The Stacks project

Proof. Note that given a chain of specializations

of points of $X$ all of the points $\xi _ i$ correspond to prime ideals of the local ring of $X$ at $\xi _0$ by Schemes, Lemma 26.13.2. Hence we see that the dimension of $X$ is the supremum of the dimensions of its local rings. In particular $\dim _ x(X) \geq \dim (\mathcal{O}_{X, x})$ as $\dim _ x(X)$ is the minimum of the dimensions of open neighbourhoods of $x$. Thus $\sup _{x \in X} \dim _ x(X) \geq \dim (X)$. On the other hand, it is clear that $\sup _{x \in X} \dim _ x(X) \leq \dim (X)$ as $\dim (U) \leq \dim (X)$ for any open subset of $X$. $\square$

Proof. By Topology, Lemma 5.11.2 we may replace $X$ by an affine open neighbourhood of $\xi $. In this case the result follows easily from Algebra, Lemma 10.26.3. $\square$

Proof. This follows from Lemma 28.10.3 for example. $\square$

Proof. A Noetherian ring of dimension $0$ is a finite product of Artinian local rings, see Algebra, Proposition 10.60.7. Hence an affine open of a locally Noetherian scheme $X$ of dimension $0$ has discrete underlying topological space. This implies that the topology on $X$ is discrete. The lemma follows easily from these remarks. $\square$

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. $\square$