The Stacks project

Proof. The implications (3) $\Rightarrow $ (1) and (1) $\Rightarrow $ (2) are immediate. Thus it suffices to prove that (2) implies (3). We urge the reader to find their own proof of this fact. Let $Z'$ and $Z''$ be closed subscheme structures on $Z$ such that $Z' \to S$ is proper. We have to show that $Z'' \to S$ is proper. Let $Z''' = Z' \cup Z''$ be the scheme theoretic union, see Morphisms, Definition 29.4.4. Then $Z'''$ is another closed subscheme structure on $Z$. This follows for example from the description of scheme theoretic unions in Morphisms, Lemma 29.4.6. Since $Z'' \to Z'''$ is a closed immersion it suffices to prove that $Z''' \to S$ is proper (see Morphisms, Lemmas 29.41.6 and 29.41.4). The morphism $Z' \to Z'''$ is a bijective closed immersion and in particular surjective and universally closed. Then the fact that $Z' \to S$ is separated implies that $Z''' \to S$ is separated, see Morphisms, Lemma 29.41.11. Moreover $Z''' \to S$ is locally of finite type as $X \to S$ is locally of finite type (Morphisms, Lemmas 29.15.5 and 29.15.3). Since $Z' \to S$ is quasi-compact and $Z' \to Z'''$ is a homeomorphism we see that $Z''' \to S$ is quasi-compact. Finally, since $Z' \to S$ is universally closed, we see that the same thing is true for $Z''' \to S$ by Morphisms, Lemma 29.41.9. This finishes the proof. $\square$

Proof. Omitted. $\square$

Proof. Observe that the statement makes sense as $f'$ is locally of finite type by Morphisms, Lemma 29.15.4. Endow $Z$ with the reduced induced closed subscheme structure. Denote $Z' = (g')^{-1}(Z)$ the scheme theoretic inverse image (Schemes, Definition 26.17.7). Then $Z' = X' \times _ X Z = (S' \times _ S X) \times _ X Z = S' \times _ S Z$ is proper over $S'$ as a base change of $Z$ over $S$ (Morphisms, Lemma 29.41.5). $\square$

Proof. Proof of (1). Assume $Y$ is separated over $S$ and $Z \subset X$ is a closed subset proper over $S$. Endow $Z$ with the reduced induced closed subscheme structure and apply Morphisms, Lemma 29.41.10 to $Z \to Y$ over $S$ to conclude.

Proof of (2). Assume $f$ is universally closed and $Z \subset X$ is a closed subset proper over $S$. Endow $Z$ and $Z' = f(Z)$ with their reduced induced closed subscheme structures. We obtain an induced morphism $Z \to Z'$. Denote $Z'' = f^{-1}(Z')$ the scheme theoretic inverse image (Schemes, Definition 26.17.7). Then $Z'' \to Z'$ is universally closed as a base change of $f$ (Morphisms, Lemma 29.41.5). Hence $Z \to Z'$ is universally closed as a composition of the closed immersion $Z \to Z''$ and $Z'' \to Z'$ (Morphisms, Lemmas 29.41.6 and 29.41.4). We conclude that $Z' \to S$ is separated by Morphisms, Lemma 29.41.11. Since $Z \to S$ is quasi-compact and $Z \to Z'$ is surjective we see that $Z' \to S$ is quasi-compact. Since $Z' \to S$ is the composition of $Z' \to Y$ and $Y \to S$ we see that $Z' \to S$ is locally of finite type (Morphisms, Lemmas 29.15.5 and 29.15.3). Finally, since $Z \to S$ is universally closed, we see that the same thing is true for $Z' \to S$ by Morphisms, Lemma 29.41.9. This finishes the proof.

Proof of (3). Assume $f$ is proper and $Z \subset Y$ is a closed subset proper over $S$. Endow $Z$ with the reduced induced closed subscheme structure. Denote $Z' = f^{-1}(Z)$ the scheme theoretic inverse image (Schemes, Definition 26.17.7). Then $Z' \to Z$ is proper as a base change of $f$ (Morphisms, Lemma 29.41.5). Whence $Z' \to S$ is proper as the composition of $Z' \to Z$ and $Z \to S$ (Morphisms, Lemma 29.41.4). This finishes the proof. $\square$

Proof. Endow $Z_ i$ with their reduced induced closed subscheme structures. The morphism

is finite by Morphisms, Lemmas 29.44.12 and 29.44.13. As finite morphisms are universally closed (Morphisms, Lemma 29.44.11) and since $Z_1 \amalg \ldots \amalg Z_ n$ is proper over $S$ we conclude by Lemma 30.26.5 part (2) that the image $Z_1 \cup \ldots \cup Z_ n$ is proper over $S$. $\square$

Proof. The support $\text{Supp}(\mathcal{F})$ of $\mathcal{F}$ is a closed subset of $X$, see Morphisms, Lemma 29.5.3. Hence we can apply Definition 30.26.2. Since the scheme theoretic support of $\mathcal{F}$ is a closed subscheme whose underlying closed subset is $\text{Supp}(\mathcal{F})$ we see that (1) and (2) are equivalent by Definition 30.26.2. It is clear that (2) implies (3). Conversely, if (3) is true, then $\text{Supp}(\mathcal{F}) \subset Z$ (an inclusion of closed subsets of $X$) and hence $\text{Supp}(\mathcal{F})$ is proper over $S$ for example by Lemma 30.26.3. $\square$

