hw3sols_143f12

hw3sols_143f12 - Math 143 Fall 2012 HW3 Solutions(UAG 2.4...

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Math 143, Fall 2012. HW3 Solutions (UAG) 2.4 Consider the cubic curve C = { ( X : Y : Z ) P 2 R | ZY 2 = X 3 + 4 XZ 2 } so that the only point with Z = 0 is the ‘point at infinity’ O = (0 : 1 : 0). This is the neutral element in the ‘simplified group law’ described on p. 39 of UAG. Considering the ‘affine’ part of C (ie, where Z 6 = 0), we have C 0 = { ( x , y ) R 2 | y 2 = x 3 + 4 x } , so that C = { O } ∪ C 0 . Then, we can consider C as an affine curve providing we remember the extra point at infinity. For P = (2, 4) C ( ie , (2 : 4 : 1) C ) we have that the tangent line to C at P is given by L = { ( x , y ) R 2 | y - 4 = 2( x - 2) } Hence, we see that Q def = (0, 0) L . Therefore, we have just shown that P + P = Q . Now, the tangent line to C at Q is just L 0 = { ( x , y ) R 2 | x = 0 } and this intersects C at O . Hence, we have just shown that O = Q + Q = P + P + P + P = 4 P , so that P is an order 4 point of the group C . Note that we do not have 2 P = O so that P is a genuine order 4 point. 2.5 Consider the nonsingular (ie, choose appropriate a , b R ) cubic curve C = { ( x , y ) R 2 | y 2 = x 3 + ax + b } We want all those points P = (
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