Let $S$ be a scheme and let $X$ be a quasi-separated algebraic space over $S$. Does there exist an open subspace of $X$ which is a scheme and which is dense in each fiber $X_s$, $s \in S$?

I am happy to make the following additional assumptions:

(1) $S$ is the spectrum of a complete dvr with algebraically closed residue field and $X$ is flat and of finite type over $S$.

(2) All fibers of $X$ are geometrically irreducible.

(3) The generic fiber of $X$ is geometrically normal and the special fiber of $X$ is geometrically integral (then $X$ is normal).

  • $\begingroup$ There is a Nagata compactification theorem for algebraic spaces due to Conrad. Then there is a version of Chow's lemma for proper algebraic spaces. Combined with the valuative criterion, that should imply that the schematic locus contains every codimension one point of $X$ (e.g., the generic point of the closed fiber). $\endgroup$ Commented May 18, 2017 at 10:34
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    $\begingroup$ Probably you do not need the Nagata compactification, this should follow just from Chow's lemma for finite type, separated algebraic spaces. $\endgroup$ Commented May 18, 2017 at 10:59
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    $\begingroup$ @JasonStarr: The paper on Nagata compactification to which you refer is written by 3 people rather than 1 (also Lieblich & Olsson). $\endgroup$
    – nfdc23
    Commented May 18, 2017 at 13:09
  • $\begingroup$ @nfdc23. Sorry about that! $\endgroup$ Commented May 18, 2017 at 13:26

2 Answers 2


There are quasi-separated counterexamples. If you assume that $X$ is separated, then this is true. The basic reference is Donald Knutson's book.

MR0302647 (46 #1791)
Knutson, Donald
Algebraic spaces.
Lecture Notes in Mathematics, Vol. 203.
Springer-Verlag, Berlin-New York, 1971. vi+261 pp.

If $X$ is separated, then by Chow's Lemma, pp. 192-193, there exists a projective birational morphism $g:\widetilde{X}\to X$ such that the composite morphism $\widetilde{X}\to S$ is a quasi-projective morphism. Since $X$ is normal, by the valuative criterion of properness and by Zariski's Main Theorem, $g$ is an isomorphism over a dense open subset $U$ of $X$ that contains all codimension one points. In particular, $U$ contains the generic point of the closed fiber. Thus, $U$ is dense in every fiber, and $U$ is isomorphic to an open subscheme of the quasi-projective $S$-scheme $\widetilde{X}$. If you use Raynaud-Gruson, then you can probably say more about $U$.

Edit. The first example that I wrote, Knutson's beautiful example from pp. 9-10, has nonreduced closed fiber. The OP asks about the case when the closed fiber is geometrically irreducible. Here is a slightly different counterexample that is quasi-separated, and even locally separated, such that the closed fiber is geometrically irreducible.

Let $S$ be $\text{Spec}\ \mathbb{C}[[s]]$. Let $U$ be $\text{Spec}\ \mathbb{C}[[s]][t,t^{-1}]$, i.e., the multiplicative group over $S$. The fiber product $U\times_S U$ is $\text{Spec}\ \mathbb{C}[[s]][t_1,t_1^{-1},t_2,t_2^{-1}]$. Define $R \subset U\times_S U$ to be the disjoint union of the closed subscheme $\text{Zero}(t_1-t_2)$, i.e., the diagonal, and the locally closed subscheme $U' = \text{Zero}(t_1+t_2)\cap D(s)$. This is an étale equivalence relation on $U$. Thus, by Proposition 1.3(b), p. 93, there is an algebraic space $X$ and an étale, surjective morphism $h:U\to X$ such that $U\times_X U$ equals $R$. Since $R\to U\times_S U$ is a quasi-compact, locally closed immersion, $X\to S$ is quasi-separated and even locally separated. The closed fiber $X_0=\text{Zero}(s)$ is just $\text{Spec}\ \mathbb{C}[t,t^{-1}]$, so it is geometrically integral.

No Zariski open subset of $X$ containing the generic point of the closed fiber is a scheme. If there were such an open, then there would be an open affine $V$. Since $X$ is integral, also $V$ is integral. Thus the ring homomorphism $\mathcal{O}_X(V)\to \mathcal{O}_X(V\cap D(s))$ is injective. But $D(s)\subset X$ is affine, namely $\text{Spec}\ \mathbb{C}[[s]][s^{-1}][t^2, t^{-2}].$ Every fraction of elements in this ring pulls back on $U$ to a rational function that identifies $t$ and $-t$ on the closed fiber $X_0$. For an open affine $V$, the restriction map $\mathcal{O}_X(V)\to \mathcal{O}_{X_0}(V\cap X_0)$ is surjective. Since there are elements in $\mathcal{O}_{X_0}(X_0)$ that do not identify $t$ and $-t$, this is a contradiction.

Second edit. My second example above is precisely the same as Count Dracula's example.


A counter example to the question in the presence of (1), (2), and (3) is to take $\mathbf{G}_{m, S}$ and divide out by the etale equivalence relation given by $(x, s) \sim (x, s)$ and for $s \not = 0$ also $(x, s) \sim (-x, s)$ (assume the residue characteristic of your dvr is not $2$). Then the resulting morphisms $$ \mathbf{G}_{m, S} \longrightarrow X \longrightarrow S $$ have the following properties: the first is surjective \'etale and the second is smooth of relative dimension 1 and both $\mathbf{G}_{m, S}$ and $X$ have geometrically integral fibres over $S$. The schematic locus of X is exactly the complement of the special fibre. It seems to me the "problem" is that the degree of the function field extensions induced between the fibres of $\mathbf{G}_{m, S}$ and $X$ drops from $2$ to $1$ from the generic to the special fibre.

PS: As Jason points out on locally Noetherian separated algebraic spaces the schematic locus contains all codimension $1$ points, see Tag 0ADD

  • $\begingroup$ Does "$s \ne 0$" refer to the open generic point of $S$ (and so it is generic characteristic 2 rather than residual characteristic 2 that is to be avoided)? $\endgroup$
    – nfdc23
    Commented May 18, 2017 at 15:24
  • $\begingroup$ Excellent point! $\endgroup$ Commented May 18, 2017 at 15:39
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    $\begingroup$ D'oh! My new example is the same as yours! I did not see your answer until I finished typing (which took a while, since I had an advisee meeting in the meantime). $\endgroup$ Commented May 18, 2017 at 15:40
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    $\begingroup$ OK, great. That should make it even more likely readers will "get it". $\endgroup$ Commented May 18, 2017 at 17:16
  • $\begingroup$ Thanks a lot! Both your counterexample and your reference are very helpful. I am sorry that it is not possible to accept two answers. $\endgroup$ Commented May 19, 2017 at 8:04

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