If given $[\mu_i,\nu_i]\in P^1$ distinct $d+1$ points where $1\leq i\leq d+1$, one can obtain $G(x_0,x_1)=\prod_{i=1}^{d+1}(\mu_ix_1-\nu_ix_0)$. Then one can define a rational curve in $P^{d}$ by $[x_0,x_1]\to [H_0(x_0,x_1),\dots,H_{d}(x_0,x_1)]=[z_0,\dots,z_d]$ where $H_i(x_0,x_1)=\frac{G(x_0,x_1)}{\mu_{i+1}x_1-\nu_{i+1}x_0}$. Call this curve $C$. I am looking for homogeneous polynomials $f_k$ in $z_0\dots z_d$ such that the variety $V(\{f_k\})=C$. I could not find quadratics though it is rational normal curve.

  • $\begingroup$ Can you please explain $f\in C[z_0,\dots,z_d]$ to me? So you have the curve, and you are looking for a polynomial in up to $d+1$ variables (which $f(C)$ will use as coordinates of points from your curve), but to me the notation suggests you'd be taking the coefficients from the curve, which doesn't make a lot of sense. Should that be $\mathbb C[\ldots]$ instead? The index shift is also confusing, having $1$ through $d+1$ in most places but $0$ through $d$ for the $z_i$ (and the $x_i$). Perhaps edit that out? $\endgroup$
    – MvG
    Nov 2, 2016 at 9:11
  • $\begingroup$ @MvG. Edited. I am looking for a polynomial describing this variety. I did not find a quadratic function. $\endgroup$
    – user45765
    Nov 2, 2016 at 12:33

1 Answer 1


Let's consider the non-projective case first: you'd have $z_i=H_i(x)$ or equivalently $H_i(x)-z_i=0$. If you have two such equations, you can eliminate $x$ from them (e.g. using resultants) and end up with a polynomial in $z_i$ and $z_j$ which characterizes the relation between these two coordinates.

Doing the same in a homogeneous coordinate setup, you have to write your equations differently. Pick three indices $i,j,k$ and then consider the equations

$$H_i(x,1)z_k - H_k(x,1)z_i = 0 \qquad H_j(x,1)z_k - H_k(x,1)z_j = 0$$

A point on your curve for parameter $(x,1)$ has to satisfy both of these. Eliminating $x$ from these equations will lead to a homogeneous equation in $z_i,z_j,z_k$ which will describe a hypersurface containing your curve. So that's one of the $f_k$ you're asking about (although the $k$ there is not the same as the $k$ I used).

In the original version of this question, where you were just asking for one polynomial that vanishes for points on the curve, I'd be done at this point. As your edited question now asks for multiple polynomials which completely define the curve, this is no longer the whole story. I would assume that if you take enough triples $i,j,k$ you should end up with a unique description of the curve.

For general position I'd assume you'll need $d-1$ such polynomials to get from a $d$-dimensional space to a $1$-dimensional curve, as each additional condition should lower the dimension by one. In that case I'd use triples $(0,1,2),(1,2,3),\dots,(d-2,d-1,d)$. But I'm sure there will be nasty corner cases, where these contain redundant information so you need extra triples. You might even have to consider linear combinations of coordinates, similar to a change of coordinates. Perhaps you can pick a random basis each time the process fails, assuming that you'll pick a basis for which the input is in general position with probability one. Haven't tried it, though.

Note that the polynomials I suggest, coming from the resultants, will likely have degree $d^2$ so if you are looking for a simple description of the curve as an implicit variety, there may be better ways. Probably using some other tools from elimination theory. Not sure.

  • $\begingroup$ Is there a consistent way to determine the minimal set of polynomials which fully determine the curve? I mean I knew I can get $\frac{(d+1)d}{2}$ relations and I only need $d$ relations to solve it as it is rational curve. $\endgroup$
    – user45765
    Nov 2, 2016 at 13:20

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