When is an affine part of an elliptic curve isomorphic to an affine part of a norm equation? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T21:21:03Z http://mathoverflow.net/feeds/question/55139 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/55139/when-is-an-affine-part-of-an-elliptic-curve-isomorphic-to-an-affine-part-of-a-nor When is an affine part of an elliptic curve isomorphic to an affine part of a norm equation? Dror Speiser 2011-02-11T16:13:40Z 2011-02-13T18:26:25Z <p>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.</p> <p>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?</p> <p>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.</p> http://mathoverflow.net/questions/55139/when-is-an-affine-part-of-an-elliptic-curve-isomorphic-to-an-affine-part-of-a-nor/55170#55170 Answer by Felipe Voloch for When is an affine part of an elliptic curve isomorphic to an affine part of a norm equation? Felipe Voloch 2011-02-11T21:27:57Z 2011-02-12T21:17:25Z <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/55139/when-is-an-affine-part-of-an-elliptic-curve-isomorphic-to-an-affine-part-of-a-nor/55206#55206 Answer by David Speyer for When is an affine part of an elliptic curve isomorphic to an affine part of a norm equation? David Speyer 2011-02-12T14:42:28Z 2011-02-13T14:09:04Z <p>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 $f(x,y,z)$.</p> <p>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.</p> <p>The space of all cubics is $\mathbb{P}^9$. (There are $10$ coefficients in a cubic, and we don't care about rescaling.) </p> <p>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 $2$.</p> <p>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 $L_1 L_2 L_3=0$. <strong>Corrected from earlier version:</strong> The map from the Segre product to $W$ is $6$ to $1$ (the $6$ orderings of the lines). So $W$ has degree $90/6=15$.</p> <p>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 <a href="http://math.stanford.edu/~vakil/245/245class17.pdf" rel="nofollow">excess intersection computation</a> which I'm not sure how to do. So I'll stop here. </p> <p>If anyone wants to complete the computation, I'll leave this answer as Community Wiki.</p> http://mathoverflow.net/questions/55139/when-is-an-affine-part-of-an-elliptic-curve-isomorphic-to-an-affine-part-of-a-nor/55318#55318 Answer by David Speyer for When is an affine part of an elliptic curve isomorphic to an affine part of a norm equation? David Speyer 2011-02-13T13:48:36Z 2011-02-13T18:26:25Z <p>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.</p> <p>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.</p> <p>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.</p> <hr> <p>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?</p> <p>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.</p>