show/hide this revision's text 3 small correction in explanation of count

I'm adding a second answer here because it doesn't appear to agree with my first answer. This second answer came from me trying to understand Felipe's arguments; it is possible that I am just rewriting what he said in more words.

Second answer: Let $X$ be a smooth cubic curve. Over $\mathbb{C}$, the ways to express $X$ as a linear combination of a product of three lines and a triple line are in bijection with triples $(P_1, P_2, P_3)$ of colinear cusps of $X$. Proof: Let $f = a L_1 L_2 L_3 + b L_4^3$. Then $L_4$ intersects $L_1$, $L_2$ and $L_3$ at one point each. Then $f$ restricted to $L_i$ vanishes to order $3$ at $P_i$. So $P_i$ is a cusp of $f$ and $L_i$ is the tangent line there. And $(P_1, P_2, P_3)$ all lie on $L_4$, so they are colinear. So, given an expression of $f$ as above, we find a triple of colinear cusps.

Conversely, given three colinear cusps $(P_1, P_2, P_3)$, let $L_4$ be the line through them and let $L_i$ be the tangent line to $P_i$. So $f$ restricted to $L_i$ is the cubic with an order $3$ root at $P_i$. So, choosing an appropriate scalar $b$, we have $f|_{L_1} = b L_4^3|_{L_1}$. Let $g = f - b L_4^3$. So $g$ vanishes on the line $L_1$; let $g = L_1 Q$, where $Q$ is a conic. Then $g$, restricted to $L_2$, vanishes to order $3$ at $P_2$. Since $L_1$ does not pass through $P_2$, this shows that $Q$ vanishes to order $3$ at $P_2$. But $Q$ is a conic, so this implies that $Q$ contains $L_2$. Similar, $Q$ contains $L_3$. So $Q=a L_2 L_3$ for some scalar $a$, and $f=a L_1 L_2 L_3 + b L_4^3$. So, given a triple of colinear cusps, we get such a linear representation.

My confusion: If I count correctly, there are $12$ such triples of colinear cusps. Namely, given any two of the $9$ cusps, there is a unique way to complete it to such a triple. There are $\binom{9}{2} = 36$ pairs of cusps, and we get each such triple $12$ 3$ ways. How do I square this with the $810$ count I got earlier?

An additional note: I'd been working over $\mathbb{C}$. For the original question, we want to impose the additional conditions that $L_4$ has coordinates in $\mathbb{Q}$, and the group $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ acts transitively on the $P_i$. In other words, that the cubic $f \cap L_4$ has no rational root. Someone better at computer algebra than I am should be able to turn that into the condition that a certain degree $12$ polynomial has a rational root, and that a certain degree $3$ polynomial does not.
show/hide this revision's text 2 small correction on third par.

I'm adding a second answer here because it doesn't appear to agree with my first answer. This second answer came from me trying to understand Felipe's arguments; it is possible that I am just rewriting what he said in more words.

Second answer: Let $X$ be a smooth cubic curve. Over $\mathbb{C}$, the ways to express $X$ as a linear combination of a product of three lines and a triple line are in bijection with triples $(P_1, P_2, P_3)$ of colinear cusps of $X$. Proof: Let $f = a L_1 L_2 L_3 + b L_4^3$. Then $L_4$ intersects $L_1$, $L_2$ and $L_3$ at one point each. Then $f$ restricted to $L_i$ vanishes to order $3$ at $P_i$. So $P_i$ is a cusp of $f$ and $L_i$ is the tangent line there. And $(P_1, P_2, P_3)$ all lie on $L_4$, so they are colinear. So, given an expression of $f$ as above, we find a triple of colinear cusps.

