= Solution
Choose a <flex> as $O=[0:1:0]$, with <tangent line> $Z=0$. The <flex coordinates for a smooth plane cubic> give
$$
F=aX^3+bY^2Z+cXYZ+dX^2Z+eYZ^2+fXZ^2+gZ^3,
\qquad ab\ne0.
$$
The coefficient $a$ is nonzero because $Z=0$ is not a component, and $b\ne0$ expresses smoothness at $O$. Since the <field characteristic> is not two, the projective linear change
$$
Y'=Y+\frac{cX+eZ}{2b}
$$
completes the square. The equation becomes $Y'^2Z=P_3(X,Z)$ after dividing by a nonzero constant, where $P_3$ is a degree-three <homogeneous polynomial> with nonzero $X^3$ coefficient.
On $Z=1$, the <polynomial> $P_3(x,1)$ has three distinct roots $r_1,r_2,r_3$ in the <algebraically closed field>. A repeated root $r$ would make $(r,0)$ a <singular point> of $y'^2=P_3(x,1)$, since both first partial derivatives vanish there. Put
$$
x'=\frac{x-r_1}{r_2-r_1},\qquad
\lambda=\frac{r_3-r_1}{r_2-r_1}.
$$
A further nonzero scaling of $y'$ absorbs the leading coefficient and $(r_2-r_1)^3$; the required square root exists in the <algebraically closed field>. These are projective linear changes of the original coordinates. Renaming the new coordinates gives the <Legendre form of an elliptic curve>
$$
\boxed{y^2=x(x-1)(x-\lambda),\qquad\lambda\notin\{0,1\}.}
$$
Its projective completion is $Y^2Z=X(X-Z)(X-\lambda Z)$, with the chosen <flex> at infinity.
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