TITLE: When is an affine part of an elliptic curve isomorphic to an affine part of a norm equation?
QUESTION [6 upvotes]: Given a cubic number field and a basis $\{\gamma_1,\gamma_2,\gamma_3\}$ for it over the rationals, we can write down the norm equation $N(x_1\gamma_1+x_2\gamma_2+x_3\gamma_3)=1$. For almost all substitutions, say $x_1=c$, the resulting affine cubic curve is an affine part of an elliptic curve.
I was wandering what can be said of the converse. If we are given an elliptic curve over the rationals, is there a cubic number field such that for some substitution (any kind) in the norm equation, we get an affine curve isomorphic to an affine part of an elliptic curve?
I've started reading Serre's Algebraic Groups and Class Fields, which seems relevant, since its main results concern rational maps $C\rightarrow G$ from a curve to a commutative algebraic group, which is the case above.

REPLY [3 votes]: 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 $3$ 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.