Finite fields. Denoted by GF(p^k), the set of all polynomials in Z_p[x] of degree at most k-1. Addition is modulo p. Multiplication reduced modulo P(x). P(x) is a monic irreducible polynomial of degree k in Z_p[x].
Properties:
Properties of GF(2^k):
E_{a,b} = \{\mathcal O\} \cup \{(x, y) \in K^2 : y^2 = x^3 + ax + b\}
For an elliptic curve to be non-singular, we have to fulfill 4a^3 + 27b^2 \ne 0. \lambda = \frac{y_Q - y_P}{x_Q - x_P} is the chord slope. \lambda = \frac{3x_P^2 + a}{2y_P} is the tangent slope.
Let P = (x_P, y_P), Q = (x_Q, y_Q), and R = (x_R, y_R)
-P = (x_P, -y_P), -\mathcal O = \mathcal O
if P = -Q then P + Q = \mathcal O
P + Q = R:
\begin{aligned} \lambda &= \begin{cases} \frac{y_Q - y_P}{x_Q - x_P} & \text{ if } x_P \ne x_Q \\ \frac{3x_P^2 + a}{2y_P} & \text{ if } x_P = x_Q \\ \end{cases} \\ x_R &= \lambda^2 - x_P - x_Q \\ x_Y &= (x_P - x_R)\lambda - y_P \\ \end{aligned}
\mathcal O is the neutral element in the group for addition
E_{a, b}(K) is an abelian group of an elliptic curve over a field x, y \in K. The characteristic should be larger than 3.
If a point P = (x, y) is of order 2, then P + P = 0, then P = -P so y = 0. So P is the solution of x^3 + ax + b = 0. As there are at most three solutions, there are at most 3 elements of order 2 in E_{a,b}. If E_{a,b} is cyclic then there can be only one or zero elements of order 2.
Same as before, but the discriminant is now \Delta = -16(4a^3 + 27b^2) \ne 0
E_{a,b} \simeq E_{u^4a,u^6b} by (x, y) \mapsto (u^2x, u^3y).
E_{a,b} and E_{v^2a,v^3b} are twist of each other if u^2 = x has no solution (ie u is not a square). They can become isomorphic in the extension field K[\theta]/(\theta^2 - v), because v = \theta^2 becomes a square.
q+1-2\sqrt{q} \le |E_{a,b}| \le q+1+2\sqrt{q} where q is the cardinality of K.
j = 1728 \frac{4a^3}{4a^3 + 27b^2}. Isomorphic groups have always the same j. Alternatively, if \frac{a^3}{b^2} are equal, then the curves are isomorphic.
E_{a_2,a_6} = \{\mathcal O\} \cup \{(x, y) \in K^2: y^2 + xy = x^3 + a_2x^2 + a_6\}
Discriminant \Delta = a_6 \ne 0
Let P = (x_P, y_P), Q = (x_Q, y_Q), and R = (x_R, y_R)
-P = (x_P, x_P + y_P), -\mathcal O = \mathcal O
if P = -Q then P + Q = \mathcal O
P + Q = R:
\begin{aligned} \lambda &= \begin{cases} \frac{y_Q + y_P}{x_Q + x_P} & \text{ if } x_P \ne x_Q \\ \frac{x_P^2 + y_P}{x_P} & \text{ if } x_P = x_Q \\ \end{cases} \\ x_R &= \lambda^2 + \lambda + a_2 + x_P + x_Q \\ x_Y &= (x_P + x_R)\lambda + y_P + x_R\\ \end{aligned}
\mathcal O is the neutral element in the group for addition
E_{a_2,a_6} \simeq E_{a_2+u^2 + y, a_6} by (x, y) \mapsto (x, ux + y).
j-invariant: 1 \over \Delta
The Pollard’s p-1 factorization method can be used on elliptic curves
For a prime field case, a single x leads to two possible y values. Given x and the parity of y we have a point (which is cheaper than storing both and x and y).
If n is prime and is coprime with h then \langle G \rangle = \{Q \in \text{group} : nQ = \mathcal O\}
For some pairs of elliptic curves G_1 and G_2 we can construct a function e: G_1 \times G_2 \to G_T which maps a pair to a group with multiplicative notation such that
Let G generate a subgroup of order p of G_1 = G_2 such that e(G, G) \ne 1