Consider a cubic $C$ of the form $$F(u, v) = u^3 + v^3 - \alpha = 0.$$ The Weierstrass equation is claimed to be $$y^2 = x^3 - 432 \alpha^2.$$ I've been trying to prove this, following the procedure outlined by Silverman--Tate's Rational Points on Elliptic Curves, p. 23. However, I cannot get it to work.
First, homogenize the equation to get $$F(U, V, W) = U^3 + V^3 - \alpha W^3 = 0.$$ We start with a rational point $P = [1, -1, 0]$, and we declare the line $Z = 0$ to be the tangent line to this rational point. Explicitly then, $$Z = \frac{\partial F}{\partial U} \bigg|_{P}(U - 1) + \frac{\partial F}{\partial V} \bigg|_{P}(V + 1) + \frac{\partial F}{\partial W}\bigg|_{P}(W - 0) = 3U + 3V.$$ We take $Y = 0$ to be any line other than $Z = 0$ going through $P$. Let me arbitrarily take $Y = U + V + W$ then. As for the line $X = 0$, usually you would look at the intersection of $C$ with $Z = 0$; you'd find that, apart from $P$, you'll get some other intersection $Q$, and $X = 0$ must be any other line through $Q$. But $P$ is an inflection point, giving us freedom to take $X = 0$ to by any line not containing $P$. So let's just take $X = U$.
Our coordinate change can be captured as $$ \begin{pmatrix} X \\ Y \\ Z \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 1 \\ 3 & 3 & 0 \end{pmatrix} \begin{pmatrix} U \\ V \\ W \end{pmatrix}.$$ We invert the matrix to obtain the equations $$\begin{split} U &= X \\ V &= -X + \frac{1}{3} Z \\ W &= Y - \frac{1}{3} Z \end{split}$$ which we substitute into $F(U, V, W)$. According to Silverman--Tate, the result should then be an equation of the form $$F'(X, Y, Z) = XY^2 + Z \cdot(\text{quadratic equation in $X,Y,Z$}) = 0,$$ which can then be simplified using algebra. But this doesn't happen. When I substitute, I instead obtain an equation of the form $$-\alpha Y^3 + Z \cdot(\text{quadratic equation in $X,Y,Z$}) = 0.$$
Question: Am I doing something wrong? If not, then how should I deal with this?
Upon inspection, I think it's not surprising that this happens. A priori, the equation $F'(X,Y,Z)$ will tautologically be of the form $$F'(X, Y, Z) = c_1 X^3 + c_2 X^2 Y + c_3 XY^2 + c_4 Y^3 + c_5 Z \cdot (\text{quadratic}) = 0.$$ But this equation must vanish on $P = [1, 0, 0]$ and $Q = [0, 1, 0]$ forcing $c_1$ and $c_4$ to be $0$, and $c_2$ must also be $0$ because $F'(X, Y, 0)$ must have a second-order root at $Y = 0$. However, the vanishing of $c_4$ fails precisely when we have an inflection point, since there won't be a point $Q$ anymore. My guess, then, is that I'm not making a mistake, but rather I'm dealing with a situation that the book doesn't cover in detail.
