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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 $e>0$?

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up vote 2 down vote accepted

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.

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Thank you very much Karl. One last question, is there anyway to define Sharp F-purity or Strongly F-regular for this kind of pairs ? –  Omprokash Das Aug 17 '13 at 2:13
    
Yes there are and this is simpler. The easiest way is a local definition. In the Sharp F-pure case, it's easily seen to be equivalent to there locally existing a boundary $\Delta' \geq \Delta$ such that $(K_X + \Delta')$ is $\mathbb{Q}$-Cartier with index not divisible by $p$ and such that $(X, \Delta')$ is sharply $F$-pure. In the strongly $F$-regular case it can be shown to be equivalent to the analogous condition, but this is more work (see my paper Test ideals in Q-Gorenstein rings or my paper with Karen Globally $F$-regular vs log Fano... Hopefully this helps. –  Karl Schwede Aug 17 '13 at 2:32
    
Thanks again, I will look at those papers. –  Omprokash Das Aug 17 '13 at 4:00
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