Tangent surface of a twisted cubic curve

Question

I am trying to describe the tangent surface to a twisted cubic curve $C$, i.e. the curve which is given parametrically by $t\mapsto(t, t^2, t^3)$. This surface $S$ is given parametrically by $(t,u)\mapsto(t+u, t^2+2tu, t^3+3t^2u)$. Denote $x=t+u$, $y=t^2+2tu$, $z=t^3+3t^2u$. What is the polynomial equation of $S$ in terms of $x,y,z$?

Answer 1

This is best solved with Gröbner bases

nmajo@laureline:~$ /usr/bin/M2
Macaulay2, version 1.7
with packages: ConwayPolynomials, Elimination, IntegralClosure, LLLBases, PrimaryDecomposition, ReesAlgebra, TangentCone

i1 : R=QQ[t,u,x,y,z]

o1 = R

o1 : PolynomialRing

i2 : I=ideal(x-(t+u), y-(t^2+2*t*u), z-(t^3+3*t^2*u))

                            2                3     2
o2 = ideal (- t - u + x, - t  - 2t*u + y, - t  - 3t u + z)

o2 : Ideal of R

i3 : gens gb I

o3 = | t+u-x u2-x2+y 2uy2+xy2-2uxz-2x2z+yz uxy-x2y+2y2-uz-xz 2ux2-2x3-2uy+3xy-z 3x2y2-4x3z-4y3+6xyz-z2 |

             1       6
o3 : Matrix R  <--- R

The last entry in the gröbner basis is independent of t and u, and is your equation $3x^2y^2-4x^3z-4y^3+6xyz-z^2$

Here is a gif.

Answer 2

I used Mathematica to eliminate $u$ and $t$, and I got $$4x^3z - 3x^2y^2 - 6xyz + 4y^3 + z^2 = 0$$

The command is simply

Eliminate[{x = t + u, y = t^2 + 2*t*u, z = t^3 + 3*t*t*u}, {t,u}]

You can use Wolfram Alpha if you don't have access to Mathematica. I don't know what algorithm is used internally, but I would guess that it's probably Groebner bases.

Answer 3

An alternative description is that (x,y,z) belongs to the tangent surface if and only if the polynomial (in the variable $s$) $1+3sx+3s^2y+s^3z$ has a double root. The corresponding M2 commands to get the equation of the tangent surface are

R=QQ[s,x,y,z]

factor (discriminant(1+3*s*x+3*s^2*y+s^3*z,t))

The reason is that the twisted cubic corresponds to polynomials $p(t)=(1+st)^3$ and the tangent line at $p(t)$ is parametrized by cubic polynomials $p(t)+up'(t)=(1+st)^2(1+(t+3u)s)$.