Lemma 110.28.1. There exists a quasi-compact and quasi-separated algebraic space which does not contain a quasi-compact dense open subscheme.
110.28 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 66.13.3. In fact, this result holds when X is reasonable, see Decent Spaces, Proposition 68.10.1. 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 65.10.5). 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 110.27 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 65.14.2 in the same manner as we used the construction of Spaces, Example 65.14.1 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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