Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 125 / 2

For an endomorphism $\varphi$ of an elliptic curve, define its trace of an elliptic-curve endomorphism byPolarizing the quadratic form $\mathrm{deg}$ shows that this is the integer for whichwhere $\hat{\varphi }$ is the dual isogeny. ConsequentlyAlsoThe left side is $[\mathrm{tr}({\varphi }^2)]$, so

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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 $\#E({\mathbb{F}}_{p^r})=p^r+1-a_r$.

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Suppose that $uP+v\alpha (P)=O$ with $u,v\in {\mathbb{F}}_{\ell }$. If $v=0$, then $u=0$ because $P$ has order $\ell$. If $v\ne 0$, thenApplying $\alpha$ again and using ${\alpha }^2=[-1]$ yields ${\lambda }^2=-1$ in ${\mathbb{F}}_{\ell }$. This is impossible when $\ell \equiv 3(\mathrm{mod}4)$, because $-1$ is then not a quadratic residue. Hence $P$ and $\alpha (P)$ are linearly independent in the two-dimensional vector space $E[\ell ]$.

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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

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