . Theorem 5.3.
Let E be an elliptic curve over K . Then E ( K ) contains a subgroupisomorphic to Z / Z ⊕ Z / Z if and only if there exists λ ∈ K \ { , ± } such that E is isomorphic to E ,λ .Proof. We already know that E ,λ ( K ) contains a subgroup isomorphic to Z / Z ⊕ Z / Z . Conversely, suppose that E is an elliptic curve over K such that E ( K )contains a subgroup isomorphic to Z / Z ⊕ Z / Z . Then E ( K ) contains all threepoints of order 2, and therefore E can be represented in the form (1). It is alsoclear that at least one of the points (2) is divisible by 2 in E ( K ). Suppose that W is divisible by 2. We may assume that α = 0. By Theorem 2.1, both nonzerodifferences − α = α − α , − α = α − α are squares in K ; in addition, they are distinct elements of K . Thus there arenonzero a, b ∈ K such that a = ± b and − α = a , − α = b . Since α = 0, theequation for E is E : y = ( x + a )( x + b ) x. If we put κ = b , then we obtain that E is isomorphic to E ( κ ) : y ′ = (cid:18) x ′ + a b (cid:19) ( x ′ + 1) x ′ , which is nothing else but E ,λ with λ = a/b . (cid:3) HE DIVISIBILITY BY 2 OF RATIONAL POINTS ON ELLIPTIC CURVES 9
Corollary 5.4.
Let E be an elliptic curve over F . The group E ( F ) is isomorphicto Z / Z ⊕ Z / Z if and only if E is isomorphic to the elliptic curve y = x − x .Proof. Suppose that E ( F ) is isomorphic to Z / Z ⊕ Z / Z . By Theorem 5.3, E isisomorphic to y = ( x + λ )( x + 1) x with λ ∈ F \ { , , − } . This implies that λ = ± , λ = −
1, and so E is isomorphic to E , : y = ( x − x + 1) = x − x. Now we need to check that E , ( F ) ∼ = Z / Z ⊕ Z / Z . By Theorem 5.3, E ( F )contains a subgroup isomorphic to Z / Z ⊕ Z / Z ; in particular, 8 divides | E ( F ) | .In order to finish the proof, it suffices to check that | E ( F ) | <
16, but this inequalityfollows from the Hasse bound (10) | E ( F ) | ≤ √ < . (cid:3) Corollary 5.5.
Let E be an elliptic curve over F . The group E ( F ) is isomorphicto Z / Z ⊕ Z / Z if and only if E is isomorphic to the elliptic curve y = ( x + 2)( x +1) x .Proof. Suppose that E ( F ) is isomorphic to Z / Z ⊕ Z / Z . It follows from Theorem5.3 that E is isomorphic to y = ( x + λ )( x + 1) x with λ ∈ F \ { , , − } . Thisimplies that λ = ± ±
3, and therefore λ = 4 or 2, i.e., E is isomorphic to oneof the two elliptic curves E , : y = ( x + 2)( x + 1) x, E , : y = ( x + 4)( x + 1) x. Since 1 / F , the elliptic curve E , coincides with E , (2); in particular, E , and E , are isomorphic.Now suppose that E = E , . We need to prove that E ( F ) is isomorphic to Z / Z ⊕ Z / Z . By Theorem 5.3, E ( F ) contains a subgroup isomorphic to Z / Z ⊕ Z / Z ; in particular, 8 divides | E ( F ) | . In order to finish the proof, it suffices tocheck that | E ( F ) | <
16, but this inequality follows from the Hasse bound (10) | E ( F ) | ≤ √ < . (cid:3) Theorem 5.6.
Suppose that K contains i = √− . Let a, b be nonzero elements of K such that a = ± b, a = ± i b . Let us consider the elliptic curve E a,b : y = ( x − α )( x − α )( x − α ) over K with α = ( a − b ) , α = ( a + b ) , α = 0 . Then all points of order on E are divisible by in E ( K ) , i.e., E ( K ) contains all twelve points of order .In particular, E a,b ( K ) contains a subgroup isomorphic to Z / Z ⊕ Z / Z .Proof. Clearly, all α i and − α j are squares in K . In addition, α − α = ( a + b ) − ( a − b ) = (2 ab ) , α − α = (2 i ab ) . This implies that all α i − α j are squares in K . It follows from Theorem 2.1 thatall points W i = ( α i ,
0) of order 2 are divisible by 2 in E ( K ), and therefore E ( K )contains all twelve (3 ×
4) points of order 4. (cid:3)
Keeping the notation and assumptions of Theorem 5.6, we describe explicitly alltwelve points of order 4, using formula (6).(1) Dividing the point W = ( α ,
0) = (cid:0) ( a + b ) , (cid:1) by 2, we have r = 0 andget four choices r = ± ab, r = ± ( a + b ). This gives us four points Q with 2 Q = W , namely, two points (cid:0) ( a + b ) + 2 ab ( a + b ) , ± ( a + b + 2 ab )2 ab ( a + b ) (cid:1) = (cid:0) ( a + b )( a + b ) , ± ab ( a + b )( a + b ) (cid:1) and two points (cid:0) ( a + b )( a − b ) , ± ab ( a + b )( a − b ) (cid:1) .(2) Dividing the point W = ( α ,
0) = (0 ,
0) by 2, we have r = 0 and get fourchoices r = ± i ( a − b ) , r = ± i ( a + b ). This gives us four points Q with 2 Q = W , namely, two points (cid:0) ( a − b )( a + b ) , ± ( i (( a − b ) + i ( a + b ))( a − b )( a + b ) (cid:1) = (cid:0) a − b , ± i a ( a − b ) (cid:1) and two points (cid:0) b − a , ± i b ( b − a ) (cid:1) .(3) Dividing the point W = ( α ,
0) = (cid:0) ( a − b ) , (cid:1) by 2, we have r = 0 andget four choices r = ± i ab, r = ± ( a − b ). This gives us four points Q with 2 Q = W , namely, two points (cid:0) ( a − b ) + 2 i ab ( a − b ) , ± (2 i ab + ( a − b ))2 i ab ( a − b ) (cid:1) = (cid:0) ( a − b )( a + i b ) , ± i ab ( a − b )( a + i b ) (cid:1) and two points (cid:0) ( a − b )( a − i b ) , ± i ab ( a − b )( a − i b ) (cid:1) . Remark 5.7.
Let λ be an element of K \ { , ± , ±√− } . We write E ,λ for theelliptic curve E ,λ : y = (cid:18) x + ( λ − ( λ + 1) (cid:19) ( x + 1) x over K . The elliptic curves E ,λ and E a,b are isomorphic if a = λb . Indeed, one hasonly to put κ = a + b and notice that E a,b ( κ ) = E ,λ . It follows from Theorem5.6 that E ,λ ( K ) contains a subgroup isomorphic to Z / Z ⊕ Z / Z .There is another family of elliptic curves with this property, namely, y = x ( x − (cid:18) x − ( u + u − ) (cid:19) ([19], [15, pp. 451–453]; see also Remark 5.9). Theorem 5.8.
Let E be an elliptic curve over K . Then E ( K ) contains a subgroupisomorphic to Z / Z ⊕ Z / Z if and only if K contains √− and there exists λ ∈ K \ { , ± , ±√− } such that E is isomorphic to E ,λ .Proof. Recall (Remark 5.7) that E ,λ ( K ) contains a subgroup isomorphic to Z / Z ⊕ Z / Z .Conversely, suppose that E is an elliptic curve over K and E ( K ) contains asubgroup isomorphic to Z / Z ⊕ Z / Z . By Theorem 5.3, there is δ ∈ K \ { , ± } such that E is isomorphic to E ,δ : y = ( x + δ )( x + 1) x. HE DIVISIBILITY BY 2 OF RATIONAL POINTS ON ELLIPTIC CURVES 11
Hence we may assume that α = − δ , α = − , α = 0. It follows from Theorem2.1 that all ± , ± ( δ −
1) are squares in K . (In particular, i = √− K .) So,there is γ ∈ K with γ = 1 − δ . Clearly, γ = 0 , ±
1. We have δ + γ = 1 . The well-known parametrization of the “unit circle” (that goes back to Euler) tellsus that there exists λ ∈ K such that λ + 1 = 0 and δ = λ − λ + 1 , γ = 2 λλ + 1 . Now one has only to plug in the formula for δ into the equation of E ,δ and get E ,λ . (cid:3) Remark 5.9.
