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Let $R\rightarrow S$ be a morphism of rings in characteristic $p$ which is formally smooth.

Is it true that $R$ is Frobenius splitting if and only if $S$ is Frobenius splitting?

One direction seems to be easy but I am not sure if both implications are true.

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  • $\begingroup$ I don't recall seeing this in the literature. Although I'd point out that Frobenius splitting isn't such a good notion in the case of non-$F$-finite rings (ie, rings where the Frobenius is not a finite map). It might be better to look at $F$-purity. Do you really want to look at Frobenius splittings? $\endgroup$ Dec 12, 2013 at 14:09
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    $\begingroup$ @KarlSchwede you can assume $F$-finite if you want. Sorry for my ignorance but why is $F$-purity better in this case? I am interested in maps coming from geometry. In my situation I have a triangle $X\stackrel{f}{\leftarrow} Z \stackrel{g}{\rightarrow} Y$ of schemes where $f$ and $g$ are formally smooth and $Y$ is Frobenius split. I want deduce weak normality of $X$. If what I asked is true then also $X$ is Frobenius split hence weakly normal. $\endgroup$
    – alfs
    Dec 12, 2013 at 14:25
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    $\begingroup$ $F$-splitting is picking elements in $\Hom_R(F_* R, R)$. When $F_* R$ is not finitely generated, that $\Hom$ set is badly behaved (say with respect to localization or completion). $\endgroup$ Dec 12, 2013 at 16:04

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Consider the following. I'm going to assume that all rings are Noetherian (some things probably generalize, but I want to be careful). First a definition,

Definition: An extension of rings $A \subseteq B$ is pure if $M \otimes_A A \to M \otimes_A B$ is injective for every $A$-module $M$. A ring $A$ is called $F$-pure if the Frobenius map $F : A \to A$ is pure. For an $F$-finite ring, being $F$-pure is the same as being split.

Hochtser-Roberts, The purity of the Frobenius and local cohomology, Proposition 5.4(b) implies that if $A \to B$ is faithfully flat, then it's pure. I'm going to then assume that in your case, the ring maps are faithfully flat.

Theorem: (Hochster-Roberts, Prop 5.4(a)) If $A \subseteq B \subseteq C$ are rings and $B$ is pure in $C$, then $A$ is pure in $B$ if and only if $A$ is pure in $C$.

Now suppose that $R \subseteq S$ is faithfully flat (I'm going to assume you can do faithful, as otherwise there isn't any hope).

Ok, so then set $A = R$, $B=S$ and $C = S$ with $B \to C$ being the Frobenius on $S$. Likewise set $B' = R$ with $R = A \to B'$ being the Frobenius on $R$. In otherwords we have $$ A \subseteq B \subseteq C $$ and $$ A \subseteq B' \subseteq C $$ Since $S$ is $F$-pure, $B \to C$ is pure, and so $A \to C$ is pure since $A \to B$ is faithfully flat. But now certainly $B' \to C$ is also pure (since it is also faithfully flat) and so $A \to B'$ is also pure so that $R = A$ is $F$-pure.

Conclusion: If $S$ is $F$-pure and $R \subseteq S$ is faithfully flat, then $R$ is $F$-pure.


For the other direction, obviously faithful flatness will not be good enough but the following is true.

Theorem: (Hochster-Huneke?) If $(R, \mathfrak{m}) \subseteq (S, \mathfrak{n})$ is a flat local extension of local rings (say excellent to be safe) such that the closed fiber is regular and $R$ is $F$-pure, then $S$ is $F$-pure.

Hochster and Huneke proved a harder result (replacing $F$-purity by $F$-regularity in F-Regularity, Test Elements, and Smooth Base Change Theorem 7.3 and adding some assumptions about generic fibers which allow one to mess about with test elements, see the theorem on page 171 of these notes by Hochster) Their methods will prove the Theorem. The case you probably want follows directly from the proof of a result of myself and Wenliang Zhang, Lemma 4.5 in Bertini Theorems for $F$-singularities, but the above theorem was certainly known long before that by all the experts.

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  • $\begingroup$ thank you very much. I have the impression I am missing something. Assume $R\rightarrow S$ is smooth. We have a factorization $R\rightarrow C:=R[x_1,\dots,x_n]\rightarrow S$ where the second map is étale. Now if $A\rightarrow B$ is étale we have $F_{*}B\cong F_{*}A\otimes_A B$ (standard theorem in étale cohomology), hence if $A$ is split so is $B$. For $R\rightarrow C$, if $R$ is split the same holds for $C$ because we can send $x_i^p$ to $x_i$. So in conclusion if $R$ is split also $S$ split. In this case we don't need Hochster-Huneke. Am I wrong? $\endgroup$
    – alfs
    Dec 13, 2013 at 19:25
  • $\begingroup$ Dear alfs you are right. The Hochster-Huneke theorem actually holds for maps with regular fibers though (not necessarily geometrically regular), so the conclusion is stronger. Best wishes, $\endgroup$ Dec 13, 2013 at 21:59

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