A variety $ X $ is $ F $-split if there exists an $ \mathcal{O}_{X} $-linear map $ \phi: F_{\ast}(\mathcal{O}_{X}) \to \mathcal{O}_{X} $ such that $ \phi \circ F^{\sharp} = \operatorname{id}_{\mathcal{O}_{X}} $. Such a map $ \phi $ is called a splitting. A closed sub-scheme $ Y $ of $ X $ is compatibly split if there exists a splitting $ \phi $ such that $ \phi(F_{\ast}(\mathcal{I}_{Y})) \subseteq \mathcal{I}_{Y} $. Note that $ \mathcal{I}_{Y} \subseteq \phi(F_{\ast}(\mathcal{I}_{Y})) $ already.
If $ X $ is a normal variety, then $ X $ if $ F $-regular if for all effective Weil divisors $ D $ there is an $ e \in \mathbb{N} $ such that $ \mathcal{O}_{X} \to F^{e}_{\ast}(\mathcal{O}_{X}) \to F^{e}_{\ast}(\mathcal{O}_{X}(D)) $ splits.
Does anyone know an example of a projective, $ F $-regular variety $ X $ and a smooth sub-variety $ Y $ such that $ \operatorname{Bl}_{Y}(X) $ is not $ F $-regular?
My initial inclination is that one should not exist. The reason is the following. The variety $ X $ is $ F $-regular if and only if there is an ample divisor $ D $ such that the section ring $ R(X,D) $ equal to $ \oplus_{m \in \mathbb{N}_{0}} H^{0}(X, \mathcal{O}_{X}(mD)) $ is $ F $-regular.
If $ E $ is the exceptional divisor of $ \operatorname{Bl}_{Y}(X) $, then there exist $ a,b \in \mathbb{N} $ such that $ aD-bE $ is ample. If $ \pi: \operatorname{Bl}_{Y}(X) \to X $ is the natural projection morphism, and we denote $ \operatorname{Bl}_{Y}(X) $ by $ Z $, then $ R(X,D) \cong \left(\oplus_{0 < i <b} \oplus_{m \in \mathbb{N}_{0}} R(Z, a\pi^{\ast}(D)-(mb+i)E)\right) \oplus R(Z, a\pi^{\ast}(D)-bE) $. If a ring inclusion $ R \subseteq S $ splits as a morphism of $ R $-modules, and if $ S $ is $ F $-regular, then so is $ R $. As a result $ R(Z, \pi^{\ast}(D)-bE) $ is $ F $-regular. Therefore, $ \operatorname{Bl}_{Y}(X) $ would be $ F $-regular as well. This seems too good to be true, so I thought I would ask fellow mathematicians if they knew of a counterexample.
