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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