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In Kollár and Mori's Birational Geometry of Algebraic Varieties, lemma2.60 claim some multiple of a big divisor induced birational morphism onto its image in a projective space. But the proof only show it can be written as a sum of an ample divisor and an effective divisor, and say the result is obvious for the latter. I try to find the details for the latter but have no clues. Thank you for any answer or comments. Furthermore the lemma assume the scheme is a projective variety, does it must be integral? Can the result be done for proper varieties?

The second question: The authors also claim a divisor is big iff its birational pullback is big. I know when the varieties are integral normal and proper, this can be done by Zariski' main theorem and projective formula. Are these conditions necessary?

Thank you!

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If $mD=A+E$, where A is wlog very ample, the dimension of the image of $\phi_{|mD|}$ is at least the dimension of the image of $\phi_{|A|}$, because $H^0(X,kA)\subset H^0(X,mkD)$ for $k\ge 0$. –  J.C. Ottem Apr 23 '12 at 2:52
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And to see that the rational map you obtain is birational, take a non-zero section $s_E$ of $E$ whose divisor is $E$, and then the rational map $x\mapsto [s_0\otimes s_E:\cdots:s_N\otimes s_E]$ defined outside of $E$ ($(s_i)$ being a basis of $H^0(X,kA)$ for $k$ large enough) is clearly birational onto its image whose dimension is $\dim X$ as John explained above. –  Henri Apr 23 '12 at 9:03
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By the way, the book you're referring to has two authors. –  Artie Prendergast-Smith Apr 24 '12 at 20:11
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Actually, the missing author is Kollár... –  Sándor Kovács Apr 25 '12 at 5:54
    
But mD may not be generated by global sections, right? So if the definition domain of $\phi_{|mD|}$ is not the entire X, while the image would be a closed subscheme with its image scheme structure, why $\phi_{|mD|}$ should be a morphism and onto its image? –  MZWang Apr 28 '12 at 6:20
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