Proof. The only if part is obvious. Assume s: 𝒪_X → 𝒪_X(D) and t: 𝒪_X → 𝒪_X(−D) are nontrivial sections. We have isomorphisms of sheaves $\mathop {\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_X}(\mathcal{O}_X, \mathcal{O}_X(-D))\cong \mathop {\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_X}(\mathcal{O}_X(D), \mathcal{O}_X)$ and $\mathop {\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_X}(\mathcal{O}_X(D), \mathcal{O}_X(D))\cong \mathop {\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_X}(\mathcal{O}_X, \mathcal{O}_X)$ . Therefore, we can interpret t as a global section of $\mathop {\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_X}(\mathcal{O}_X(D), \mathcal{O}_X)$ . The compositions s ∘ t and t ∘ s are both nonzero and give elements in H ⁰(X,𝒪_X), thus they are isomorphisms. Hence, 𝒪_X ≅ 𝒪_X(D), which shows that D is trivial.

Proof. Let ω _Z^• := p ^! ω _X^• = q ^! ω _Y^•. Recall, e.g., from [Sta24, Tag 0AU3], that the functor ${\rm R}\mathop {\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_Z}(-, \omega _Z^\bullet )$ defines an involution of D_Coh(𝒪_Z) and that we have the duality formula

$${\rm R}\mathop {\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_Z}(p^! M, \omega _Z^\bullet ) = {\rm L}p^*{\rm R}\mathop {\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_X}(M, \omega _X^\bullet ), $$

which holds naturally in M ∈ D_Coh⁺(𝒪_X) and similarly for q: Z → Y. Setting M = 𝒪_X or M = 𝒪_Y yields

$$ {\rm R}\mathop {\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_Z}(p^! \mathcal{O}_X, \omega _Z^\bullet ) = {\rm L}p^* \omega _X^\bullet \quad {\rm and} \quad {\rm R}\mathop {\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_Z}(q^! \mathcal{O}_Y, \omega _Z^\bullet ) = {\rm L}q^* \omega _Y^\bullet . $$

Therefore, p ^!𝒪_X ≅ q ^!𝒪_Y if and only if Lp*ω _X^• ≅ Lq*ω _Y^•.

Proof. Without loss of generality we can assume U = X. Let Spec A ⊆ X be an affine open subset and let Spec B := π ⁻¹(SpecA) ⊆ Y be the pre-image under π. Let s ∈ S := H ⁰(Y,𝒪_Y) be nonzero. Since A → B is integral, there exists a monic irreducible polynomial P _A ∈ A[T] such that P _A(s) = 0 in B. By normality of X, A is integrally closed in K(X), and it follows that P _A is irreducible as a polynomial in K(X)[T]. Indeed, if P _A = QR in K(X)[T] with Q and R monic, then denoting C the integral closure of A in a splitting field of P _A over K(X), we have Q, R ∈ C[T] ∩ K(X)[T] = A[T]. Let now Spec A′ ⊆ X be another affine open subset and let P ∈ K(X)[T] be the greatest common divisor (gcd) of P _A and P _A′. Then P(s) = 0, so P is not constant as s ≠ 0 in K(Y) by integrality of Y. As P _A and P _A′ are irreducible, we obtain P _A = P = P _A′. It follows that P = P _A has coefficients in R = H ⁰(X,𝒪_X) = ⋂A, where the intersection runs over all affine open subsets of X. Thus s is integral over R.

If Y is normal, then H ⁰(V,𝒪_Y) = ⋂_y ∈ V𝒪_Y, y ⊆ K(Y) is integrally closed for every open subset V ⊆ Y. Hence, for every open subset U ⊆ X, H ⁰(π ⁻¹(U),𝒪_Y) is the integral closure of H ⁰(U,𝒪_X) in K(Y). This coincides with the relative normalization in [Sta24, Tag 0BAK] if U is affine.

Proof. If L/K is not separable, then tr_L/K is zero. Thus we can and do assume that L/K is separable. Let M be the Galois closure of L over K, denote by G = Gal(M/K) the Galois group, and let H ⊆ G be the stabilizer of L. Let T ⊆ M be the integral closure of R in M. Since R ↪ S is integral, S ⊆ T. Note that, for any s ∈ L, we have ${\rm tr}_{L/ K} (s) = \sum _{g \in G/H} g(s)$ , where g runs over a set of coset representatives of G/H. By definition of the integral closure, any automorphism g ∈ G satisfies g(T) = T ⊆ M, so tr_L/K(s) lies in T ∩ K = R for any s ∈ S. Similarly, any g ∈ G preserves $\sqrt {T \mathfrak{p}} \subseteq T$ . Hence, for any $s \in \sqrt {S \mathfrak{p}} \subseteq \sqrt {T \mathfrak{p}}$ , tr_L/K(s) lies in $\sqrt {T \mathfrak{p}}\cap R = \mathfrak{p}$ .

