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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

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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).

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  • $\begingroup$ Right, but what about the converse? Is something along those lines true? $\endgroup$ – user101036 May 4 '14 at 11:01
  • $\begingroup$ @Tedar: I added something. $\endgroup$ – Brenin May 4 '14 at 11:47
  • $\begingroup$ Ok, So I am having some trouble parsing the latter part you wrote. Are you saying that it is not neccesarily true that if $O_X(D)$ has a global section, $D$ is effective? I have a counterquestion then, if you have the time: If we have an effective Cartier divisor - does the dual have a global section? If not, why? $\endgroup$ – user101036 May 4 '14 at 12:23
  • $\begingroup$ As for your counterquestion: An effective (nontrivial) divisor has positive degree. The dual has then negative degree. An invertible sheaf of negative degree has no nonzero sections. $\endgroup$ – Brenin May 4 '14 at 18:51
  • $\begingroup$ What do you mean with degree for a Cartier divisor? I have only seen degree defined for curves. $\endgroup$ – user101036 May 4 '14 at 19:00
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Effective cartier divisors are just closed subschemes which locally are cut out by the vanishing of a non-zero divisor.

I don't know what you mean by $\mathcal{O}_X(D)$, but if you mean $\mathcal{O}_D(D)$, then certainly it need not have a nonconstant global section. Take for example a family of elliptic curves over $\mathbb{A}^1$. Each elliptic curve is a cartier divisor, but being a projective curve, doesn't have any nonconstant global sections.

See, for example, section 9 of http://stacks.math.columbia.edu/download/divisors.pdf

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  • $\begingroup$ With $\mathcal{O}_X(D)$ I mean the associated invertible sheaf $\mathcal{O}_X(D)$ coming from the injective homomorphism from the Cartier divisors modulo linear equivalence to the Picard group of X. $\endgroup$ – user101036 May 3 '14 at 23:35
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I know this question has been answered long ago... but I feel that the question is settled more completely by the following simple fact:

$\mathcal O_X(D)$ has a global section if and only if $D$ is linearly equivalent to an effective divisor. (see See this MS question)

Given this, is clear that effectivity implies the existence of a global section; while a global section doesn't imply effectivity (because an effective divisor can be linearly equivalent to a noneffective one).

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