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).



Schematic locus of algebraic spaces in fibers
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. 


