So in Ravi Vakil's notes Ex 8.2C, I have to prove that if $\pi:X\hookrightarrow\text{Proj}\ S_{\cdot}$ is a closed subscheme (here $S_{\cdot}$ is a graded ring finitely generated by elements of degree 1), then there exists a homogeneous ideal $I$ such that $X=\text{Proj}\ S_{\cdot}/I$.
I have gone as far as proving the following:
Suppose $S_\cdot$ is generated by $x_i$'s for $i=0,…,n$. Then by definition of closed subscheme, on $\mathcal{O}_{\text{Proj}\ S_{\cdot}}(D(x_i ))$ we have an ideal $I_i$ of $((S_{\cdot})_{x_{i}})_{0}$. Then I let $J_i$ to be the homogeneous ideal in $S_{\cdot}$ such that
$$ J_{i}=\bigoplus_{d}\{a\in S_d: a/x_{i}^{d}\in I_{i}\}.$$
Then I let $I=\cap J_{i}$. The reason why I arrive at this is as follows and in fact I am not even sure it is correct: that each of the $I_i$ in $\mathcal{O}_{\text{Proj}\ S_{\cdot}}(D(x_i ))$ corresponds to a closed subscheme of $D(x_i )$. Then $X$ is supposed to be the union of these closed subschemes which translates to the intersection of ideals, and hence the result. Surprisingly, I got the correct homogeneous ideal by checking the sources, but I have a problem of proving that this corresponds to the desired $X$. I have searched several sources but it requires using coherent modules and some other techniques which are only proven in the later chapters. Is there a way to prove this using elementary way? Thanks!
Best Answer
The ideal in the post is the correct one. It suffices to show that the sections of $\widetilde{I\ }$ are the same as those of $\mathcal{I}_X$, the ideal sheaf of $X$ on each open affine $D(x_i)$.
First, we note that $\widetilde{J_i}(D(x_i))=(J_i)_{(x_i)}=\mathcal{I}_{X}(D(x_i))$. So it's enough to show that $I_{(x_i)}=(J_i)_{(x_i)}$ for all $i$. As $I\subset J_i$ implies $I_{(x_i)}\subset (J_i)_{(x_i)}$, it's enough to show that for any $f\in J_i$, we can find $fx_i^e\in J_j$ for all $j\neq i$ for some $e\geq 0$, as then $\frac{fx_i^e}{x_i^{e+\deg f}}=\frac{f}{x_i^{\deg f}}$ and we see the equality $I_{(x_i)}=(J_i)_{(x_i)}$.
Fix $i$. Pick some $f\in S_d$ so that $f\in J_i$. Then on $D(x_j)$ for $j\neq i$, $\frac{f}{x_j^d}=\frac{f}{x_i^d}\cdot\frac{x_i^d}{x_j^d}$ is in $\mathcal{I}_X(D(x_j) \cap D(x_i))$, so after adjustment by a factor of $\frac{x_i^{d_j}}{x_j^{d_j}}$, we have that $\frac{fx_i^{d_j}}{x_j^{d+d_j}}\in (J_j)_{(x_j)}$ and therefore $fx_i^{d_j}\in J_j$. So by picking $e=\max_j(d_j)$, we get that $fx_i^e\in J_j$ for all $j$, and thus $fx_i^e\in I$. This exactly shows what we wanted.