Lemma 27.13.7. Let $R$ be a ring. Let $Z \subset \mathbf{P}^ n_ R$ be a closed subscheme. Let

$I_ d = \mathop{\mathrm{Ker}}\left( R[T_0, \ldots , T_ n]_ d \longrightarrow \Gamma (Z, \mathcal{O}_{\mathbf{P}^ n_ R}(d)|_ Z)\right)$

Then $I = \bigoplus I_ d \subset R[T_0, \ldots , T_ n]$ is a graded ideal and $Z = \text{Proj}(R[T_0, \ldots , T_ n]/I)$.

Proof. It is clear that $I$ is a graded ideal. Set $Z' = \text{Proj}(R[T_0, \ldots , T_ n]/I)$. By Lemma 27.11.5 we see that $Z'$ is a closed subscheme of $\mathbf{P}^ n_ R$. To see the equality $Z = Z'$ it suffices to check on an standard affine open $D_{+}(T_ i)$. By renumbering the homogeneous coordinates we may assume $i = 0$. Say $Z \cap D_{+}(T_0)$, resp. $Z' \cap D_{+}(T_0)$ is cut out by the ideal $J$, resp. $J'$ of $R[T_1/T_0, \ldots , T_ n/T_0]$. Then $J'$ is the ideal generated by the elements $F/T_0^{\deg (F)}$ where $F \in I$ is homogeneous. Suppose the degree of $F \in I$ is $d$. Since $F$ vanishes as a section of $\mathcal{O}_{\mathbf{P}^ n_ R}(d)$ restricted to $Z$ we see that $F/T_0^ d$ is an element of $J$. Thus $J' \subset J$.

Conversely, suppose that $f \in J$. If $f$ has total degree $d$ in $T_1/T_0, \ldots , T_ n/T_0$, then we can write $f = F/T_0^ d$ for some $F \in R[T_0, \ldots , T_ n]_ d$. Pick $i \in \{ 1, \ldots , n\}$. Then $Z \cap D_{+}(T_ i)$ is cut out by some ideal $J_ i \subset R[T_0/T_ i, \ldots , T_ n/T_ i]$. Moreover,

$J \cdot R\left[ \frac{T_1}{T_0}, \ldots , \frac{T_ n}{T_0}, \frac{T_0}{T_ i}, \ldots , \frac{T_ n}{T_ i} \right] = J_ i \cdot R\left[ \frac{T_1}{T_0}, \ldots , \frac{T_ n}{T_0}, \frac{T_0}{T_ i}, \ldots , \frac{T_ n}{T_ i} \right]$

The left hand side is the localization of $J$ with respect to the element $T_ i/T_0$ and the right hand side is the localization of $J_ i$ with respect to the element $T_0/T_ i$. It follows that $T_0^{d_ i}F/T_ i^{d + d_ i}$ is an element of $J_ i$ for some $d_ i$ sufficiently large. This proves that $T_0^{\max (d_ i)}F$ is an element of $I$, because its restriction to each standard affine open $D_{+}(T_ i)$ vanishes on the closed subscheme $Z \cap D_{+}(T_ i)$. Hence $f \in J'$ and we conclude $J \subset J'$ as desired. $\square$

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