LECTURE 7, WEDNESDAY

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1 LECTURE 7, WEDNESDAY FRANZ LEMMERMEYER 1. Singular Weierstrass Curves Consider cubic curves in Weierstraß form (1) E : y 2 + a 1 xy + a 3 y = x 3 + a 2 x 2 + a 4 x + a 6, the coefficients a i lie in some field K; we say that E is defined over K and occasionally denote this by E/K. Note that (1) is irreducible; this is clear if a 1 = a 3 = 0, since Y 2 = f(x) can only be reducible if deg f is even, and can be proved with a little bit more effort in the general case. We next observe that the point O = [0 : 1 : 0] at infinity is always smooth since F Z (O) = 1. It thus remains to study affine points. Before we do that, we will show how to transform long into short Weierstrass forms over fields of characteristic 2, 3. If K is a field of characteristic 2, we can put η = y + (a 1 x + a 3 )/2 (we are completing the square) and find (2) η 2 = x 3 + b 2 4 x2 + b 4 2 x + b 6 4 with b 2 = a a 2, b 4 = a 1 a 3 + 2a 4, and b 6 = a a 6. If, moreover, char K 3, then we can put ξ = x + b 2 /12 and find the short Weierstraß normal form (3) η 2 = ξ 3 c 4 48 ξ c 6 864, c 4 = b b 4, c 6 = b b 2 b 4 216b 6. We also introduce and define the discriminant of E by b 8 = a 2 1a 6 a 1 a 3 a 4 + 4a 2 a 6 + a 2 a 2 3 a 2 4 = b 2 2b 8 8b b b 2 b 4 b 6. We will see below that 0 for elliptic (i.e., nonsingular) cubics, hence we can define the j-invariant of E by j = c 3 4/. We have collected all these definitions in table 1. Recall that the discriminant of a cubic polynomial f(x) = x 3 + ax 2 + bx + c = (x α)(x α )(x α ) is defined to be disc f = [(α α )(α α )(α α)] 2. 1

2 2 FRANZ LEMMERMEYER Table 1 b 2 = a a 2, b 4 = a 1 a 3 + 2a 4, b 6 = a a 6, b 8 = a 2 1a 6 a 1 a 3 a 4 + 4a 2 a 6 + a 2 a 2 3 a 2 4, c 4 = b b 4, c 6 = b b 2 b 4 216b 6, = b 2 2b 8 8b b b 2 b 4 b 6, 4b 8 = b 2 b 6 b 2 4, 1728 = c 3 4 c 2 6, j = c 3 4/ = c 2 6/. Thus disc f = 0 if and only if f has multiple roots. Moreover, for the polynomial f(x) = x b 2x b 4x+ 1 4 b 6 on the right hand side of (2) we have 16 disc f =. When should we consider two elliptic curves to be essentially the same? Consider the transformation (4) x x + r, y y + sx + t for coefficients r, s, t, u K. Substituting these equations into (1) gives a new equation E : y 2 + a 1x y + a 3y = x 3 + a 2x 2 + a 4x + a 6, (with a little help from pari) a 1 = a 1 + 2s, a 2 = a 2 a 1 s + 3r s 2, a 3 = a 3 + a 1 r + 2t, a 4 = a 4 a 3 s + 2ra 2 a 1 (rs + t) 2st + 3r 2, a 6 = a 6 + a 4 r a 3 t + a 2 r 2 a 1 rt t 2 + r 3, b 2 = b r, b 4 = b 4 + rb 2 + 6r 2, b 6 = b 6 + 2rb 4 + r 2 b 2 + 4r 3, c 4 = c 4, c 6 = c 6, =. hence Thus changing the coordinate system via (4) does not change c 4, c 6, and the discriminant. These transformations allow us to move any affine point P into the origin. Note that the point at infinity as well as the line at infinity are fixed by (4). In addition to these translations we can also rescale the equation by substituting (5) x = u 2 x, y = u 3 y for some u K. After getting rid of denominators we find E : y 2 + a1 ux y + a 3 u 3 y = x 3 + a2 u 2 x 2 + a4 u 4 x + a 6 u 6.

