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The Hasse theorem for elliptic curves states that, for an elliptic curve over ${\mathbb{F}}_q$,Let $\pi$ be the Frobenius isogeny of an elliptic curve and put $a=q+1-\#E({\mathbb{F}}_q)$. The fixed points of $\pi$ are $E({\mathbb{F}}_q)$, and $1-\pi$ is separable, soHence the trace of an elliptic-curve endomorphism iswhile $\mathrm{deg}\pi =q$.

The degree on $\mathrm{End}(E)$ is a nonnegative quadratic form. Polarization and the identities for the dual isogeny give, for integers $m,n$,If $a^2>4q$, this real binary quadratic form is indefinite. An open cone on which it is negative contains a nonzero rational point and therefore a nonzero integer point, contradicting nonnegativity of isogeny degree. Thus $a^2\le 4q$, which is exactly the claimed inequality.

The Hasse theorem for elliptic curves states thatLet $\pi$ be the Frobenius isogeny of an elliptic curve and put $a=p+1-\#E({\mathbb{F}}_p)$. The degree on $\mathrm{End}(E)$ is a positive-definite quadratic form, its associated bilinear form gives $\mathrm{tr}(\pi )=a$, and $\mathrm{deg}\pi =p$. Consequentlyfor all integers $m,n$. If $a^2-4p>0$, this real quadratic form is indefinite, so by density of rational slopes it is negative at some nonzero integer pair $(m,n)$, contradicting nonnegativity of the degree. Hence $a^2\le 4p$, which is the claimed bound.

Let $\pi$ be the Frobenius isogeny of an elliptic curve and putPart (a) givesLet $\alpha ,\beta$ be the two roots of $T^2-aT+p$. The points over ${\mathbb{F}}_{p^r}$ are exactly $\ker (1-{\pi }^r)$. Since the differential of $1-{\pi }^r$ is the identity, this isogeny is separable, and thereforeUsing $\mathrm{deg}\psi =(1-\psi )(1-\hat{\psi })$ in the endomorphism algebra gives the elliptic-curve point count over a finite fieldEquivalently, if $a_r={\alpha }^r+{\beta }^r$, thenand $\#E({\mathbb{F}}_{p^r})=p^r+1-a_r$.

For $E:y^2=x^3-x$ over ${\mathbb{F}}_3$, each of $x=0,1,2$ gives the single affine point with $y=0$. Including $O$ gives $\#E({\mathbb{F}}_3)=4$, so the Frobenius trace is $a=3+1-4=0$. Therefore the Frobenius isogeny of an elliptic curve $\pi$ satisfiesOn the $11$-torsion, this reads ${\pi }^2=[-3]=[8]$. The element $8\in {\mathbb{F}}_{11}^{\times }$ has order ten and $8^5=-1$. Henceand no smaller positive power of $\pi$ is the identity on $E[11]$. A division field of an elliptic curve over a finite field has degree equal to the order of Frobenius on the torsion module, so