IRREDUCIBILITY OF ELLIPTIC CURVES AND INTERSECTION WITH LINES.

Transcription

1 IRREDUCIBILITY OF ELLIPTIC CURVES AND INTERSECTION WITH LINES. IAN KIMING 1. Non-singular points and tangents. Suppose that k is a field and that F (x 1,..., x n ) is a homogeneous polynomial in n variables (n 2) with coefficients in k, and of degree d, say. The equation: F (x 1,..., x n ) = 0 defines a projective variety V defined over k. If (ξ 1,..., ξ n ) P n 1 ( k) is a point on V ( k), we say that (ξ 1,..., ξ n ) is nonsingular if the partial derivatives x i do not all vanish at the point. If so, we have a well-defined hyperplane in P n given by the equation: ( ) x x n = 0 ; x 1 x 1=ξ 1 x n x n=ξ n this hyperplane is called the tangent (space) at the point (ξ 1,..., ξ n ). This tangent space does indeed pass through the point (ξ 1,..., ξ n ): For F is a sum with coefficients in k of monomials x i1 1 x in n where i i n = d; such a monomial is seen to give a contribution (i i n ) x i1 1 x in n = d x i1 1 x in n to the left hand side of the tangent equation ( ), whence: ξ ξ n = d F (ξ 1,..., ξ n ) = 0. x 1 x 1=ξ 1 x n x n=ξ n We call V smooth if all points in V ( k) are non-singular. It is clear that the question of whether V contains any singular points is independent of whether we apply a linear change of variables in the equation of V. 2. Elliptic curves. We define (in this course...) an elliptic curve over a field k as a non-singular projective curve given by a general Weierstraß equation: y 2 z + a 1 xyz + a 3 yz 2 = x 3 + a 2 x 2 z + a 4 xz 2 + a 6 z 3. Proposition 1. (i). Assume char(k) 2. Then, using a linear change of variables, a general Weierstraß equation may be put in shape: ( ) y 2 z = x 3 + ax 2 z + bxz 2 + cz 3. 1

2 2 IAN KIMING If additionally char(k) 3, we can also attain a = 0. A Weierstraß equation ( ) defines an elliptic curves if and only if the polynomial x 3 + ax 2 + bx + c does not have any multiple roots. (ii). Assume char(k) = 2. Then, using a linear change of variables, a general Weierstraß equation may be put in shape ( ) y 2 z + a 1 xyz = x 3 + ax 2 z + bxz 2 + cz 3, a 1 0, or ( ) y 2 z + a 3 yz 2 = x 3 + ax 2 z + bxz 2 + cz 3, a 3 0. An equation ( ) defines an elliptic curve if and only if b 2 + a 1 c 0. equation ( ) always defines an elliptic curve. Proof. (i). Since char(k) 2 we can consider the linear change of variables: (x, y, z) (x, y a 1 2 x a 3 2, z) that will bring the Weierstraß equation in shape ( ). If also char(k) 3, we can further consider the change x x a 3, y y, z z. By definition, a Weierstraß equation ( ) defines an elliptic curve if and only if the projective curve defined by it is without singularities. Writing the equation ( ) as F (x, y, z) = 0 where: F (x, y, z) := y 2 z x 3 ax 2 z bxz 2 cz 3, we find that: x = 3x2 2axz bz 2, y = 2yz, z = y2 ax 2 2bxz 3cz 2. Hence, we see that ( x, y, z )(0, 1, 0) = (0, 0, 1), and so O = (0, 1, 0) is never a singular point. Suppose that (x 0, y 0, 1) is an affine singular point. Then 2y 0 = 0, and hence y 0 = 0 since char(k) 2. We conclude that x 0 is a root of x 3 + ax 2 + bx + c. Since we have 3x 2 0 2ax 0 b = 0 we conclude that x 0 is a multiple root of x 3 + ax 2 + bx + c. If conversely the polynomial x 3 +ax 2 +bx+c has a multiple root x 0 then (x 0, 0, 1) is a point on the projective curve defined by ( ). We also see that x and y both vanish at this point. The same holds for the partial derivative z : ax 2 0 2bx 0 3c = (3x ax 0 + b)x 0 3(x ax bx 0 + c) = 0, so that (x 0, 0, 1) is in fact a singular point. (ii). If a 1 0 the linear change of variables (x, y, z) (x + a 3 a 1, y, z) An

