Divisor of a global section of a line bundle associated to a Weil divisor 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.
 A: 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.
A: 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 $H^0(V,\mathcal{O}(D))\subset H^0(V,\mathcal{O}(D'))$?
A: Actually I found in Hindry & Silverman, Diophantine Geometry And Introduction, Springer 2000, page 63, that
$$\{div(s)|\;s\in\Gamma(V,\mathcal{O}(D))\}=|D|,$$
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 
$$D+div(f)\geq 0$$
and
$$div(s)=D+div(f).$$
This reassures me of the fact that a global section has no pole.
