# Elliptic curves - Weierstrass normal form

Are there any reasons why we transform elliptic curves with the equation $Y^2Z+a_1XYZ+a_3YZ^2=X^3+a_2X^2Z+a_4XZ²+a_6Z³$ over $\mathbb{R}$ or a finite field $F_q$ ($q$ is prime) to the shorter Weierstrass forms?

Are the shorter forms more often used in cryptography?

• As a rule of thumb in proofs and programming, the fewer parameters the better. For one thing fewer parameters makes it easier to recognize when two such elliptic curves may be "equivalent". The "right" notion of equivalent here is a birational transformation of variables. Some references for this are given in answer to this previous Question. – hardmath Jun 10 '17 at 14:26
• It is much easier to check if the curve is singular, the $j$-invariant and the group law with $Z Y^2 = X^3+aXZ^2+bZ^3$ (this form is always possible when the characteristic is not $2,3$ ?). And $Y^2 = X^3+aX+b$ means that the function field is $K(X, \sqrt{X^3+aX+b})$ – reuns Jun 10 '17 at 14:44