The Stacks project

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. $\square$


Comments (2)

Comment #6661 by Jonas Ehrhard on

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

Comment #6879 by on

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


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 01Z2. Beware of the difference between the letter 'O' and the digit '0'.