I have a brief question – I seem to have a vague recollection that if we have a Cartier divisor $D$ on a scheme $X$ , then we can determine whether $D$ is effective by saying whether $\mathcal{O}_X(D)$ has a global section or not. I have tried to prove this fact, but can't seem to do it (is it true?) . If anyone could confirm / deny this I would be very happy, and I would also be glad if someone could give me a reference or a proof of the fact!
Sincerely
Tedar
Best Answer
To give an effective Cartier divisor $D\subset X$ is to give its invertible ideal sheaf $\mathcal O_X(-D)\to \mathcal O_X$. Tensoring this inclusion with the dual $\mathcal O_X(D)$, you get a canonical section $\mathcal O_X\to \mathcal O_X(D)$ attached to $D\subset X$. (Actually, $D$ is the zero scheme of this section).
So, if $D$ is effective, $\mathcal O_X(D)$ always has a section. A possible reference is Vakil's course notes, the chapter on invertible sheaves.
Added. Conversely, if $\mathcal O_X(D)$ has a section $s$ which does not restrict to a zero divisor over any open set $U\subset X$, then $D$ is an effective Cartier divisor. I think the existence of such a section is not automatic.
In order for $D$ to be effective, it is not enough to have a section: what you need a section $s\in H^0(X,\mathcal O_X(D))$ restricting to $s|_{U_i}=f_i$, where $\{(U_i,f_i)\}$ are the local data defining $D$ (which we assumed Cartier).
Another possible translation of effectivity is: the $f_i$'s actually live in $\mathcal O_X(U_i)$ (subring of $\mathscr M(U_i)$, the meromorphic functions on $U_i$, where they a priori lie).