Let $(R,\mathfrak m)$ be a reduced Noetherian local ring of prime characteristic $p$. For integer $e>0$, let $F^e_* R$ denote the $R$-module which is $R$ as an abelian group, but the $R$-module structure is given by
$r\cdot s=r^{p^e}s, \forall r \in R, s\in F^e_* R$. Assume that $F^1_* R$ is a finitely generated $R$-module (hence so is $F^e_* R$ for all $e>0$).
My question is: When is it true that $R$ is a direct summand of $F^e_*R$ for infinitely many $e>0$? When is it true that $R$ is a direct summand of $F^e_*R$ for all large enough $e \gg 0$?
If $R$ is strongly $F$-regular, then it is clear from Theorem 0.2 of https://doi.org/10.4310/MRL.2003.V10.N1.A6 that $R$ is a direct summand of $F^e_*R$ for all large enough $e\gg 0$. But I would like to believe that more of class of rings satisfies one of the properties I want in my question.
Best Answer
If there exists one $e > 0$ so that $R \to F^e_* R$ splits, then by composing splittings one sees that $R \to F^{ne}_* R$ splits for all $n > 0$. Ie, if $\phi : F^e_* R \to R$ is a splitting (sends $F^e_* 1 \mapsto 1$), then $\phi \circ (F^e_* \phi) : F^{2e}_* R \to R$ also sends $F^{2e}_* 1 \mapsto 1$.
Next suppose that $0 < d < e$ and $R \to F^e_* R$ splits. Then $R \to F^d_* R \to F^e_* R$ splits and so $R \to F^d_* R$ splits as well. Thus by the previous observation, you have splitting for all $e > 0$ if you have it for $1$.
I don't think you need $F$-finite or even Noetherian for this stuff. However, $F$-splitting is probably not the right notion without $F$-finiteness (assuming $R$ is not complete Noetherian). $F$-purity is better behaved and agrees with $F$-splitting for $F$-finite rings.
Regardless, in summary:
If you have a specific ring that you want to check whether or not is $F$-split, you can use Fedder's criterion F-purity and rational singularity. The Macaulay2 package TestIdeals also checks $F$-splitting/$F$-purity.