What is the transformation such that a general cubic polynomial to be made a cube,

$$ax^3+bx^2+cx+d = y^3\tag{1}$$

can be transformed to Weierstrass form,

$$x^3+Ax+B = t^2\tag{2}$$

(The special case $b = c = 0$ is easier.) Given an initial rational point, I know a method how to find subsequent ones, but it would be nice to know the general transformation. For example,

$$3x^3+9x^2+15x+9 = y^3\tag{3}$$

I find that,

$$x_1 = 3$$

$$x_2 = -1839/1871$$

$$x_3 = -13898941449153/12222218425537$$

and so on. (I may have skipped some points.) But how do you transform $(3)$ to $(2)$?

Postscript (After Jyrki's answer)

For those interested, the cubic $(3)$ in two-variable form is equivalent to,

$$3p^3+9p^2q+15pq^2+9q^3 =\; p^3 + (p+q)^3+(p+2q)^3 = t^3$$

or three cubes (not necessarily positive) in arithmetic progrees. Thus, $p,q = 3,1$ gives the well-known,

$$3^3+4^3+5^3 = 6^3$$

and $p,q = -1839, 1871$ yields,

$$(-1839)^3+(-1839+1871)^3+(-1839+2\cdot1871)^3 = 876^3$$

  • $\begingroup$ The general case of cubics is dealed with in this article. $\endgroup$ – Cantlog Sep 10 '13 at 7:29

The point $P=(x,y)=(-1,0)$ is a point of inflection in the sense that the line $x=-1$ makes 3-fold contact there. This makes it a prime candidate to be moved to the point at infinity. Recalling that in Weierstrass form the two coordinates have poles of orders $2$ and $3$ respectively at the point of infinity (following the known process outlined e.g. in the proof of Proposition 3.1. of chapter 3 in The Arithmetic of Elliptic Curves, GTM#106 by J.Silverman) this suggests the pair $$ u=\frac1{x+1},\qquad v=\frac{y}{x+1} $$ as new coordinates. These are chosen to have poles of orders $3$ and $2$ at $P$, and no other poles. This is clear in the case of $u$. To see that $v$ has no other poles we observe that your curve has three points at infinity with homogeneous coordinates $$[X_0:X_1:X_2]=[1,\omega^k\root3\of3:0],$$ $\omega=e^{2\pi i/3}, k=0,1,2$, where $v$ might potentially go to infinity. But we see that the ratio $v=y/(x+1)=X_1/(X_0+X_2)$ is well defined and finite at all those points, so none of these points are poles of $v$. Furthermore, $y$ is a local parameter at $P$, so the pole of $v$ at $P$ is of order $2$.

Indeed, the calculation $$ v^3=\frac{y^3}{(x+1)^3}=\frac{3(x+1)[(x+1)^2+2]}{(x+1)^3}=3+\frac{6}{(x+1)^2}=3+6u^2 $$ then shows that you can write your curve in the form $$ 6u^2=v^3-3. $$ To get to the Weierstrass form we should multiply this equation by $6^3=216$. This gives us $$ (36u)^2=6^4u^2=216v^3-348=(6v)^3-3\cdot216. $$ The final coordinates should thus be $X=6v$ and $Y=36u$, and the Weierstrass form $$ Y^2=X^3-648. $$ Altogether the transformation was $$ X=\frac{6y}{x+1},\qquad Y=\frac{36}{x+1}. $$ So for example the point $Q=(x,y)=(3,6)$ in the original coordinates becomes $(X,Y)=(9,9).$ As the curve is now in Weierstrass form, $-Q=(X,Y)=(9,-9)$ is another rational point with original coordinates $(x,y)=(-5,-6).$ Of course, you could equally well do the curve's group operation in the original coordinates using the point $P$ as the neutral element. The three points at infinity in the original coordinates have now become the three points with $Y=0$.

If you start with a point that is not an inflection point, then the functions with the prescribed poles still exist by Riemann-Roch. Finding them takes a bit more work though. With an inflection point you can start with the tangent, and life is simpler.

  • 1
    $\begingroup$ This was ad hoc. I'm sure that a general algorithm is given in some (if not all) books dedicated to elliptic curves. $\endgroup$ – Jyrki Lahtonen Aug 28 '13 at 8:25
  • 1
    $\begingroup$ I checked Silverman's AEC (GTM#106). He proves the existence of Weierstrass form only after Riemann-Roch. I learned ECs mostly from that book, so that's why I followed the route of locating functions with given pole divisors supported at a single point. I have a vague recollection that at least Menezes' book (targeting cryptography people, so may not be everybody's cup of coffee) discusses the steps needed to bring a generic non-singular cubic to Weierstrass form. $\endgroup$ – Jyrki Lahtonen Aug 28 '13 at 8:49
  • 2
    $\begingroup$ Thanks, Jyrki. I was hoping for the general transformation, but this detailed answer and references will do. As far back as Dickson's History of the Theory of Numbers he gives a transformation, but it is only for the general monic cubic (with leading coefficient $a=1$) as $x^3+bx^2+cx+d = y^3$. $\endgroup$ – Tito Piezas III Aug 28 '13 at 23:56

General algorithm for transforming a cubic into Weierstrass form, with several worked examples, can be found in Section 1.4 of Ian Connell's Elliptic Curve Handbook


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.