3 IRREDUCIBILITY OF ELLIPTIC CURVES AND INTERSECTION WITH LINES. 3 transforms ( ) to an equation of type ( ). The remaining statements are left as exercises. Proposition 2. Suppose that E is an elliptic curve defined over a field k. Then E is irreducible, i.e., does not contain a whole line in P 2 ( k). Proof. We use the previous proposition and leave the proof as an exercise in case char(k) = 2. So assume char(k) 2. According to the previous proposition we may then assume the equation for E in form: E : y 2 z = x 3 + ax 2 z + bxz 2 + cz 3. By definition E is a curve without singularities, and by the previous proposition we know that this is equivalent to the polynomial not having any multiple roots. x 3 + ax 2 + bx + c Now suppose that E contains the line: l: ux + vy + wz = 0, where (u, v, w) k\f(0, 0, 0)g. Suppose that w = 0. Then (u, v) (0, 0). The equation ux + vy = 0 has a solution (x 0, y 0 ) (0, 0). Then l would be the set of points (x 0, y 0, t), t k. But then we would have: 0 = ct 3 + bx 0 t 2 + (ax 2 0 y 2 0)t + x 3 0 for all t k; but that implies first x 3 0 = 0 hence x 0 = 0, and then y 2 0 = ax 2 0 y 2 0 = 0 so y 0 = 0 and contradiction. So we may suppose that l is given by an equation of form z = αx + βy. If now l is contained in E we deduce that the polynomial: g(x, y) := y 2 (αx + βy) x 3 ax 2 (αx + βy) bx(αx + βy) 2 c(αx + βy) 3 vanishes identically. Looking at the coefficients of x 3, xy 2 and x 2 y, respectively, we deduce: (1) cα 3 + bα 2 + aα + 1 = 0, (2) α bβ 2 3cαβ 2 = 0, (3) aβ 2bαβ 3cα 2 β = 0. Then (1) implies α 0; then (2) gives β 0. Then (3) implies a+2bα+3cα 2 = 0. Together with (1) this implies that α 1 is (at least) a double root of the polynomial x 3 + ax 2 + bx + c, contradiction.

4 4 IAN KIMING 3. Intersection between lines and elliptic curves. Let k be a field and let E be an elliptic curve over k given by a Weierstrass equation: We shall write this as: where: y 2 z + a 1 xyz + a 3 yz 2 = x 3 + a 2 x 2 z + a 4 xz 2 + a 6 z 3. F (x, y, z) = 0 F (x, y, z) := y 2 z + a 1 xyz + a 3 yz 2 (x 3 + a 2 x 2 z + a 4 xz 2 + a 6 z 3 ) so that F is a homogeneous polynomial of degree 3. Consider a projective line l given by an equation: αx + βy + γz = 0. Thus, (α, β, γ) (0, 0, 0). Notice that we do not necessarily assume that l is k-rational in the sense that α, β, γ k. The statement we will now discuss is the following: With an appropriately defined notion of multiplicity of intersection points, the line l intersects E in precisely 3 points. Furthermore, if l is k-rational, and if 2 of the intersection points are k-rational then so is the 3 rd point of intersection. Actually, what we describe below is not only a proof of this statement but also an algorithm for computing these intersection points. With the above statement we will not need any reference to Bezout s theorem as in chap. 1 of [1]. The statement above holds of course in much greater generality: Without going into Bezout s theorem, you can at least easily see from the following how to generalize the statement to a statement about points of intersection between a line and a projective curve of degree n (n N); you do need some kind of assumption on the curve though, specifically an irreducibility condition. For us, in connection with elliptic curves it is enough that we showed that an elliptic curve does not contain any line. Now to the argument in favor of the above statement. As we said, the notion of multiplicity will be defined in course of the analysis. One needs to split the discussion up into 3 cases depending on which of α, β, γ is 0. We do only 1 case, assuming that: γ 0. We may thus assume an equation of form: for our line l. z = rx + sy

