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Jan 28, 2017 at 15:03 comment added nfdc23 @JasonStarr: Ah, whoops indeed. So with $k$ perfect of characteristic $p>0$ any regular connected $K(Y)$-scheme $X_{\eta}$ that is generically $K(Y)$-smooth but not everywhere $K(Y)$-smooth spreads out to a counterexample over a dense open in $Y$, and such do exist over $K(Y)$ whenever $\dim Y>0$ since $K(Y)$ is then imperfect. That is, we can find $t \in K(Y) - K(Y)^p$ and then adapt your examples to also allow $p=2$: for any $m>1$ not divisible by $p$ (such as $m=2$ if $p>2$) the integral $K(Y)$-curve $z^m = w^p+t$ is Dedekind and so spreads out to such an $X$ over some dense open in $Y$.
Jan 28, 2017 at 10:08 comment added Jason Starr @nfdc23. The wording tricked you. Munchlax wants the open subset to be in the target not in the source. The quasi-elliptic fibrations are counterexamples. I realize you already know those, but for others: let $Y$ be $\text{Spec}\ k[t]$, and let $X$ be $\text{Spec}\ k[t,z,w]/\langle z^2 - (w^p+t) \rangle$, $p\neq 2$. Then $X$ is isomorphic to $\text{Spec}\ k[z,w]$, and so it is smooth. There is a dense open, $V=X\setminus \text{Zero}(\langle z \rangle)$, over which $\pi$ is smooth. However, the singular locus of $\pi$ surjects to $Y$.
Jan 28, 2017 at 3:45 comment added nfdc23 It is necessary and sufficient for $K(X)/K(Y)$ to be separable, since that is exactly the condition for the existence of a separating transcendence basis, which in turn is the ``generic'' effect of the Zariski-local description of a smooth morphism as the composition of an etale map to an affine space over the base (i.e., an etale $Y$-map $U \rightarrow \mathbf{A}^n_Y$ for Zariski-open $U \subset X$).
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Jan 28, 2017 at 3:24 history asked Munchlax CC BY-SA 3.0