The Stacks project

Proof. In case each of the schemes $S_ i$ is affine, and we consider only affine schemes of finite presentation over $S_ i$, resp. $S$ this lemma is equivalent to Algebra, Lemma 10.127.8. We claim that the affine case implies the lemma in general.

Let us prove (3). Suppose given an index $i \in I$, schemes $X_ i$, $Y_ i$ of finite presentation over $S_ i$ and a pair of morphisms $\varphi _ i, \psi _ i : X_ i \to Y_ i$. Assume that the base changes are equal: $\varphi _{i, S} = \psi _{i, S}$. We will use the notation $X_{i'} = X_{i, S_{i'}}$ and $Y_{i'} = Y_{i, S_{i'}}$ for $i' \geq i$. We also set $X = X_{i, S}$ and $Y = Y_{i, S}$. Note that according to Lemma 32.2.3 we have $X = \mathop{\mathrm{lim}}\nolimits _{i' \geq i} X_{i'}$ and similarly for $Y$. Additionally we denote $\varphi _{i'}$ and $\psi _{i'}$ (resp. $\varphi$ and $\psi$) the base change of $\varphi _ i$ and $\psi _ i$ to $S_{i'}$ (resp. $S$). So our assumption means that $\varphi = \psi$. Since $Y_ i$ and $X_ i$ are of finite presentation over $S_ i$, and since $S_ i$ is quasi-compact and quasi-separated, also $X_ i$ and $Y_ i$ are quasi-compact and quasi-separated (see Morphisms, Lemma 29.21.10). Hence we may choose a finite affine open covering $Y_ i = \bigcup V_{j, i}$ such that each $V_{j, i}$ maps into an affine open of $S$. As above, denote $V_{j, i'}$ the inverse image of $V_{j, i}$ in $Y_{i'}$ and $V_ j$ the inverse image in $Y$. The immersions $V_{j, i'} \to Y_{i'}$ are quasi-compact, and the inverse images $U_{j, i'} = \varphi _ i^{-1}(V_{j, i'})$ and $U_{j, i'}' = \psi _ i^{-1}(V_{j, i'})$ are quasi-compact opens of $X_{i'}$. By assumption the inverse images of $V_ j$ under $\varphi$ and $\psi$ in $X$ are equal. Hence by Lemma 32.4.11 there exists an index $i' \geq i$ such that of $U_{j, i'} = U_{j, i'}'$ in $X_{i'}$. Choose an finite affine open covering $U_{j, i'} = U_{j, i'}' = \bigcup W_{j, k, i'}$ which induce coverings $U_{j, i''} = U_{j, i''}' = \bigcup W_{j, k, i''}$ for all $i'' \geq i'$. By the affine case there exists an index $i''$ such that $\varphi _{i''}|_{W_{j, k, i''}} = \psi _{i''}|_{W_{j, k, i''}}$ for all $j, k$. Then $i''$ is an index such that $\varphi _{i''} = \psi _{i''}$ and (3) is proved.

Let us prove (2). Suppose given an index $i \in I$, schemes $X_ i$, $Y_ i$ of finite presentation over $S_ i$ and a morphism $\varphi : X_{i, S} \to Y_{i, S}$. We will use the notation $X_{i'} = X_{i, S_{i'}}$ and $Y_{i'} = Y_{i, S_{i'}}$ for $i' \geq i$. We also set $X = X_{i, S}$ and $Y = Y_{i, S}$. Note that according to Lemma 32.2.3 we have $X = \mathop{\mathrm{lim}}\nolimits _{i' \geq i} X_{i'}$ and similarly for $Y$. Since $Y_ i$ and $X_ i$ are of finite presentation over $S_ i$, and since $S_ i$ is quasi-compact and quasi-separated, also $X_ i$ and $Y_ i$ are quasi-compact and quasi-separated (see Morphisms, Lemma 29.21.10). Hence we may choose a finite affine open covering $Y_ i = \bigcup V_{j, i}$ such that each $V_{j, i}$ maps into an affine open of $S$. As above, denote $V_{j, i'}$ the inverse image of $V_{j, i}$ in $Y_{i'}$ and $V_ j$ the inverse image in $Y$. The immersions $V_ j \to Y$ are quasi-compact, and the inverse images $U_ j = \varphi ^{-1}(V_ j)$ are quasi-compact opens of $X$. Hence by Lemma 32.4.11 there exists an index $i' \geq i$ and quasi-compact opens $U_{j, i'}$ of $X_{i'}$ whose inverse image in $X$ is $U_ j$. Choose an finite affine open covering $U_{j, i'} = \bigcup W_{j, k, i'}$ which induce affine open coverings $U_{j, i''} = \bigcup W_{j, k, i''}$ for all $i'' \geq i'$ and an affine open covering $U_ j = \bigcup W_{j, k}$. By the affine case there exists an index $i''$ and morphisms $\varphi _{j, k, i''} : W_{j, k, i''} \to V_{j, i''}$ such that $\varphi |_{W_{j, k}} = \varphi _{j, k, i'', S}$ for all $j, k$. By part (3) proved above, there is a further index $i''' \geq i''$ such that

