Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 125 / 1

The Weierstrass equation of an elliptic curve over a field $K$ has the general formwith nonzero elliptic-curve discriminant. Admissible change of Weierstrass coordinates relates two such equations defining isomorphic pointed curves over $K$:where $u\in K^{\times }$ and $r,s,t\in K$.

If $\mathrm{char}K\ne 2,3$, completing the square and translating $x$ simplify every equation. Short Weierstrass form isTwo short equations are $K$-isomorphic precisely when, for some $u\in K^{\times }$,the isomorphism from the first curve to the second is $(x,y)\mapsto (u^{2}x,u^{3}y)$.

A twist of an elliptic curve $E/\mathbb{Q}$ is an elliptic curve that becomes isomorphic to $E$ over $\overline{\mathbb{Q}}$. WriteFor $d\in {\mathbb{Q}}^{\times }$, its quadratic twist may be writtenOver $\mathbb{Q}(\sqrt{d})$, the map $(x,y)\mapsto (d^{-1}x,d^{-3/2}y)$ identifies $E^d$ with $E$.

The hypothesis $j(E)\ne 0,1728$ impliesConsequently twists are classified bywhere the last identification is Kummer theory. In the explicit equations, $E^d{\simeq }_{\mathbb{Q}}{E}^e$ exactly when $d/e$ is a rational square. Every nonzero rational square class has a unique square-free integer representative, including its sign, so the twists are parametrized by the nonzero square-free integers.

If every point of $E[n]$ is rational, the Weil pairingand its nondegeneracy imply that every $n$th root of unity belongs to $\mathbb{Q}$. The only roots of unity in $\mathbb{Q}$ are $\pm 1$, so $n\le 2$. Since the question assumes $n\ge 2$, only $n=2$ is possible. It does occur: any nonsingular equationwith distinct $a,b,c\in \mathbb{Q}$ has all four points of $E[2]$ rational.

Mod-three Galois representation of an elliptic curve is the representationin this case. Twisting by the quadratic character ${\chi }_d$ replaces it by ${\chi }_d{\rho }_3$. Thus $E^d$ has a rational point of order three exactly when $E[3]$ contains a nonzero vector $v$ satisfyingfor every $\sigma \in G_{\mathbb{Q}}$: the line ${\mathbb{F}}_{3}v$ is a one-dimensional Galois subrepresentation with character ${\chi }_d$.

The two-dimensional representation $E[3]$ has at most two distinct one-dimensional characters among its Jordan–Hölder factors. Therefore at most two quadratic characters ${\chi }_d$, and hence at most two rational isomorphism classes of twists, can have a rational point of order three.

Write $K=\mathbb{Q}(\sqrt{d})$ and let $\sigma$ be its nontrivial automorphism. On the rational vector space $E(K)\otimes \mathbb{Q}$, the involution $\sigma$ gives the eigenspace decompositionThe positive eigenspace is $E(\mathbb{Q})\otimes \mathbb{Q}$. Choose the quadratic twist $E^{\prime }=E^d$ and an isomorphism $\iota :E^{\prime }\to E$ over $K$ for which ${\iota }^{\sigma }=-\iota$. It maps $E^{\prime }(\mathbb{Q})\otimes \mathbb{Q}$ isomorphically onto the negative eigenspace. Taking dimensions gives