The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

Email from Ofer Gabber dated June 4, 2016

Lemma 27.10.6. Let $X$ be a scheme of dimension zero. The following are equivalent

  1. $X$ is quasi-separated,

  2. $X$ is separated,

  3. $X$ is Hausdorff,

  4. every affine open is closed.

In this case the connected components of $X$ are points.

Proof. As the dimension of $X$ is zero, we see that for any affine open $U \subset X$ the space $U$ is profinite and satisfies a bunch of other properties which we will use freely below, see Algebra, Lemma 10.25.5. We choose an affine open covering $X = \bigcup U_ i$.

If (4) holds, then $U_ i \cap U_ j$ is a closed subset of $U_ i$, hence quasi-compact, hence $X$ is quasi-separated, by Schemes, Lemma 25.21.6, hence (1) holds.

If (1) holds, then $U_ i \cap U_ j$ is a quasi-compact open of $U_ i$ hence closed in $U_ i$. Then $U_ i \cap U_ j \to U_ i$ is an open immersion whose image is closed, hence it is a closed immersion. In particular $U_ i \cap U_ j$ is affine and $\mathcal{O}(U_ i) \to \mathcal{O}_ X(U_ i \cap U_ j)$ is surjective. Thus $X$ is separated by Schemes, Lemma 25.21.6, hence (2) holds.

Assume (2) and let $x, y \in X$. Say $x \in U_ i$. If $y \in U_ i$ too, then we can find disjoint open neighbourhoods of $x$ and $y$ because $U_ i$ is Hausdorff. Say $y \not\in U_ i$ and $y \in U_ j$. Then $y \not\in U_ i \cap U_ j$ which is an affine open of $U_ j$ and hence closed in $U_ j$. Thus we can find an open neighbourhood of $y$ not meeting $U_ i$ and we conclude that $X$ is Hausdorff, hence (3) holds.

Assume (3). Let $U \subset X$ be affine open. Then $U$ is closed in $X$ by Topology, Lemma 5.12.4. This proves (4) holds.

We omit the proof of the final statement. $\square$

Comments (0)

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 0CKV. Beware of the difference between the letter 'O' and the digit '0'.