for all $j_1, j_2, k_1, k_2$. Then $i'''$ is an index such that there exists a morphism $\varphi _{i'''} : X_{i'''} \to Y_{i'''}$ whose base change to $S$ gives $\varphi$. Hence (2) holds.

Let us prove (1). Suppose given a scheme $X$ of finite presentation over $S$. Since $X$ is of finite presentation over $S$, and since $S$ is quasi-compact and quasi-separated, also $X$ is quasi-compact and quasi-separated (see Morphisms, Lemma 29.21.10). Choose a finite affine open covering $X = \bigcup U_ j$ such that each $U_ j$ maps into an affine open $V_ j \subset S$. Denote $U_{j_1j_2} = U_{j_1} \cap U_{j_2}$ and $U_{j_1j_2j_3} = U_{j_1} \cap U_{j_2} \cap U_{j_3}$. By Lemmas 32.4.11 and 32.4.13 we can find an index $i_1$ and affine opens $V_{j, i_1} \subset S_{i_1}$ such that each $V_ j$ is the inverse of this in $S$. Let $V_{j, i}$ be the inverse image of $V_{j, i_1}$ in $S_ i$ for $i \geq i_1$. By the affine case we may find an index $i_2 \geq i_1$ and affine schemes $U_{j, i_2} \to V_{j, i_2}$ such that $U_ j = S \times _{S_{i_2}} U_{j, i_2}$ is the base change. Denote $U_{j, i} = S_ i \times _{S_{i_2}} U_{j, i_2}$ for $i \geq i_2$. By Lemma 32.4.11 there exists an index $i_3 \geq i_2$ and open subschemes $W_{j_1, j_2, i_3} \subset U_{j_1, i_3}$ whose base change to $S$ is equal to $U_{j_1j_2}$. Denote $W_{j_1, j_2, i} = S_ i \times _{S_{i_3}} W_{j_1, j_2, i_3}$ for $i \geq i_3$. By part (2) shown above there exists an index $i_4 \geq i_3$ and morphisms $\varphi _{j_1, j_2, i_4} : W_{j_1, j_2, i_4} \to W_{j_2, j_1, i_4}$ whose base change to $S$ gives the identity morphism $U_{j_1j_2} = U_{j_2j_1}$ for all $j_1, j_2$. For all $i \geq i_4$ denote $\varphi _{j_1, j_2, i} = \text{id}_ S \times \varphi _{j_1, j_2, i_4}$ the base change. We claim that for some $i_5 \geq i_4$ the system $((U_{j, i_5})_ j, (W_{j_1, j_2, i_5})_{j_1, j_2}, (\varphi _{j_1, j_2, i_5})_{j_1, j_2})$ forms a glueing datum as in Schemes, Section 26.14. In order to see this we have to verify that for $i$ large enough we have

and that for large enough $i$ the cocycle condition holds. The first condition follows from Lemma 32.4.11 and the fact that $U_{j_2j_1j_3} = U_{j_1j_2j_3}$. The second from part (1) of the lemma proved above and the fact that the cocycle condition holds for the maps $\text{id} : U_{j_1j_2} \to U_{j_2j_1}$. Ok, so now we can use Schemes, Lemma 26.14.2 to glue the system $((U_{j, i_5})_ j, (W_{j_1, j_2, i_5})_{j_1, j_2}, (\varphi _{j_1, j_2, i_5})_{j_1, j_2})$ to get a scheme $X_{i_5} \to S_{i_5}$. By construction the base change of $X_{i_5}$ to $S$ is formed by glueing the open affines $U_ j$ along the opens $U_{j_1} \leftarrow U_{j_1j_2} \rightarrow U_{j_2}$. Hence $S \times _{S_{i_5}} X_{i_5} \cong X$ as desired. $\square$

Proof. We sketch two proofs, but we omit the details.

First proof. If $S$ and $S_ i$ are affine schemes, then this lemma is equivalent to Algebra, Lemma 10.127.6. In the general case, use Zariski glueing to deduce it from the affine case.

Second proof. We use

1. there is an equivalence of categories between quasi-coherent $\mathcal{O}_ S$-modules and vector bundles over $S$, see Constructions, Section 27.6, and

2. a vector bundle $\mathbf{V}(\mathcal{F}) \to S$ is of finite presentation over $S$ if and only if $\mathcal{F}$ is an $\mathcal{O}_ S$-module of finite presentation.

