Lemma 109.27.1. There exists a quasi-compact and quasi-separated algebraic space which does not contain a quasi-compact dense open subscheme.

## 109.27 Nonexistence of quasi-compact dense open subscheme

Let $X$ be a quasi-compact and quasi-separated algebraic space over a field $k$. We know that the schematic locus $X' \subset X$ is a dense open subspace, see Properties of Spaces, Proposition 65.13.3. In fact, this result holds when $X$ is reasonable, see Decent Spaces, Proposition 67.10.1. A natural question is whether one can find a quasi-compact dense open subscheme of $X$. It turns out this is not possible in general.

Assume the characteristic of $k$ is not 2. 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 $U = \mathop{\mathrm{Spec}}(B)$. Denote $0 \in U$ the closed point all of whose coordinates are zero. Set

where $\Delta $ is the image of the diagonal morphism of $U$ over $k$ and

It is clear that $s, t : R \to U$ are étale, and hence $j$ is an étale equivalence relation. The quotient $X = U/R$ is an algebraic space (Spaces, Theorem 64.10.5). Note that $j$ is not an immersion because $(0, 0) \in \Delta $ is in the closure of $\Gamma $. Hence $X$ is not a scheme. On the other hand, $X$ is quasi-separated as $R$ is quasi-compact. Denote $0_ X$ the image of the point $0 \in U$. We claim that $X \setminus \{ 0_ X\} $ is a scheme, namely

(details omitted). On the other hand, we have seen in Section 109.26 that the scheme on the right hand side does not contain a quasi-compact dense open.

**Proof.**
See discussion above.
$\square$

Using the construction of Spaces, Example 64.14.2 in the same manner as we used the construction of Spaces, Example 64.14.1 above, one obtains an example of a quasi-compact, quasi-separated, and locally separated algebraic space which does not contain a quasi-compact dense open subscheme.

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