Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 29 / 1 / ii / Solution

The given relation impliesTake isogeny degrees and cancel the positive integer $\mathrm{deg}\pi$. Multiplicativity of isogeny degree givesAs above, $\mathrm{deg}(1-{\varphi }_i)=\#E_i({\mathbb{F}}_q)$ and $\mathrm{deg}{\varphi }_i=q$. The degree parallelogram law on $\mathrm{End}(E_2)$ now yieldsSubstitution provesEquivalently, their Frobenius traces are negatives of one another. The isogeny degree argument works over the algebraic closure, so it does not wrongly assume that $\pi$ is defined over ${\mathbb{F}}_q$; the stated quadratic-extension hypothesis is compatible with the anticommutation relation.