Suppose $(X,\Delta\ge 0)$ is a pair such that $(p^g-1)(K\_X+\Delta)$ is an Integral Weil Divisor for some $g>0$ and $X$ is a normal variety. Define $\mathcal{L}\_{e,\Delta} = \mathcal{O}\_X( (1-p^g)(K\_X + \Delta) )$ and note it is a reflexive sheaf.


We can still define the map $\phi^e: \mathcal{L}\_{e, \Delta} \to \mathcal{O}\_X$ for $g|e$, following the Grothendieck Trace map, Reflexivity of Weil divisors and normality of $X$. Now my question is, can I define the *non-$F$-pure* ideal $\sigma(X, \Delta)$ of $(X, \Delta$) as $\sigma(X, \Delta)=\bigcap\_{e\ge 0} \phi^{eg}F^{eg}\_\*\mathcal{L}\_{e, \Delta}$ ?


What I mean is that, will this definition have all the good properties of the $\mathbb{Q}$-Cartier case? For example, I need to know whether $\phi^{eg}F^{eg}\_\*\mathcal{L}\_{e, \Delta}=\sigma(X, \Delta)$ for all $e\gg 0$ or not?


Also, is $\sigma(X, \Delta)$ the Unique Largest ideal $\mathcal{I}$ such that 
$\phi^{eg}F^{eg}\_\*(\mathcal{L}\_{e, \Delta}\cdot \mathcal{I})=\mathcal{I}$ for all $e>0$?



On Non F-pure ideal and Sharp F-Purity for a pair $(X, \Delta)$ where $K_X+\Delta$ is NOT $\mathbb{Q}$-Cartier
I think the answer is that we don't know.


You can certainly make the definition of $\sigma(X, \Delta)$ as that intersection. It seems a reasonable one to me. The intersection is still descending. 


However, I don't think we know whether that intersection stabilizes (if I recall correctly, the usual proof doesn't work, but I haven't checked it recently). I wouldn't be surprised if the image stabilized though. Here's a subtle point that we'd need to run the usual proof.


Suppose $\phi^e(F^e\_\* \mathcal{L}\_{e, \Delta}) = \phi^{e+g}(F^{e+g}\_\* \mathcal{L}\_{e+g, \Delta})$. Is it automatically true that 
$$
\phi^e(F^e\_\* \mathcal{L}\_{e, \Delta}) = \phi^{e+g}(F^{e+g}\_\* \mathcal{L}\_{e+g, \Delta}) = \phi^{e+g}(F^{e+2g}\_\* \mathcal{L}\_{e+2g, \Delta}) = \ldots?
$$
In the case that $(p^g -1 )(K\_X + \Delta)$ is Cartier, this is easy (work locally, then use the projection formula). However, in the more general case I don't see how to do it. Just because $\phi^e$ and $\phi^{e+g}$ have the same image in $O\_X$, I don't see why 
$$
F^{e+g}\_\* \mathcal{L}\_{e+g, \Delta} \to \mathcal{L}\_{g, \Delta} \text{ and } F^{e+2g}\_\* \mathcal{L}\_{e+2g, \Delta} \to \mathcal{L}\_{g, \Delta} 
$$
should have the same image. These maps are obtained from the maps you wrote by tensoring and then reflexifying (and reflexifying could break things a priori).


Still, I wouldn't be surprised if these images stabilized for some more subtle reason (there are other things worth trying too in this direction too).


I would be more surprised if the second statement about $\sigma(X, \Delta)$ being the unique largest ideal satisfying that property was true, but I don't know.