5 IRREDUCIBILITY OF ELLIPTIC CURVES AND INTERSECTION WITH LINES. 5 and: A point P = (x, y, z) is a point of intersection between l and E precisely if: Thus, if we put: (x, y, z) = (x, y, rx + sy) 0 = F (x, y, z) = F (x, y, rx + sy). g(x, y) := F (x, y, rx + sy), the points of intersection are precisely the points (x, y, rx+sy) where (x, y) (0, 0) is such that g(x, y) = 0 (notice that (x, y, rx + sy) is a point in the projective plane exactly if (x, y) (0, 0)). Now, g(x, y) is clearly a homogeneous polynomial of degree 3, so we can write: ( ) g(x, y) = ax 3 + bx 2 y + cxy 2 + dy 3. Notice that, as the coefficients of F are in k, we can conclude that a, b, c, d are also in k if l is k-rational, i.e., if r, s k. We first claim that g splits into linear factors: ( ) g(x, y) = (u 1 x + v 1 y)(u 2 x + v 2 y)(u 3 x + v 3 y) with u i, v i in an algebraic closure of k and (u i, v i ) (0, 0) for each i = 1, 2, 3. To see this, notice first that g is not identically 0: For otherwise, we would conclude that E contained the whole line l. So if a = d = 0 then (b, c) (0, 0), and also: so that things are clear in this case. g(x, y) = xy(bx + cy) Now, in principle we would have to discuss both of the cases a 0 and d 0. Let us take just the second case: If d 0 the polynomial: ( ) g(1, y) = a + by + cy 2 + dy 3 is a polynomial of degree 3 in y. Hence: g(1, y) = d(y θ 1 )(y θ 2 )(y θ 3 ) with θ 1, θ 2, θ 3 in an algebraic closure of k. Then: d( θ 1 θ 2 θ 3 ) = c, d(θ 1 θ 2 + θ 1 θ 3 + θ 2 θ 3 ) = b, d( θ 1 θ 2 θ 3 ) = a. Using these formulas we see that: ( ) g(x, y) = d(y θ 1 x)(y θ 2 x)(y θ 3 x) simply by multiplying the right hand side out. Thus we have established ( ) in all cases. Using the form ( ) we can now say that the points of intersection between l and E are the points (x 0, y 0, rx 0 + sy 0 ) where (x 0, y 0 ) (0, 0) is a solution to an equation: ( ) u i x 0 + v i y 0 = 0

6 6 IAN KIMING for some i f1, 2, 3g. Now, if (x 0, y 0 ), (x 0, y 0) (0, 0) are 2 solutions to ( ) then (x 0, y 0) = (λx 0, λy 0 ) for some non-zero constant λ: We can deduce this since we had (u i, v i ) (0, 0). Consequently, these 2 solutions to ( ) (for a fixed i) give us the same intersection point (x 0, y 0, rx 0 + sy 0 ) in the projective plane. Or put in another way: We get 3 intersection points by considering the 3 linear factors of g in ( ) one after the other. Of course we may get the same point 2 or even 3 times, but it is now clear how to define multiplicity of an intersection point: It is the number of times the point arises when we consider the linear factors of g one after the other. Concerning the last remark (about k-rational intersection points): Suppose that our line l is k-rational. Then r, s above are in k. Since F has coefficients in k (the elliptic curve E being defined over k) we deduce that a, b, c, d in ( ) and ( ) are in k. Let the 3 points of intersection be (x i, y i, rx i + sy i ), i = 1, 2, 3, corresponding as in the above to solutions to ( ) for i = 1, 2, 3, respectively. Assume now further that the first 2 of these points are k-rational, i.e., that x 1, x 2, y 1, y 2 k. We claim that then the u i, v i in the decomposition ( ) can be chosen in k (the decomposition ( ) is not unique, the factors are only determined up to multiplication with a non-zero constant; clearly we could multiply one of the factors with some non-zero λ and then some other factor with λ 1 ). Let us argue in the case we considered above: If g(x, y) = xy(bx + cy) the claim is clear as b, c k. Suppose then that d 0. Considering ( ) we may choose (say) (u i, v i ) = ( θ i, 1) for i = 1, 2, and (u 3, v 3 ) = ( dθ 3, d). Since (x i, y i ) is a solution (0, 0) to u i x + v i y = 0 we see that we can not possibly have x i = 0. Thus, θ i = y i /x i k for i = 1, 2 by our assumption x 1, x 2, y 1, y 2 k. We deduce: θ 3 = c/d θ 1 θ 2 k and hence the claim u i, v i k for i = 1, 2, 3. But since now u 3, v 3 k the equation u 3 x + v 3 y = 0 has a solution (x 3, y 3) (0, 0) with x 3, y 3 k. As r, s k the point (x 3, y 3, rx 3 + sy 3) is a k-rational point. But this point is the 3 rd point of intersection between l and E. References [1] J. H. Silverman, J. Tate: Rational points on elliptic curves, Undergraduate Texts in Mathematics. Springer, Department of Mathematics, University of Copenhagen, Universitetsparken 5, DK Copenhagen Ø, Denmark. address: kiming@math.ku.dk