 Solution

The degree of an isogeny $\varphi :E\to E^{\prime }$ is the degree of the induced finite extension of function fields. Saying that degree is a quadratic form on $\mathrm{End}(E)$ meansand thatis bilinear, equivalentlyThe trace of an elliptic-curve endomorphism isThe relation ${\varphi }^2-[\mathrm{tr}\varphi ]\varphi +[\mathrm{deg}\varphi ]=0$ impliesthe trace of the square of an elliptic-curve endomorphism.

The Hasse theorem for elliptic curves states that for $E/{\mathbb{F}}_q$,Let $\pi$ be the Frobenius isogeny of an elliptic curve, put $a=\mathrm{tr}(\pi )$, and note that $\mathrm{deg}\pi =q$ andFor integers $m,n$, quadraticity givesThis binary quadratic form cannot have positive discriminant, since rational numbers $m/n$ are dense, so $a^2-4q\le 0$. Substitution proves the bound. This is the degree-form proof of the Hasse bound.

Both endpoints occur. The curve $E:y^2=x^3-x$ over ${\mathbb{F}}_3$ is supersingular with trace zero. Over ${\mathbb{F}}_9$, its Frobenius is $[-3]$, soIts nontrivial quadratic twist over ${\mathbb{F}}_9$ has the opposite trace and therefore haspoints.