Proof. Assume π is quasi-étale. Let U ⊆ X be an open subset with codim_X(X\U) ≥ 2 such that $\pi \vert _U \colon \pi ^{-1}(U) \to U$ is finite étale. The map $\mathcal{O}_Y \to \pi ^!\mathcal{O}_X = \mathop {\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_X}(\pi _*\mathcal{O}_Y, \mathcal{O}_X), 1 \mapsto {\rm tr}_{Y/X}$ is a map of quasi-coherent sheaves on Y that restricts to an isomorphism on π ⁻¹(U), and hence is an isomorphism on Y since both sheaves are reflexive after applying π _*. Conversely, we claim that, if π: Y → X is a finite surjective morphism with Y integral, and X normal and universally catenary, then there exists an open subset U ⊆ X with codim_X(X\U) ≥ 2 such that $\pi \vert _U \colon V:= \pi ^{-1}(U) \to U$ is finite flat. This suffices to conclude since by [Sta24, Tag 0BVH] the classical trace for $\pi \vert _U$ is nonsingular if and only if $\pi \vert _U$ is étale. Let V ⊆ Y be the flat locus of π, which by [Sta24, Tag 0398] is open, and let Z := Y \ V. Since X is universally catenary, any codimension-1 point of Y is mapped to a codimension-1 point of X by [Sta24, Tag 02JT], and by [Reference HartshorneHar77, Chapter III, Proposition 9.7], V contains all codimension-1 points of Y. Hence Z has codimension at least 2 in Y, and so π(Z) defines a closed subscheme of codimension at least 2 in X. We may then choose U := X \ π(Z).

Proof. For a proper variety X over a field of positive characteristic, there exists, by [Reference BhattBha12, Proposition 7.2], for any i > 0 a finite surjective morphism π: Y → X such that the induced map H ⁱ(X,ℒ) → H ⁱ(Y,π*ℒ) is zero. If now X is a splinter, the pullback map 𝒪_X → π _*𝒪_Y admits a splitting s, i.e., we have

$$ \operatorname {id} \colon \mathcal{O}_X \to \pi _*\mathcal{O}_Y \stackrel {s}{\to } \mathcal{O}_X.$$

Tensoring with ℒ and using the projection formula, we obtain

$$ \operatorname {id} \colon H^i(X,\mathcal{L}) \xrightarrow {0} H^i(Y,\pi ^*\mathcal{L}) = H^i(X,\pi _* \pi ^*\mathcal{L}) \to H^i(X,\mathcal{L}),$$

where the equality in the middle uses that π is finite, in particular affine. We conclude that H ⁱ(X,ℒ) = 0.

Proof. Since a splinter is normal, by working on each connected component of X separately, we can assume that X is of pure positive dimension, say n. By Proposition 3.5, X is Cohen–Macaulay, so ω _X^• = ω _X[n]. By Proposition 3.6 and Serre duality for Cohen–Macaulay schemes, we obtain H ⁰(X,ω _X)^∨ ≅ H ⁿ(X,𝒪_X) = 0. This shows that the Weil divisor K _X is not effective.

Proof. Fix a point x ∈ X. If c ∈ 𝒪_X, x is a nonzero element, c defines an effective divisor in the neighborhood of x. By taking the Zariski closure, this divisor extends to a Weil divisor D on X such that the map 𝒪_X → 𝒪_X(D) localizes to the map 𝒪_X, x → 𝒪_X, x[c ⁻¹] sending 1 ↦ 1. Since X is globally F-regular, there exists e > 0 such that 𝒪_X → F _*^e𝒪_X(D) splits. Thus, localizing yields a splitting of the map 𝒪_X, x → F _*^e(𝒪_X, x[c ⁻¹]) sending 1 ↦ 1. Multiplying by c yields a splitting of the map 𝒪_X, x → F _*^e𝒪_X, x sending 1 ↦ F _*^e c.

Proof. We only prove (iii) since (i) and (ii) admit similar simpler proofs. First note that the condition that 𝒪_X → π _*𝒪_Y splits forces π to be dominant. We can assume that X and Y are integral. Thus, the pullback π*Δ is well defined if Δ is ℚ-Cartier [Sta24, Tag 02OO], or if π is finite. Let f: Z → X be a finite surjective morphism with Z normal. Let Z′ → Y×_XZ be the normalization of the fiber product, which is finite over Y since Y is excellent. (This step is not needed for the proofs of (i) and (ii) where one simply takes Z′ = Y×_XZ.) We obtain the following commutative square.

