Model of a scheme regular over the generic point

Question

Let all schemes below be excellent. Let $X_0$ be a regular (not necessarily smooth, projective) non-empty scheme of finite type over the generic point $\eta$ of a regular connected scheme $S$. As the answers to my question For a morphism f from a regular scheme, should there exist an open subscheme U of the target such that fibre of f at each point of U is regular show, there does not have to exist a dense open $U\subset S$ such that $X_0$ possesses a fibrewise regular model over $U$. Yet, should there always exist a pseudo-finite dominant morphism $j:U\to S$ and some model $X$ of $X_0$ over $S$ such that $U$ is regular and the reduced scheme associated to $X_U$ is smooth over $U$? Can we assume that the morphism $U\to S$ is radiciel?

Upd.: it is probably sufficient for my purposes to find such a $U$ for a variety $X_0$ that is smooth over the perfect closure of $\eta$. It seems that (the first) BCnrd comment solves the problem; thanks!!

Answer 1

To synthersize a bit: let $S$ be an excellent (quasi-excellent is enough) integral scheme, let $X_0$ be a regular scheme of finite type over $K=K(S)$. I will suppose $X_0$ separated to avoid possible pathologies.

1. There exists a separated integral noetherian scheme $X$ over $S$ with generic fiber isomorphic to $X_0$.

2. The singular locus of $X$ is closed (excellence), its projection in $S$ is constructible (Chevalley) and does not contain the generic point of $S$, therefore is contained in a proper closed subset $F$ of $S$. The scheme $X_U$, where $U=S \setminus F$, is a regular model of $X_0$ over $U$ (or $S$).

3. If $(\overline{X_0})_{\rm red}$ is smooth, then, as explained by BCnrd, there exits a radicial finite flat (hence dominant) morphism $U\to S$ with $U$ integral such that $(X_U)_{\rm red}\to U$ is smooth. [Edit] The assumption of smoothness is also necessary if $(X_U)_{\rm red}\to U$ is smooth for some dominant morphism $U\to S$ (just consider the generic fiber).

4. The above assumption of smoothness is essentially sharp [Edit] if we want to have a quasi-finite (i.e. finite type with finite fibers) and dominant base change $U\to S$ such that $(X_U)_{\rm red}\to U$ is fiberwise regular. Consider the example $S={\rm Spec} (k[t])$ with $k$ perfect of characteristic $p>2$, and let $X_0$ be the regular affine curve over $K=k(t)$ defined by the equation $y^2=x^p-t$. Then $X_0$ is geometrically integral but not smooth over $K$. Let $X\to S$ be a model of $X_0$ (i.e. finite type separated morphism with generic fiber isomorphic to $X_0$). Shrinking $S$ if necessary, we can suppose that $X\to S$ is flat with geometrically integral fibers ([EGA], IV.9 or directly compare with the obvious projective model associated to the equation $y^2=x^p-t$ over $S$) [Edit] and that the Zariski closure in $X$ of the non-smooth point $(y, x^p-t)\in X_0$ meets every fiber. Let $U\to S$ be any quasi-finite dominant morphism $U\to S$ (with $U$ integral). Then $X_U$ is integral. But none of the closed fibers of $X_U\to U$ is regular because it would be smooth (any finite extension of $k$ is perfect) and then $X_0$ would be smooth at $(y, x^p-t)$, contradiction.