2.9. If $Y\subseteq \mathbb{A} ^{n}$ is an affine variety, we identify $\mathbb{A} ^{n}$ with an open set $U_{0} \subseteq \mathbb{P} ^{n}$ by homeomorphism $\varphi _{0}$. Then we can speak of $\bar{Y}$,the closure of $Y$ in $\mathbb{P} ^{n}$, which is called the projective closure of $Y$.
(i) Show that $ I(\bar{Y})$ is the ideal generated by $\beta (I(Y))$, using the notation of the proof of (2.2).
(ii) Let $Y\subseteq \mathbb{A} ^{3}$ be the twisted of exercise 1.2. Its projective closure $\bar{Y} \subseteq \mathbb{P} ^{3}$ is called the twisted cubic curve in $\mathbb{P} ^{3}$. Find generators for $ I(\bar{Y} )$.
It's not hard to see that $Y= V(Y-X^{2},Z-X^{3})$ and $I(Y)= (Y-X^{2},Z-X^{3})$.
Notation: Let $S:=k[W,X,Y,Z]$ and let $A:=k[X,Y,Z]$. Identify $\mathbb{A}^n$ with the open subset $U:=\mathbb{P}^n\setminus Z(W)$ via the obvious map $\varphi$. Define a map $\beta\colon A\to S$ by $$\beta(g)=W^{\deg{g}}g\left(\frac{X}{W},\frac{Y}{W},\frac{Z}{W}\right),$$ so $\beta (Y-X^{2})=WY-X^{2}$ and $\beta (Z-X^{3})=W^{2}Z-X^{3}$. Because of (i), $I(\bar{Y})$ is the ideal generated by $\beta(I(Y))$.
My question is how to find the generators for $I(\bar{Y})$? I guess $I(\bar{Y})=(WY-X^{2},W^{2}Z-X^{3},XY-WZ)$, but I don't know how to prove this. Can someone please help me? Thanks.
