Timeline for answer to Isotrivial families with non-zero Kodaira spencer map by Jason Starr

9 events

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Aug 17, 2015 at 15:57	comment	added	Ariyan Javanpeykar		@JasonStarr How can one get rid of the assumption on the characteristic of $k$? It seems to me that if, for instance, the morphism $\pi$ were finite type, we don't need $k$ to be of characteristic zero. Does that seem right to you?
Aug 17, 2015 at 15:04	history	edited	Jason Starr	CC BY-SA 3.0	added 80 characters in body
Aug 17, 2015 at 15:01	comment	added	Jason Starr		@AriyanJavanpeykar: "Did you mean to write ..." Yes, you are absolutely right. Thank you for catching the error. I will correct it now.
Aug 17, 2015 at 14:58	comment	added	Ariyan Javanpeykar		@JasonStarr You wrote "Since k is algebraically closed, there exists a dense open subscheme of I that is k-scheme." Did you mean to write "Since $k$ is algebraically closed and of characteristic zero, there exists a smooth dense open subscheme of $I$ over $k$"?
Jul 30, 2015 at 18:54	comment	added	Pancho		This is beautifully clear. Thank you. One small comment: Somewhere one should mention that the Isom-scheme satisfies some finiteness properties, i.e., the morphism $I\to S$ is of finite type. If not, one can't use all the "generic" arguments, right?
Jul 30, 2015 at 18:14	history	edited	Jason Starr	CC BY-SA 3.0	added 279 characters in body
Jul 30, 2015 at 18:04	vote	accept	Pancho
S Jul 30, 2015 at 17:31	history	answered	Jason Starr	CC BY-SA 3.0
S Jul 30, 2015 at 17:31	history	made wiki			Post Made Community Wiki by Jason Starr