2.6. Elliptic curves over finite fields 35 2.6. Elliptic curves over finite fields Let p 2 be a prime and let Fp be the finite field with p elements, i.e., Fp = Z/pZ = {a mod p : a = 0,1,2,...,p 1}. Fp is a field and we may consider elliptic curves defined over Fp. As for elliptic curves over Q, there are two conditions that need to be satisfied: the curve needs to be given by a cubic equation, and the curve needs to be smooth. Example 2.6.1. For instance, E : y2 x3 + 1 mod 5 is an ellip- tic curve defined over F5. It is clearly given by a cubic equation (zy2 x3+z3 mod 5 in the projective plane P2(F5)) and it is smooth, because for F zy2 x3 z3 mod 5, the partial derivatives are: ∂F ∂x −3x2, ∂F ∂y 2yz, ∂F ∂z y2 3z2 mod 5. Thus, if the partial derivatives are congruent to 0 modulo 5, then x 0 mod 5 and yz 0 mod 5. The latter congruence implies that y or z 0 mod 5, and ∂F/∂z 0 implies that y z 0 mod 5. Since [0,0,0] is not a point in the projective plane, we conclude that there are no singular points on E/F5. However, C/F3 : y2 x3 + 1 mod 3 is not an elliptic curve be- cause it is not smooth. Indeed, the point P = (2 mod 3,0 mod 3) C(F3) is a singular point: ∂F ∂x (P) −3 · 22 0, ∂F ∂y (P) 2 · 0 · 1 0, and ∂F ∂z (P) 02 3 · 12 0 mod 3. Let E/Q be an elliptic curve given by a Weierstrass equation y2 = x3 + Ax + B with integer coefficients A, B Z, and let p 2 be a prime number. If we reduce A and B modulo p, then we obtain the equation of a curve E given by a cubic curve and defined over the field Fp. Even though E is smooth as a curve over Q, the curve E may be singular over Fp. In the previous example, we saw that E/Q : y2 = x3 + 1 is smooth over Q and F5 but it has a singularity over F3. If the reduction curve E is smooth, then it is an elliptic curve over Fp.
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