Lemma 31.4.14. In Situation 31.4.5 if $S$ is separated, then for some $i_0 \in I$ the schemes $S_ i$ for $i \geq i_0$ are separated.

**Proof.**
Choose a finite affine open covering $S_0 = U_{0, 1} \cup \ldots \cup U_{0, m}$. Set $U_{i, j} \subset S_ i$ and $U_ j \subset S$ equal to the inverse image of $U_{0, j}$. Note that $U_{i, j}$ and $U_ j$ are affine. As $S$ is separated the intersections $U_{j_1} \cap U_{j_2}$ are affine. Since $U_{j_1} \cap U_{j_2} = \mathop{\mathrm{lim}}\nolimits _{i \geq 0} U_{i, j_1} \cap U_{i, j_2}$ we see that $U_{i, j_1} \cap U_{i, j_2}$ is affine for large $i$ by Lemma 31.4.13. To show that $S_ i$ is separated for large $i$ it now suffices to show that

is surjective for large $i$ (Schemes, Lemma 25.21.7).

To get rid of the annoying indices, assume we have affine opens $U, V \subset S_0$ such that $U \cap V$ is affine too. Let $U_ i, V_ i \subset S_ i$, resp. $U, V \subset S$ be the inverse images. We have to show that $\mathcal{O}(U_ i) \otimes \mathcal{O}(V_ i) \to \mathcal{O}(U_ i \cap V_ i)$ is surjective for $i$ large enough and we know that $\mathcal{O}(U_) \otimes \mathcal{O}(V) \to \mathcal{O}(U \cap V)$ is surjective. Note that $\mathcal{O}(U_0) \otimes \mathcal{O}(V_0) \to \mathcal{O}(U_0 \cap V_0)$ is of finite type, as the diagonal morphism $S_ i \to S_ i \times S_ i$ is an immersion (Schemes, Lemma 25.21.2) hence locally of finite type (Morphisms, Lemmas 28.14.2 and 28.14.5). Thus we can choose elements $f_{0, 1}, \ldots , f_{0, n} \in \mathcal{O}(U_0 \cap V_0)$ which generate $\mathcal{O}(U_0 \cap V_0)$ over $\mathcal{O}(U_0) \otimes \mathcal{O}(V_0)$. Observe that for $i \geq 0$ the diagram of schemes

is cartesian. Thus we see that the images $f_{i, 1}, \ldots , f_{i, n} \in \mathcal{O}(U_ i \cap V_ i)$ generate $\mathcal{O}(U_ i \cap V_ i)$ over $\mathcal{O}(U_ i) \otimes \mathcal{O}(V_0)$ and a fortiori over $\mathcal{O}(U_ i) \otimes \mathcal{O}(V_ i)$. By assumption the images $f_1, \ldots , f_ n \in \mathcal{O}(U \otimes V)$ are in the image of the map $\mathcal{O}(U) \otimes \mathcal{O}(V) \to \mathcal{O}(U \cap V)$. Since $\mathcal{O}(U) \otimes \mathcal{O}(V) = \mathop{\mathrm{colim}}\nolimits \mathcal{O}(U_ i) \otimes \mathcal{O}(V_ i)$ we see that they are in the image of the map at some finite level and the lemma is proved. $\square$

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## Comments (1)

Comment #4527 by awllower on