3 added 420 characters in body

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 $S={\rm Spec} k[t]$ (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$is regular, geometrically integral but not smooth over$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$). S$)  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 (i.e. finite type with finite fibers) dominant morphism $U\to S$ (with $U$ integral). Then $X_U$ is integral. But $X_U\to U$ is not smooth because its generic fiber is not smooth. Even worse, 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 the generic fiber $X_0$ would be smooth at $(y, x^p-t)$, contradiction.

2 edited body

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.

4. The above assumption of smoothness is essentially shapesharp. Consider the example $S={\rm Spec} k[t]$ with $k$ perfect of characteristic $p>2$, and let $X_0$ be the affine curve over $K=k(t)$ defined by the equation $y^2=x^p-t$. Then $X_0$ is regular, 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 model $y^2=x^p-t$ over $S$). Let $U\to S$ be any quasi-finite (i.e. finite type with finite fibers) dominant morphism $U\to S$ (with $U$ integral). Then $X_U$ is integral. But $X_U\to U$ is not smooth because its generic fiber is not smooth. Even worse, none of the closed fibers is regular because it would be smooth (any finite extension of $k$ is perfect) and then the generic fiber would be smooth.

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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.

4. The above assumption of smoothness is essentially shape. Consider the example $S={\rm Spec} k[t]$ with $k$ perfect of characteristic $p>2$, and let $X_0$ be the affine curve over $K=k(t)$ defined by the equation $y^2=x^p-t$. Then $X_0$ is regular, 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 model $y^2=x^p-t$ over $S$). Let $U\to S$ be any quasi-finite (i.e. finite type with finite fibers) dominant morphism $U\to S$ (with $U$ integral). Then $X_U$ is integral. But $X_U\to U$ is not smooth because its generic fiber is not smooth. Even worse, none of the closed fibers is regular because it would be smooth (any finite extension of $k$ is perfect) and then the generic fiber would be smooth.