Using a different parametrization of the unit circle in the proof ofTheorem 5.8, we obtain the family of elliptic curves E : y = (cid:18) x + (2 λ ) ( λ + 1) (cid:19) ( x + 1) x with the same property as the family E ,λ . Notice that, for each λ ∈ K \ { , ± } ,the elliptic curve E is isomorphic to the elliptic curve y = x ( x − (cid:0) x − ( u + u − ) / (cid:1) mentioned in Remark 5.7. Indeed, the latter differs from E ( κ ), where κ = 2 λ √− / ( λ +1), only with the change of the parameter λ by u . Corollary 5.10.
Let E be an elliptic curve over F q , where q = 9 , , . The group E ( F q ) is isomorphic to Z / Z ⊕ Z / Z if and only if E is isomorphic to one of ellipticcurves E ,λ . If q = 9 , then E ( F q ) is isomorphic to Z / Z ⊕ Z / Z if and only if E isisomorphic to y = x − x .Proof. First, F q contains √−
1. Suppose that E ( F q ) is isomorphic to Z / Z ⊕ Z / Z .It follows from Theorem 5.8 that E is isomorphic to E ,λ .Conversely, suppose that E is isomorphic to one of these curves. We need toprove that E ( F q ) is isomorphic to Z / Z ⊕ Z / Z . By Theorem 5.8, E ( F q ) containsa subgroup isomorphic to Z / Z ⊕ Z / Z ; in particular, 16 divides | E ( F q ) | . In orderto finish the proof, it suffices to check that | E ( F q ) | <
32, but this inequality followsfrom the Hasse bound (10) | E ( F q ) | ≤ q + 2 √ q + 1 ≤
17 + 2 √
17 + 1 < . Now assume that q = 9. Then λ is one of four ± (1 ± i ). For all such λ we have λ = ± i = ∓ i , ( λ − ( λ + 1) = (1 ∓ i ) ( − ∓ i ) = ∓ i ± i = − . Therefore the equation for E ,λ is y = ( x − x + 1) x = x − x. (cid:3) Corollary 5.11.
Let E be an elliptic curve over F . The group E ( F ) is iso-morphic to Z / Z ⊕ Z / Z if and only if E is isomorphic to one of elliptic curves E ,λ . Proof.
First, F contains √−
1. Suppose that E ( F ) is isomorphic to Z / Z ⊕ Z / Z .Then E ( F ) contains a subgroup isomorphic to Z / Z ⊕ Z / Z . It follows fromTheorem 5.8 that E is isomorphic to E ,λ .Conversely, suppose that E is isomorphic to one of these curves. We need toprove that E ( F ) is isomorphic to Z / Z ⊕ Z / Z . By Theorem 5.8, E ( F ) containsa subgroup isomorphic to Z / Z ⊕ Z / Z ; in particular, 16 divides | E ( F ) | . TheHasse bound (10) tells us that29 + 1 − √ ≤ | E ( F q ) | ≤
29 + 1 + 2 √ < | E ( F ) | < . It follows that | E ( F ) | = 32; in particular, E ( F ) is a finite 2-group. Clearly, E ( F ) is isomorphic to a product of two cyclic 2-groups, each of which has orderdivisible by 4. It follows that E ( F ) is isomorphic to Z / Z ⊕ Z / Z . (cid:3) Theorem 5.12.
Let K = Q ( √− and E be an elliptic curve over Q ( √− . Thenthe torsion subgroup E ( Q ( √− t of E ( Q ( √− is isomorphic to Z / Z ⊕ Z / Z ifand only if there exists λ ∈ K \ { , ± , ±√− } such that E is isomorphic to E ,λ .Proof. By Theorem 4.3, if E ( Q ( √− Z / Z ⊕ Z / Z then E ( Q ( √− t is isomorphic to Z / Z ⊕ Z / Z . Now the desired resultfollows from Theorem 5.3. (cid:3) Points of order 8
Let us return to the curve E ,λ and consider Q ∈ E ,λ ( K ) with 2 Q = W . Let ustry to divide Q by 2 in E ( K ). By Remark 5.1, x ( Q ) = ± λ . First, we assume that x ( Q ) = λ (such a Q does exist). Lemma 6.1.
Let Q be a point of E ,λ ( K ) with x ( Q ) = λ . Then Q is divisible by in E ,λ ( K ) if and only if there exists c ∈ K \ { , ± , ± ± √ , ±√− } such that λ = (cid:20) c − c (cid:21) . Proof.
We have λ − α = λ − ( − λ ) = λ + λ , λ − α = λ − ( −
1) = λ + 1 , λ − α = λ − λ. By Theorem 2.1, Q ∈ E ,λ ( K ) if and only if all three λ + λ , λ + 1 , λ are squaresin K . The latter means that both λ and λ + 1 are squares in K , i.e., there exist a, b ∈ K such that a = λ + 1 , λ = b . This implies that the pair ( a, b ) is a K -pointon the hyperbola u − v = 1 . Recall that λ = 0 , ±
1. Using the well-known parametrization u = t + t , v = t − t λ and λ + 1 are squares in K if and only ifthere exists a nonzero c ∈ K such that λ = (cid:20) c − c (cid:21) . HE DIVISIBILITY BY 2 OF RATIONAL POINTS ON ELLIPTIC CURVES 13
If this is the case, then a = ± c + c , b = ± c − c λ + 1 = (cid:20) c + c (cid:21) . Recall that λ = 0 , ±
1. This means that c − c = 0 , ± , ±√− , i.e., c = 0 , ± , ± ± √ , ±√− . (cid:3) Now let us assume that x ( Q ) = − λ (such a Q does exist). Lemma 6.2.
Let Q be a point of E ,λ ( K ) with x ( Q ) = − λ . Then Q is divisible by in E ,λ ( K ) if and only if there exists c ∈ K \ { , ± , ± ± √ , ±√− } such that λ = − (cid:20) c − c (cid:21) . Proof.
Applying Lemma 6.1 to − λ (instead of λ ) and the curve E , − λ = E ,λ , weobtain that Q ∈ E , − λ ( K ) = 2 E ,λ ( K ) if and only if there exists c ∈ K \ { , ± , ± ± √ , ±√− } such that − λ = (cid:20) c − c (cid:21) . (cid:3) Lemmas 6.1 and 6.2 give us the following statement.
Proposition 6.3.
The point W = (0 , is divisible by in E ,λ ( K ) if and only ifthere exists c ∈ K such that c = 0 , ± , ± ± √ , ±√− and λ = ± (cid:20) c − c (cid:21) , i.e., λ = (cid:20) c − c (cid:21) . Proposition 6.4.
The following conditions are equivalent. (i) If Q ∈ E ,λ ( K ) is any point with Q = W , then Q lies in E ,λ ( K ) . (ii) If R is any point of E ,λ with R = W , then R lies in E ,λ ( K ) . (iii) There exist c, d ∈ K \ { , ± , ± ± √ , ±√− } such that λ = (cid:20) c − c (cid:21) , − λ = (cid:20) d − d (cid:21) . If these equivalent conditions hold, then K contains √− and E ,λ ( K ) contains all(twelve) points of order 4. Proof.
