unramified morphisms

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Feb 10, 2010 at 6:08	history	made wiki			Post Made Community Wiki by Harry Gindi
Feb 9, 2010 at 22:24	history	edited	Harry Gindi	CC BY-SA 2.5	deleted 130 characters in body
Feb 9, 2010 at 22:22	comment	added	Harry Gindi		sure, but I was under the impression that you're really only supposed to vote down answers that are wrong.
Feb 9, 2010 at 22:21	comment	added	Emerton		Incidentally, Qing Liu (the author of the book) has already stated that the exercise was in error.
Feb 9, 2010 at 21:57	comment	added	Emerton		Your answer addresses part of the problem of showing that if $Y \to X$ smooth/etale/unram., then $Z\to X$ is smooth/etale/unram. if and only if $Y\times_X Z \to X$ is smooth/etale/unram., namely it addresses a part of the only if direction. (This is the revised exercise suggested in Qing Liu's answer.) It doesn't fully address that revised question, though (e.g. the converse is slightly more subtle, I think) and it doesn't address the original question at all. (Without wanting to speak for whoever downvoted your answer, this may go some way to explaining the downvote).
Feb 9, 2010 at 21:44	comment	added	Harry Gindi		Without this fact, etale morphisms don't form a grothendieck topology.
Feb 9, 2010 at 21:42	comment	added	Harry Gindi		This follows formally from the fact that {smooth, unramified, etale} morphisms are stable under base change by the tensor product in the opposite category of commutative rings. The etale topology on Sch is merely the extension of the etale topology from CommRing^op. This follows from SGA 4.1.ii.2.5 and 4.1.ii.5
Feb 9, 2010 at 21:40	history	edited	Harry Gindi	CC BY-SA 2.5	deleted 3 characters in body
Feb 9, 2010 at 21:33	history	answered	Harry Gindi	CC BY-SA 2.5