Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 125 / 1 / c / Solution

Because $\varphi$ is a separable isogeny,The degree-zero divisorcorresponds under $E_1\simeq {\mathrm{Pic}}^0(E_1)$ to the sum of all elements of the finite abelian group $E_1[\varphi ]$. Pairing every $T$ with $-T$ shows that this sum is the sum of the elements in $E_1[\varphi ]\cap E_1[2]$. It vanishes when that intersection has one element and also when it has four elements, since the sum of the four elements of $(\mathbb{Z}/2\mathbb{Z}{)}^2$ is zero. The principal divisor criterion on an elliptic curve therefore gives a rational function $g$ with $\mathrm{div}(g)=D$.

The divisor $D$ is invariant under $[-1]$, so $[-1{]}^{\ast }g/g$ has zero divisor and is constant. Applying $[-1]$ twice shows that this constant has square one; hence $[-1{]}^{\ast }g=\pm g$.

For a short Weierstrass equation of an elliptic curvethe multiplication-by-two isogeny hasThus one may take $g=y$, and $[-1{]}^{\ast }y=-y$. For multiplication by three, the third division polynomial of an elliptic curvevanishes simply at the eight nonzero points of $E[3]$ and has a pole of order eight at $O_E$. Hence $g={\psi }_3(x)$ has divisor $[3{]}^{\ast }(O_E)-9(O_E)$ and satisfies $[-1{]}^{\ast }g=g$. Both signs occur.