I am trying to solve exercise 2.9 b from section 1.2 of Hartshorne algebraic geometry. It asks to find out the generator of $I(\bar Y)$ where $\bar Y$ is the projective closure of the twisted cubic curve in $\mathbb P^3$ given by $\{(t,t^2,t^3)| t\in k\}\subset \mathbb A^3$.
My claim is that $I(\bar Y)= \langle x_1^2-x_0x_2, x_2^2-x_1x_3,x_1x_2-x_0x_3\rangle$. Let’s call the right hand side ideal as $\alpha$.
I have shown using the first part of the exercise which says $I(\bar Y)=\langle \beta I(Y)\rangle$ where $\beta $ is homogenisation map, that, $\alpha \subset I(\bar Y)$.
For the reverse inclusion, I am trying to show $\alpha $ is a homogeneous radical prime ideal.
I defined the map, $\theta :k[x_0,x_1,x_2,x_3] \rightarrow k[u,v], \theta$ sends, $x_0$ to $u^3$, $x_1$ to $u^2v$, $x_2$ to $uv^2$ and $x_3$ to $v^3$.
My claim is that $\alpha = \ker \theta$. $\alpha \subset \ker\theta$ is clear. I am stuck with the second part.
If I’m done with this part I can prove $Z(\alpha) \cap U_0 \subset \phi_0^{-1} (Y) $ where $U_0 = \mathbb P^3- Z(x_0)$ and $\phi_0$ is the homeomorphism from $U_0$ to $\mathbb A^3$. Then using irreducibility of $Z(\alpha)$ and taking closures both the sides I can say that $Z(\alpha) \subset \bar Y$. Which, by homogeneity and radical condition will tell us $I(\bar Y) \subset \alpha$. Also it would tell us that the 3-IPLs embedding of $\mathbb P^1 $ into $\mathbb P^3$ is the twisted cubic curve in $\mathbb P^3$.
I need help in completing the proof of the claim regarding the kernel of $\theta$.