The equivalence of (i) and (ii) is obvious. It is also clear that (ii) impliesthat all points of order (dividing) 4 lie in E ,λ ( K ).Recall (Remark 5.1) that Q with 2 Q = W are exactly the points of E ,λ with x ( Q ) = ± λ . Now the equivalence of (ii) and (iii) follows from Lemmas 6.1 and 6.2.In order to finish the proof, we notice that λ = 0 and − − λλ = h d − d ih c − c i . (cid:3) Suppose that λ = (cid:20) c − c (cid:21) with c ∈ K \ { , ± , ± ± √ , ±√− } and consider Q = ( λ, ( λ + 1) λ ) ∈ E ,λ ( K ) of order 4 with 2 Q = W . Let us find apoint R ∈ E ,λ ( K ) of order 8 with 2 R = Q . First, notice that Q = ( λ, ( λ + 1) λ ) = (cid:20) c − c (cid:21) , (cid:20) c + c (cid:21) · (cid:20) c − c (cid:21) ! = (cid:18) ( c − c , ( c − c (cid:19) . We have r = p λ + λ = p ( λ + 1) λ, r = √ λ + 1 , r = √ λ ; r r r = − ( λ + 1) λ. This means that r = ± c − c · c + c , r = ± c + c , r = ± c − c , and the signs should be chosen in such a way that the product r r r coincideswith − (cid:20) c − c (cid:21) · (cid:20) c + c (cid:21) . For example, we may take r = − c − c · c + c − c − c − c − c , r = c + c , r = c − c r + r = c and r r = ( c − / c )) r + r + r = − c − c + c = − c + 4 c + 14 c ,r r + r r + r r = cr + r r = − c ( c − c + c − c = (1 − c )( c − c . Now (4) and (7) tell us that the coordinates of the corresponding R with 2 R = Q are as follows: x ( R ) = x ( Q ) + r r + r r + r r = ( c − c + (1 − c )( c − c = (1 − c ) ( c + 1)4 c ,y ( R ) = − ( r + r )( r + r )( r + r ) = HE DIVISIBILITY BY 2 OF RATIONAL POINTS ON ELLIPTIC CURVES 15 − (cid:18) − c − c · c + c c + c (cid:19) c (cid:18) − c − c · c + c c − c (cid:19) = − (cid:18) − c − c (cid:19) · c + c · c · (cid:18) − c + c (cid:19) c − c − c − c · (cid:18) c − − c (cid:19) (cid:18) c − c (cid:19) c = − (cid:0) c − c (cid:1) (cid:0) ( c − − c (cid:1) c . So, we get the K -point of order 8 R = (1 − c ) ( c + 1)4 c , − (cid:0) c − c (cid:1) (cid:0) ( c − − c (cid:1) c ! on the elliptic curve E ,c := E , (cid:18) ± c − c (cid:19) : y = " x + (cid:18) c − c (cid:19) ( x + 1) x for any c ∈ K \ { , ± , ± ± √ , ±√− } . The group E ,c ( K ) contains the subgroupgenerated by R and W , which is isomorphic to Z / Z ⊕ Z / Z . Theorem 6.5.
Let E be an elliptic curve over K . Then E ( K ) contains a subgroupisomorphic to Z / Z ⊕ Z / Z if and only if there exists c ∈ K \{ , ± , ± ±√ , ±√− } such that E is isomorphic to E ,c .Proof. We know that E ,c ( K ) contains a subgroup isomorphic to Z / Z ⊕ Z / Z .Conversely, suppose that E ( K ) contains a subgroup isomorphic to Z / Z ⊕ Z / Z .This implies that E ( K ) contains all three points of order 2, i.e., E may be repre-sented in the form (1). Clearly, one of the points (2) is divisible by 4 in E ( K ).We may assume that W is divisible by 4. We may also assume that α = 0, i.e., W = (0 , a, b ∈ K such that α = − a , α = − b , i.e., the equation of E is y = ( x + a )( x + b ) x. Replacing E by E ( b ) and putting λ = a/b , we may assume that E = E ,λ : y = ( x + λ )( x + 1) x. Since W is divisible by 4 in E ,λ ( K ), the desired result follows from Proposition6.3. (cid:3) Remark 6.6.
There is another family of elliptic curves ([9, Table 3 on p. 217],[11, Appendix E])) y + (1 − a ( t )) xy − b ( t ) y = x − b ( t ) x with a ( t ) = (2 t + 1)(8 t + 4 t + 1)2(4 t + 1)(8 t − t , b ( t ) = (2 t + 1)(8 t + 4 t + 1)(8 t − , whose group of rational points contains a subgroup isomorphic to Z / Z ⊕ Z / Z .Let us assume that t is an element of an arbitrary field K (with char( K ) = 2)such that t = 0 , t − = 0 , t + 1 = 0and put U ( t ) := (2 t + 1)(8 t + 4 t + 1) , A ( t ) = 2(4 t + 1)(8 t − t = 0 , B ( t ) = (8 t − = 0 , a ( t ) = U ( t ) A ( t ) , b ( t ) = U ( t ) B ( t ) . Let us consider the cubic curve E ,t over K defined by the same equation E ,t : y + (1 − a ( t )) xy − b ( t ) y = x − b ( t ) x as above. In light of Theorem 6.5, if E ,t is an elliptic curve over K , then E ,t is isomorphic to E ,c for a certain c ∈ K . Let us find the corresponding λ (as arational function of t ). First, rewrite the equation for E ,t as (cid:18) y + (1 − a ( t ) x ) − b ( t )2 (cid:19) = x − b ( t ) x + (cid:18) (1 − a ( t )) x − b ( t )2 (cid:19) , i.e., (cid:18) y + (1 − a ( t ) x ) − b ( t )2 (cid:19) = x − U ( t ) B ( t ) · x + (cid:16) − U ( t ) A ( t ) (cid:17) x − U ( t ) B ( t ) , Second, multiplying the last equation by ( A ( t ) B ( t )) and introducing new variables y = ( A ( t ) B ( t )) · (cid:18) y + (1 − a ( t )) x − b ( t )2 (cid:19) , x = ( A ( t ) B ( t )) · x, we obtain (with help of magma ) the following equation for an isomorphic cubiccurve ˜ E ,t : y = x + − U ( t ) A ( t ) B ( t ) + (( U ( t ) − A ( t )) B ( t ) x + ( U ( t ) − A ( t )) U ( t ) A ( t ) B ( t ) x + A ( t ) B ( t ) U ( t )
4= ( x − α )( x − α )( x − α ) , where α = − ( − t − t − t + 2162688 t + 753664 t − t − t − t + 14336 t + 2304 t − t − t − t ) ,α = − (4194304 t + 4194304 t − t − t − t +491520 t + 163840 t − t − t + 1792 t + 192 t − t − t ,α = − ( − t − t − t + 2424832 t + 1015808 t − t − t − t + 30720 t + 8960 t − t − t − t + 16 t + 4 t + 1 / . Using magma , we obtain that α − α = − t ( t + 1 / ( t − / , α − α = − ( t + 1 / ( t − / . This implies that ˜ E ,t (and therefore E ,t ) is an elliptic curve over K (i.e., all three α , α , α are distinct elements of K ) if and only if t = 0 , − , − , ± √ HE DIVISIBILITY BY 2 OF RATIONAL POINTS ON ELLIPTIC CURVES 17 and α − α α − α = (cid:18) t ( t + 1 / t + 1 / (cid:19) = 1 . Assume that all these conditions hold. Then the change of variable x = x + α transforms ˜ E ,t to the elliptic curve E : y = x ( x − ( α − α ))( x − ( α − α )) = x (cid:0) x + 2 t ( t + 1 / ( t − / (cid:1) (cid:0) x + 2 ( t + 1 / ( t − / (cid:1) . If we put κ = 2 ( t + 1 / ( t − / , then κ = − ( α − α )and E is isomorphic to the elliptic curve E ( κ ) : y ′ = x ′ (cid:18) x ′ + α − α α − α (cid:19) ( x ′ + 1) = x ′ x ′ + (cid:18) t ( t + 1 / t + 1 / (cid:19) ! ( x ′ + 1) . Notice that2 t ( t + 1 / t + 1 / t (4 t + 2)(4 t + 1) = 4 t (4 t + 2)2(4 t + 1) = (4 t + 1) − t + 1) = (4 t + 1) − t +1) , and therefore E ( κ ) = E ,c with c = (4 t + 1). This implies that E ,t is isomorphicto E ,c with c = (4 t + 1). Remark 6.7.
Suppose that K = F q with q = 3 , , F q \ { , , − , ± ± √ , ±√− } = ∅ . Corollary 6.8.
