The rational normal curve is described by $\nu_4:[x_0,x_1]\to [x_0^4,x_0^3x_1,x_0^2x_1^2,x_0x_1^3,x_1^4]=[z_0,z_1,z_2,z_3,z_4]$. Now it is the intersection of 4 quadratics. $F_{11}=z_1^2-z_0z_2,F_{22}=z_2^2-z_1z_3,F_{33}=z_3^2-z_2z_4,F_{13}=z_1z_3-z_0z_4$. By Bezout theorem, $F_{11}\cap F_{22}$ should have a degree of 4. Now we know the curve is of degree 4 and it is the solution to above 4 quadratics. I should get nothing else. However, $F_{11}\cap F_{22}$ clearly contains $V(z_1,z_2)$.

The other thing is that $F_{11}\cap F_{22}\cap F_{33}$ should have degree 8. However, I got the curve of degree 4 and a line which is degree 1. I should have a degree 3 left. Can someone clarify this?

  • 2
    $\begingroup$ $F_{11}\cap F_{22}$ is a surface, not a curve. $\endgroup$ – Mohan Nov 2 '16 at 2:03
  • $\begingroup$ @Mohan I see. In wiki, en.wikipedia.org/wiki/B%C3%A9zout%27s_theorem, there is a generalized statement on $P^n$. If that is the case, I need n equations. However, product of degrees of $F_{11}, F_{22},F_{33}$ and $F_{13}$ is 16. This is even more confusing now. What am I missing here? $\endgroup$ – user45765 Nov 2 '16 at 2:09
  • $\begingroup$ And there are two more quadrics --- $z_0z_3 - z_1z_2$ and $z_1z_4 - z_2z_3$. $\endgroup$ – Sasha Nov 2 '16 at 13:14

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