Since (Y,π*Δ) is globally +-regular, we have a splitting

$$\mathcal{O}_Y\to f_*' \mathcal{O}_{Z'}(\lfloor f'^*\pi ^* \Delta \rfloor ) \xrightarrow {t}\mathcal{O}_Y.$$

Fixing a splitting $\mathcal{O}_X \to \pi _*\mathcal{O}_Y \xrightarrow {s} \mathcal{O}_X$ , we obtain a factorization

$$\operatorname {id}_{\mathcal{O}_X}\colon \mathcal{O}_X \to \pi _* \mathcal{O}_Y \to \pi _* f_*' \mathcal{O}_{Z'}(\lfloor f'^*\pi ^* \Delta \rfloor ) = f_* \pi '_* \mathcal{O}_{Z'}(\lfloor \pi '^* f^* \Delta \rfloor ) \xrightarrow {\pi _* t} \pi _* \mathcal{O}_Y \xrightarrow {s} \mathcal{O}_X.$$

Note that π′*⌊f*Δ⌋ ≤ ⌊π′*f*Δ⌋. Hence, 𝒪_X → f _*π′_*𝒪_Z′(⌊π′*f*Δ⌋) factors through 𝒪_X → f _*𝒪_Z(⌊f*Δ⌋) → f _*π′_*π′*𝒪_Z(⌊f*Δ⌋) and the lemma follows.

Proof. Let f: Z → Y be a finite surjective morphism. The splitting of f ^♯: 𝒪_Y → f _*𝒪_Z is equivalent to the surjectivity of

$$\operatorname {Hom}_{\mathcal{O}_Y}(f_*{\mathcal{O}_Z}, \mathcal{O}_Y) \xrightarrow {-\circ f^\sharp } \operatorname {Hom}_{\mathcal{O}_Y}(\mathcal{O}_Y, \mathcal{O}_Y).$$

The adjunction π _* ⊣ π ^! provides the following commutative diagram.

(5.4)

Here, Tr_Y/X is simply defined as the composite of the right vertical arrows. In case (a), we choose the isomorphism π ^!𝒪_X ≅ 𝒪_Y to be the one induced by the classical trace map tr_Y/X; under the identifications Hom_𝒪Y (𝒪_Y,𝒪_Y) = H ⁰(Y,𝒪_Y) and Hom_𝒪X (𝒪_X,𝒪_X) = H ⁰(X,𝒪_X), the map Tr_Y/X agrees with tr_Y/X. Since X is a splinter, the bottom-left vertical arrow  −  ∘ (π∘f)^♯ is surjective. This implies on the one hand that  −  ∘ f ^♯ is nonzero, and on the other hand that  −  ∘ π ^♯ is surjective. This is already sufficient to establish Lemma 5.3 under condition (b). Assuming that H ⁰(Y,𝒪_Y) is a field, that  −  ∘ f ^♯ is nonzero yields that  −  ∘ f ^♯ is surjective since it is a map of H ⁰(Y,𝒪_Y)-modules. Assuming that H ⁰(X,𝒪_X) = H ⁰(Y,𝒪_Y), that  −  ∘ π ^♯ is surjective yields that the composition of the right vertical arrows, which is H ⁰(X,𝒪_X)-linear, is bijective and hence that  −  ∘ f ^♯ is surjective.

We now prove Lemma 5.3 under conditions (a) or (c). The proof works uniformly for both cases, but we refer to Remark 5.8 for a direct proof in case π is étale. We write R = H ⁰(X,𝒪_X) and S = H ⁰(Y,𝒪_Y). Recall that the map $\coprod \operatorname {Spec} R_z^h \to \operatorname {Spec} R$ , where the coproduct runs through all closed points z of Spec R and where R _z^h is the henselization of R _z, is faithfully flat [Sta24, Tag 07QM]. We thus have to prove that, for all closed points of Spec R, the map  −  ∘ f ^♯ is surjective after base change to R _z^h. Write π ^h: Y ^h → X ^h and f ^h: Z ^h → Y ^h for the pullbacks of π and f along R → R _z^h, respectively. It is a priori not clear that R is Noetherian; however, since Spec R _z^h → Spec R is affine, X ^h and Y ^h are quasi-compact and quasi-separated. By base change for the exceptional inverse image [Reference Görtz and WedhornGW23, Theorem 25.31], we have h*π ^!𝒪_X = (π ^h)^×𝒪_{X ^h} in D_QCoh(𝒪_{Y ^h} ), where h: Y ^h → Y is the natural map and (π ^h)^× is the right adjoint to Rπ _*^h: D_QCoh⁺(𝒪_Y) → D_QCoh⁺(𝒪_X). Since h is flat, we have ℋ⁰(h*π ^!𝒪_X) = h*ℋ⁰(π ^!𝒪_X), and so ℋ⁰((π ^h)^×𝒪_{X ^h} ) ≅ 𝒪_{Y ^h} . As outlined in § 2.3, (π ^h)^× is the right derived functor of $\mathop {\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_{X^h}}(\pi ^h_* \mathcal{O}_{Y^h}, -)$ and we thus have $\mathop {\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_{X^h}}(\pi ^h_* \mathcal{O}_{Y^h}, \mathcal{O}_{X^h}) \cong \mathcal{O}_{Y^h}$ as 𝒪_{Y ^h} -modules. On the other hand, for sheaves ℱ, 𝒢 ∈ Coh (X) we have $\operatorname {Hom}_{\mathcal{O}_X}(\mathcal{F}, \mathcal{G}) = H^0(X, {\rm R} \mathop {\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_X}( \mathcal{F}, \mathcal{G}))$ [Sta24, Tag 08DK]. Therefore, by flat base change of quasi-coherent cohomology and the compatibility of ${\rm R}\mathop {\mathcal{H}\! \mathit{om}}\nolimits (-, -)$ with flat pullback [Reference Görtz and WedhornGW23, Proposition 22.70], we obtain the same diagram as (5.4) with X replaced by X ^h, Y by Y ^h, Z by Z ^h, π ^! by (π ^h)^×, and where  −  ∘ (π ^h∘f ^h)^♯ is surjective. We are thus reduced to showing that  −  ∘ (f ^h)^♯ is surjective for every closed point z ∈ Spec R.

