When is an affine part of an elliptic curve isomorphic to an affine part of a norm equation?

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.

• "the resulting affine cubic curve is an affine part of an elliptic curve": I agree that we get a plane affine cubic, and that the projective completion is "generally" smooth. But an elliptic curve has a rational point by definition. Why should this be the case here? Of course you can force it by requiring the element 1 of the cubic field to be on the curve, but this is no longer "general". – Laurent Moret-Bailly Feb 12 '11 at 9:10

If you have $C$ a curve of genus one and $P,Q,R$ points on it such that $P+Q \sim 2R, Q+R \sim 2P, P+R \sim 2Q$ (any two implies the third, btw), then $P-Q,R-P,Q-R$ have order three and, if you embed $C$ in the plane by the linear system $P+Q+R$ and choose coordinates in the affine plane such that the (inflectional) tangents at $P,Q,R$, which are collinear, meet at the origin, then the equation for $C$ will be of the form $f(x,y)=1$ where $f$ is a homogeneous cubic. The converse is also true. Your question has a further issue of rationality, as you want things defined over $\mathbb{Q}$. My guess is that, as long as $C$ is defined over $\mathbb{Q}$ and $P+Q+R$ is also defined over $\mathbb{Q}$ as a divisor, it should work out.

Edit: David's comment below correctly points out that not all curves of genus 1 can be put on the form $f(x,y)=1$. The mistake, as usual, is in the unjustified assertion. The inflectional tangents are not collinear. I think my method leads to an equation in the form $L_1L_2L_3=1$, notation as in David's answer. Note that, in this kind of equation, the lines $L_i=0$ are inflectional tangents at points of the cubic on the line at infinity.

• Nice argument... but it doesn't seem like this is the form the OP's elliptic curves take. If you specialize $x_1$ he suggests, you don't generally get a homogeneous cubic in the two remaining variables. – Ramsey Feb 11 '11 at 21:58
• @Ramsey: Right! Only when $c=0$ in his notation. I think it's a matter again of moving the origin by an affine transformation. – Felipe Voloch Feb 11 '11 at 22:54
• Felipe: I don't think this works. I don't think a generic cubic curve can put be in the form $f(x,y)=1$, where $f$ is a homogeneous cubic. More specifically, I claim that the only elliptic curve (over $\mathbb{C}$) which can be put in this form is the one with complex multiplication by a cube root of unity. Proof: Consider the automorphism $(x,y) \mapsto (\zeta x, \zeta y)$, where $\zeta$ is a primitive cube root of $1$. This automorphism has order $3$, and fixes $3$ points (the points at infinity). Taking any of those points as the origin, the curve has complex multiplication as required. – David E Speyer Feb 12 '11 at 16:46
• Second proof by counting dimensions. You are trying to write the curve as a linear combination of a triple line ($z^3=0$) and three concurrent lines ($f(x,y)=0$). The space of triple lines is $2$-dimensional; the space of triples of concurrent lines is $2+3=5$ dimensional (choose the place of concurrence, choose three slopes). So the pairs (triple line, three concurrent lines) gives you 7 dimensions of choices. Each pair gives you a one dimensional pencil, so there are 8 dimensions of cubic curves which can be written this way. The space of cubics is 9 dimensional. – David E Speyer Feb 12 '11 at 16:50

I'll assume you are happy changing coordinates in $\mathbb{P}^2$ to whatever you want. I'll approach the complex geometry problem and ignore the number theory. Let you cubic be $f(x,y,z)$.

You write $f(x,y,1) = L_1(x,y,1) L_2(x,y,1) L_3(x,y,1) - 1$, for three linear forms $L_i$. Working homogeneously, you want $f(x,y,z) = L_1(x,y,z) L_2(x,y,z) L_3(x,y,z) - z^3$. In other words, you want to write $f$ as a linear combination of a product of three lines, and a triple line.

The space of all cubics is $\mathbb{P}^9$. (There are $10$ coefficients in a cubic, and we don't care about rescaling.)

The space of cubics which are triple lines is the $3$-uple Veronese embedding of $\mathbb{P}^2$ into $\mathbb{P}^9$. Calling the space of triple lines $V$; it has degree $3^2=9$ and dimension $2$.

The space of cubics which are a product of three lines is a finite ($6$ to $1$) projection of the Segre embedding of $\mathbb{P}^2 \times \mathbb{P}^2 \times \mathbb{P}^2$ into $\mathbb{P}^{26}$. Call the spaces of products of three lines $W$. The Segre embedding has degree $6!/(2! 2! 2!) = 90$, if I recall correctly. The Segre product stays away from the base points of the projection -- explicitly, you can't have three lines $(L_1, L_2, L_3)$ such that $L_1 L_2 L_3=0$. Corrected from earlier version: The map from the Segre product to $W$ is $6$ to $1$ (the $6$ orderings of the lines). So $W$ has degree $90/6=15$.

Morally, we have a point $x$ (our given cubic) in $\mathbb{P}^9$, and we want to know how many lines through $x$ meet $V$ and $W$. However, there is a problem. In fact, $W$ contains $V$! So there is a huge excess intersection contribution. The space of lines through $V$ and $W$ thus splits into two components: Lines which meet $V$ at one point and $W$ at another point; and just the space of all lines that meet $V$. We want to understand the first space. But separating out the second component is going to require an excess intersection computation which I'm not sure how to do. So I'll stop here.

If anyone wants to complete the computation, I'll leave this answer as Community Wiki.

• This count may be wrong; see my other count below. – David E Speyer Feb 13 '11 at 13:49
• I've found some errors, but I haven't found all of them yet. I'm waiting to update until I'm confident all the data lines up. – David E Speyer Feb 13 '11 at 13:58
• OK, I think I've found the errors. I think it will be hard to get a correct count so I am updating now. – David E Speyer Feb 13 '11 at 14:01

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.