Legendre form of an elliptic curve ($Y^{2}Z=X(X-Z)(X-\lambda Z),\lambda \ne 0,1$)

Over an algebraically closed field of characteristic different from two, a cubic Weierstrass model with distinct roots becomes this form by sending two roots to zero and one, then scaling $Y$ by a square root of the resulting nonzero coefficient. The third root becomes $\lambda \notin \{0,1\}$. The point $[0:1:0]$ is a flex and the curve is a genus one curve.

For $Y^{2}Z=X(X-Z)(X-\lambda Z)$ over $\mathbb{C}$, order its three finite roots as $(r_0,r_1,r_2)$ and put $h=r_1-r_0$. The projective transformation $X^{\prime }=(X-r_{0}Z)/h$, $Y^{\prime }=Y/s$, $Z^{\prime }=Z$, with $s^2=h^3$, gives Legendre form of an elliptic curve with parameter $(r_2-r_0)/(r_1-r_0)$. The six root orderings give the six displayed values, all different from zero and one. These exhaust changes that act affinely on $X/Z$ and scale $Y/Z$, because such a change must send the finite root set onto the new finite root set. There are six distinct values except that $\lambda \in \{-1,2,1/2\}$ gives three and ${\lambda }^2-\lambda +1=0$ gives two.

For the Legendre form of an elliptic curve $y^2=x(x-1)(x-\lambda )$, a change in the $x$ coordinate that permutes its branch values lifts with either of two opposite signs in the $y$ coordinate. These lifts differ by the involution $(x,y)\mapsto (x,-y)$. At $\lambda =e^{\pi i/3}$ the parameter transformation $x\mapsto 1-x/\lambda$ has order three. Its lift fixing $y$ has order three, whereas its lift negating $y$ has order six, since its cube is the nontrivial involution. Thus the order must be computed for a specified lift; both imply the same hexagonal shape of the associated period lattice.