Since a splinter is normal, we have in both cases (a) and (c) that X and Y are normal. Working componentwise, we may and do assume that X and Y are both integral. In particular, as discussed in § 2.4, both R and S are integrally closed domains in K(X) and K(Y), respectively. Normality of rings is preserved under smooth base change [Sta24, Tag 033C] and under filtered colimits [Sta24, Tag 037D], and hence X ^h and Y ^h are normal. Since Y ^h is quasi-compact, Y ^h has only finitely many connected components. Let Y ^h = Y ₁ ∐ … ∐ Y _r be the decomposition into connected components and let S⊗_RR _z^h = H ⁰(Y ^h,𝒪_{Y ^h} ) = S ₁ × … × S _r, where S _i = H ⁰(Y _i,𝒪_{Y i} ). By Lemma 2.7, R ↪ S is integral, hence each S _i is integral over R _z^h.

Recall that R _z^h is the filtered colimit of the pointed étale ring maps (R _z,z) → (T,𝔮) such that κ(z) = κ(𝔮). Each projector S⊗_RR _z^h → S _i is defined by an idempotent in S⊗_RR _z^h. Thus, there exists a pointed étale ring map (R _z,z) → (T,𝔮) such that each projector S⊗_RR _z^h → S _i is already defined in S⊗_RT, i.e., S⊗_RT = S ₁′ × … × S _r′ with S _i = S _i′⊗_TR _z^h. We then have Y _T := Y×_Spec RSpec T = Y ₁′ ∐ … ∐ Y _r′ with Y _i = Y _i′×_Spec TS _i. Note that Y _T is normal Noetherian so that each Y _i′ is an integral normal Noetherian scheme. Using that for any normal R-algebra A and any étale ring morphism T → T′ the base change map A⊗_RT → A⊗_RT′ is injective (by openness of flat morphisms locally of finite presentation [Sta24, Tag 01UA]), and that a filtered colimit of integral domains is an integral domain, we see that any affine open cover of Y′_i yields an affine open cover of Y _i by integral domains and hence that Y _i isintegral.

In particular, the rings S _i are normal domains. We claim that each S _i is a local domain. Indeed, R _z^h → S _i is an integral ring map and S _i contains no nontrivial idempotent elements. Since R _z^h is henselian, any finite R _z^h-subalgebra of S _i is a finite product of local rings [Sta24, Tag 04GG (10)]. Since any such subalgebra contains no nontrivial idempotents it is a local ring. Hence, S _i is a filtered colimit of local rings and because S _i is integral over R _z^h, each morphism in the directed system is a morphism of local rings and so S _i is a local ring.

Moving forward, since we are assuming X to be Nagata, after possibly normalizing Z and working on each component separately, we may and do assume that Z is integral and normal. By further taking the normalization of Z in a normal closure of K(Z) over K(X) we may and do assume that K(Z)/K(X) is normal. We summarize the situation in the commutative diagram

where f _i: Z _i → Y _i is the base change of Z ^h along S⊗_RR _z^h → S _i and g _i = π _i ∘ f _i: Z _i → X ^h.

Proof of the Claim. Let G be the Galois group of K(Z) over K(X) and let T := H ⁰(Z,𝒪_Z) be the integral closure of R in K(Z). The field extension K(Z)^G/K(X) is purely inseparable and the ring homomorphism R → T ^G to the invariant subring T ^G, which is the integral closure of R in K(Z)^G, is a homeomorphism on spectra. By a straightforward variant of [Sta24, Tag 0BRK] for domains that are integrally closed in a field, G acts transitively on the maximal ideals 𝔭₁, …, 𝔭_s of T lying over z. As discussed previously in the case of the integral extension R → S, T⊗_RR _z^h splits into a product of local rings T ₁ × … × T _s. Let Z _j′ be the connected component of Z ^h with H ⁰(Z _j′,𝒪_{Z j′}) = T _j. The induced action of G on Z ^h then permutes the morphisms g _j′: Z _j′ → X ^h and so induces isomorphisms (g _j′)_*𝒪_{Z j′} ≅ (g _k′)_*𝒪_{Z k′} of 𝒪_{X ^h} -algebras. Since X is a splinter, the pullback 𝒪_X → π _*f _*𝒪_Z splits. By base change, such a splitting induces a splitting

$$\mathcal{O}_{X^{h}} \to \bigoplus _{j=1}^r (g_j')_*\mathcal{O}_{Z_j'} \xrightarrow {\sum _j s_j} \mathcal{O}_{X^{h}}\!.$$

