Question: How can we show algebraically that three cubic homogeneous polynomials in two variables satisfy a cubic polynomial of three variables?

More specifically, let $f_1(x_0,x_1),f_2(x_0,x_1),f_3(x_0,x_1)$ be cubic and homogeneous having no common roots. How can we show the existence of a cubic homogeneous $p(z_0,z_1,z_2)$ such that $p(f_1,f_2,f_3)=0$?

I can handle the case where $f_1,f_2,f_3$ are monomials by inpsection, but how about the general case?


Your polynomials define a non-constant morphism $$f:\mathbb P^1\to \mathbb P^2:(x_0:x_1)\mapsto (f_1(x_0,x_1):f_2(x_0,x_1):f_3(x_0,x_1))$$ Its image is closed and hence is an algebraic curve $f(\mathbb P^1)=C\subset\mathbb P^2$ .
Cutting that curve $C$ with a line $az_1+bz_2+cz_3=0$ shows that $C$ has degree $3$.
Thus $C=V(p(z_1,z_2,z_3))$ is the zero locus of some homogeneous third degree polynomial $p$: the one you desired!

  • $\begingroup$ So here is how i interpret your argument: the curve C intersects a generic line at 3 points (counting multiplicities) and this number must be the degree of C by Bezout's theorem. Is that how you think about it? $\endgroup$ – Manos Apr 5 '15 at 21:21
  • $\begingroup$ Yes${}{}{}{}$ ! $\endgroup$ – Georges Elencwajg Apr 5 '15 at 21:36
  • 1
    $\begingroup$ You are welcome, dear Manos! $\endgroup$ – Georges Elencwajg Apr 5 '15 at 21:38

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