## Model of a scheme regular over the generic point

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 http://mathoverflow.net/questions/32083/for-a-morphism-f-from-a-regular-scheme-should-there-exist-an-open-subscheme-u-of 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!!

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Mikhail, I edited the question (e.g., "generic fiber $S_0$" of $S$ seemed likely to be a generic pt of $S$). The answer to the modified question (which I hope is what you intended) is "yes". By generic flatness can assume $X$ is $S$-flat, and by descent from perfect closure of $k(\eta)$ (over which regular implies smooth) there's a finite purely insep $K/k(\eta)$ such that $((X_0)_K)_ {\rm{red}}$ is $K$-smooth. Spread $K$ to reduced $U$ finite flat radiciel over dense open in $S$. Then $(X_U)_ {\rm{red}}$ is $U$-smooth. Replace $U$ with regular dense open (excellence!), so you win. QED – BCnrd Jul 16 2010 at 16:30
Thank you very much!! – Mikhail Bondarko Jul 16 2010 at 16:38
Whoops!! Mea culpa, the step where I did descent from perfect closure is wrong, since the underlying reduced scheme of the base change to the perfect closure is merely generic smooth, perhaps not smooth if $X_ {\eta}$ wasn't smooth. I'm now a bit doubtful. Is there a reason why alterations aren't enough for your needs? – BCnrd Jul 16 2010 at 16:51
@B.: For the conclusion to be true, the reduced geometric generic fiber $((X_0)_{\bar{k(\eta)}})_{\rm red}$ must be smooth. Your proof shows that this condition is sufficient. I think that just assuming $X_0$ regular is not enough (consider the curve $y^2=x^p-a$ with $a$ not a $p$-th power in characteristic $p>0$, it is even geometrically reduced). – Qing Liu Jul 16 2010 at 16:54
Oops, the condition is not necessary. – Qing Liu Jul 16 2010 at 16:56

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.  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  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 $S$)  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.

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 Item 4 concerns with models with regular fibers, which a priori is a weaker condition then smoothness. – Qing Liu Jul 17 2010 at 13:36