Since $\sum _j s_j$ sends the constant section 1_{X ^h}  ↦ 1_{X ^h}  ∈ R _z^h, there exists j ₀ such that s _{j 0} (1_{X ^h} ) ∉ 𝔪_z^h, where 𝔪_z^h denotes the maximal ideal of R _z^h. Thus, 𝒪_{X ^h}  → (g _{j 0} ′)_*𝒪_{Z j 0 ′} splits. Using the isomorphisms (g _j′)_*𝒪_{Z j′} ≅ (g _k′)_*𝒪_{Z k′}, we see that every 𝒪_{X ^h}  → (g _j′)_*𝒪_{Z j′} splits. The claim then follows, as each Z _i is a finite disjoint union of some connected components Z _j′.

Now the map  −  ∘ (f ^h)^♯ decomposes as the direct sum

$$ - \circ \oplus _i f_i^\sharp \colon \bigoplus _{i=1}^r \operatorname {Hom}_{\mathcal{O}_{Y_i}}((f_i)_* \mathcal{O}_{Z_i}, \mathcal{O}_{Y_i}) = \operatorname {Hom}_{\mathcal{O}_{Y^{h}}} (f^{h}_* \mathcal{O}_{Z^{h}}, \mathcal{O}_{Y^{h}}) \to \operatorname {Hom}_{\mathcal{O}_{Y^{h}}}\left (\mathcal{O}_{Y^{h}},\mathcal{O}_{Y^{h}}\right )= \bigoplus _{i=1}^r S_i. $$

By the claim above, 𝒪_{X ^h}  → (g _i)_*𝒪_{Z i} splits for every i. Thus, the left-vertical arrow in (5.4) base changed to R _z^h remains surjective after restricting to a component Y _i of Y ^h. Hence, it is enough to show that any morphism of rings ϕ _i: T _i → S _i such that the composite Tr_{Y i/X ^h}  ∘ ϕ _i: T _i → R _z^h is surjective is also surjective. Denote 𝔪 ⊆ R and 𝔫_𝔦 ⊆ S _i the maximal ideals.

Under condition (c), we still have that R _z^h → S _i has reduced closed fibers, and hence 𝔪S _i = 𝔫_i. By R _z^h-linearity of the map Tr_{Y i/X ^h} , we get Tr_{Y i/X ^h} (𝔫_i) ⊆ 𝔪. Thus, any element s ∈ S _i such that Tr_{Y i/X ^h} (s) = 1 is a unit in S _i. Since ϕ _i is S _i-linear, we conclude that ϕ _i is surjective.

Assume now condition (a). Note that the restriction of the classical trace for Y/X is the classical trace for Y _i/X ^h; indeed, they both agree on the base change along R → R _z^h of the flat locus U ⊆ X of π and then use that X ^h is integral. Note that for any integral extension of local rings ϕ: (A,𝔞) ↪ (B,𝔟) we have $\mathfrak b = \sqrt {\mathfrak{a}B}$ ; this is indeed clear if ϕ is finite and the general case follows by a limit argument. By Lemma 2.8 and the discussion in § 2.4, the trace tr_{Y i/X ^h} : S _i → R _z^h sends $\mathfrak{n}_i = \sqrt {\mathfrak{m}S}$ into 𝔪. We can then conclude as before.

Proof. This is proved as in Lemma 5.3(b) by using the adjunction Rπ _* ⊣ π ^! for a proper morphism π: Y → X instead of the adjunction π _* ⊣ π ^! for a finite morphism π: Y → X.

Proof. If X is a derived splinter, then it is normal so that π _*𝒪_Y = 𝒪_X and hence π ^♯: H ⁰(X,𝒪_X) → H ⁰(Y,𝒪_Y) is an isomorphism. We can then apply Lemma 5.10. The statement (i) follows from the fact [Reference BhattBha12, Theorem 1.4] that splinters in positive characteristic agree with derived splinters.

