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Sep 2, 2011 at 14:45 answer added David E Speyer timeline score: 4
Aug 19, 2011 at 14:30 comment added Karl Schwede It can be used to prove things like that, but for this particular question, I'm curious what Shrawan Kumar says. Certain Frobenius splittings (ie, global F-regularity) is enough to prove $f_* \omega_X = \omega_Y$ but it doesn't give you the higher $R^i f_* \omega_X = 0$ without knowing the existence of a resolution with a certain form. Perhaps I should point out that anything that is Frobenius split in char. $p > 0$ has something like Du Bois singularities (which can be viewed as a limit of rational singularities).
Aug 19, 2011 at 2:43 comment added Alexander Woo Does Frobenius splitting say something about rational singularities?
Aug 18, 2011 at 21:10 comment added David E Speyer Ah, a good question, I recently got this wrong. The right definition (see for example Brion and Kumar's book, section 3.4) is that there is a proper birational map $f: X \to Y$, with $X$ nonsingular, $f_{\ast} \mathcal{O}_X =\mathcal{O}_Y$ and $R^i f_{\ast} \mathcal{O}_X = R^i f_{\ast} \omega_X=0$ for $i>0$.
Aug 18, 2011 at 21:04 comment added Karl Schwede David, just curious, how do you define rational singularities in characteristic $p > 0$.
Aug 18, 2011 at 19:49 history asked David E Speyer CC BY-SA 3.0