The <principal divisor criterion on an elliptic curve> says that $D=\sum n_P(P)$ is principal exactly when $\deg D=0$ and $\sum[n_P]P=O$.

On $E':y^2+dy=x^3$, take
$$
T=(0,0),\qquad f=y.
$$
The line $y=0$ meets the cubic three times at $T$, while $y$ has a triple pole at the point at infinity. Hence
$$
\operatorname{div}(f)=3(T)-3(O_{E'}).
$$
With $g=u/v\in\mathbb Q(E)$, part (a) gives
$$
\phi^*f=\frac{u^3}{v^3}=g^3.
$$
Taking divisors and cancelling the factor three yields
$$
\operatorname{div}(g)=\phi^*((T)-(O_{E'})).
$$
Pullback on degree-zero divisor classes is the <dual isogeny>, so the pulled-back class is represented by $\widehat\phi(T)$. It is principal, and therefore
$$
\boxed{T\in\ker\widehat\phi.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 125 / 5 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-125/5/b/solution