Skip to main content

Timeline for Log resolutions of linear series

Current License: CC BY-SA 2.5

8 events
when toggle format what by license comment
Jun 18, 2010 at 15:52 comment added Angelo No problem; you are learning.
Jun 18, 2010 at 9:49 vote accept Gianni Bello
Jun 18, 2010 at 9:48 comment added Gianni Bello Ok. Thank you for your answers and forgive me for the triviality of the question.
Jun 18, 2010 at 5:25 comment added Angelo This seems pretty obvious to me. The linear system corresponds to a line bundle $L$ with a number of independent sections $s_i$, whose zero scheme is the base locus. When you pullback the $s_i$ the base scheme pulls back.
Jun 17, 2010 at 21:11 comment added Gianni Bello Ok, this is what I had in mind. The point I lack is why the inverse image of the base ideal is the base ideal of the pullback of the linear series. I suppose this is trivial but it is not clear to me.
Jun 17, 2010 at 20:38 comment added Angelo I guess we are using different definitions. With the definitions I seem to understand you have in mind, the result seems trivial to me. The point is that when you subtract the inverse image of the base ideal, which is supported on a divisor with normal crossing, the linear system becames base point free, so this is a log-resolution, by definition. Is there something wrong with this argument? If so, can you spell out your defitions exactly.
Jun 17, 2010 at 19:07 comment added Gianni Bello Maybe we are using two different definitions. When I speak about a log resolution of the base ideal, I mean that I want the inverse image of the ideal that defines the scheme-theoretical base locus to become a line bundle (defining a divisor with snc support). In particular I'm using the definitions of Lazarsfeld's "Positivity in algebraic geometry", chap 9.1. Using this definition, in the same chapter, Lazarsfeld remarks that for smooth varieties the claim holds.
Jun 17, 2010 at 18:08 history answered Angelo CC BY-SA 2.5