Let E be an elliptic curve over F q , where q = 11 , , , . Thegroup E ( F q ) is isomorphic to Z / Z ⊕ Z / Z if and only if E is isomorphic to oneof elliptic curves E ,c .Proof. Suppose that E ( F q ) is isomorphic to Z / Z ⊕ Z / Z . It follows from Theorem6.5 that E is isomorphic to one of the elliptic curves E ,c : y = " x + (cid:18) c − c (cid:19) ( x + 1) x with c ∈ K \ { , ± , ±√− , ±√− } . Conversely, suppose that E is isomorphic toone of these curves. We need to prove that E ( F q ) is isomorphic to Z / Z ⊕ Z / Z . ByTheorem 6.5, E ( F q ) contains a subgroup isomorphic to Z / Z ⊕ Z / Z ; in particular,16 divides | E ( F q ) | . In order to finish the proof, it suffices to check that | E ( F q ) | < | E ( F q ) | ≤ q + 2 √ q + 1 ≤
19 + 2 √
19 + 1 < . (cid:3) Corollary 6.9.
Let E be an elliptic curve over F . The group E ( F ) is iso-morphic to Z / Z ⊕ Z / Z if and only if E is isomorphic to one of elliptic curves E ,c . Proof.
Suppose that E ( F ) is isomorphic to Z / Z ⊕ Z / Z . Then it containsa subgroup isomorphic to Z / Z ⊕ Z / Z . It follows from Theorem 6.5 that E isisomorphic to one of elliptic curves E ,c : y = " x + (cid:18) c − c (cid:19) ( x + 1) x with c ∈ K \ { , ± , ± ± √ , ±√− } .Conversely, suppose that E is isomorphic to one of these curves. We need to provethat E ( F ) is isomorphic to Z / Z ⊕ Z / Z . By Theorem 6.5, E ( F ) contains asubgroup isomorphic to Z / Z ⊕ Z / Z ; in particular, 16 divides | E ( F ) | . The Hassebound tells us that 47 + 1 − √ ≤ | E ( F ) | ≤
47 + 1 + 2 √ < | E ( F ) | <
62. This implies that | E ( F ) | = 48; in particular, E ( F ) contains a point of order 3. This implies that E ( F ) contains a subgroupisomorphic to ( Z / Z ⊕ Z / Z ) ⊕ Z / Z ∼ = Z / Z ⊕ Z / Z . Since this subgroup has the same order 48 as the whole group E ( F ), we get thedesired result. (cid:3) Theorem 6.10.
Let K = Q and E be an elliptic curve over Q . Then the torsionsubgroup E ( Q ) t of E ( Q ) is isomorphic to Z / Z ⊕ Z / Z if and only if there exists c ∈ Q \ { , ± } such that E is isomorphic to E ,c .Proof. By Theorem 4.2 applied to m = 4, if E ( Q ) contains a subgroup isomorphicto Z / Z ⊕ Z / Z then E ( Q ) t is isomorphic to Z / Z ⊕ Z / Z . Now the desired resultfollows from Theorem 6.5, since neither √ √− Q . (cid:3) Theorem 6.11.
Let E be an elliptic curve over K . Then E ( K ) contains a subgroupisomorphic to Z / Z ⊕ Z / Z if and only if K contains i = √− and there exist c, d ∈ K \ { , ± , ± ± √ , ±√− } such that c − c = i (cid:18) d − d (cid:19) and E is isomorphic to E ,c . Remark 6.12.
The above equation defines an open dense set in the plane affinecurve(11) M , : ( c − d = i ( d − c. It is immediate that the corresponding projective closure is a nonsingular cubic¯ M , with a K -point, i.e., an elliptic curve. To obtain a Weierstrass normal form of¯ M , , we first slightly simplify equation(11) by the change of variables d = s, i c = t and get s t + ts + s − t = 0. Then, using the birational transformation s = ηξ + ξ , t = η ξ , we obtain η = ξ − ξ . See [16, Example 1.4.2 on p. 88] for an explicit description of the (finite) set of all Q ( i )-pointson this elliptic curve; none of them corresponds to ( c, d ) that satisfy the conditions of Theorem6.11. HE DIVISIBILITY BY 2 OF RATIONAL POINTS ON ELLIPTIC CURVES 19
Proof of Theorem 6.11.
We have already seen that E ,c ( K ) contains an order 8point R with 4 R = W . It follows from Proposition 6.4 that E ,c ( K ) contains allpoints of order 4. In particular, it contains an order 4 point Q with 2 Q = W .Clearly, R and Q generate a subgroup isomorphic to Z / Z ⊕ Z / Z .Conversely, suppose that E ( K ) contains a subgroup isomorphic to Z / Z ⊕ Z / Z .This implies that E ( K ) contains all twelve points of order 4. In particular, E maybe represented in the form (1). Clearly, one of the points of order 2 is divisible by4 in E ( K ). We may assume that W is divisible by 4. The same arguments as inthe proof of Theorem 6.5 allow us to assume that E = E ,λ : y = ( x + λ )( x + 1) x. Since W is divisible by 4 in E ,λ ( K ) and all points of order dividing 4 lie in E ,λ ( K ),every point R of E ,λ with 4 R = W also lies in E ,λ ( K ). It follows from Proposition6.3 that K contains i = √− c, d ∈ K \ { , , − , ± ± √ , ±√− } such that λ = (cid:20) c − c (cid:21) , − λ = (cid:20) d − d (cid:21) . This implies that c − c = ± i (cid:18) d − d (cid:19) . Replacing if necessary d by − d , we obtain the desired c − c = i (cid:18) d − d (cid:19) . (cid:3) Points of order 3
The following assertion gives a simple description of points of order 3 on ellipticcurves.
Proposition 7.1.
A point P = ( x , y ) ∈ E ( K ) has order if and only if one canchoose three square roots r i = √ x − α i in such a way that r r + r r + r r = 0 . Proof.
Indeed, let P be a point of order 3. Then 2( − P ) = P . Hence, all x − α i are squares in K . By (4), x ( − P ) = x + ( r r + r r + r r )for a suitable choice of r , r , r . Since x ( − P ) = x ( P ) = x , we get r r + r r + r r = 0.Conversely, suppose that there exists a triple of square roots r i = √ x − α i suchthat r r + r r + r r = 0. Since P ∈ E ( K ),( r r r ) = ( x − α )( x − α )( x − α ) = y , i.e., r r r = ± y . Replacing r , r , r by − r , − r , − r if necessary, we may assumethat r r r = − y . Then there exists a point Q = ( x ( Q ) , y ( Q )) ∈ E ( K ) such that Q = P , and x = x ( Q ) , y = y ( Q ) are expressed in terms of r , r , r as in (6).Therefore x ( Q ) = x + ( r r + r r + r r ) = x ,y ( Q ) = − y − ( r + r + r )( r r + r r + r r ) = − y , i.e., Q = − P, − P ) = P , and so P has order 3. (cid:3) Theorem 7.2.
Let a , a , a be elements of K such that all a , a , a are distinct.Let us consider the elliptic curve E = E a ,a ,a : y = ( x + a )( x + a )( x + a ) over K . Let P = (0 , a a a ) . Then P enjoys the following properties. (i) P is divisible by in E ( K ) . More precisely, there are four points Q ∈ E ( K ) with Q = P , namely, ( a a − a a − a a , ( a − a )( a + a )( a − a )) , ( a a − a a − a a , ( a − a )( a − a )( a + a )) , ( a a − a a − a a , ( a + a )( a − a )( a − a ) , ( a a + a a + a a , ( a + a )( a + a )( a + a )) . (ii) The following conditions are equivalent. (1) P has order . (2) None of a i vanishes, i.e., ± a , ± a , ± a are six distinct elements of K , and one of the following four equalities holds: a a = a a + a a , a a = a a + a a ,a a = a a + a a , a a + a a + a a = 0 . (iii) Suppose that equivalent conditions ( i ) − ( ii ) hold. Then one of four points Q coincides with − Q and has order 3, while the three other points are of order . In addition, E ( K ) contains a subgroup isomorphic to Z / Z ⊕ Z / Z . Remark 7.3.
Clearly, E a ,a ,a = E ± a , ± a , ± a . Proof of Theorem 7.2.