3 LECTURE 7, WEDNESDAY Thus each a i gets multiplied by u i, and the same thing happens to the b s and the c s. In particular c 4 = u 4 c 4 and = u 12, hence j = c 43 / = j remains the same: the j-invariant is invariant under both types of transformations (whence the name). Weierstrass curves that can be transformed into each other using (4) or (5) are called isomorphic. Isomorphic elliptic curves have the same j-invariant. It can be proved (quite easily) that elliptic curves with the same j-invariant are isomorphic over the algebraic closure of K (that is, the transformations (4) and (5) might involve coefficients from some extension of K). Now we are ready for Theorem 1. The cubic E in (1) defined over some field K is singular if and only if = 0 in K. In this case there exists a unique singular point P which is determined as follows: If char K = 2 and E in given by (1), then ( a 4, a 2 a 4 + a 6 ) if c 4 = 0 ( a 1 = 0) (a 3 /a 1, (a a 2 1a 4 )/a 3 1) if c 4 0 ( a 1 0) If char K = 3 and E is given by (2), then ( 3 b 6, 0) if c 4 = 0 ( b 2 = 0) ( b 4 /b 2, 0) if c 4 0 ( b 2 0) Finally, if char K 2, 3 and E is given by (3), then (0, 0) if c 4 = 0 ( c 6 /12c 4, 0) if c 4 0. In particular, the unique singularity of E/K is always K-rational if K has characteristic 0 or is a finite field. Proof. Let us first consider the case char K 2, 3. Then we may assume that E is given in short Weierstrass form. Here we use the fact that invertible affine transformations x = ax + by + c, y = dx + ey + f map singular points to singular points, since the derivatives of g(x, y ) = f(x, y) with respect to x and y are linear combinations of f 1 and f 2 and therefore vanish; we also use that translations x = x + c, y = y + c do not change the discriminant. The derivatives we have to look at are f x = 3x 2 + c 4 /48 and f y = 2y. The first equation gives x = ± c 4 /12, the second y = 0; plugging this into (3) we get c 6 = c 4 3 (in particular we have c 4 K). This implies that = 0. If c 4 = 0, then x = 0 and y = 0; if c 4 0, then x = ± c 4 /12 = c 6 /12c 4. Thus we have seen: if there is a singular point, then it is the one given above, and we have = 0. Conversely, if = 0, then the derivatives f x and f y vanish in the given point, and the equation = 0 guarantees that the point lies on E. Now assume that char K = 2. Then b 2 = a 2 1, b 4 = a 1 a 3, c 4 = a 4 1, hence c 4 = 0 a 1 = 0. The coordinates of a singular point in the affine plane satisfy the three equations f = y 2 + a 1 xy + a 3 y + x 3 + a 2 x 2 + a 4 x + a 6, f x = a 1 y + x 2 + a 4, f y = a 1 x + a 3.

4 4 FRANZ LEMMERMEYER Note that signs are irrelevant in light of 1 = +1. If a 1 = 0, then f y = 0 is equivalent to a 3 = 0, and in this case we have = 0 as well as x 0 = a 4 and y 0 = a 2 a 4 + a 6. If a 1 0, then x = a 3 /a 1 (since f y = 0) and y = (a a 2 1a 4 )/a 3 1 (since f x = 0). The condition f = 0 now yields = 0: on the one hand we have (observe that 2 = 0 and 1 = 1) = b 2 2b 8 + b b 2 b 4 b 6 = a 6 1a 6 + a 5 1a 3 a 4 + a 4 1a 2 a a 4 1a a a 3 1a 3 3, on the other hand we find a 6 1f(x, y) = a 6 1(y 2 + a 1 xy + a 3 y + x 3 + a 2 x 2 + a 4 x + a 6 ) = a a 4 1a a 3 1a a 3 1a a 5 1a 3 a 4 + a 3 1a a 4 1a 2 a a 6 a 6 1, and since 3a 3 1a 3 3 = a 3 1a 3 3, we see that = 0 is in fact equivalent to f(x, y) = 0. The case char K = 3 is left as an exercise. Finally some remarks on the question whether the singular point is defined over K (i.e. has coordinates from K): if K is a finite field of characteristic 2, then x x 2 is an automorphism, hence every element of K is a square. Corollary 2. Cubic curves in Weierstrass form are irreducible. Proof. Reducible cubic curves consist of three lines or a conic and a line; every point of intersection of components is singular, hence reducible cubics have at least two singular points. If a Weierstrass curve E defined over a field K has a singular point P, then E ns = E(K) \ P } is called the nonsingular part of E. 2. Addition Formulas Let E be an elliptic curve in long Weierstrass form (1) defined over some field K. For P E(K) we define P as the third point of intersection of the line through P and O = [0 : 1 : 0] with E. If [x 1 : y 1 : 1], the line P O is given by x = x 1 ; intersection gives y 2 + (a 1 x 1 + a 3 )y (x a 2 x a 4 x 1 + a 6 ) = 0, that is, the sum of the two roots of this quadratic equation is (a 1 x 1 + a 3 ); note that this equation describes the affine points only. Since one root is given by y = y 1, the other one must be (a 1 x 1 + a 3 ) y 1. Given two points P 1 = (x 1, y 1 ) and P 1 = (x 2, y 2 ) in E(K) with x 1 x 2, let P 1 P 2 = P 3 = (x 3, y 3 ) be the third point of intersection of the line P 1 P 2 with E. The line P 1 P 2 has the equation y = y 1 + m(x x 1 ) with m = (y 2 y 1 )/(x 2 x 1 ); plugging this into (1) yields a cubic equation in x with the roots x 1, x 2 and x 3. The sum of these roots is the coefficient of x 2 (observe that the coefficient of x 3 is 1), hence x 3 = x 1 x 2 a 2 + m(a 1 + m). Plugging this into the line equation we find the y-coordinate of P 3, hence x 3 = x 1 x 2 + m(a 1 + m), y 3 = [y 1 + m(x 3 x 1 ) + a 1 x 3 + a 3 ]. If x 1 = x 2, then y 1 = ±y 2. If y 1 = y 2, then we put P 1 + P 2 = O; if y 1 = y 2, that is, P 1 = P 2, then we let 2P 1 be the third point of intersection of the tangent to E in P 1 with E; a simple calculation then gives