For (iii), choose a splitting $\mathcal{O}_X \to F_*\mathcal{O}_X \xrightarrow {s} \mathcal{O}_X$ . Let s̃ be the image of s under

\begin{align*} \operatorname {Hom}_{\mathcal{O}_X}(F_* \mathcal{O}_X, \mathcal{O}_X) = &\operatorname {Hom}_{\mathcal{O}_X}(\mathcal{O}_X,F^! \mathcal{O}_X) \to \operatorname {Hom}_{\mathcal{O}_Y}(\pi ^!\mathcal{O}_X, \pi ^!F^!\mathcal{O}_X) \\ = &\operatorname {Hom}_{\mathcal{O}_Y}(\pi ^!\mathcal{O}_X, F^! \pi ^!\mathcal{O}_X) \cong \operatorname {Hom}_{\mathcal{O}_Y}(\mathcal{O}_Y,F^! \mathcal{O}_Y) = \operatorname {Hom}_{\mathcal{O}_Y}(F_*\mathcal{O}_Y, \mathcal{O}_Y). \end{align*}

Then $\mathcal{O}_Y \to F_* \mathcal{O}_Y \xrightarrow {\tilde {s}} \mathcal{O}_Y$ is the identity. Indeed, since Y is integral, this can be verified after restricting to the dense open subset where π is an isomorphism so that the claim follows from the fact that s is a splitting of 𝒪_X → F _*𝒪_X.

Proof. Let f: Z → Y be a finite surjective morphism with Z normal. We have to show that the map f _π*Δ^♯: 𝒪_Y → f _*𝒪_Z(⌊f*π*Δ⌋) splits, or equivalently, that

$$\operatorname {Hom}_{\mathcal{O}_Y}(f_*{\mathcal{O}_Z}(\lfloor f^* \pi ^*\Delta \rfloor ), \mathcal{O}_Y) \xrightarrow {-\circ f^\sharp _{\pi ^*\Delta }} \operatorname {Hom}_{\mathcal{O}_Y}(\mathcal{O}_Y, \mathcal{O}_Y)$$

is surjective. Note that the map (π∘f)_Δ^♯: 𝒪_X → π _*f _*𝒪_Z(⌊f*π*Δ⌋) factors through 𝒪_X → π _*𝒪_Y. Hence, the following diagram commutes.

We then proceed as in the proof of Lemma 5.3.

Proof. The ‘only if’ part of the proposition follows from Lemma 5.1. Let K := H ⁰(X,𝒪_X). First assume L is a finite extension of K and let h: Spec L → Spec K. Then

$$h^!\mathcal{O}_{\operatorname {Spec} K} =\operatorname {Hom}_K(L, K) \cong L = \mathcal{O}_{\operatorname {Spec} L}$$

as L-vector spaces. By base change for the exceptional inverse image [Sta24, Tag 0E9U], there exists an isomorphism π ^!𝒪_X ≅ 𝒪_{X L} , where π: X _L → X is the base change of h along X → Spec K. We conclude by Lemma 5.3 that X _L is a splinter, since

$$H^0(X_L, \mathcal{O}_{X_L})=H^0(X, \mathcal{O}_X) \times _K L=L$$

by flat base change. Now assume K → L is any algebraic field extension and take a finite cover f: Y → X _L. Since f is defined by finitely many equations and L is algebraic over K, we can find a finite cover f′: Y′ → X _L′ such that K ⊆ L′ ⊆ L is an intermediate extension, finite over K, and f = f′×_L′L is the base change of f′. By the previous argument X _L′ is a splinter, thus 𝒪_{X L′}  → f′_*𝒪_Y′ admits a section s: f′_*𝒪_Y′ → 𝒪_{X L′} . Now, pulling back to X _L and using flat base change, we obtain the desired section of 𝒪_{X L}  → f _*𝒪_Y.

Proof. The ‘only if’ part of the corollary follows from Lemma 5.1. Let k̄ be an algebraic closure of k. By the assumption on H ⁰(X,𝒪_X), X _k̄ := X×_kk̄ is isomorphic to the disjoint union of dim_kH ⁰(X,𝒪_X) copies of X×_{H ⁰(X,𝒪X)}k̄. By the ‘if’ part of Proposition 5.14, if X is a splinter, then X _k̄ is a splinter. By Lemma 5.1, we get that X _K is a splinter.

Proof. By Proposition 3.5, a splinter is normal. In particular, H ⁰(X,𝒪_X) is a field. The corollary then follows from Proposition 5.14.

Proof. By Proposition 3.10, Y is a splinter and, by Lemma 5.1, X is a splinter. Hence, by Proposition 3.5, X is normal. We can now apply [Reference Schwede and SmithSS10, Corollary 6.4].

Proof. We first prove (a), (b), and (c). By Proposition 3.10, X is a splinter, and it follows from Lemma 5.3 that Y is a splinter and hence is normal. By normality of X, the complement of X _reg has codimension at least 2, and by finiteness of π, the complement of π ⁻¹(X _reg) in Y has codimension at least 2. Therefore, by Lemma 4.4, Y is globally F-regular if and only if π ⁻¹(X _reg) ⊆ Y is globally F-regular. Replacing X by X _reg we can and do assume that X is regular. Since Y is normal, Y _sing is a proper closed subset of Y and since π is finite, π(Y _sing) ⊆ X is also a proper closed subset. Let U ⊆ X be an affine open subset in the complement of π(Y _sing). By [Sta24, Tag 0BCU], D := X \ U has codimension 1, so defines a Weil divisor on X. Since X is regular, D is further Cartier and π*𝒪_X(D) is invertible. Note that the pullback π*D = π ⁻¹(D) of D is a Cartier divisor [Sta24, Tag 02OO]. Let σ _D: 𝒪_X → 𝒪_X(D) be the global section defined by the divisor D. Then the pullback π*σ _D defines a global section of π*𝒪_X(D) whose zero locus is precisely π ⁻¹(D). Thus π*σ _D is the global section of π*𝒪_X(D) = 𝒪_Y(π*D) defined by the divisor π*D [Sta24, Tag 0C4S].

