Lemma 32.4.14. In Situation 32.4.5 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 32.4.13. To show that S_ i is separated for large i it now suffices to show that
is surjective for large i (Schemes, Lemma 26.21.7).
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 26.21.2) hence locally of finite type (Morphisms, Lemmas 29.15.2 and 29.15.5). 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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Comment #4527 by awllower on
Comment #4743 by Johan on