5 LECTURE 7, WEDNESDAY Theorem 3. Let E/K be an elliptic curve given in long Weierstrass form (1). The chord-tangent method defines an addition on the set E(K) of K-rational points on E; the addition formulas are given by and (x 1, y 1 ) + (x 2, y 2 ) = (x 3, y 3 ), x 3 = x 1 x 2 a 2 + a 1 m + m 2 y 3 = y 1 (x 3 x 1 )m a 1 x 3 a 3 y 2 y 1 if x 1 x 2 x m = 2 x 1 3x 1 + 2a 2 x 1 + a 4 a 1 y 1 if x 1 = x 2 2y 1 + a 1 x 1 + a 3 In particular, for (x, y) the x-coordinate of 2P is given by (6) x 2 x4 b 4 x 2 2b 6 x b 8 4x 3 + b 2 x 2 + 2b 4 x + b 6. Specializing this to curves in short Weierstrass form, we get x 3 = x 1 x 2 + m 2, y 3 = y 1 m(x 3 x 1 ), y2 y 1 x m = 2 x 1 if x 2 x 1, 3x 2 +a 2y if x 2 = x 1, y 0. The cases not covered by these formulas are a) x 1 = x 2 and y 1 = y 2 : here P 1 = P 2, hence P 1 + P 2 = O; b) 2(x, y) with y = 0: here (x, 0) is a point of order 2, that is, 2(x, y) = O. In order to see these formulas in action take the curve E : y 2 +xy = x 3 18x+27. Here a 1 = 1, a 2 = a 3 = 0, a 4 = 18 and a 6 = 27. We find b 2 = 1, b 4 = 36, b 6 = 108, b 8 = 324, hence c 4 = 865, c 6 = and = = Thus E is an elliptic curve over F p for all p 3, 5, 7. The point (1, 1) is in E(F 2 ) : y 2 + xy = x 3 + 1; let us compute a few multiples of P. Bu the addition law we have m = 3x a 2 x 1 + a 4 a 1 y 1 2y 1 + a 1 x 1 + a 3 = x2 1 y 1 x 1 = 0, hence x 2 2x P + m + m 2 = 0 and y 2 y P (x 2P x P )m x 2 1, i.e., 2 (0, 1). We get 3P by adding P and 2P ; here m = (y 2 y 1 )/(x 2 x 1 ) = 0, hence x 3 x P x 2 1, as well as y 3 y P x 3 0, hence 3 (1, 0). Finally 4 P + 3P, but here m is not defined because x x 3P. This implies (since P 3P ) that 4 O. In fact we have seen that (x, y) = (x, a 1 x a 3 y), so in our case we have (x, y) = (x, x y), in particular (1, 1) = (1, 0). Thus P generates a group of order 4. Listing all points in E(F 2 ) we find that these are all, and we have proved that E(F 2 ) Z/4Z.