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

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

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. $\square$

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## Comments (2)

Comment #6658 by Jonas Ehrhard on

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