Let $X$ be a normal projective variety over a field of characteristic $p>0$ and $(X, \Delta\geq 0)$ be a pair such that $K_X+\Delta$ is $\mathbb{Q}$-Cartier whose index is not divisible by $p$. Also assume that $A$ is an effective Cartier divisor, then $\tau(X, \Delta)=\tau(X, \Delta+\varepsilon A)$ for $0<\varepsilon\ll 1$. Now my question is, if $(X, \Delta)$ is Strongly $F$-regular does that imply $S^0(X, \tau(X, \Delta)\otimes {\mathcal{O}}_X(M))=S^0(X, \tau(X, \Delta ')\otimes {\mathcal{O}}_X(M))$, where $\Delta ' = \Delta+\varepsilon A$ and $M$ is a Cartier divisor ?

I know one of the inclusion is true for the obvious reason $\Delta '\geq \Delta$, how to prove the other inclusion, if it is true at all!



1 Answer 1


It isn't true as stated unfortunately. For example, take $X$ to be an ordinary elliptic curve, $\Delta = 0$ and $M = 0$. Then $S^0(X, \tau(X) \otimes O(M)) = H^0(X, O_X)$. However, for any effective Cartier $A > 0$ and any $\varepsilon > 0$, we have $S^0(X, \tau(X, \varepsilon A) \otimes O_X(M)) = 0$ (this can be checked easily with a direct computation).

However: Probably it is true for something like $P^0$, for a definition see Test ideals of non-principal ideals: Computations, Jumping Numbers, Alterations and Division Theorems (there are some modifications one can make to that definition too which might make this easier).

Definitely it is true for $P^0$ under suitable positivity assumptions. What can you assume about $M - K_X - \Delta$?.

  • $\begingroup$ Thank you very much Karl for your reply! Now I know that it's not gonna work the way I was trying to make it work. To be honest I was expecting that you would answer something when I asked this problem yesterday, :-) Anyway, I looked at your paper and it seems $P^0$ might be a good candidate for me! I have the appropriate assumption on $M-K_X-\Delta$ that it's ample. So, under this assumption is the relation true replacing $S^0$ by $P^0$ on both sides? I will read the paper carefully. Thanks again. $\endgroup$
    – Omprokash
    Mar 1, 2013 at 19:28
  • $\begingroup$ Happy to help. Yes, it should be true under the assumption that $M - K_X - \Delta$ is ample. At the very least, you should be able to take some affine cone and just argue that way for $\tau$. It's probably true more generally though. Let me know if I can be of any other assistance. $\endgroup$ Mar 1, 2013 at 20:16

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