Divisor proof of the degree parallelogram law ($\mathrm{deg}(\varphi +\psi )+\mathrm{deg}(\varphi -\psi )=2\mathrm{deg}\varphi +2\mathrm{deg}\psi$)

On the product of an elliptic curve with itself, let $D_{+}$ and $D_{-}$ be the diagonal and the graph of negation. The principal divisor of $x(P)-x(Q)$ is $D_{+}+D_{-}-2(\{O\}\times E)-2(E\times \{O\})$: equality of the coordinates means $Q=\pm P$, while $x$ has a double pole at $O$. Pulling back along $(\varphi ,\psi )$ and taking degrees proves the parallelogram law for the degree of an isogeny when $\varphi ,\psi$ are nonzero and $\varphi \ne \pm \psi$. The exceptional cases follow from $\mathrm{deg}[2]=4$. The same argument uses the quotient by negation in characteristic two. Thus degree, with degree zero assigned to the zero map, is a positive quadratic form on $\mathrm{Hom}(E_1,E_2)$.