Lemma 31.31.1. Let $S$ be a scheme. Let $\mathcal{A}$ be a quasi-coherent graded $\mathcal{O}_ S$-algebra. Let $p : X = \underline{\text{Proj}}_ S(\mathcal{A}) \to S$ be the relative Proj of $\mathcal{A}$. Let $i : Z \to X$ be a closed subscheme. Denote $\mathcal{I} \subset \mathcal{A}$ the kernel of the canonical map

$\mathcal{A} \longrightarrow \bigoplus \nolimits _{d \geq 0} p_*\left((i_*\mathcal{O}_ Z)(d)\right).$

If $p$ is quasi-compact, then there is an isomorphism $Z = \underline{\text{Proj}}_ S(\mathcal{A}/\mathcal{I})$.

Proof. The morphism $p$ is separated by Constructions, Lemma 27.16.9. As $p$ is quasi-compact, $p_*$ transforms quasi-coherent modules into quasi-coherent modules, see Schemes, Lemma 26.24.1. Hence $\mathcal{I}$ is a quasi-coherent $\mathcal{O}_ S$-module. In particular, $\mathcal{B} = \mathcal{A}/\mathcal{I}$ is a quasi-coherent graded $\mathcal{O}_ S$-algebra. The functoriality morphism $Z' = \underline{\text{Proj}}_ S(\mathcal{B}) \to \underline{\text{Proj}}_ S(\mathcal{A})$ is everywhere defined and a closed immersion, see Constructions, Lemma 27.18.3. Hence it suffices to prove $Z = Z'$ as closed subschemes of $X$.

Having said this, the question is local on the base and we may assume that $S = \mathop{\mathrm{Spec}}(R)$ and that $X = \text{Proj}(A)$ for some graded $R$-algebra $A$. Assume $\mathcal{I} = \widetilde{I}$ for $I \subset A$ a graded ideal. By Constructions, Lemma 27.8.9 there exist $f_0, \ldots , f_ n \in A_{+}$ such that $A_{+} \subset \sqrt{(f_0, \ldots , f_ n)}$ in other words $X = \bigcup D_{+}(f_ i)$. Therefore, it suffices to check that $Z \cap D_{+}(f_ i) = Z' \cap D_{+}(f_ i)$ for each $i$. By renumbering we may assume $i = 0$. Say $Z \cap D_{+}(f_0)$, resp. $Z' \cap D_{+}(f_0)$ is cut out by the ideal $J$, resp. $J'$ of $A_{(f_0)}$.

The inclusion $J' \subset J$. Let $d$ be the least common multiple of $\deg (f_0), \ldots , \deg (f_ n)$. Note that each of the twists $\mathcal{O}_ X(nd)$ is invertible, trivialized by $f_ i^{nd/\deg (f_ i)}$ over $D_{+}(f_ i)$, and that for any quasi-coherent module $\mathcal{F}$ on $X$ the multiplication maps $\mathcal{O}_ X(nd) \otimes _{\mathcal{O}_ X} \mathcal{F}(m) \to \mathcal{F}(nd + m)$ are isomorphisms, see Constructions, Lemma 27.10.2. Observe that $J'$ is the ideal generated by the elements $g/f_0^ e$ where $g \in I$ is homogeneous of degree $e\deg (f_0)$ (see proof of Constructions, Lemma 27.11.3). Of course, by replacing $g$ by $f_0^ lg$ for suitable $l$ we may always assume that $d | e$. Then, since $g$ vanishes as a section of $\mathcal{O}_ X(e\deg (f_0))$ restricted to $Z$ we see that $g/f_0^ d$ is an element of $J$. Thus $J' \subset J$.

Conversely, suppose that $g/f_0^ e \in J$. Again we may assume $d | e$. Pick $i \in \{ 1, \ldots , n\}$. Then $Z \cap D_{+}(f_ i)$ is cut out by some ideal $J_ i \subset A_{(f_ i)}$. Moreover,

$J \cdot A_{(f_0f_ i)} = J_ i \cdot A_{(f_0f_ i)}.$

The right hand side is the localization of $J_ i$ with respect to $f_0^{\deg (f_ i)}/f_ i^{\deg (f_0)}$. It follows that

$f_0^{e_ i}g/f_ i^{(e_ i + e)\deg (f_0)/\deg (f_ i)} \in J_ i$

for some $e_ i \gg 0$ sufficiently divisible. This proves that $f_0^{\max (e_ i)}g$ is an element of $I$, because its restriction to each affine open $D_{+}(f_ i)$ vanishes on the closed subscheme $Z \cap D_{+}(f_ i)$. Hence $g/f_0^ e \in J'$ and we conclude $J \subset J'$ as desired. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).