 Solution

For $E:y^2=x^3+1$, let $\alpha (x,y)=(\zeta x,y)$. The three points $P,\alpha P,{\alpha }^{2}P$ lie on the horizontal line through $P$, so their sum is zero. Thusin the endomorphism ring of an elliptic curve. Since $\mathrm{deg}\alpha =1$ and complex conjugation sends $\alpha$ to ${\alpha }^2$, degree on $\mathbb{Z}[\alpha ]$ is the Eisenstein-integer norm:

A separable isogeny is determined by its kernel up to unique isomorphism of its target, and every finite Galois-stable subgroup of an elliptic curve is the kernel of the corresponding quotient isogeny. Put $T=(-1,0)$. The three nonzero points of $E[2]$ are $T,\alpha T,{\alpha }^{2}T$, andHence $\phi (\alpha -{\alpha }^2)$ kills $\ker \phi =\{O,T\}$ and factors uniquely through $\phi$:for an isogeny $\psi :E^{\prime }\to E^{\prime }$. Degrees giveMoreover,Composing the factorization twice and using the surjectivity of $\phi$ gives ${\psi }^2=[-3]$.