The image of the point at infinity under birational equivalence from an elliptic curve

Question

To find rational points on the curve $C:y^2=P(x)$ where $P$ is a cubic or quartic with rational coefficients and no repeated roots, I can derive a birational equivalence between it and an elliptic curve $E$, and work with the elliptic curve. For example, $y^2=x^4-18x^2+36x-27$ is birationally equivalent to $w^2=z^3-432$ with $x=\frac{w-36}{2(z-12)}$ and $y=x^2-3-\frac z2$.

However, transforming points through birational equivalences may be a little tricky. For example, if the circle is parametrised as $x=\frac{2t}{1+t^2},y=\frac{1-t^2}{1+t^2}$, the point at infinity ($t=\infty$) corresponds to the finite point $(0,-1)$. This question is about the genus-$1$ case.

If $E$ is in Weierstrass normal form and hence has the point at infinity $\mathcal O$ that serves as the identity of its Mordell–Weil group, do I need to consider $\mathcal O$ when transforming rational points on $E$ back to $C$? If so, what will $\mathcal O$ map to, given the map?

I was left with this question after editing my answer to a question that asked for all rational points on $y^2=x^4-18x^2+36x-27$. I was a little worried that I might have missed a few details, because I did not consider $\mathcal O$.

Answer 1

Not sure if these are the same maps that you found, but we have rational maps \begin{align*} \varphi: C &\dashrightarrow E\\ (x,y) &\mapsto \left(\frac{12x}{x-3}, -\frac{36y}{(x-3)^2}\right) \end{align*} and \begin{align*} \psi: E &\dashrightarrow C\\ (z,w) &\mapsto \left(\frac{3z}{z-12}, -\frac{36w}{(z-12)^2} \right) \end{align*} that are mutually inverse where they are defined. To compute $\psi(\infty)$, we change coordinates to a chart containing $\infty$. Homogenizing the equations for $C$ and $E$, we have \begin{align*} C: Y^2 Z^2 = X^4 - 18X^2 Z^2 + 36 X Z^3 - 27 Z^4 \end{align*} where $x = X/Z$ and $y = Y/Z$ on the affine open where $Z \neq 0$, and \begin{align*} E: U^2 V = T^3 - 432 V^3 \end{align*} where $z = T/V$ and $w = U/V$ on the affine open where $V \neq 0$. The homogenized version of $\psi$ is $$ \psi = (3T(T-12V) : -36UV : (T-12V)^2) \, . $$ The affine open where $U \neq 0$ contains the point $\infty = (0 : 1 : 0)$, so we move to this chart with coordinates $t = T/U$ and $v = V/U$. In this chart, $E$ is given by \begin{align} \label{eqn:tv} \tag{1} E: v = t^3 - 432 v^3 \, , \end{align} the point $(0:1:0)$ is $(t,v) = (0,0)$, and $\psi$ becomes $$ \psi = \left(3t(t-12v) : -36v : (t - 12v)^2 \right) \, . $$ From (\ref{eqn:tv}) we see that $v=0$ is the tangent line to $E$ at $(0,0)$ and $v$ vanishes to order $3$. Then $t=0$ intersects transversely at $(0,0)$, so $t$ is a uniformizer, i.e., it vanishes to order $1$. Multiplying the expression for $\psi$ by $1/t^2$, we have \begin{align*} \psi &= \left(3 \frac{t-12v}{t} : -36 \frac{v}{t^2} : \frac{(t - 12v)^2}{t^2} \right) = \left(3\left(1 - 12\frac{v}{t}\right) : -36 \frac{v}{t^2} : \left(1 - 12 \frac{v}{t} \right)^2 \right) \, . \end{align*} Since $v$ vanishes to order $3$ at $(0,0)$ and $t$ only vanishes to order $1$, then this expression shows that $$ \psi(0,0) = (3 : 0 : 1) \, . $$ (One can also use (\ref{eqn:tv}) to see directly that $v/t$ vanishes at $(0,0)$: $$ \frac{v}{t} \cdot \frac{t^2}{t^2} = \frac{t^2 v}{t^3} = \frac{t^2 v}{v + 432 v^3} = \frac{t^2}{1 + 432v^2}. ) $$ Thus $\psi(0:1:0) = (3:0:1)$, or in terms of the original affine coordinates, $\psi(\infty) = (3,0)$.