Flexes and projective equivalence of smooth cubics

Question

I am trying to study Kock and Vainsencher's book "An invitation to Quantum Cohomology", working my way through the exercises. One of them ($0$th chapter) asks two prove that two elliptic curves $Y^2 Z=X(X-Z)(X-\lambda Z)$ are projectively equivalent if and only if they share the same set $\Lambda(\lambda)$, where such set is defined as $$ \Lambda(\lambda)=\left\lbrace \lambda,1-\lambda,\frac 1\lambda,\frac\lambda{\lambda-1},\frac 1{1-\lambda},\frac{\lambda-1}\lambda\right\rbrace $$ (that is, the orbit of the cross-ratio of 4 points under permutations of the 4 points).

What puzzles me is that they also suggest a Hint: first show that if $P,Q$ are flexes of a smooth plane cubic $C$, then there exists $\phi\in\mathrm{Aut}(\mathbb P^2)$ such that $\phi(C)=C$, $\phi(P)=Q$.

I am a bit rusty in Algebraic Geometry, but think I managed to prove the Hint. However, I cannot see how to apply the hint to solve the exercise. Any ideas on how to solve the exercise using the suggested hint would be greatly appreciated!

Answer 1

Let me unravel the kind suggestion by Sasha in the comments section.

1. The point $F=[0:1:0]$ is a flex of the cubic $Y^2Z=X(X-Z)(X-\lambda Z)$. Let us call $t_0$ the tangent line through $F$, which is actually $Z=0$. There are exactly three more tangents $t_1,t_2,t_3$ to the cubic which pass through $F$, which are $$ X=0,\ X=Z,\ X=\lambda Z. $$ The cross ratio of the four lines $t_0,\dots,t_3$ (meeting in a single point) in $\mathbb P^2$ is, up to ordering the four lines, $\lambda$.

2. Next, let $F'$ be any flex of the cubic, $t_0'$ the corresponding tangent. It can be shown that there are exactly three more tangents $t_i'$ ($i=1,2,3$) to the cubic through $F'$. Let $\lambda'$ be the cross ratio of the four lines $t_0',\dots,t_3'$ (meeting in a single point) - in some order. By the HINT, there is an automorphism of $\mathbb P^2$ fixing the cubic and sending $F'$ to $F$, thus sending $t_i'$ to $t_i$ (up to reordering), whence the cross-ratios $\lambda$ and $\lambda'$ share the same set $\Lambda(\lambda)=\Lambda(\lambda')$. This shows that one can really associate the set $\Lambda(\lambda)=:\Lambda(\mathcal C)$ to any cubic $\mathcal C$ by applying this recipe to any flex of $\mathcal C$.

3. Finally, given two cubic curves $\mathcal C:Y^2Z=X(X-Z)(X-\lambda)$ and $\mathcal C':Y^2Z=X(X-Z)(X-\lambda'Z)$, we can compute $\Lambda(\mathcal C)$ and $\Lambda(\mathcal C')$ according to 1 above as cross-ratios of four lines through a point in $\mathbb P ^2$. If $\mathcal C$ and $\mathcal C'$ are projectively equivalent, then these cross-ratios must be equal, i.e. $\Lambda(\lambda)=\Lambda(\lambda')$.

(Conversely, if $\Lambda(\lambda)=\Lambda(\lambda')$ it is easy to construct a projective equivalence of the two curves.)