We have α = − a , α = − a , α = − a . Let us try to divide P by 2 in E ( K ). We have r = ± a , r = ± a , r = ± a . Since all r i lie in K , the point P = (0 , a a a ) is divisible by 2 in E ( K ). Let Q bea point on E with 2 Q = P . By (4) and (7), x ( Q ) = r r + r r + r r , y ( Q ) = − ( r + r )( r + r )( r + r )with r r r = − a a a . Plugging in r i = ± a i into the formulas for x ( Q ) and y ( Q ),we get explicit formulas for points Q as in the statement of the theorem. Thisproves (i).Let us prove (ii). Suppose that P has order 3. Since P is not of order 2, we have0 = x ( P ) = α i for all i = 1 , ,
3. Since { α , α , α } = {− a , − a , − a } , none of a i vanishes. It follows from Proposition 7.1 that one may choose the signsfor r i in such a way that r r + r r + r r = 0. Plugging in r i = ± a i into thisformula, we get four relations between a , a , a as in (ii)(2). HE DIVISIBILITY BY 2 OF RATIONAL POINTS ON ELLIPTIC CURVES 21
Now suppose that one of the relations as in (ii)(2) holds. This means that onemay choose the signs of r i = ± a i in such a way that r r + r r + r r = 0. Itfollows from Proposition 7.1 that P has order 3. This proves (ii).Let us prove (iii). Since P has order 3, 2( − P ) = P , i.e., − P is one of the four Q ’s. Suppose that Q is a point of E with 2 Q = P, Q = − P . Clearly, the order of Q is either 3 or 6. Assume that Q has order 3. Then P = 2 Q = − Q and therefore Q = − P , which is not the case. Hence Q has order 6. Then 3 Q has order 2, i.e.,coincides with W i = ( − a i ,
0) for some i ∈ { , , } . Pick j ∈ { , , } \ { i } andconsider the point W j = ( − a j , = W i . Then the subgroup of E ( K ) generated by Q and W j is isomorphic to Z / Z ⊕ Z / Z . This proves (iii). (cid:3) Remark 7.4.
In Theorem 7.2 we do not assume that char( K ) = ! Corollary 7.5.
Let a , a , a be elements of K such that a , a , a are distinct.Then the following conditions are equivalent. (i) The point P = (0 , a a a ) ∈ E a ,a ,a ( K ) has order . (ii) None of a i vanishes, and one may choose signs for a = ± a , b = ± a , c = ± a in such a way that c = ab/ ( a + b ) .If these conditions hold, then E a ,a ,a = E λ,b : y = (cid:0) x + ( λb ) (cid:1) (cid:0) x + b (cid:1) x + (cid:18) λλ + 1 b (cid:19) ! , where λ = a/b ∈ K \ { , ± , − , − } . Proof.
Suppose that condition (ii) of the corollary holds, i.e., none of a i vanishes,and one may choose signs for a = ± a , b = ± a , c = ± a in such a way that c = ab/ ( a + b ). Then none of a, b, c vanishes and ab = ac + bc .By Theorem 7.2(ii), P = (0 , abc ) is a point of order 3 on the elliptic curve E λ,b = E a ,a ,a . Since abc = ± a a a , either P = P or P = − P . In both cases P has order 3.Notice that ± a , ± a , ± a are six distinct elements of K . This means that ± a, ± b, ± c are also six distinct elements of K . If we put λ = a/b , then ± λb, ± b, ± λ + 1 λ b are six distinct elements of K . This means (in light of the inequalities a = 0 , b = 0)that λ = 0 , ± , − , − . Suppose P has order 3. By Theorem 7.2(ii), none of a i vanishes and one of thefollowing four equalities holds: a a = a a + a a , a a = a a + a a ,a a = a a + a a , a a + a a + a a = 0 . Here are the corresponding choices of a, b, c with c = ab/ ( a + b ): a = a , b = − a , c = a ; a = a , b = − a , c = a ; a = a , b = a , c = a ; a = a , b = a , c = − a . In order to finish the proof, we just need to notice that a = λb and c = aba + b = λb · bλb + b = λλ + 1 b. (cid:3) Theorem 7.6.
Let E be an elliptic curve over K . Then E ( K ) contains a subgroupisomorphic to Z / Z ⊕ Z / Z if and only if there exists λ ∈ K \ { , ± , − , − } suchthat E is isomorphic to E ,λ : y = (cid:0) x + λ (cid:1) ( x + 1) x + (cid:18) λλ + 1 (cid:19) ! . Proof of Theorem 7.6.
Let λ ∈ K \ { , ± , − , − / } and put a = λ, a = 1 , a = λ/ ( λ + 1). Then all a i do not vanish, a , a , a are three distinct elements of K , a a = a a + a a , and E ,λ = E a ,a ,a . It follows from Theorem 7.2 that E ,λ contains a subgroup isomorphic to Z / Z ⊕ Z / Z .Conversely, suppose that E is an elliptic curve over K such that E ( K ) containsa subgroup isomorphic to Z / Z ⊕ Z / Z . It follows that all three points of order 2lie in E ( K ), and therefore E can be represented in the form (1). It is also clear that E ( K ) contains a point of order 3. Let us choose a point P = ( x ( P ) , y ( P )) ∈ E ( K )of order 3. We may assume that x ( P ) = 0. We have P = 2( − P ), and therefore P is divisible by 2 in E ( K ). By Theorem 2.1, all x ( P ) − α i = − α i are squares in K .This implies that there exist elements a , a , a ∈ K such that α i = − a i . Clearly,all three a , a , a are distinct. Since P lies on E , y ( P ) = ( x ( P ) + a )( x ( P ) + a )( x ( P ) + a ) = a a a = ( a a a ) , and therefore y ( P ) = ± a a a . Replacing P by − P if necessary, we may assumethat y ( P ) = a a a , i.e., P = (0 , a a a ) is a K -point of order 3 on E = E a ,a ,a : y = ( x + a ) ( x + a )( x + a ) . It follows from Corollary 7.5 that there exist nonzero b ∈ K and λ ∈ K \{ , ± , − , − / } such that E = E a ,a ,a = E λ,b : y = (cid:0) x + ( λb ) (cid:1) (cid:0) x + b (cid:1) x + (cid:20) λλ + 1 b (cid:21) ! . But E λ,b is isomorphic to E λ,b ( b ) : y ′ = ( x ′ + λ )( x ′ + 1) x ′ + (cid:20) λλ + 1 (cid:21) ! while the latter coincides with E ,λ . (cid:3) Remark 7.7.
