Proof.
Choose 0 \in I and an affine open covering S_0 = \bigcup _{j \in J} U_{0, j}. For i \geq 0 let U_{i, j} = f_{i, 0}^{-1}(U_{0, j}) and set U_ j = f_0^{-1}(U_{0, j}). Here f_{i'i} : S_{i'} \to S_ i is the transition morphism and f_ i : S \to S_ i is the projection. For j \in J the following are equivalent: (a) s \in U_ j, (b) s_0 \in U_{0, j}, (c) s_ i \in U_{i, j} for all i \geq 0. Let J' \subset J be the set of indices for which (a), (b), (c) are true. Then \overline{\{ s\} } = \bigcup _{j \in J'} (\overline{\{ s\} } \cap U_ j) and similarly for \overline{\{ s_ i\} } for i \geq 0. Note that \overline{\{ s\} } \cap U_ j is the closure of the set \{ s\} in the topological space U_ j. Similarly for \overline{\{ s_ i\} } \cap U_{i, j} for i \geq 0. Hence it suffices to prove the lemma in the case S and S_ i affine for all i. This reduces us to the algebra question considered in the next paragraph.
Suppose given a system of rings (A_ i, \varphi _{ii'}) over I. Set A = \mathop{\mathrm{colim}}\nolimits _ i A_ i with canonical maps \varphi _ i : A_ i \to A. Let \mathfrak p \subset A be a prime and set \mathfrak p_ i = \varphi _ i^{-1}(\mathfrak p). Then
V(\mathfrak p) = \mathop{\mathrm{lim}}\nolimits _ i V(\mathfrak p_ i)
This follows from Lemma 32.4.1 because A/\mathfrak p = \mathop{\mathrm{colim}}\nolimits A_ i/\mathfrak p_ i. This equality of rings also shows the final statement about reduced induced scheme structures holds true. The equality \kappa (\mathfrak p) = \mathop{\mathrm{colim}}\nolimits \kappa (\mathfrak p_ i) follows from the statement as well.
\square
Comments (2)
Comment #6660 by Jonas Ehrhard on
Comment #6878 by Johan on