## 71.16 Closed subspaces of relative proj

Some auxiliary lemmas about closed subspaces of relative proj. This section is the analogue of Divisors, Section 31.31.

Lemma 71.16.1. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $\mathcal{A}$ be a quasi-coherent graded $\mathcal{O}_ X$-algebra. Let $\pi : P = \underline{\text{Proj}}_ X(\mathcal{A}) \to X$ be the relative Proj of $\mathcal{A}$. Let $i : Z \to P$ be a closed subspace. Denote $\mathcal{I} \subset \mathcal{A}$ the kernel of the canonical map

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

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

Proof. The morphism $\pi$ is separated by Lemma 71.11.6. As $\pi$ is quasi-compact, $\pi _*$ transforms quasi-coherent modules into quasi-coherent modules, see Morphisms of Spaces, Lemma 67.11.2. Hence $\mathcal{I}$ is a quasi-coherent $\mathcal{O}_ X$-module. In particular, $\mathcal{B} = \mathcal{A}/\mathcal{I}$ is a quasi-coherent graded $\mathcal{O}_ X$-algebra. The functoriality morphism $Z' = \underline{\text{Proj}}_ X(\mathcal{B}) \to \underline{\text{Proj}}_ X(\mathcal{A})$ is everywhere defined and a closed immersion, see Lemma 71.12.3. Hence it suffices to prove $Z = Z'$ as closed subspaces of $P$.

Having said this, the question is étale local on the base and we reduce to the case of schemes (Divisors, Lemma 31.31.1) by étale localization. $\square$

In case the closed subspace is locally cut out by finitely many equations we can define it by a finite type ideal sheaf of $\mathcal{A}$.

Lemma 71.16.2. Let $S$ be a scheme. Let $X$ be a quasi-compact and quasi-separated algebraic space over $S$. Let $\mathcal{A}$ be a quasi-coherent graded $\mathcal{O}_ X$-algebra. Let $\pi : P = \underline{\text{Proj}}_ X(\mathcal{A}) \to X$ be the relative Proj of $\mathcal{A}$. Let $i : Z \to P$ be a closed subscheme. If $\pi$ is quasi-compact and $i$ of finite presentation, then there exists a $d > 0$ and a quasi-coherent finite type $\mathcal{O}_ X$-submodule $\mathcal{F} \subset \mathcal{A}_ d$ such that $Z = \underline{\text{Proj}}_ X(\mathcal{A}/\mathcal{F}\mathcal{A})$.

Proof. The reader can redo the arguments used in the case of schemes. However, we will show the lemma follows from the case of schemes by a trick. Let $\mathcal{I} \subset \mathcal{A}$ be the quasi-coherent graded ideal cutting out $Z$ of Lemma 71.16.1. Choose an affine scheme $U$ and a surjective étale morphism $U \to X$, see Properties of Spaces, Lemma 66.6.3. By the case of schemes (Divisors, Lemma 31.31.4) there exists a $d > 0$ and a quasi-coherent finite type $\mathcal{O}_ U$-submodule $\mathcal{F}' \subset \mathcal{I}_ d|_ U \subset \mathcal{A}_ d|_ U$ such that $Z \times _ X U$ is equal to $\underline{\text{Proj}}_ U(\mathcal{A}|_ U/\mathcal{F}'\mathcal{A}|_ U)$. By Limits of Spaces, Lemma 70.9.2 we can find a finite type quasi-coherent submodule $\mathcal{F} \subset \mathcal{I}_ d$ such that $\mathcal{F}' \subset \mathcal{F}|_ U$. Let $Z' = \underline{\text{Proj}}_ X(\mathcal{A}/\mathcal{F}\mathcal{A})$. Then $Z' \to P$ is a closed immersion (Lemma 71.12.5) and $Z \subset Z'$ as $\mathcal{F}\mathcal{A} \subset \mathcal{I}$. On the other hand, $Z' \times _ X U \subset Z \times _ X U$ by our choice of $\mathcal{F}$. Thus $Z = Z'$ as desired. $\square$

Lemma 71.16.3. Let $S$ be a scheme. Let $X$ be a quasi-compact and quasi-separated algebraic space over $S$. Let $\mathcal{A}$ be a quasi-coherent graded $\mathcal{O}_ X$-algebra. Let $\pi : P = \underline{\text{Proj}}_ X(\mathcal{A}) \to X$ be the relative Proj of $\mathcal{A}$. Let $i : Z \to X$ be a closed subspace. Let $U \subset X$ be an open. Assume that

1. $\pi$ is quasi-compact,

2. $i$ of finite presentation,

3. $|U| \cap |\pi |(|i|(|Z|)) = \emptyset$,

4. $U$ is quasi-compact,

5. $\mathcal{A}_ n$ is a finite type $\mathcal{O}_ X$-module for all $n$.