Conversely, given three colinear cusps $(P_1, P_2, P_3)$, let $L_4$ be the line through them and let $L_i$ be the tangent line to $P_i$. So $f$ restricted to $L_i$ is the cubic with an order $3$ root at $P_i$. So, choosing an appropriate scalar $b$, we have $f|_{L_1} = b L_4^3|_{L_1}$. Let $g = f - a b L_4^3$. So $g$ vanishes on the line $L_1$; let $g = L_1 Q$, where $Q$ is a conic. Then $g$, restricted to $L_2$, vanishes to order $3$ at $P_2$. Since $L_1$ does not pass through $P_2$, this shows that $Q$ vanishes to order $3$ at $P_2$. But $Q$ is a conic, so this implies that $Q$ contains $L_2$. Similar, $Q$ contains $L_3$. So $Q=a L_2 L_3$ for some scalar $a$, and $f=a L_1 L_2 L_3 + b L_4^3$. So, given a triple of colinear cusps, we get such a linear representation.

My confusion: If I count correctly, there are $12$ such triples of colinear cusps. Namely, given any two of the $9$ cusps, there is a unique way to complete it to such a triple. There are $\binom{9}{2} = 36$ pairs of cusps, and we get each such triple $12$ ways. How do I square this with the $810$ count I got earlier?

An additional note: I'd been working over $\mathbb{C}$. For the original question, we want to impose the additional conditions that $L_4$ has coordinates in $\mathbb{Q}$, and the group $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ acts transitively on the $P_i$. In other words, that the cubic $f \cap L_4$ has no rational root. Someone better at computer algebra than I am should be able to turn that into the condition that a certain degree $12$ polynomial has a rational root, and that a certain degree $3$ polynomial does not.
show/hide this revision's text 1

I'm adding a second answer here because it doesn't appear to agree with my first answer. This second answer came from me trying to understand Felipe's arguments; it is possible that I am just rewriting what he said in more words.

Second answer: Let $X$ be a smooth cubic curve. Over $\mathbb{C}$, the ways to express $X$ as a linear combination of a product of three lines and a triple line are in bijection with triples $(P_1, P_2, P_3)$ of colinear cusps of $X$. Proof: Let $f = a L_1 L_2 L_3 + b L_4^3$. Then $L_4$ intersects $L_1$, $L_2$ and $L_3$ at one point each. Then $f$ restricted to $L_i$ vanishes to order $3$ at $P_i$. So $P_i$ is a cusp of $f$ and $L_i$ is the tangent line there. And $(P_1, P_2, P_3)$ all lie on $L_4$, so they are colinear. So, given an expression of $f$ as above, we find a triple of colinear cusps.

Conversely, given three colinear cusps $(P_1, P_2, P_3)$, let $L_4$ be the line through them and let $L_i$ be the tangent line to $P_i$. So $f$ restricted to $L_i$ is the cubic with an order $3$ root at $P_i$. So, choosing an appropriate scalar $b$, we have $f|_{L_1} = b L_4^3|_{L_1}$. Let $g = f - a L_4^3$. So $g$ vanishes on the line $L_1$; let $g = L_1 Q$, where $Q$ is a conic. Then $g$, restricted to $L_2$, vanishes to order $3$ at $P_2$. Since $L_1$ does not pass through $P_2$, this shows that $Q$ vanishes to order $3$ at $P_2$. But $Q$ is a conic, so this implies that $Q$ contains $L_2$. Similar, $Q$ contains $L_3$. So $Q=a L_2 L_3$ for some scalar $a$, and $f=a L_1 L_2 L_3 + b L_4^3$. So, given a triple of colinear cusps, we get such a linear representation.

My confusion: If I count correctly, there are $12$ such triples of colinear cusps. Namely, given any two of the $9$ cusps, there is a unique way to complete it to such a triple. There are $\binom{9}{2} = 36$ pairs of cusps, and we get each such triple $12$ ways. How do I square this with the $810$ count I got earlier?

An additional note: I'd been working over $\mathbb{C}$. For the original question, we want to impose the additional conditions that $L_4$ has coordinates in $\mathbb{Q}$, and the group $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ acts transitively on the $P_i$. In other words, that the cubic $f \cap L_4$ has no rational root. Someone better at computer algebra than I am should be able to turn that into the condition that a certain degree $12$ polynomial has a rational root, and that a certain degree $3$ polynomial does not.