Timeline for Hilbert polynomial of $X\times P^1$

8 events

when toggle format	what		by	license	comment
Dec 21, 2012 at 21:43	vote	accept	Mike Lowrey
Dec 21, 2012 at 20:46	comment	added	Mike Lowrey		Isn't $K_{X\times \mathbf{P}^1}\|_{X\times a} = K_X\oplus K_{\mathbf{P}^1,a}$ for all $a$ in $\mathbf{P}^1$. Thus, this would show that the canonical sheaf is never ample because it contains a direct factor.
Dec 21, 2012 at 20:45	answer	added	Will Sawin		timeline score: 6
Dec 21, 2012 at 20:40	comment	added	Will Sawin		Is the canonical divisor of $X \times \mathbb P^1$ ever ample?
Dec 21, 2012 at 20:37	comment	added	Daniel Litt		Yes--but the canonical divisor on $X\times \mathbb{P}^1$ might not be ample.
Dec 21, 2012 at 20:26	comment	added	Mike Lowrey		Yes. I thought that when one says "canonically polarized" this means we choose some power of the canonical divisor to define the Hilbert polynomial, no?
Dec 21, 2012 at 20:24	comment	added	Daniel Litt		A Hilbert polynomial depends on a choice of ample line bundle...
Dec 21, 2012 at 20:18	history	asked	Mike Lowrey	CC BY-SA 3.0