reason to change elliptic curves to normal form

Question

I would like to know about the concept of why we change the curve into Weierstrass normal form. I mean for a specific example, if we have $x^3 +7y^3 + 64z^3=0$ and I went through all the steps and I changed it to $y^2= x^3 +ax^2 +bx + c$ . I did this and I got very huge number like in trillions for $a, b, c$. so we have $P$($2,2,-1$) belong to our first curve, what is this point considered w.r.t. the normal form? I understood the process but I don't know why we do that and how can we find the rational points. Thanks for your help in advance.

Answer 1

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I would like to know about the concept of why we change the curve into Weierstrass normal form.

One reason might be that the curves entered into the LMFDB are given in terms of the Weierstrass normal form. However, depending on context, you don't have to change to the Weierstrass normal form. It is one of a few standard forms of elliptic curve equations.

For example, $\,y^2 + a_1 xy + a_3 y = x^3 + a_2 x^2 + a_4 x + a_6,\,$ is a popular one for number theory. The original Weierstrass equation $\,y^2 = 4x^3 - g_2x - g_3\,$ has Weierstrass elliptic function solutions. Some other forms include Edwards curve, Montgomery curve, Mordell curve, and Hessian curve.

Some integer points on your homogeneous curve $\,C\!: 0 = x^3 +7y^3 + 64z^3\,$ are: $$\,(2,2,-1),\,(4,0,-1),\,(5,-3,1),\,(8,-4,-1),\,(16,-12,5),\,(68,152,-73)$$ and infinitely many others. You can use a standard method to find more points. For example, if $\,P\,$ and $\,Q\,$ are two points on the curve $\,C$, then let $\,R := a P+b Q\,$ where $\,ab\ne 0\,$ and solve the equation $\,0 = x^3 +7y^3 + 64z^3\,$ to find a new point.

For example, let $\,P := (2,2,-1),\,$ $Q := (4,0,-1).\,$ Then $$R := a P+b Q = (2a+4b, 2a, -a-b).$$ The cubic equation applied to $\,R\,$ becomes $$0 = (2a+4b)^3 + 7(2a)^3 + 64(-a-b)^3 = -48ab(3a + 2b). $$ The nonzero solution is $\,a=-2,\, b=3,\, R=(8,-4,-1).\,$

The elliptic curve group law is given by any line that intersects the curve at three points which are now considered to have a group sum of zero. The point $\,R_0 := (4,0,-1)\,$ is the group identity and the point $\,R_1 := (2,2,-1)\,$ is a group generator. Now, find that $\,R_{-1} := (8,-4,-1)\,$ is the additive inverse of $\,R_1\,$ because $\,R_{-1} = 3 R_0 - 2 R_1\,$ is a solution of the cubic equation. The line $\,0 = x + 7y + 16z\,$ is tangent to $\,C\,$ at $\,R_1\,$ and it intersects $\,C\,$ at $\,R_{-2} := (5,-3,1).\,$ Now $\,R_2 := (16,-12,5)\,$ is the additive inverse of $\,R_{-2}\,$ because $\,R_2 = -R_0 + 4R_{-2}.\,$ Continue this chord and tangent process to find any integer multiple $\,R_n\,$ of the point $\,R_1\,$ on the curve $\,C.$

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However, it is possible to change your curve to Weierstrass normal form with a few simple replacement steps.

The original curve is $$ 0 = x^3 +7 y^3 +64 z^3. \tag1 $$ Scale the variables with the replacements $(x\to x,\,y\to -z,\,z\to y/4)$ to get $$ 0 = x^3 +y^3 -7 z^3. \tag2 $$ This curve appears in the solution to MSE question 4298049. Apply the replacements $(x\to z-y,\,y\to z+y,\,z\to x)$ to get $$ 0 = -6 y^2 z +7 x^3 - 2 z^3. \tag3 $$ This kind of step appears in the solution to MO question 145877 by Noam Elkies. Scale the variables again with the replacements $(x\to 6x,\,y\to y,\,z\to 252z)$ and remove a common factor of $1512$ to get $$ 0 = -y^2 z +x^3 -21168 z^3. \tag4 $$ This is the homogeneous equation of a elliptic curve. Dehomogenize by setting $z=1$ to get $$ y^2 = x^3 -21168. \tag5 $$ which is in Weierstrass normal form, However, it is not listed in the LMFDB in this form. Now use the replacements $(x\to 4x,\,y\to 8y+4z,\,z\to z)$ and remove a common factor of $64$ to get $$ 0 = -y^2 z -y z^2 +x^3 -331 z^3. \tag6 $$ Dehomogenize by setting $z=1$ to get $$ y^2 +y = x^3 -331. \tag7 $$ This is the curve LMFDB 441.d1. The LMFDB gives the infinite order generator $P=(21,94).$

To conversion of points on this curve $(7)$ to the original curve is given by the mapping $$ (x,y)\mapsto(-8(y-31), -24 x, 2 (y+32)). \tag8 $$ For example, the generator $P$ is mapped to $(-504,-504,252)$ which is projectively equivalent to $(2,2,-1).$ Another example, the double of the generator $2P=(7,3)$ is mapped to $(224,-168,70)$ which is projectively equivalent to $(16,-12,5).$

Answer 2

First of all, ultimately you need to deal with such curves without using explicit equations. The higher-dimensional analogues of elliptic curves (called abelian varieties) do not easily submit to equation-theoretic reasoning.

Even for elliptic curves, the group law is best thought of geometrically, not in terms of equations for the curve.

If we do use equations, a reason for working with Weierstrass equations is related to the definition of an elliptic curve over a field: it is not just given by a smooth cubic, but also must have a point defined over the field. A famous example of your type of cubic that has rational coefficients but no point with rational coordinates is $3x^3 + 4y^3 + 5z^3 = 0$ (viewed in $\mathbf P^2$, so $[0:0:0]$ does not count). The projective curve defined by this equation (in algebraic geometry) is smooth of genus 1 but it is not an elliptic curve over $\mathbf Q$.

Every curve over a field defined by a Weierstrass equation has a rational point over that field: $[0:1:0]$. This point is also rather special geometrically among all points on a curve defined by a Weierstrass equation: the tangent line to the curve at that point is the line at infinity $Z = 0$ and this tangent line meets the curve at $[0:1:0]$ with multiplicity 3 (it is called a flex point), not just multiplicity at least 2. This can help you figure out how to find the Weierstrass equation for an elliptic curve given by a different equation: make a change of coordinates to move the flex point to $[0:1:0]$ so that the tangent line to the curve at the flex point turns into $Z = 0$.