## 110.27 Nonexistence of suitable opens

This section complements the results of Properties, Section 28.29.

Let $k$ be a field and let $A = k[z_1, z_2, z_3, \ldots ]/I$ where $I$ is the ideal generated by all pairwise products $z_ iz_ j$, $i \not= j$, $i, j \in \mathbf{N}$. Set $S = \mathop{\mathrm{Spec}}(A)$. Let $s \in S$ be the closed point corresponding to the maximal ideal $(z_ i)$. We claim there is no quasi-compact open $V \subset S \setminus \{ s\}$ which is dense in $S \setminus \{ s\}$. Note that $S \setminus \{ s\} = \bigcup D(z_ i)$. Each $D(z_ i)$ is open and irreducible with generic point $\eta _ i$. We conclude that $\eta _ i \in V$ for all $i$. However, a principal affine open of $S \setminus \{ s\}$ is of the form $D(f)$ where $f \in (z_1, z_2, \ldots )$. Then $f \in (z_1, \ldots , z_ n)$ for some $n$ and we see that $D(f)$ contains only finitely many of the points $\eta _ i$. Thus $V$ cannot be quasi-compact.

Let $k$ be a field and let $B = k[x, z_1, z_2, z_3, \ldots ]/J$ where $J$ is the ideal generated by the products $xz_ i$, $i \in \mathbf{N}$ and by all pairwise products $z_ iz_ j$, $i \not= j$, $i, j \in \mathbf{N}$. Set $T = \mathop{\mathrm{Spec}}(B)$. Consider the principal open $U = D(x)$. We claim there is no quasi-compact open $V \subset S$ such that $V \cap U = \emptyset$ and $V \cup U$ is dense in $S$. Let $t \in T$ be the closed point corresponding to the maximal ideal $(x, z_ i)$. The closure of $U$ in $T$ is $\overline{U} = U \cup \{ t\}$. Hence $V \subset \bigcup _ i D(z_ i)$ is a quasi-compact open. By the arguments of the previous paragraph we see that $V$ cannot be dense in $\bigcup D(z_ i)$.

Lemma 110.27.1. Nonexistence quasi-compact opens of affines:

1. There exist an affine scheme $S$ and affine open $U \subset S$ such that there is no quasi-compact open $V \subset S$ with $U \cap V = \emptyset$ and $U \cup V$ dense in $S$.

2. There exists an affine scheme $S$ and a closed point $s \in S$ such that $S \setminus \{ s\}$ does not contain a quasi-compact dense open.

Proof. See discussion above. $\square$

Let $X$ be the glueing of two copies of the affine scheme $T$ (see above) along the affine open $U$. Thus there is a morphism $\pi : X \to T$ and $X = U_1 \cup U_2$ such that $\pi$ maps $U_ i$ isomorphically to $T$ and $U_1 \cap U_2$ isomorphically to $U$. Note that $X$ is quasi-separated (by Schemes, Lemma 26.21.6) and quasi-compact. We claim there does not exist a separated, dense, quasi-compact open $W \subset X$. Namely, consider the two closed points $x_1 \in U_1$, $x_2 \in U_2$ mapping to the closed point $t \in T$ introduced above. Let $\tilde\eta \in U_1 \cap U_2$ be the generic point mapping to the (unique) generic point $\eta$ of $U$. Note that $\tilde\eta \leadsto x_1$ and $\tilde\eta \leadsto x_2$ lying over the specialization $\eta \leadsto s$. Since $\pi |_ W : W \to T$ is separated we conclude that we cannot have both $x_1$ and $x_2 \in W$ (by the valuative criterion of separatedness Schemes, Lemma 26.22.2). Say $x_1 \not\in W$. Then $W \cap U_1$ is a quasi-compact (as $X$ is quasi-separated) dense open of $U_1$ which does not contain $x_1$. Now observe that there exists an isomorphism $(T, t) \cong (S, s)$ of schemes (by sending $x$ to $z_1$ and $z_ i$ to $z_{i + 1}$). Hence by the first paragraph of this section we arrive at a contradiction.

Lemma 110.27.2. There exists a quasi-compact and quasi-separated scheme $X$ which does not contain a separated quasi-compact dense open.

Proof. See discussion above. $\square$

## Comments (2)

Comment #3232 by Laurent Moret-Bailly on

There are similar examples which I think are more "visual", and avoid using coordinates. Start with a compact space $\vert S\vert$ consisting of a sequence $(s_n)$ converging to a limit $s$. Fix a field $k$ and turn $\vert S\vert$ into an affine scheme $S=\mathrm{Spec}(A)$ where $A$ is the ring of locally constant functions $\vert S\vert\to k$. As a topological space, $S\smallsetminus\{s\}$ is just $\mathbb{N}$, so property (2) is clear.

From this $S$ you get a new $T$ by attaching $\mathbb{A}^1_k$ to $S$ via identification of $s$ and the origin of $\mathbb{A}^1_k$. Of course, you take for $U$ the complement of the origin in $\mathbb{A}^1_k$.

Comment #3331 by on

Yes, you are correct. We already discuss this type of ring and its spectrum very briefly in Section 110.63, but we should add a discussion of spectra of rings of continuous functions somewhere and then use that to make all kinds of examples like this. Anybody who is interested?

On the other hand, there is a method in commutative algebra of making counter examples by just taking a bunch of variables and imposing conditions... so I think it is worthwhile to do this if you can.

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