There is a family of elliptic curves over Q [9, Table 3 on p. 217] (seealso [11, Appendix E]), E ,t : y + (1 − a ( t )) xy − b ( t ) y = x − b ( t ) x , HE DIVISIBILITY BY 2 OF RATIONAL POINTS ON ELLIPTIC CURVES 23 with a ( t ) = 10 − tt − , b ( t ) = − t − ( t − t − (with t ∈ Q \ { , , ± , } ), whose group of rational points contains a subgroupisomorphic to Z / Z ⊕ Z / Z . (The point (0 ,
0) of E ,t has order 6, ibid.) Let usassume that t = ± K (with char( K ) = 2) andconsider the cubic curve E ,t over K defined by the same equation as above.In light of Theorem 7.6, if E ,t is an elliptic curve over K , then E ,t is isomorphicto E ,λ for a certain λ ∈ K . Let us find the corresponding λ (as a rational functionof t ). First, rewrite the equation for E ,λ as (cid:18) y + (1 − a ( t ) x ) − b ( t )2 (cid:19) = x − b ( t ) x + (cid:18) (1 − a ( t )) x − b ( t )2 (cid:19) . Second, multiplying the last equation by ( t − and introducing new variables y = ( t − · (cid:18) y + (1 − a ( t )) x − b ( t )2 (cid:19) , x = ( t − · x, we obtain (with help of magma ) the equation for an isomorphic cubic curve˜ E ,t : y = ( x − α )( x − α )( x − α ) , where α = − (2 t − t − t + 90) = − t − t − t + 3) ,α = − (2 t − t + 14 t −
6) = − t − t − ,α = − (cid:18) t − t − t + 7 t − (cid:19) = −
14 ( t − t + 3)( t − . We have α − α = − ( t − , α − α = 14 · ( t − ( t − , α − α = − · ( t − ( t + 3) . This implies that ˜ E ,t (and therefore E ,t ) is an elliptic curve over K if and only if t ∈ K \ { , ± , , } . Further we assume that this condition holds and therefore ˜ E ,t and E ,t are ellipticcurves over K . Clearly, all three points of order 2 on ˜ E ,t are defined over K andthe K -point Q = ( x ( Q ) , y ( Q )) = (0 , − ( t − t − t + 3)( t − )lies on ˜ E ,t . We prove that Q has order 6. Let us consider the point P = 2 Q ∈ E ( K )with coordinates x ( P ) , y ( P ) ∈ K . (Since y ( P ) = 0, P = ∞ .) According toformulas of Section 1, there exists a unique triple r , r , r of distinct elements of K such that( r + r )( r + r )( r + r ) = − y ( Q ) = ( t − t − t + 3)( t − and for all i = 1 , , x ( P ) − α i = r i , = − α i = x ( Q ) − α i = ( r i + r j )( r i + r k ) , where ( i, j, k ) is a permutation of (1 , , r + r = ( t − t − t + 3)( t − − a = ( t − t − t + 3)( t − ( t − t + 3)( t − = 4( t − , r + r = ( t − t − t + 3)( t − − a = ( t − t − t + 3)( t − t − t − t + 3) = 12 · ( t − ,r + r = ( t − t − t + 3)( t − − a = ( t − t − t + 3)( t − t − t − = 12 · ( t − t +3) . Hence r + r = 4( t − , r + r = ( t − , r + r = ( t + 3)( t − , and therefore r + r + r = 12 · (( r + r ) + ( r + r ) + ( r + r )) = 12 · ( t + 2 t − , which, in turn, implies that r = 2 t −
10 = 2( t − , r = 2 t − t − , r = 12 · ( t − t −
5) = 18 r r . One may easily check that c ( t ) := − t + 14 t − t + 10 = r i + α i for all i = 1 , , . This implies that x ( P ) = c ( t ) , c ( t ) − α i = r i for all i = 1 , , E ,t is isomorphic to the elliptic curve E r ,r ,r : y = ( x + r )( x + r )( x + r )with x = x − c ( t ). In addition, y ( P ) = − r r r = − t − ( t − . We have r r = 8 r , r − r = 8 . This implies that ( r − r ) r = r r , which means that( − r ) r + r r + ( − r ) r = 0 . It follows from Proposition 7.1 that P has order 3 in ˜ E ,t ( K ). (In particular, all r i = 0.) Since 2 Q = P , the order of Q in ˜ E ,t is 6.Notice that − r = ( − r ) r ( − r ) + r and E r ,r ,r = E − r ,r , − r . It follows from Corollary 7.5 and the end of the proof of Theorem 7.6 that E r ,r ,r is isomorphic to E ,λ with λ = − r r = − (2 t − t − − t − t − . This implies that E ,t is isomorphic to E ,λ with λ = − ( t − / ( t − Corollary 7.8.
Let E be an elliptic curve over F q , where q = 7 , , , . Thegroup E ( F q ) is isomorphic to Z / Z ⊕ Z / Z if and only if E is isomorphic to oneof elliptic curves E ,λ . HE DIVISIBILITY BY 2 OF RATIONAL POINTS ON ELLIPTIC CURVES 25
Proof.
Suppose that E ( F q ) is isomorphic to Z / Z ⊕ Z / Z . By Theorem 7.6, E isisomorphic to one of elliptic curves E ,λ .Conversely, suppose that E is isomorphic to one of these curves. We need toprove that E ( F q ) is isomorphic to Z / Z ⊕ Z / Z . By Theorem 7.6, E ( F q ) containsa subgroup isomorphic to Z / Z ⊕ Z / Z ; in particular, 12 divides | E ( F q ) | . In orderto finish the proof, it suffices to check that | E ( F q ) | <
24, but this inequality followsfrom the Hasse bound (10) | E ( F q ) | ≤ q + 2 √ q + 1 ≤
13 + 2 √
13 + 1 < . (cid:3) Corollary 7.9.
Let E be an elliptic curve over F . The group E ( F ) is iso-morphic to Z / Z ⊕ Z / Z if and only if E is isomorphic to one of elliptic curves E ,λ .Proof. Suppose that E ( F ) is isomorphic to Z / Z ⊕ Z / Z . Then it containsa subgroup isomorphic to Z / Z ⊕ Z / Z . It follows from Theorem 7.6 that E isisomorphic to one of elliptic curves E ,λ .Conversely, suppose that E is isomorphic to one of these curves. We need to provethat E ( F ) is isomorphic to Z / Z ⊕ Z / Z . By Theorem 7.6, E ( F ) contains asubgroup isomorphic to Z / Z ⊕ Z / Z ; in particular, 12 divides | E ( F ) | . The Hassebound (10) tells us that23 + 1 − √ ≤ | E ( F ) | ≤
23 + 1 + 2 √ < | E ( F ) | <
34. It follows that | E ( F ) | = 24; in particularthe 2-primary component E ( F )(2) of E ( F ) has order 8. On the other hand, E ( F )(2) is isomorphic to a product of two cyclic groups, each of which has evenorder. This implies that E ( F )(2) is isomorphic to Z / Z ⊕ Z / Z . Taking intoaccount that E ( F ) contains a point of order 3, we conclude that it contains asubgroup isomorphic to( Z / Z ⊕ Z / Z ) ⊕ Z / Z ∼ = Z / Z ⊕ Z / Z . This subgroup has the same order 24 as the whole group E ( F ), which ends theproof. (cid:3) Theorem 7.10.
Let K = Q and E an elliptic curve over Q . Then the torsionsubgroup E ( Q ) t of E ( Q ) is isomorphic to Z / Z ⊕ Z / Z if and only if there exists λ ∈ Q \ { , ± , − , − } such that E is isomorphic to E ,λ .Proof. By Theorem 4.2 applied to m = 3, if E ( Q ) contains a subgroup isomorphicto Z / Z ⊕ Z / Z then E ( Q ) t is isomorphic to Z / Z ⊕ Z / Z . Now the desired resultfollows from Theorem 7.6. (cid:3) Points of order
Proposition 8.1.
Let P = ( x , y ) ∈ E ( K ) . The point P has order if and onlyif, for any permutation i, j, k of , , , one can choose square roots r i = √ x − α i and r (1) i = p ( r i + r j )( r i + r k ) in such a way that (12) ( r r + r r + r r ) + ( r (1)1 r (1)2 + r (1)2 r (1)3 + r (1)3 r (1)1 ) = 0 ,r r + r r + r r = 0 . Remark 8.2.
Notice that if we drop the condition r r r = − y in formulas (4)and (7), then we get 8 points Q such that 2 Q = ± P . Similarly, if we drop theconditions r r r = − y , r (1)1 r (1)2 r (1)3 = ( r + r )( r + r )( r + r ) in the formulas(9), then we obtain all points R for which 4 R = ± P . Proof of Proposition 8.1.
Suppose that P has order 5. Then − P is a 1 / P .Therefore there exist r i and r (1) j such that x ( − P ) = x ( P ) + ( r r + r r + r r ) + ( r (1)1 r (1)2 + r (1)2 r (1)3 + r (1)3 r (1)1 ) . Since x ( P ) = x ( − P ), we have( r r + r r + r r ) + ( r (1)1 r (1)2 + r (1)2 r (1)3 + r (1)3 r (1)1 ) = 0 . On the other hand, if r r + r r + r r , then the corresponding Q (with 2 Q = P )satisfies x ( Q ) = x ( P ) + ( r r + r r + r r ) = x ( P )and therefore Q = P or − P . Since 2 Q = P , either P = 2 P or Q = − P = − Q has order 5. Clearly, P = 2 P . If Q = − Q then Q has order dividing 3,which is not true, because its order is 5. The contradiction obtained proves that r r + r r + r r = 0.Conversely, suppose there exist square roots r i = √ x − α i and r (1) i = q ( r i + r j )( r i + r k )that satisfy (12). Replacing if necessary all r i by − r i , we may and will assume that r r r = − y ( P ). Let Q = ( x ( Q ) , y ( Q )) be the corresponding half of P with x ( Q ) = x ( P ) + ( r r + r r + r r ). Since r r + r r + r r = 0, we have x ( Q ) = x ( P );in particular, Q = − P . Replacing if necessary all r (1) i by r (1) i , we may and willassume that r (1)1 r (1)2 r (1)3 = ( r + r )( r + r )( r + r ) = − y ( Q ) . Let R = ( x ( R ) , y ( R )) be the corresponding half of Q . Then 4 R = 2(2 R ) = 2 Q = P and x ( R ) = x ( P ) + ( r r + r r + r r ) + ( r (1)1 r (1)2 + r (1)2 r (1)3 + r (1)3 r (1)1 ) = x ( P ) . This means that either R = P or R = − P . If R = P , then R = 4 R and R has order3. This implies that both Q = 2 R and P = 4 R also have order 3. It follows that P = − Q , which is not the case. Therefore R = − P . This means that R = − R ,i.e., R has order 5 and therefore P = − R also has order 5. (cid:3) In what follows we will use the following identities in the polynomial ring Z [ t , t , t ]that could be checked either directly or by using magma .(13) ( − t + t + t )( t − t + t ) + ( t − t + t )( t + t − t )+( t + t − t )( − t + t + t ) = − ( t + t + t )( − t + t + t )( t − t + t )( t + t − t ) , HE DIVISIBILITY BY 2 OF RATIONAL POINTS ON ELLIPTIC CURVES 27 (14) ( − t + t + t )( t − t + t ) + ( t − t + t )( t + t − t )+( t + t − t )( − t + t + t ) + 4 t t t + 4 t t t + 4 t t t = t + t + t − t t − t t − t t − t t t − t t t − t t t = ( t + t + t ) (cid:0) t + t + t − t t − t t − t t − t t − t t − t t − t t t (cid:1) . Theorem 8.3.
