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22.11. Zariski topology of schemes

See Topology, Section 5.1 for some basic material in topology adapted to the Zariski topology of schemes.

Lemma 22.11.1. Let $X$ be a scheme. Any irreducible closed subset of $X$ has a unique generic point. In other words, $X$ is a sober topological space, see Topology, Definition 5.5.4.

Proof. Let $Z \subset X$ be an irreducible closed subset. For every affine open $U \subset X$, $U = \mathop{\rm Spec}(R)$ we know that $Z \cap U = V(I)$ for a unique radical ideal $I \subset R$. Note that $Z \cap I$ is either empty or irreducible. In the second case (which occurs for at least one $U$) we see that $I = \mathfrak p$ is a prime ideal, which is a generic point $\xi$ of $Z \cap U$. It follows that $Z = \overline{\{\xi\}}$, in other words $\xi$ is a generic point of $Z$. If $\xi'$ was a second generic point, then $\xi' \in Z \cap U$ and it follows immediately that $\xi' = \xi$. $\square$

Lemma 22.11.2. Let $X$ be a scheme. The collection of affine opens of $X$ forms a basis for the topology on $X$.

Proof. This follows from the discussion on open subschemes in Section 22.9. $\square$

Remark 22.11.3. In general the intersection of two affine opens in $X$ is not affine open. See Example 22.14.3.

Lemma 22.11.4. The underlying topological space of any scheme is locally quasi-compact, see Topology, Definition 5.20.1.

Proof. This follows from Lemma 22.11.2 above and the fact that the spectrum of ring is quasi-compact, see Algebra, Lemma 9.16.10. $\square$

Lemma 22.11.5. Let $X$ be a scheme. Let $U, V$ be affine opens of $X$, and let $x \in U \cap V$. There exists an affine open neighbourhood $W$ of $x$ such that $W$ is a standard open of both $U$ and $V$.

Proof. Write $U = \mathop{\rm Spec}(A)$ and $V = \mathop{\rm Spec}(B)$. Say $x$ corresponds to the prime $\mathfrak p \subset A$ and the prime $\mathfrak q \subset B$. We may choose a $f \in A$, $f \not \in \mathfrak p$ such that $D(f) \subset U \cap V$. Note that any standard open of $D(f)$ is a standard open of $\mathop{\rm Spec}(A) = U$. Hence we may assume that $U \subset V$. In other words, now we may think of $U$ as an affine open of $V$. Next we choose a $g \in B$, $g \not \in \mathfrak q$ such that $D(g) \subset U$. In this case we see that $D(g) = D(g_A)$ where $g_A \in A$ denotes the image of $g \in A$. Thus the lemma is proved. $\square$

Lemma 22.11.6. Let $X$ be a scheme. Let $X = \bigcup_i U_i$ be an affine open covering. Let $V \subset X$ be an affine open. There exists a standard open covering $V = \bigcup_{j = 1, \ldots, m} V_j$ (see Definition 22.5.2) such that each $V_j$ is a standard open in one of the $U_i$.

Proof. Pick $v \in V$. Then $v \in U_i$ for some $i$. By Lemma 22.11.5 above there exists an open $v \in W_v \subset V \cap U_i$ such that $W_v$ is a standard open in both $V$ and $U_i$. Since $V$ is quasi-compact the lemma follows. $\square$

Lemma 22.11.7. Let $X$ be a scheme whose underlying topological space is a finite discrete set. Then $X$ is affine.

Proof. Say $X = \{x_1, \ldots, x_n\}$. Then $U_i = \{x_i\}$ is an open neighbourhood of $x_i$. By Lemma 22.11.2 it is affine. Hence $X$ is a finite disjoint union of affine schemes, and hence is affine by Lemma 22.6.8. $\square$

Example 22.11.8. There exists a scheme without closed points. Namely, let $R$ be a local domain whose spectrum looks like $(0) = \mathfrak p_0 \subset \mathfrak p_1 \subset \mathfrak p_2 \subset \ldots \subset \mathfrak m$. Then the open subscheme $\mathop{\rm Spec}(R) \setminus \{\mathfrak m\}$ does not have a closed point. To see that such a ring $R$ exists, we use that given any totally ordered group $(\Gamma, \geq)$ there exists a valuation ring $A$ with valuation group $(\Gamma, \geq)$, see [Krull]. See Algebra, Section 9.48 for notation. We take $\Gamma = \mathbf{Z}x_1 \oplus \mathbf{Z}x_2 \oplus \mathbf{Z}x_3 \oplus \ldots$ and we define $\sum_i a_i x_i \geq 0$ if and only if the first nonzero $a_i$ is $> 0$, or all $a_i = 0$. So $x_1 \geq x_2 \geq x_3 \geq \ldots \geq 0$. The subsets $x_i + \Gamma_{\geq 0}$ are prime ideals of $(\Gamma, \geq)$, see Algebra, notation above Lemma 9.48.13. These together with $\emptyset$ and $\Gamma_{\geq 0}$ are the only prime ideals. Hence $A$ is an example of a ring with the given structure of its spectrum, by Algebra, Lemma 9.48.13.

