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Let us denote the Frobenius endomorphism of a variety $ X $ by $ F $. A variety $ X $ over a field $ k $ of positive characteristic is globally $ F $-regular if for every effective Weil divisor $ D $, there is an $ e \in \mathbb{N}_{0} $ such that the morphism $ \mathcal{O}_{X} \to F^{e}_{\ast}(\mathcal{O}_{X}(D)) $ splits as a morphism of $ \mathcal{O}_{X} $ modules.

There is a theorem that in characteristic zero if $ X $ is a non-singular variety over an algebraically closed field of characteristic zero and $ f: X \to Y $ is a morphism, then there is an open sub-variety $ V \subseteq Y $ such that $ f: f^{-1}(V) \to V $ is generically smooth of relative dimension $ \dim(X)-\dim(Y) $. This is commonly known as generic smoothness.

Does anyone know of a globally $ F $-regular, smooth variety $ X $ over a field $ k $ of positive characteristic such that there exists a morphism $ f: X \to Y $ such that there is no $ V \subseteq Y $ with the property that $ f: f^{-1}(V) \to V $ is generically smooth of relative dimension $ \dim(X)-\dim(Y) $? Namely, does anyone know of a globally $ F $-regular, smooth variety $ X $ over a field $ k $ of positive characteristic for which generic smoothness does not hold?

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    $\begingroup$ Of course, take any smooth $F$-regular $X$ and $f\colon X\to Y$ equal to the Frobenius $F\colon X\to X$. $\endgroup$
    – Piotr Achinger
    Commented Jun 21, 2022 at 6:32
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    $\begingroup$ P.S. Generic smoothness of $f$ is equivalent to geometric regularity of its generic fiber. With $X$ smooth, the generic fiber of any $f$ is regular, which in char. 0 implies geometrically regular. I hope this helps. $\endgroup$
    – Piotr Achinger
    Commented Jun 21, 2022 at 6:44
  • $\begingroup$ There is something I don't understand about your example @PiotrAchinger. If $ F_{\ast}(\mathcal{O}_{X}) $ is locally free, then $ f: X \to Y $ is generically flat. So the question of smoothness is local. We may assume that $ Y $ is equal to $ \operatorname{Spec}(k) $ by restricting to any point in the flat locus. If $ F_{\ast}(\mathcal{O}_{X}) $ is locally free then Kunz's theorem suggests that the fibre is regular. What is wrong with this argument? This seems to imply that for regular $ X $ that $ F: X \to X $ is generically regular. What am I missing? $\endgroup$
    – Schemer1
    Commented Jun 27, 2022 at 9:48
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    $\begingroup$ It's the difference between smoothness and regularity. Let's contemplate $X=\mathbf{A}^1_k$ ($k=\mathbf{F}_p$) with coordinate $t$, then the Frobenius is the inclusion $k[t^p]\to k[t]$ and the base change to the fraction field is $k(t^p)\to k(t)$. And indeed $\operatorname{Spec} k(t)$ is regular. But it is not smooth over $k(t^p)$, as it is not geometrically regular as a $k(t^p)$-scheme. In fact $k(t)\otimes_{k(t^p)} k(t) \simeq k(t)[s]/(s^p)$ is not reduced. $\endgroup$
    – Piotr Achinger
    Commented Jun 27, 2022 at 13:20
  • $\begingroup$ Thank you @Piotr Achinger!! $\endgroup$
    – Schemer1
    Commented Jun 27, 2022 at 14:04

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