I'm coming from a projective setting of a smooth cubic plane curve over a field $K$ and want to show that I can bring it to the Weierstrass long form. The usual method is to start with a point and bring it to $(1:0:0)$ and then take the tangent line of this point and bring it to $z=0$ without changing $(1:0:0)$. But when doing this, one wants that the tangent line of the chosen point meet the curve at another point. I want to argue why this can happen at all.

In other words, I want to argue why a smooth cubic curve in the projective plane will have a point whose tangent line does not meet the point with multiplicity $3$. So if $C$ is the curve and $P$ any point and I denote $L_P$ as the tangent line at that point, do I really need to use commutative algebra and show that there is a $P$ such that $I(C\cap L_P , P) =2$ or is there a more elementary method of showing this? Here, when I meant commutative algebra, I view the intersection number as the $K$ vector space dimension of the polynomial ring localized at $P$ modulo the defining equations for the curve and the tangent line.

Edit: From the answer and comments, I realize that I should probably use case distinctions if I want to show that I can bring it into Weierstrass long form (by some isomorphisms). Namely, cases when I have only a flex (how to show in this case the curve consist of only finite number of points?) and cases when there is a non-flex. I suppose I can argue that there is a non-flex over $\bar K$ (algebraic closure), but I think I should resist this because I still want to have this curve defined over $K$ and fix the $K$-rational points.

  • $\begingroup$ How about Bezout's Theorem? Over a non-algebraically closed field $K$, for two homogeneous polynomials $f,g$ defining two curves without common intersection, it states that most number of intersections is at most $\text{deg}f\cdot \text{deg} g$ times (if $K$ is algebraically closed then it is an equality). Then as a tangent line $L_{P}$ to a point $P$ has $\text{deg}L_{P} = 1$, it will have at most 3 points of intersections as the elliptic curve has degree 3. It is only equal to 3 if $P$ is a point of inflection, so just choose another point. Hopefully this aids you at least geometrically $\endgroup$ – BenCWBrown Nov 10 '18 at 21:47
  • $\begingroup$ Yes, but how do you argue that there are points which are not points of inflection? $\endgroup$ – quantum Nov 10 '18 at 21:54
  • $\begingroup$ Guess it depends on the field $K$ (I'm more used to working over one that's closed), but the intersection of the cubic with its Hessian gives you the 9 points of inflexion. That is the maximum you can have by Bezout's, then see if the curve has more? $\endgroup$ – BenCWBrown Nov 10 '18 at 22:08
  • $\begingroup$ But isn't this intersection with Hessian only a statement that works with characteristic 0? What do I do in other characteristics (even if the field is alg. closed)? $\endgroup$ – quantum Nov 10 '18 at 22:16

There are elliptic curves over $\mathbb{Q}$ which have only one rational point. At this unique point the tangent line meet the curve three times (so it is a flex).

For example, the curve given by the Weierstrass equation $$y^2 + y = x^3 - x^2 - 7820x - 263580$$ has only the point at infinity as a rational point.

There is also other examples of curves with exactly three points, all of them flexes. For example the one given by the equation $$y^2 + y = x^3 + x^2 + x$$ with points $(0,0),(0,-1)$ and the point at infinity.

All this can be shown for example by using some software like pari or sage or magma.

Note that any genus 1 curve in Weierstrass form has at least one flex point: the point at infinity.

Moreover, any genus 1 curve over an algebraically closed field of characteristic $\ne 3$ has exactly 9 flexes. If the characteristic is $=3$ it has either 1 or 3 flexes. They correspond to the 3-torsion points in the group structure of the (rational points of the) elliptic curve.

Concretely, if you put your genus 1 curve in simplified Weierstrass form, $y^2=x^3+Ax+B$ for some $A$ and $B\in K$, which is always possible if the characteristic of $K$ is $>3$, the flexes are the 3-torsion points, which have $x$-coordinate a root of $$\psi_{3} = 3x^{4} + 6Ax^{2} + 12Bx - A^{2}$$ (which is the $3$-Division polynomial), or it is the point at infinity. This polynomial has 4 distinct roots, hence we get eight points (two for each root), plus the point at infinity. If the characteristic is $=2$ a similar polynomial works for the simplifed form $y^2 + xy = x^3 + ax^2 + b$.

  • $\begingroup$ Two things would be interesting: 1. Could you give a more concrete example of this elliptic curve. 2. And if I require algebraic closedness (but leave the field characteristic arbitrary)? $\endgroup$ – quantum Nov 10 '18 at 22:14
  • $\begingroup$ Thank you for the detailed answer. I'm interested to know why 9 flexes if char$\ne$3, do you have any references I could look at (or maybe this is easy to explain?). $\endgroup$ – quantum Nov 11 '18 at 0:28

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