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Sep 23, 2013 at 14:15 vote accept Dan Petersen
Sep 23, 2013 at 14:14 vote accept Dan Petersen
Sep 23, 2013 at 14:15
Sep 23, 2013 at 14:04 comment added naf @Dan Petersen: Choose a closed point on the fibre of $f$ over each (scheme theoretic) generic point of $Y$ and let $Z$ be the Zariski closure of these points.
Sep 23, 2013 at 13:50 comment added Dan Petersen @ulrich: Oh, even better! Because any proper variety $X$ has a modification $X' \to X$ which is projective? Or is there a simpler argument?
Sep 23, 2013 at 11:26 comment added naf @Dan Petersen: Proper and surjective is all one needs to get such a subvariety.
Sep 23, 2013 at 8:53 comment added Dan Petersen That's a nice reference. In fact the only point of the argument that uses projectivity is in order to find a closed subvariety $X' \subset X$ mapping generically finitely onto $Y$.
Sep 23, 2013 at 8:31 vote accept Dan Petersen
Sep 23, 2013 at 8:32
Sep 22, 2013 at 13:35 review First posts
Sep 22, 2013 at 13:39
Sep 22, 2013 at 13:16 history answered novice CC BY-SA 3.0