Let a , a , a be elements of K such that ± a , ± a , ± a are sixdistinct elements of K and none of three elements β = − a + a + a , β = a − a + a , β = a + a − a vanishes. Then the following conditions hold. (i) None of a i vanishes and β , β , β are three distinct elements of K . (ii) Let us consider an elliptic curve E a ,a ,a : y = (cid:18) x + β (cid:19) (cid:18) x + β (cid:19) (cid:18) x + β (cid:19) with P = (0 , − β β β / ∈ E a ,a ,a ( K ) .Then P enjoys the following properties. (1) P ∈ E a ,a ,a ( K ) . (2) Assume that (15) a + a + a − a a − a a − a a − a a − a a − a a − a a a = 0 , ( a + a + a )( a − a − a )( a + a − a )( a − a + a ) = 0 . Then P has order . In addition, E a ,a ,a ( K ) contains a subgroup iso-morphic to Z / Z ⊕ Z / Z .Proof. (i) Since a i = − a i , none of a i vanishes. Let i, j ∈ { , , } be two distinctindices and k ∈ { , , } be the third one. Then β i − β j = a j − a i = 0 , β i + β j = 2 a k = 0 . This implies that β i = β j .(ii) Keeping our notation, we obtain that r = ± β ± − a + a + a , r = ± β a − a + a , r = ± β ± a + a − a ,r (1) i = ± q ( r i + r j )( r i + r k )where i, j, k is any permutation of 1 , ,
3. Thanks to Proposition 8.1, it sufficesto check that one may choose the square roots r i and r (1) i in such a way that r r + r r + r r = 0 and(16) ( r r + r r + r r ) + ( r (1)1 r (1)2 + r (1)2 r (1)3 + r (1)3 r (1)1 ) = 0 . Let us put r i = β i − a i + a j + a k . We have r + r = a , r + r = a , r + r = a . It follows that ( r (1)1 ) = a a , ( r (1)2 ) = a a , ( r (1)3 ) = a a . Let us put r (1)1 = a a , r (1)2 = a a , r (1)3 = a a . Then the condition (16) may be rewritten as follows.( − a + a + a )( a − a + a ) + ( a − a + a )( a + a − a )+( a + a − a )( − a + a + a ) + 4 a a a + 4 a a a + 4 a a a = 0 . In light of (14), the condition (16) may be rewritten as( a + a + a )( a + a + a − a a − a a − a a − a a − a a − a a − a a a ) = 0 . The latter equality follows readily from the assumption (15) of Theorem. By Propo-sition 8.1, it suffices to check that r r + r r + r r = 0. In other words, we needto prove that(17) ( − a + a + a )( a − a + a ) + ( a − a + a )( a + a − a )+( a + a − a )( − a + a + a ) = 0 . In light of (13), this inequality is equivalent to( a + a + a )( a − a − a )( a + a − a )( a − a + a ) = 0 . But the latter inequality holds, by the assumption (15) of the theorem. Hence, P has order 5. Clearly, P and all points of order 2 generate a subgroup that isisomorphic to Z / Z ⊕ Z / Z . (cid:3) Theorem 8.4.
Let E be an elliptic curve over K . Then the following conditionsare equivalent. (i) E ( K ) contains a subgroup isomorphic to Z / Z ⊕ Z / Z . (ii) There exists a triple { a , a , a } ⊂ K that satisfies all the conditions ofTheorem 8.3, including (15), and such that E is isomorphic to E a ,a ,a .Proof. (i) follows from (ii), thanks to Theorem 8.3.Suppose (i) holds. In order to prove (ii) it suffices to check that E is isomorphicto a certain E a ,a ,a over K . We may assume that E is defined by an equation ofthe form (1). Suppose that P = (0 , y ( P )) ∈ E ( K ) has order 5. Then P = 4( − P ) isdivisible by 4 in E ( K ). This implies the existence of square roots r i = √− α i ∈ K and r (1) i = p ( r i + r j )( r i + r k ) ∈ K in such a way that x ( − P ) = x ( P ) + ( r r + r r + r r ) + ( r (1)1 r (1)2 + r (1)2 r (1)3 + r (1)3 r (1)1 ) ,r (1)1 r (1)2 r (1)3 = ( r + r )( r + r )( r + r ) . Since x ( − P ) = x ( P ) = 0,(18) ( r r + r r + r r ) + ( r (1)1 r (1)2 + r (1)2 r (1)3 + r (1)3 r (1)1 ) = 0 . Since the order of P is not r r + r r + r r = 0 . Recall that none of r i + r j vanishes. Let us choose square roots b = √ r + r , b = √ r + r , b = √ r + r in such a way that r (1)1 = b b , r (1)2 = b b . Since r (1)1 r (1)2 r (1)3 = b b b = ( b b b ) , we conclude that r (1)3 = r (1)1 r (1)2 r (1)3 r (1)2 r (1)3 = ( b b b ) ( b b )( b b ) = b b . We obtain that(20) r (1)1 = b b , r (1)2 = b b , r (1)3 = b b . Unfortunately, b i do not have to lie in K . However, all the ratios b i /b j lie in K ∗ .We have r + r = b , r + r = b , r + r = b and therefore(21) r = − b + b + b , r = b − b + b , r = b + b − b ,α = − r = ( − b + b + b ) , α = − r = − ( b − b + b ) ,α = − r = − ( b + b − b ) ,P = (0 , − ( r + r )( r + r )( r + r )) = (0 , − b b b ) ∈ E ( K ) . Since none of r i vanishes, we get − b + b + b = 0 , b − b + b = 0 , b + b − b = 0 . Let us put γ = − b + b + b , γ = b − b + b , γ = b + b − b . It follows from Theorem 8.3(i) that all β i are distinct nonzero elements of K . Theinequality (19) combined with first formula of (21) gives us( − b + b + b )( b − b + b ) + ( b − b + b )( b + b − b )+( b + b − b )( − b + b + b ) = 0 , which is equivalent (thanks to (13)) to( b + b + b )( b − b − b )( b + b − b )( b − b + b ) = 0 . In particular, b + b + b = 0 . The equality (18) gives us (thanks to (14))( b + b + b )( b + b + b − b b − b b − a b − b b − b b − b b − b b b ) = 0 , i.e., ( b + b + b − b b − b b − a b − b b − b b − b b − b b b ) = 0 . Let us put a = b b , a = b b , a = b b = 1 . All a i lie in K . Clearly, the triple { a , a , a } satisfies all the conditions of Theorem8.3 including (15). Let us put β = − a + a + a = γ b = γ r + r ,β = a − a + a = γ b = γ r + r ,β = a + a − a = γ b = γ r + r . The equation of E is y = (cid:18) x + γ (cid:19) (cid:18) x + γ (cid:19) (cid:18) x + γ (cid:19) . Then E is isomorphic to E ( r + r ) : y ′ = (cid:18) x ′ + γ r + r ) (cid:19) (cid:18) x ′ + γ r + r ) (cid:19) (cid:18) x ′ + γ r + r ) (cid:19) = (cid:18) x ′ + β (cid:19) (cid:18) x ′ + γ (cid:19) (cid:18) x ′ + γ (cid:19) . Clearly, E ( r + r ) coincides with E a ,a ,a . (cid:3) Remark 8.5.