Having said this, we can use Lemma 32.10.1 to show that the category of vector bundles of finite presentation over $S$ is the colimit over $I$ of the categories of vector bundles over $S_ i$. $\square$

Proof. Let $\mathcal{E}$ be a finite locally free $\mathcal{O}_ S$-module. Since finite locally free modules are of finite presentation we can find an $i$ and an $\mathcal{O}_{S_ i}$-module $\mathcal{E}_ i$ of finite presentation such that $f_ i^*\mathcal{E}_ i \cong \mathcal{E}$, see Lemma 32.10.2. After increasing $i$ we may assume $\mathcal{E}_ i$ is a flat $\mathcal{O}_{S_ i}$-module, see Algebra, Lemma 10.168.1. (Using this lemma is not necessary, but it is convenient.) Then $\mathcal{E}_ i$ is finite locally free by Algebra, Lemma 10.78.2.

If $\mathcal{L}$ is an invertible $\mathcal{O}_ S$-module, then by the above we can find an $i$ and finite locally free $\mathcal{O}_{S_ i}$-modules $\mathcal{L}_ i$ and $\mathcal{N}_ i$ pulling back to $\mathcal{L}$ and $\mathcal{L}^{\otimes -1}$. After possible increasing $i$ we see that the map $\mathcal{L} \otimes _{\mathcal{O}_ X} \mathcal{L}^{\otimes -1} \to \mathcal{O}_ X$ descends to a map $\mathcal{L}_ i \otimes _{\mathcal{O}_{S_ i}} \mathcal{N}_ i \to \mathcal{O}_{S_ i}$. And after increasing $i$ further, we may assume it is an isomorphism. It follows that $\mathcal{L}_ i$ is an invertible module (Modules, Lemma 17.25.2) and the proof of (2) is complete.

Given $\mathcal{I}$ as in (3) we see that $\mathcal{O}_ S \to \mathcal{O}_ S/\mathcal{I}$ is a map of finitely presented $\mathcal{O}_ S$-modules. Hence by Lemma 32.10.2 this is the pullback of some map $\mathcal{O}_{S_ i} \to \mathcal{F}_ i$ of finitely presented $\mathcal{O}_{S_ i}$-modules. After increasing $i$ we may assume this map is surjective (details omitted; hint: use Algebra, Lemma 10.127.5 on affine open cover). Then the kernel of $\mathcal{O}_{S_ i} \to \mathcal{F}_ i$ is a finite type quasi-coherent ideal in $\mathcal{O}_{S_ i}$ whose pullback gives $\mathcal{I}$. $\square$

Proof. (This lemma is the analogue of Lemma 32.8.7 for modules.) For $i' \geq i$ denote $X_{i'} = S_{i'} \times _{S_ i} X_ i$, $\mathcal{F}_{i'} = (X_{i'} \to X_ i)^*\mathcal{F}_ i$ and similarly for $Y_{i'}$. Denote $\varphi _{i'}$ the base change of $\varphi _ i$ to $S_{i'}$. Also set $X = S \times _{S_ i} X_ i$, $Y =S \times _{S_ i} X_ i$, $\mathcal{F} = (X \to X_ i)^*\mathcal{F}_ i$ and $\varphi$ the base change of $\varphi _ i$ to $S$. Let $Y_ i = \bigcup _{j = 1, \ldots , m} V_{j, i}$ be a finite affine open covering such that each $V_{j, i}$ maps into some affine open of $S_ i$. For each $j = 1, \ldots m$ let $\varphi _ i^{-1}(V_{j, i}) = \bigcup _{k = 1, \ldots , m(j)} U_{k, j, i}$ be a finite affine open covering. For $i' \geq i$ we denote $V_{j, i'}$ the inverse image of $V_{j, i}$ in $Y_{i'}$ and $U_{k, j, i'}$ the inverse image of $U_{k, j, i}$ in $X_{i'}$. Similarly we have $U_{k, j} \subset X$ and $V_ j \subset Y$. Then $U_{k, j} = \mathop{\mathrm{lim}}\nolimits _{i' \geq i} U_{k, j, i'}$ and $V_ j = \mathop{\mathrm{lim}}\nolimits _{i' \geq i} V_ j$ (see Lemma 32.2.2). Since $X_{i'} = \bigcup _{k, j} U_{k, j, i'}$ is a finite open covering it suffices to prove the lemma for each of the morphisms $U_{k, j, i} \to V_{j, i}$ and the sheaf $\mathcal{F}_ i|_{U_{k, j, i}}$. Hence we see that the lemma reduces to the case that $X_ i$ and $Y_ i$ are affine and map into an affine open of $S_ i$, i.e., we may also assume that $S$ is affine.

