Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 125 / 2

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.

For $E:y^2=x^3+3$, direct counting givesNeither group order is divisible by $17$, so neither group contains a point of order $17$.

At $p=17$ the Frobenius trace is zero. The elliptic-curve point count over a finite field has trace recurrenceFor every $n\ge 1$, this order is congruent to one modulo $17$. Consequently $E({\mathbb{F}}_{17^n})$ has no point of order $17$ for any $n$.

At $p=31$, the trace is $a=31+1-43=-11$. On $E[17]$, Frobenius has characteristic polynomialIts discriminant is $14$, a nonsquare in ${\mathbb{F}}_{17}$, so its two distinct eigenvalues lie in ${\mathbb{F}}_{17^2}^{\times }$. Their orders divide $17^2-1=288$, whence ${\pi }^{288}=1$ on $E[17]$. Thus all of $E[17]$ is rational over ${\mathbb{F}}_{31^{288}}$, and in particular a point of order $17$ exists over some extension with $n\ge 2$.