Proposition

Let $C$ be the twisted cubic curve in $k^3$.

1. Show that $C$ is a subvariety of the surface $S = V(xz - y^2)$.
2. Find an ideal $J \subset k[S]$ such that $C = V_S(J)$.

Solution

1

Let $f(x, y, z) = xz - y^2$. $C = \{ (a, a^2, a^3) \mid a \in k \}$. For every $(a, a^2, a^3) \in C$, $f(a, a^2, a^3) = 0$. Therefore, $C \subset S$.

2

The ideal $J$ generated by $y - x^2 + \ev{xz - y^2}$ and $z - x^3 + \ev{xz - y^2}$ gives $C = V_S(J)$.