Let E a ,a ,a be as in Theorem 8.3. Clearly, E a ,a ,a ( a ) = E a /a ,a /a , . Let us put λ = a /a , µ = a /a . Then(22) E a /a ,a /a , = E λ,µ, : y = " x + (cid:18) − λ + µ + 12 (cid:19) x + (cid:18) λ − µ + 12 (cid:19) x + (cid:18) λ + µ − (cid:19) . The equation of (isomorphic) E λ,µ, (cid:16) λ + µ − (cid:17) is as follows.(23) E λ,µ, (cid:18) λ + µ − (cid:19) : y = " x + (cid:18) − λ + µ λ + µ − (cid:19) x + (cid:18) λ − µ + 1 λ + µ − (cid:19) ( x +1) . The conditions on a , a , a may be rewritten in terms of λ, µ as follows.(24) λ + µ − λ µ − λµ − λ − λµ − µ − λ − µ + 1 = 0 ,λ ± µ = ± , λ = 0 , µ = 0 , λ = ± µ,λ + µ = 1 , λ − µ = ± . The equality (24) is equivalent to(25) ( λ + µ )( λ − µ ) − ( λ + µ ) − ( λ + µ ) + 1 = 0 . Multiplying (25) by (non-vanishing) ( λ + µ ), we get the equivalent equation(26) ( λ − µ ) − ( λ + µ ) − ( λ + µ ) + ( λ + µ ) = 0 . Let us make the change of variables ξ = λ + µ, η = λ − µ . HE DIVISIBILITY BY 2 OF RATIONAL POINTS ON ELLIPTIC CURVES 31
Then (26) may be rewritten as(27) η = ξ ( ξ + ξ − , which is an (affine model of an) elliptic curve if char( K ) = 5 and a singular rationalplane cubic (Cartesian leaf) if char( K ) = 5. Since(28) λ + µ = ( λ + µ ) + ( λ − µ ) ξ + η ξ ξ + ξ + ξ − ξ ξ + ξ + ξ − ξ , the only restrictions on ( ξ, η ) besides the equality (27) are the inequalities ξ ( ξ + ξ − = 0 , ± ξ + ξ + ξ − = 2 ξ, ± = ηξ = s ξ ( ξ + ξ − ξ , i.e.(29) ξ = 0 , ± , − ± √ . This means that(30) ( ξ, η )
6∈ { (0 , , ( ± , ± , ( − ± √ , } . In light of (28), the equation (22) may be rewritten with coefficients being rationalfunctions in ξ, η (rather than ( λ, µ )) as follows. E ,ξ,η : y = " x + (cid:18) − η ) ξ + ξ + ξ − (cid:19) x + (cid:18) η + 1) ξ + ξ + ξ − (cid:19) ( x + 1) . Theorem 8.6.
Let E be an elliptic curve over K . Then the following conditionsare equivalent. (i) E ( K ) contains a subgroup isomorphic to Z / Z ⊕ Z / Z . (ii) There exist ( ξ, η ) ∈ K that satisfy the equation (27) and inequalities (30)and such that E is isomorphic to E ,ξ,η .Proof. The result follows from Theorem 8.4 combined with Remark 8.5. (cid:3)
Remark 8.7.
In Theorem 8.6 we do not assume that char( K ) = ! Corollary 8.8.
Let E be an elliptic curve over F q with q = 13 , , , , , .Then E ( F q ) is isomorphic to Z / Z ⊕ Z / Z if and only if E is isomorphic to oneof E ,ξ,η .Proof. Suppose that E ( F q ) is isomorphic to Z / Z ⊕ Z / Z . By Theorem 8.6, E isisomorphic to one of elliptic curves E ,ξ,η .Conversely, suppose that E is isomorphic to one of these curves. We need toprove that E ( F q ) is isomorphic to Z / Z ⊕ Z / Z . By Theorem 8.6, E ( F q ) containsa subgroup isomorphic to Z / Z ⊕ Z / Z ; in particular, 20 divides | E ( F q ) | . In orderto finish the proof, it suffices to check that | E ( F q ) | <
40, but this inequality followsfrom the Hasse bound (10) | E ( F q ) | ≤ q + 2 √ q + 1 ≤
27 + 2 √
27 + 1 < . (cid:3) Corollary 8.9.
Let E be an elliptic curve over F q with q = 31 , , , . Then E ( F q ) is isomorphic to Z / Z ⊕ Z / Z if and only if E is isomorphic to one of E ,ξ,η .Proof. Suppose that E ( F q ) is isomorphic to Z / Z ⊕ Z / Z ; the latter contains asubgroup isomorphic to Z / Z ⊕ Z / Z . By Theorem 8.6, E is isomorphic to one ofelliptic curves E ,ξ,η .Conversely, suppose that E is isomorphic to one of these curves. We need toprove that E ( F q ) is isomorphic to Z / Z ⊕ Z / Z . By Theorem 8.6, E ( F q ) containsa subgroup isomorphic to Z / Z ⊕ Z / Z ; in particular, 20 divides | E ( F q ) | . It followsfrom the Hasse bound (10) that20 < − √
31 + 1 ≤ | E ( F q ) | ≤
43 + 2 √
43 + 1 < . This implies that | E ( F q ) | = 40, and therefore E ( F q ) is isomorphic to a direct sumof Z / Z and the order 8 abelian group E ( F q )(2); in addition, the latter group isisomorphic to a direct sum of two cyclic groups of even order (because it containsa subgroup isomorphic to Z / Z ⊕ Z / Z ). This implies that E ( F q )(2) is isomorphicto Z / Z ⊕ Z / Z . It follows that E ( F q ) is isomorphic to a direct sum Z / Z ⊕ Z / Z ⊕ Z / Z ∼ = Z / Z ⊕ Z / Z . (cid:3) Corollary 8.10.
Let E be an elliptic curve over F q with q = 59 or . Then E ( F q ) is isomorphic to Z / Z ⊕ Z / Z if and only if E is isomorphic to one of E ,ξ,η .Proof. Suppose that E ( F q ) is isomorphic to Z / Z ⊕ Z / Z ; the latter contains asubgroup isomorphic to Z / Z ⊕ Z / Z . By Theorem 8.6, E is isomorphic to one ofelliptic curves E ,ξ,η .Conversely, suppose that E is isomorphic to one of these curves. We need toprove that E ( F q ) is isomorphic to Z / Z ⊕ Z / Z . By Theorem 8.6, E ( F q ) containsa subgroup isomorphic to Z / Z ⊕ Z / Z ; in particular, 20 divides | E ( F q ) | . It followsfrom the Hasse bound (10) that40 < − √
59 + 1 ≤ | E ( F q ) | <
61 + 2 √
61 + 1 < . This implies that | E ( F q ) | = 60; in particular, E ( F q ) contains a subgroup isomorphicto Z / Z . This implies that E ( F q ) contains a subgroup isomorphic to( Z / Z ⊕ Z / Z ) ⊕ Z / Z ∼ = Z / Z ⊕ Z / Z ;the order of this subgroup is 60, i.e., it coincides with the order of the whole group E ( F q ). (cid:3) Theorem 8.11.
Let K be a quadratic field and E be an elliptic curve over K .Then the following conditions are equivalent. (i) The torsion subgroup E ( K ) t of E ( K ) is isomorphic to Z / Z ⊕ Z / Z . (ii) There exist ( ξ, η ) ∈ K that satisfy the equation (27) and inequalities (30)and such that E is isomorphic to E ,ξ,η .Proof. By Theorem 4.3, if E ( K ) contains a subgroup isomorphic to Z / Z ⊕ Z / Z then E ( K ) t is isomorphic to Z / Z ⊕ Z / Z . Now the desired result follows fromTheorem 8.6. (cid:3) HE DIVISIBILITY BY 2 OF RATIONAL POINTS ON ELLIPTIC CURVES 33
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