Determining generators for vanishing ideal of projective closure

This question is along the lines of 2.9 from Hartshorne.

Notation: Let $S:=k[x_0,\ldots,x_n]$ be the coordinate ring of $\mathbb{P}^n$, and let $A:=k[y_1,\ldots,y_n]$ be the coordinate ring of $\mathbb{A}^n$. Identify $\mathbb{A}^n$ with the open subset $U:=\mathbb{P}^n\setminus Z(x_0)$ via the obvious map $\varphi$. Define a map $\beta\colon A\to S$ by $$\beta(g)=x_0^{\deg{g}}g\left(\frac{x_1}{x_0},\ldots,\frac{x_n}{x_0}\right).$$ Let $Y$ be an affine variety in $\mathbb{A}^n$, and let $\overline{Y}$ be its projective closure.

In 2.9(a), it's established that $I(\overline{Y})=(\beta(I(Y)))$. However, part (b) shows that if $I(Y)=(f_1,\ldots,f_r)$, then $I(\overline{Y})$ isn't necessarily $(\beta(f_1),\ldots,\beta(f_r))$. The particular example given is $Y=Z(f_1,f_2)$, where $f_1=x_2-x_1^2$ and $f_2=x_3-x_1^3$.

I'm pretty sure that $\overline{Y}=\{[s^3:s^2t:st^2:t^3]\}$, in which case I know how to show that $\beta(f_1)=x_2x_0-x_1^2$ and $\beta(f_2)=x_3x_0^2-x_1^3$ don't generate $I(\overline{Y})$: The point $[0:1:1:0]$ vanishes at both $\beta(f_1)$ and $\beta(f_2)$ yet is not in $\overline{Y}$. (I'm having trouble proving that this is in fact $\overline{Y}$, though).

Here's some initial work I've done to kind of get my head around this in the general case: In order for $\beta(f_1),\ldots,\beta(f_r)$ to not generate $I(\overline{Y})$, there needs to be an element $F\in \beta(I(Y))$ which is not an $S$-linear combination of $\beta(f_1),\ldots,\beta(f_r)$. Since $F\in \beta(I(Y))$, there are $h_1,\ldots,h_r\in A$ such that $F=\beta(h_1f_1+\cdots+h_rf_r)$. Let $d_i:=\deg(f_i),e_i:=\deg(h_i)$, and let $d:=\max_i(d_i+e_i)$. Then $$F=\sum_{i=1}^r x_0^{d-d_i-e_i}\beta(h_i)\beta(f_i).$$ But then $F$ is a linear combination of the $\beta(f_i)$'s. What have I missed?

1 Answer

If you let $f_3 = x_1 f_1 - f_2 = x_1x_2 - x_3$, then $\beta(f_3) = x_1x_2 -x_3t$ if one uses $t$ as the homogenizing variable. This element is not in the ideal generated by $\beta(f_1), \beta(f_2)$.

One way to see this is the follwoing: $F = \beta(x_1 f_1 - f_2) = \beta(f_3) = x_1x_2 - x_3t \neq x_1 \cdot (x_2t - x_1^2) - (x_3t^3 -x_1^3) = \beta(x_1)\beta(f_1) - \beta(f_2)$. Hence $\beta(x_1 f_1 - f_2) \neq \beta(x_1)\beta(f_1) - \beta(f_2)$ in general. Also you can not multiply by any power of $t$ to any term of the RHS to make it equal since the deg of the LHS is $2$, the lowest degree on the RHS is $2$, and these degree $2$ guys are not equal.