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The Stacks project

Lemma 32.4.1. Let S = \mathop{\mathrm{lim}}\nolimits S_ i be the limit of a directed inverse system of schemes with affine transition morphisms (Lemma 32.2.2). Then S_{set} = \mathop{\mathrm{lim}}\nolimits _ i S_{i, set} where S_{set} indicates the underlying set of the scheme S.

Proof. Pick i \in I. Take U_ i \subset S_ i an affine open. Denote U_{i'} = f_{i'i}^{-1}(U_ i) and U = f_ i^{-1}(U_ i). Here f_{i'i} : S_{i'} \to S_ i is the transition morphism and f_ i : S \to S_ i is the projection. By Lemma 32.2.2 we have U = \mathop{\mathrm{lim}}\nolimits _{i' \geq i} U_ i. Suppose we can show that U_{set} = \mathop{\mathrm{lim}}\nolimits _{i' \geq i} U_{i', set}. Then the lemma follows by a simple argument using an affine covering of S_ i. Hence we may assume all S_ i and S affine. 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. Then

\mathop{\mathrm{Spec}}(A) = \mathop{\mathrm{lim}}\nolimits _ i \mathop{\mathrm{Spec}}(A_ i)

Namely, suppose that we are given primes \mathfrak p_ i \subset A_ i such that \mathfrak p_ i = \varphi _{ii'}^{-1}(\mathfrak p_{i'}) for all i' \geq i. Then we simply set

\mathfrak p = \{ x \in A \mid \exists i, x_ i \in \mathfrak p_ i \text{ with }\varphi _ i(x_ i) = x\}

It is clear that this is an ideal and has the property that \varphi _ i^{-1}(\mathfrak p) = \mathfrak p_ i. Then it follows easily that it is a prime ideal as well. \square


Comments (2)

Comment #6658 by Jonas Ehrhard on

Typo in the third sentence of the proof: it should be transition instead of transtion.


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