The Stacks project

Email from Ofer Gabber dated June 4, 2016

Lemma 28.10.6. 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 and every quasi-compact open of $X$ is affine. In particular, if $X$ is quasi-compact, then $X$ is affine.

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 10.26.5. 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 26.21.6, 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 26.21.7, 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.

Assume $X$ satisfies the equivalent conditions (1) – (4). We prove the final statements of the lemma. Say $x, y \in X$ with $x \not= y$. Since $y$ does not specialize to $x$ we can choose $U \subset X$ affine open with $x \in U$ and $y \not\in U$. Then we see that $X = U \amalg (X \setminus U)$ is a decomposistion into open and closed subsets which shows that $x$ and $y$ do not belong to the same connected component of $X$. Next, assume $U \subset X$ is a quasi-compact open. Write $U = U_1 \cup \ldots \cup U_ n$ as a union of affine opens. We will prove by induction on $n$ that $U$ is affine. This immediately reduces us to the case $n = 2$. In this case we have $U = (U_1 \setminus U_2) \amalg (U_1 \cap U_2) \amalg (U_2 \setminus U_1)$ and the arguments above show that each of the pieces is affine. $\square$

Comments (4)

Comment #7004 by Laurent Moret-Bailly on

Suggested addition to the final statement: "and if is quasicompact, then it is affine".

Comment #7546 by Fred Vu on

In the third paragraph of the proof, I believe you accidentally cite tag 01KO instead of 01KP.

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'.