Lemma 32.4.3. 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). If all the schemes $S_ i$ are nonempty and quasi-compact, then the limit $S = \mathop{\mathrm{lim}}\nolimits _ i S_ i$ is nonempty.

Proof. Choose $0 \in I$. Note that $I$ is nonempty as the limit is directed. Choose an affine open covering $S_0 = \bigcup _{j = 1, \ldots , m} U_ j$. Since $I$ is directed there exists a $j \in \{ 1, \ldots , m\}$ such that $f_{i0}^{-1}(U_ j) \not= \emptyset$ for all $i \geq 0$. Hence $\mathop{\mathrm{lim}}\nolimits _{i \geq 0} f_{i0}^{-1}(U_ j)$ is not empty since a directed colimit of nonzero rings is nonzero (because $1 \not= 0$). As $\mathop{\mathrm{lim}}\nolimits _{i \geq 0} f_{i0}^{-1}(U_ j)$ is an open subscheme of the limit we win. $\square$

Comment #6661 by Jonas Ehrhard on

Do we need all the $S_i$ to be quasi-compact or is it sufficient to assume that $S_0$ is quasi-compact?

Comment #6879 by on

Yes, it is enough to have one of the schemes $S_i$ to be quasi-compact. But they do all need to be assumed nonempty of course. I'm going to leave this as is.

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