Then there exists a $d > 0$ and a quasi-coherent finite type $\mathcal{O}_ X$-submodule $\mathcal{F} \subset \mathcal{A}_ d$ with (a) $Z = \underline{\text{Proj}}_ X(\mathcal{A}/\mathcal{F}\mathcal{A})$ and (b) the support of $\mathcal{A}_ d/\mathcal{F}$ is disjoint from $U$.

Proof. We use the same trick as in the proof of Lemma 71.16.2 to reduce to the case of schemes. Let $\mathcal{I} \subset \mathcal{A}$ be the quasi-coherent graded ideal cutting out $Z$ of Lemma 71.16.1. Choose an affine scheme $W$ and a surjective étale morphism $W \to X$, see Properties of Spaces, Lemma 66.6.3. By the case of schemes (Divisors, Lemma 31.31.5) there exists a $d > 0$ and a quasi-coherent finite type $\mathcal{O}_ W$-submodule $\mathcal{F}' \subset \mathcal{I}_ d|_ W \subset \mathcal{A}_ d|_ W$ such that (a) $Z \times _ X W$ is equal to $\underline{\text{Proj}}_ W(\mathcal{A}|_ W/\mathcal{F}'\mathcal{A}|_ W)$ and (b) the support of $\mathcal{A}_ d|_ W/\mathcal{F}'$ is disjoint from $U \times _ X W$. By Limits of Spaces, Lemma 70.9.2 we can find a finite type quasi-coherent submodule $\mathcal{F} \subset \mathcal{I}_ d$ such that $\mathcal{F}' \subset \mathcal{F}|_ W$. Let $Z' = \underline{\text{Proj}}_ X(\mathcal{A}/\mathcal{F}\mathcal{A})$. Then $Z' \to P$ is a closed immersion (Lemma 71.12.5) and $Z \subset Z'$ as $\mathcal{F}\mathcal{A} \subset \mathcal{I}$. On the other hand, $Z' \times _ X W \subset Z \times _ X W$ by our choice of $\mathcal{F}$. Thus $Z = Z'$. Finally, we see that $\mathcal{A}_ d/\mathcal{F}$ is supported on $X \setminus U$ as $\mathcal{A}_ d|_ W/\mathcal{F}|_ W$ is a quotient of $\mathcal{A}_ d|_ W/\mathcal{F}'$ which is supported on $W \setminus U \times _ X W$. Thus the lemma follows. $\square$

Lemma 71.16.4. Let $S$ be a scheme and let $X$ be an algebraic space over $S$. Let $\mathcal{E}$ be a quasi-coherent $\mathcal{O}_ X$-module. There is a bijection

$\left\{ \begin{matrix} \text{sections }\sigma \text{ of the } \\ \text{morphism } \mathbf{P}(\mathcal{E}) \to X \end{matrix} \right\} \leftrightarrow \left\{ \begin{matrix} \text{surjections }\mathcal{E} \to \mathcal{L}\text{ where} \\ \mathcal{L}\text{ is an invertible }\mathcal{O}_ X\text{-module} \end{matrix} \right\}$

In this case $\sigma$ is a closed immersion and there is a canonical isomorphism

$\mathop{\mathrm{Ker}}(\mathcal{E} \to \mathcal{L}) \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes -1} \longrightarrow \mathcal{C}_{\sigma (X)/\mathbf{P}(\mathcal{E})}$

Both the bijection and isomorphism are compatible with base change.

Proof. Because the constructions are compatible with base change, it suffices to check the statement étale locally on $X$. Thus we may assume $X$ is a scheme and the result is Divisors, Lemma 31.31.6. $\square$

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