Now π ⁻¹(U) = Y \ π*D is an affine open subset contained in the regular locus of Y, so is strongly F-regular. By Theorem 6.1, it is enough to show that there exists an e > 0 such that the map 𝒪_Y → F _*^e𝒪_Y(π*D) splits. Since X is globally F-regular, there exists an integer e > 0 and a splitting s such that

$$ \operatorname {id}_{\mathcal{O}_X} \colon \mathcal{O}_X \to F_*^e\mathcal{O}_X \xrightarrow {F_*^e(\sigma _D)} F_*^e \mathcal{O}_X(D) \xrightarrow {s} \mathcal{O}_X.$$

Since X is a splinter, we have a splitting

$$ \operatorname {id}_{\mathcal{O}_X} \colon \mathcal{O}_X \xrightarrow {\pi ^\sharp } \pi _* \mathcal{O}_Y \xrightarrow {t} \mathcal{O}_X.$$

Note that by the projection formula, t induces a splitting of 𝒪_X(D) → π _*π*𝒪_X(D) = π _*𝒪_Y(π*D) which by abuse we still denote by t. The composite map

(6.5) $$ \mathcal{O}_X \xrightarrow {\pi ^\sharp } \pi _* \mathcal{O}_Y \xrightarrow {\pi _*F^{e\sharp } } \pi _*F^e_*\mathcal{O}_Y \xrightarrow { \pi _* F_*^e(\pi ^*\sigma _D) } \pi _* F_*^e \mathcal{O}_Y (\pi ^*D) $$

then admits

$$ \pi _* F_*^e \mathcal{O}_Y (\pi ^*D) = F_*^e \pi _* \mathcal{O}_Y (\pi ^*D) \xrightarrow {F^e_*(t)} F^e_*\mathcal{O}_X(D) \xrightarrow {s} \mathcal{O}_X$$

as a splitting. As in Lemma 5.3, we consider the following diagram.

Since the splitting of (6.5) is equivalent to the left-vertical map  −  ∘ π _*F _*^e(π*σ _D) ∘ π _*F ^e♯ ∘ π ^♯ being surjective, in case (b) we conclude, as in the proof of Lemma 5.3, that the map  −  ∘ π*σ _D is surjective, or equivalently, that 𝒪_Y → F _*^e𝒪_Y(π*D) splits. In cases (a) and (c), we argue again as in the proof of Lemma 5.3, whose notation we take up. Let R = H ⁰(X,𝒪_X) and z ∈ Spec R be a closed point. Since henselization is a filtered colimit of étale ring maps, the base change of the absolute Frobenius F: Y → Y along Y ^h → Y identifies with the absolute Frobenius of Y ^h. Let π′: Y′ → Y be the normalization of Y in a normal closure of K(Y) over K(X). Arguing as above we obtain a splitting of the map 𝒪_X → π _*π _*′F _*^e𝒪_Y′(π′*π*D). This provides a splitting of the base change $\mathcal{O}_{X^h}\to \pi _*^h\pi '^h_* F_*^e\mathcal{O}_{Y'^h} (\pi '^{h*} \pi ^{h*} D\vert _{X^h})$ . We have a decomposition into connected components Y ^h = Y ₁ ∐ … ∐ Y _r and Y′^h = Y′₁ ∐ … ∐ Y′_s, where the induced action of Gal(K(Y′)/K(X)) permutes the morphisms Y _j′ → X ^h. Hence, we obtain a splitting of $\mathcal{O}_{X^h}\to \pi _*^h\pi '^h_* F_*^e\mathcal{O}_{{Y'}^h} (\pi '^{h*} \pi ^{h*} D\vert _{X^h})\vert _{Y'_j}$ for every j. As every Y _i is dominated by one Y _j′, we obtain that $\mathcal{O}_{X^h}\to \pi _*^h F_*^e\mathcal{O}_{Y^h} ( \pi ^{h*} D\vert _{X^h})\vert _{Y_i}$ splits for every i. We then conclude as in the proof ofLemma 5.3.

