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Mar 6, 2014 at 7:06 history edited Roberto Pignatelli CC BY-SA 3.0
removed useless $c$
Mar 4, 2014 at 20:20 comment added rita Ciao Pigna! welcome to MO!
Mar 4, 2014 at 13:18 comment added rita No. For instance if $A$ is a a rational curve with $A^2<0$, then $h^0(\mathcal O_A(-A))>0$.
Mar 4, 2014 at 12:08 comment added Heitor Ah OK, I see. Thank you! I thought that $h^0(\mathcal{O}_B(−A))=0$ would follow from $A$ being effective. Is this not enough?
Mar 4, 2014 at 11:01 comment added Roberto Pignatelli You are right, the expression for $\chi({\mathcal O}_B(-A))$ is Riemann-Roch. The injectivity gives me $h^0({\mathcal O}_B(-A))=0$, else $\chi$ could be positive.
Mar 3, 2014 at 19:11 comment added Heitor Just a question: I don't really see where do you use the injection $H^0(O_D)\subset H^0(O_A)$. I guess the expression for $\chi(O_B(-A))$ is just Riemann-Roch for singular curves, no?
Mar 3, 2014 at 17:31 vote accept Heitor
Mar 3, 2014 at 17:28 history edited Roberto Pignatelli CC BY-SA 3.0
deleted 123 characters in body
Mar 3, 2014 at 14:49 history answered Roberto Pignatelli CC BY-SA 3.0