Lemma 26.11.1. 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 5.8.5.

## 26.11 Zariski topology of schemes

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

**Proof.**
Let $Z \subset X$ be an irreducible closed subset. For every affine open $U \subset X$, $U = \mathop{\mathrm{Spec}}(R)$ we know that $Z \cap U = V(I)$ for a unique radical ideal $I \subset R$. Note that $Z \cap U$ 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 26.11.2. 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 26.9.
$\square$

Remark 26.11.3. In general the intersection of two affine opens in $X$ is not affine open. See Example 26.14.3.

Lemma 26.11.4. The underlying topological space of any scheme is locally quasi-compact, see Topology, Definition 5.13.1.

**Proof.**
This follows from Lemma 26.11.2 above and the fact that the spectrum of ring is quasi-compact, see Algebra, Lemma 10.16.10.
$\square$

Lemma 26.11.5. 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 $U = \mathop{\mathrm{Spec}}(A)$ and $V = \mathop{\mathrm{Spec}}(B)$. 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 $\mathop{\mathrm{Spec}}(A) = U$. 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 $g$ by the map $B \to A$. Thus the lemma is proved.
$\square$

Lemma 26.11.6. 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 26.5.2) 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 26.11.5 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 26.11.7. Let $X$ be a scheme. Let $\mathcal{B}$ be the set of affine opens of $X$. Let $\mathcal{F}$ be a presheaf of sets on $\mathcal{B}$, see Sheaves, Definition 6.30.1. The following are equivalent

$\mathcal{F}$ is the restriction of a sheaf on $X$ to $\mathcal{B}$,

$\mathcal{F}$ is a sheaf on $\mathcal{B}$, and

$\mathcal{F}(\emptyset )$ is a singleton and whenever $U = V \cup W$ with $U, V, W \in \mathcal{B}$ and $V, W \subset U$ standard open (Algebra, Definition 10.16.3) the map

\[ \mathcal{F}(U) \longrightarrow \mathcal{F}(V) \times \mathcal{F}(W) \]is injective with image the set of pairs $(s, t)$ such that $s|_{V \cap W} = t|_{V \cap W}$.

**Proof.**
The equivalence of (1) and (2) is Sheaves, Lemma 6.30.7. It is clear that (2) implies (3). Hence it suffices to prove that (3) implies (2). By Sheaves, Lemma 6.30.4 and Lemma 26.5.1 it suffices to prove the sheaf condition holds for standard open coverings (Definition 26.5.2) of elements of $\mathcal{B}$. Let $U = U_1 \cup \ldots \cup U_ n$ be a standard open covering with $U \subset X$ affine open. We will prove the sheaf condition for this covering by induction on $n$. If $n = 0$, then $U$ is empty and we get the sheaf condition by assumption. If $n = 1$, then there is nothing to prove. If $n = 2$, then this is assumption (3). If $n > 2$, then we write $U_ i = D(f_ i)$ for $f_ i \in A = \mathcal{O}_ X(U)$. Suppose that $s_ i \in \mathcal{F}(U_ i)$ are sections such that $s_ i|_{U_ i \cap U_ j} = s_ j|_{U_ i \cap U_ j}$ for all $1 \leq i < j \leq n$. Since $U = U_1 \cup \ldots \cup U_ n$ we have $1 = \sum _{i = 1, \ldots , n} a_ i f_ i$ in $A$ for some $a_ i \in A$, see Algebra, Lemma 10.16.2. Set $g = \sum _{i = 1, \ldots , n - 1} a_ if_ i$. Then $U = D(g) \cup D(f_ n)$. Observe that $D(g) = D(gf_1) \cup \ldots \cup D(gf_{n - 1})$ is a standard open covering. By induction there is a unique section $s' \in \mathcal{F}(D(g))$ which agrees with $s_ i|_{D(gfi)}$ for $i = 1, \ldots , n - 1$. We claim that $s'$ and $s_ n$ have the same restriction to $D(gf_ n)$. This is true by induction and the covering $D(gf_ n) = D(gf_ nf_1) \cup \ldots \cup D(gf_ nf_{n - 1})$. Thus there is a unique section $s \in \mathcal{F}(U)$ whose restriction to $D(g)$ is $s'$ and whose restriction to $D(f_ n)$ is $s_ n$. We omit the verification that $s$ restricts to $s_ i$ on $D(f_ i)$ for $i = 1, \ldots , n - 1$ and we omit the verification that $s$ is unique.
$\square$

Lemma 26.11.8. 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 26.11.2 it is affine. Hence $X$ is a finite disjoint union of affine schemes, and hence is affine by Lemma 26.6.8.
$\square$

Example 26.11.9. 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{\mathrm{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 10.49 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 10.49.17. 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 10.49.17.

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