\section{Zariski topology of schemes}
\label{section-topology}

\noindent
See Topology, Section \ref{topology-section-introduction}
for some basic material in topology adapted to the Zariski
topology of schemes.

\begin{lemma}
\label{lemma-scheme-sober}
Let $X$ be a scheme.
Any irreducible closed subset of $X$ has a unique generic point.
In other words, $X$ is a sober topological space, see
Topology, Definition \ref{topology-definition-generic-point}.
\end{lemma}

\begin{proof}
Let $Z \subset X$ be an irreducible closed subset.
For every affine open $U \subset X$, $U = \Spec(R)$
we know that $Z \cap U = V(I)$ for a unique
radical ideal $I \subset R$. Note that $Z \cap I$ is either
empty or irreducible. In the second case (which occurs
for at least one $U$) we see that $I = \mathfrak p$
is a prime ideal, which is a generic point $\xi$ of $Z \cap U$.
It follows that $Z = \overline{\{\xi\}}$, in other words
$\xi$ is a generic point of $Z$. If $\xi'$ was a second
generic point, then $\xi' \in Z \cap U$ and it follows
immediately that $\xi' = \xi$.
\end{proof}

\begin{lemma}
\label{lemma-basis-affine-opens}
Let $X$ be a scheme. The collection of affine opens
of $X$ forms a basis for the topology on $X$.
\end{lemma}

\begin{proof}
This follows from the discussion on open subschemes
in Section \ref{section-schemes}.
\end{proof}

\begin{remark}
\label{remark-intersection-affine-opens}
In general the intersection of two affine opens in $X$
is not affine open. See Example \ref{example-affine-space-zero-doubled}.
\end{remark}

\begin{lemma}
\label{lemma-locally-quasi-compact}
The underlying topological space of any scheme is
locally quasi-compact, see
Topology, Definition \ref{topology-definition-locally-quasi-compact}.
\end{lemma}

\begin{proof}
This follows from Lemma \ref{lemma-basis-affine-opens} above
and the fact that the spectrum of ring is quasi-compact, see
Algebra, Lemma \ref{algebra-lemma-quasicompact}.
\end{proof}

\begin{lemma}
\label{lemma-standard-open-two-affines}
Let $X$ be a scheme.
Let $U, V$ be affine opens of $X$, and let $x \in U \cap V$.
There exists an affine open neighbourhood $W$ of $x$
such that $W$ is a standard open of both $U$ and $V$.
\end{lemma}

\begin{proof}
Write $U = \Spec(A)$ and $V = \Spec(B)$.
Say $x$ corresponds to the prime $\mathfrak p \subset A$
and the prime $\mathfrak q \subset B$.
We may choose a $f \in A$, $f \not \in \mathfrak p$ such that
$D(f) \subset U \cap V$. Note that any standard open of $D(f)$
is a standard open of $\Spec(A) = U$. Hence we may assume
that $U \subset V$. In other words, now we may think of $U$
as an affine open of $V$. Next we choose a
$g \in B$, $g \not \in \mathfrak q$ such that
$D(g) \subset U$. In this case we see that $D(g) = D(g_A)$
where $g_A \in A$ denotes the image of $g \in A$. Thus the lemma
is proved.
\end{proof}

\begin{lemma}
\label{lemma-good-subcover}
Let $X$ be a scheme.
Let $X = \bigcup_i U_i$ be an affine open covering.
Let $V \subset X$ be an affine open.
There exists a standard open covering
$V = \bigcup_{j = 1, \ldots, m} V_j$ (see
Definition \ref{definition-standard-covering})
such that each $V_j$ is a standard open in one of the $U_i$.
\end{lemma}

\begin{proof}
Pick $v \in V$. Then $v \in U_i$ for some $i$.
By Lemma \ref{lemma-standard-open-two-affines} above there exists an open
$v \in W_v \subset V \cap U_i$ such that
$W_v$ is a standard open in both $V$ and $U_i$.
Since $V$ is quasi-compact the lemma follows.
\end{proof}

\begin{lemma}
\label{lemma-scheme-finite-discrete-affine}
Let $X$ be a scheme whose underlying topological space
is a finite discrete set.
Then $X$ is affine.
\end{lemma}

\begin{proof}
Say $X = \{x_1, \ldots, x_n\}$. Then $U_i = \{x_i\}$ is an open neighbourhood
of $x_i$. By
Lemma \ref{lemma-basis-affine-opens}
it is affine. Hence $X$ is a finite disjoint union of affine schemes, and
hence is affine by
Lemma \ref{lemma-disjoint-union-affines}.
\end{proof}

\begin{example}
\label{example-scheme-without-closed-points}
There exists a scheme without closed points.
Namely, let $R$ be a local domain whose spectrum
looks like
$(0) = \mathfrak p_0 \subset \mathfrak p_1 \subset \mathfrak p_2 \subset \ldots \subset \mathfrak m$. Then the open subscheme
$\Spec(R) \setminus \{\mathfrak m\}$ does not have
a closed point. To see that such a ring $R$ exists, we use
that given any totally ordered group $(\Gamma, \geq)$
there exists a valuation ring $A$ with valuation group $(\Gamma, \geq)$,
see \cite{Krull}. See Algebra, Section \ref{algebra-section-valuation-rings}
for notation. We take
$\Gamma = \mathbf{Z}x_1 \oplus \mathbf{Z}x_2 \oplus \mathbf{Z}x_3 \oplus \ldots$ and we define $\sum_i a_i x_i \geq 0$ if and only
if the first nonzero $a_i$ is $> 0$, or all $a_i = 0$.
So $x_1 \geq x_2 \geq x_3 \geq \ldots \geq 0$.
The subsets $x_i + \Gamma_{\geq 0}$ are prime ideals
of $(\Gamma, \geq)$, see Algebra, notation above
Lemma \ref{algebra-lemma-ideals-valuation-ring}. These together
with $\emptyset$ and $\Gamma_{\geq 0}$ are the only prime ideals.
Hence $A$ is an example of a ring with the given structure of
its spectrum, by
Algebra, Lemma \ref{algebra-lemma-ideals-valuation-ring}.
\end{example}


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