We now prove (d). By Proposition 3.10, X is a splinter, so a derived splinters since k is of positive characteristic [Reference BhattBha12, Theorem 1.4]. By Remark 2.4, π is generically finite. Let U ⊆ X be an affine regular dense open subset such that the restriction of π to U is finite. After possibly further shrinking U we can assume that π ⁻¹(U) is also regular. Since restriction to open subschemes commutes with exceptional inverse image functors [Sta24, Tag 0G4J], we have $\pi \vert _U^!\mathcal{O}_U \cong \mathcal{O}_{\pi ^{-1}(U)}$ . Since X is quasi-projective, any Weil divisor on X is dominated by a Cartier divisor. Thus, by possibly further shrinking U, we may and do assume that D := X \ U is a Cartier divisor. Since π ⁻¹(U) is affine an regular, it is enough to show, by Theorem 6.1, that there exists an e > 0 such that the map 𝒪_Y → F _*^e𝒪_Y(π*D) splits. By arguing as in case (b) with π _* replaced by Rπ _*, we conclude that a splitting of 𝒪_X → F _*^e𝒪_X(D) induces a splitting of 𝒪_Y → F _*^e𝒪_Y(π*D).

Proof. Denote by π _X and π _Y the natural projections from X×_kY to X and Y, respectively. Since X and Y are normal and k is perfect, their product X×_kY is normal, see [Sta24, Tag 038L]. Moreover, X \ X _reg and Y \ Y _reg both have codimension  ≥ 2 and thus X×_kY \ X _reg×_kY _reg has codimension  ≥ 2. Therefore, by Lemma 4.4, we can and do assume without loss of generality that X and Y are smooth over k.

Assume first that X×_kY is globally F-regular. As in Remark 5.2, since π _*𝒪_X×kY = 𝒪_X⊗_kH ⁰(Y,𝒪_Y), any splitting of the k-linear map k → H ⁰(Y,𝒪_Y), 1 ↦ 1_Y provides a splitting to the natural map 𝒪_X → π _*𝒪_X×kY. By Lemma 6.3, it follows that X is globally F-regular.

For the converse, we first note that there exist effective Cartier divisors D on X and E on Y such that X \ D and Y \ E are affine. Indeed, since X and Y are normal and k is perfect, they admit dense affine open subsets, and then use the fact that the complement of a dense affine open subset is a divisor by [Sta24, Tag 0BCU]. The divisors obtained this way are a priori Weil divisors, but since X and Y are smooth, they are actually Cartier divisors. Since k is assumed to be perfect

$$X \setminus D \times _k Y \setminus E=X\times _k Y \setminus \pi _X^*D \cup \pi _Y^*E$$

is smooth and affine, so in particular strongly F-regular. By Theorem 6.1 it is enough to show that the map

$$ \mathcal{O}_{X\times Y} \to F_*^e \mathcal{O}_{X\times Y} \to F_*^e \mathcal{O}_{X\times Y}(\pi _X^*D + \pi _Y^*E)$$

splits for some e > 0. Since X is globally F-regular, we can find an e > 0 such that 𝒪_X → F _*^e𝒪_X(D) splits and similarly for Y. As remarked in [Reference SmithSmi00, p. 558], if 𝒪_X → F _*^e𝒪_X(D) splits for some e > 0, then 𝒪_X → F _*^e′𝒪_X(D) splits for all e′ ≥ e. Thus, there exists an integer e > 0 such that both 𝒪_X → F _*^e𝒪_X(D) and 𝒪_Y → F _*^e𝒪_Y(E) split. The morphism

$$\sigma _{\pi _X^*D + \pi _Y^* E} \colon \mathcal{O}_{X\times Y} \to \mathcal{O}_{X\times Y}(\pi _X^*D + \pi _Y^* E)$$

can be identified with the tensor product π _X*σ _D ⊗ π _Y*σ _E, where

$$ \sigma _D\colon \mathcal{O}_X \to \mathcal{O}_X(D)\quad {\rm and} \quad \sigma _E \colon \mathcal{O}_Y \to \mathcal{O}_Y(E) $$

denote the corresponding morphisms on X and Y. Pushing forward along Frobenius, we obtain

$$F^e_*\sigma _{\pi _X^*D + \pi _Y^* E} = \pi _X^* F^e_*\sigma _D \otimes \pi _Y^* F^e_*\sigma _E.$$

We conclude, by taking the tensor product of the sections of 𝒪_X → F _*^e𝒪_X(D) and 𝒪_Y → F _*^e𝒪_Y(E), that

$$\mathcal{O}_{X\times Y} \to F_*^e \mathcal{O}_{X\times Y}(\pi _X^*D + \pi _Y^* E)$$

splits. Hence X×_kY is globally F-regular.

Proof. Let us provide an alternate proof. We may and do assume X is connected. Since any divisor on X _L is defined over a finitely generated field extension of the field H ⁰(X,𝒪_X), we may assume that L is a simple extension of H ⁰(X,𝒪_X). If L is algebraic, we can apply Proposition 6.4, while if L is purely transcendental, we can apply Proposition 6.7 (or Remark 6.8 in case Δ ≠ 0) and Lemma 4.5 to X×_{H ⁰(X,𝒪X)}𝔸¹ → 𝔸¹. Note that in this situation it is not necessary to assume that H ⁰(X,𝒪_X) is perfect as X _reg×_{H ⁰(X,𝒪X)}𝔸¹ is regular.