Proof. Observe that the statement makes sense because $(g')*\mathcal{F}$ is of finite type by Modules, Lemma 17.9.2. We have $\text{Supp}((g')^*\mathcal{F}) = (g')^{-1}(\text{Supp}(\mathcal{F}))$ by Morphisms, Lemma 29.5.3. Thus the lemma follows from Lemma 30.26.4. $\square$

Proof. Proof of (1). Let $Z$, $Z'$ be the support of $\mathcal{F}$ and $\mathcal{G}$. Then all the sheaves mentioned in (1) have support contained in $Z \cup Z'$. Thus the assertion itself is clear from Lemmas 30.26.3 and 30.26.6 provided we check that these sheaves are finite type and quasi-coherent. For quasi-coherence we refer the reader to Schemes, Section 26.24. For “finite type” we suggest the reader take a look at Modules, Section 17.9.

Proof of (2). The proof is the same as the proof of (1). Note that the assertions make sense as $X$ is locally Noetherian by Morphisms, Lemma 29.15.6 and by the description of the category of coherent modules in Section 30.9. $\square$

Proof. By Lemma 30.26.7 there exists a closed immersion $i : Z \to X$ and a finite type, quasi-coherent $\mathcal{O}_ Z$-module $\mathcal{G}$ such that (a) $g = f \circ i : Z \to S$ is proper, and (b) $i_*\mathcal{G} = \mathcal{F}$. We see that $R^ pg_*\mathcal{G}$ is coherent on $S$ by Proposition 30.19.1. On the other hand, $R^ qi_*\mathcal{G} = 0$ for $q > 0$ (Lemma 30.9.9). By Cohomology, Lemma 20.13.8 we get $R^ pf_*\mathcal{F} = R^ pg_*\mathcal{G}$ which concludes the proof. $\square$

Proof. We will use the criterion of Homology, Lemma 12.10.2. Moreover, we will use that if $0 \to (\mathcal{G}_ n) \to (\mathcal{F}_ n) \to (\mathcal{H}_ n) \to 0$ is a short exact sequence of $\textit{Coh}(X, \mathcal{I})$, then (a) $\mathcal{G}_ n \to \mathcal{F}_ n \to \mathcal{H}_ n \to 0$ is exact for all $n \geq 1$ and (b) $\mathcal{G}_ n$ is a quotient of $\mathop{\mathrm{Ker}}(\mathcal{F}_ m \to \mathcal{H}_ m)$ for some $m \geq n$. See proof of Lemma 30.23.2.

Proof of (1). Let $(\mathcal{F}_ n)$ be an object of $\textit{Coh}(X, \mathcal{I})$. Then $\text{Supp}(\mathcal{F}_ n) = \text{Supp}(\mathcal{F}_1)$ for all $n \geq 1$. Hence by remarks (a) and (b) above we see that for any short exact sequence $0 \to (\mathcal{G}_ n) \to (\mathcal{F}_ n) \to (\mathcal{H}_ n) \to 0$ of $\textit{Coh}(X, \mathcal{I})$ we have $\text{Supp}(\mathcal{G}_1) \cup \text{Supp}(\mathcal{H}_1) = \text{Supp}(\mathcal{F}_1)$. This proves that the category defined in (1) is a Serre subcategory of $\textit{Coh}(X, \mathcal{I})$.

Proof of (2). Here we argue the same way. Let $0 \to (\mathcal{G}_ n) \to (\mathcal{F}_ n) \to (\mathcal{H}_ n) \to 0$ be a short exact sequence of $\textit{Coh}(X, \mathcal{I})$. If $Z \subset X$ is a closed subscheme and $\mathcal{I}_ Z$ annihilates $\mathcal{F}_ n$ for all $n$, then $\mathcal{I}_ Z$ annihilates $\mathcal{G}_ n$ and $\mathcal{H}_ n$ for all $n$ by (a) and (b) above. Hence if $Z \to S$ is proper, then we conclude that the category defined in (2) is closed under taking sub and quotient objects inside of $\textit{Coh}(X, \mathcal{I})$. Finally, suppose that $Z \subset X$ and $Y \subset X$ are closed subschemes proper over $S$ such that $\mathcal{I}_ Z \mathcal{G}_ n = 0$ and $\mathcal{I}_ Y \mathcal{H}_ n = 0$ for all $n \geq 1$. Then it follows from (a) above that $\mathcal{I}_{Z \cup Y} = \mathcal{I}_ Z \cdot \mathcal{I}_ Y$ annihilates $\mathcal{F}_ n$ for all $n$. By Lemma 30.26.6 (and via Definition 30.26.2 which tells us we may choose an arbitrary scheme structure used on the union) we see that $Z \cup Y \to S$ is proper and the proof is complete. $\square$