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.



Frobenius splitting and smoothness
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](http://www.math.lsa.umich.edu/~hochster/711F07/fndtc.pdf)) 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. 


