Timeline for divisors and powers of line bundles

9 events

when toggle format	what		by	license	comment
Dec 29, 2012 at 9:42	comment	added	Francesco Polizzi		Right. I was tacitly assuming $k=\mathbb{C}$.
Dec 28, 2012 at 22:59	comment	added	Felipe Voloch		There was no assumption on the ground field in the original question.
Dec 28, 2012 at 22:25	history	edited	Francesco Polizzi	CC BY-SA 3.0	added 477 characters in body; added 9 characters in body
Dec 28, 2012 at 21:59	history	edited	Francesco Polizzi	CC BY-SA 3.0	added 1 characters in body
Dec 28, 2012 at 21:44	comment	added	Francesco Polizzi		Yes, because $\textrm{Pic}^0(X)$ is a complex torus, hence a divisible group. In fact, assume that the degree of $D$ is divisible by $m$ and take any effective divisor $F$ such that $\deg F = (\deg D)/m$. Then $D-mF \in \textrm{Pic}^0(X)$, so there exists a degree $0$ divisor $\psi$ such that $D-mF=m \psi$, that is $D=m(F+ \psi)$.
Dec 28, 2012 at 21:36	comment	added	div		Thanks for your answer, Francesco! What about curves? Is it sufficient in that case?
Dec 28, 2012 at 21:34	vote	accept	div
Dec 28, 2012 at 21:34	vote	accept	div
Dec 28, 2012 at 21:34
Dec 28, 2012 at 21:33	history	answered	Francesco Polizzi	CC BY-SA 3.0