In the affine case we reduce to the following algebra result. Suppose that $R = \mathop{\mathrm{colim}}\nolimits _{i \in I} R_ i$. For some $i \in I$ suppose given a map $A_ i \to B_ i$ of finitely presented $R_ i$-algebras. Let $N_ i$ be a finitely presented $B_ i$-module. Then, if $R \otimes _{R_ i} N_ i$ is flat over $R \otimes _{R_ i} A_ i$, then for some $i' \geq i$ the module $R_{i'} \otimes _{R_ i} N_ i$ is flat over $R_{i'} \otimes _{R_ i} A$. This is exactly the result proved in Algebra, Lemma 10.168.1 part (3). $\square$

Proof. Fully faithfulness. Suppose we have $i \in I$ and objects $X_ i$, $Y_ i$ of $\mathcal{C}_{S_ i}$. Denote $X = X_ i \times _{S_ i} S$ and $Y = Y_ i \times _{S_ i} S$. Suppose given a morphism $f : X \to Y$ over $S$. We can choose a finite affine open covering $Y_ i = V_{i, 1} \cup \ldots \cup V_{i, m}$ such that $V_{i, j} \to Y_ i \to S_ i$ maps into an affine open $W_{i, j}$ of $S_ i$. Denote $Y = V_1 \cup \ldots \cup V_ m$ the induced affine open covering of $Y$. Since $f : X \to Y$ is quasi-compact (Schemes, Lemma 26.21.14) after increasing $i$ we may assume that there is a finite open covering $X_ i = U_{i, 1} \cup \ldots \cup U_{i, m}$ by quasi-compact opens such that the inverse image of $U_{i, j}$ in $Y$ is $f^{-1}(V_ j)$, see Lemma 32.4.11. By Lemma 32.10.1 applied to $f|_{f^{-1}(V_ j)}$ over $W_ j$ we may assume, after increasing $i$, that there is a morphism $f_{i, j} : V_{i, j} \to U_{i, j}$ over $S$ whose base change to $S$ is $f|_{f^{-1}(V_ j)}$. Increasing $i$ more we may assume $f_{i, j}$ and $f_{i, j'}$ agree on the quasi-compact open $U_{i, j} \cap U_{i, j'}$. Then we can glue these morphisms to get the desired morphism $f_ i : X_ i \to Y_ i$. This morphism is unique (up to increasing $i$) because this is true for the morphisms $f_{i, j}$.

To show that the functor is essentially surjective we argue in exactly the same way. Namely, suppose that $X$ is an object of $\mathcal{C}_ S$. Pick $i \in I$. We can choose a finite affine open covering $X = U_1 \cup \ldots \cup U_ m$ such that $U_ j \to X \to S \to S_ i$ factors through an affine open $W_{i, j} \subset S_ i$. Set $W_ j = W_{i, j} \times _{S_ i} S$. This is an affine open of $S$. By Lemma 32.10.1, after increasing $i$, we may assume there exist $U_{i, j} \to W_{i, j}$ of finite presentation whose base change to $W_ j$ is $U_ j$. After increasing $i$ we may assume there exist quasi-compact opens $U_{i, j, j'} \subset U_{i, j}$ whose base changes to $S$ are equal to $U_ j \cap U_{j'}$. Claim: after increasing $i$ we may assume the image of the morphism $U_{i, j, j'} \to U_{i, j} \to W_{i, j}$ ends up in $W_{i, j} \cap W_{i, j'}$. Namely, because the complement of $W_{i, j} \cap W_{i, j'}$ is closed in the affine scheme $W_{i, j}$ it is affine. Since $U_ j \cap U_{j'} = \mathop{\mathrm{lim}}\nolimits U_{i, j, j'}$ does map into $W_{i, j} \cap W_{i, j'}$ we can apply Lemma 32.4.9 to get the claim. Thus we can view both

as schemes over $W_{i, j'}$ whose base changes to $W_{j'}$ recover $U_ j \cap U_{j'}$. Hence after increasing $i$, using Lemma 32.10.1, we may assume there are isomorphisms $U_{i, j, j'} \to U_{i, j', j}$ over $W_{i, j'}$ and hence over $S_ i$. Increasing $i$ further (details omitted) we may assume these isomorphisms satisfy the cocycle condition mentioned in Schemes, Section 26.14. Applying Schemes, Lemma 26.14.1 we obtain an object $X_ i$ of $\mathcal{C}_{S_ i}$ whose base change to $S$ is isomorphic to $X$; we omit some of the verifications. $\square$