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Apr 23, 2020 at 15:52 comment added k.j. @JasonStarr Thank you for pointing out. So, for a separable $f$, does $f_*$ factor through $\Gamma(X, \Omega)$? I want its (elementary) proof.
Apr 23, 2020 at 15:50 history edited k.j. CC BY-SA 4.0
added 49 characters in body
Apr 23, 2020 at 15:40 comment added Jason Starr @k.j. You left out a hypothesis that $f$ is generically smooth.
Apr 23, 2020 at 10:53 history edited k.j. CC BY-SA 4.0
edited body
Apr 23, 2020 at 10:53 comment added k.j. @JasonStarr So does not the map $f_*$ factor through $\Gamma(X, \Omega_X)?$
Apr 23, 2020 at 7:24 comment added David Loeffler It's a pity that neither the journal web-page for Zannier's paper, nor the MathSciNet review, has any kind of note attached acknowledging the error.
Apr 22, 2020 at 19:59 comment added Jason Starr The article of Zannier is incorrect.
Apr 22, 2020 at 19:12 comment added k.j. @LSpice It's because of lack of my English skills. I edited it. And thank you for editing.
Apr 22, 2020 at 19:10 history edited k.j. CC BY-SA 4.0
added 22 characters in body
Apr 22, 2020 at 18:37 comment added LSpice You say "The definition of $f_*$ is: Now $\Omega_{Z, \zeta} = \Omega_{X, \xi} \otimes_{k(X)} k(Z)$." Is that intentional?
Apr 22, 2020 at 18:37 history edited LSpice CC BY-SA 4.0
Name of paper and link; proofreading and TeX
Apr 22, 2020 at 17:46 history asked k.j. CC BY-SA 4.0