= Divisor proof of the degree parallelogram law
{title2=$\deg(\phi+\psi)+\deg(\phi-\psi)=2\deg\phi+2\deg\psi$}
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 $(\phi,\psi)$ and taking degrees proves the <parallelogram law> for the <degree of an isogeny> when $\phi,\psi$ are nonzero and $\phi\ne\pm\psi$. The exceptional cases follow from $\deg[2]=4$. 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 $\operatorname{Hom}(E_1,E_2)$.
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