This is a simple question.

Let $D$ be some Weil divisor on a non-singular projective variety $V$, $\mathcal{O}(D)$ the associated line bundle. Suppose $s\in H^0(V,\mathcal{O}(D))$ is a global section.

How $div(s)$ and $D$ are related?

For example on an elliptic curve an elliptic curve $E/K$ over an number field $K$ with have the line bundle $\mathcal{O}(O_E)$ associated to the neutral point of the elliptic curve. Then $H^0(E,\mathcal{O}(O_E))$ is of dimension $1$ generated by a global section of divisor $O_E$.

Certainly in full generality we cannot have $div(s)=D$ especially if the divisor is not effective because a global section is regular as it seems to me.

  • $\begingroup$ 1) Only Cartier divisors have associated line bundles. 2) The dimension of $H^0(E,\mathcal{O}(O_E))$ is irrelevant. $\endgroup$ Commented Feb 14, 2016 at 21:51

3 Answers 3


First, note that you can only associate a line bundle to a Cartier divisor, not a Weil divisor.
Fortunately on non-singular varieties they coincide (up to isomorphism) so that in your case this poses no problem.
And now for the real problem: given a section $s\in H^0(V,\mathcal{O}(D))$ there are two divisors associated to it!

a) Looking at $s$ as a rational function, it has a divisor $\operatorname {div}(s)=\sum n_iH_i$, consisting in formally summing its zeros and poles counted with suitable multiplicities $n_i$, positive or negative according as $s$ has a zero or a pole along the irreducible hypersurface $H_i$.
The requirement for $s$ to be section of $\mathcal O(D)$ is of course $\operatorname {div}(s)+D\geq 0$.

b) Given a line bundle $\mathcal L$, like $\mathcal O(D)$ for example, a non zero global rational section $t$ of $\mathcal L$ has a divisor $\operatorname {div}^{\mathcal L}(t)=\sum \nu_i H_i$ obtained by the following recipe:
on a trivializing open set $U$ for $\mathcal L$ choose a nowhere zero section $u\in \Gamma(U,\mathcal L) $ and write $t\vert U=fu$ with $f\in \operatorname {Rat}(U)$ rational on $U$.
The recipe is then to write $$\operatorname {div}^{\mathcal L}(t)\vert U=\operatorname {div}(f)$$ and by covering $V$ by such $U$'s to obtain the divisor $\operatorname {div}^{\mathcal L}(t)\in Div(V)$.

c) The fundamental formula
for a rational section $t$ of $\mathcal O(D)$ is $$\operatorname {div}^{\mathcal L}(t)= \operatorname {div}(t)+D $$

No standard book or text I'm aware of seems to introduce this distinction between $\operatorname {div}^{\mathcal L}$ and $\operatorname {div}$, nor consequently the relation between them.
This is unfortunate since this seems to cause recurring misunderstandings, as witnessed by your post.

  • $\begingroup$ Dear Georges, first of all I wanted to thank you for your enlightening comment. Having said that, I would like to ask if there is an easy way to prove that $\mathrm{div}^{\mathcal{L}}(t)\cong \mathrm{div}(t)+D$ $\endgroup$
    – Howard
    Commented May 4, 2017 at 17:36
  • 1
    $\begingroup$ @Howard: If $D$ is given by the family $(U_i,z_i)$, where $(U_i)$ is an open covering of $V$ and the $t_i$'s are rational functions, the formula follows from the local isomorphisms $\mathcal O(D)\vert U_i\to \mathcal O\vert U_i$ sending $t\vert U_i$ to $z_i . t \vert U_i$ $\endgroup$ Commented May 5, 2017 at 18:16
  • $\begingroup$ Dear @GeorgesElencwajg, in (a) of your post, if we consider $s$ as a rational function then could you please explain to me why $s$ has a pole at $D$? $\endgroup$ Commented Dec 30, 2017 at 12:47
  • $\begingroup$ Dear @An Khuong Doan, Wikipedia explains this here $\endgroup$ Commented Dec 30, 2017 at 16:56
  • 1
    $\begingroup$ Dear @Alekos Robotis, yours is a subtle and absolutely valid remark. I have replaced my correct but less precise congruence by your much more satisfying equality. I am very grateful to you for this improvement. $\endgroup$ Commented Sep 25, 2020 at 15:20

Does it mean that a global section can have poles? I knew the isomorphism

$$\mathcal{L}(D)=\{f\in k^*(V)|\;div(f)+D\geq 0\}\cong H^0(V,\mathcal{O}(D))$$

but is this really an identification?

Furthermore if $D\leq D'$ we have $\mathcal{L}(D)\subset\mathcal{L}(D')$ but then do we have also $H^0(V,\mathcal{O}(D))\subset H^0(V,\mathcal{O}(D'))$?


Actually I found in Hindry & Silverman, Diophantine Geometry And Introduction, Springer 2000, page 63, that


where $|D|$ is the linear system of effective divisors linearly equivalent to $D$.

Therefore it seems that for any global section $s\in\Gamma(V,\mathcal{O}(D))$ there exist a meromorphic function $f\in K^*(V)$ such that

$$D+div(f)\geq 0$$



This reassures me of the fact that a global section has no pole.

  • $\begingroup$ Just a remark, in the preceeding you have to make the assuptiion that s is not identically 0... $\endgroup$
    – Benji01
    Commented Feb 16, 2016 at 14:01

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