I'm reading Hartshorne (II) Proposition 7.7.
Proposition 7.7. Let $X$ be a nonsingular projective variety over the algebraically closed field $k$. Let $D_0$ be a divisor on $X$ and let $\mathscr L\cong \mathscr L(D_0)$ be the corresponding invertible sheaf. Then
(a) for each nonzero $s\in \Gamma(X,\mathscr L)$, the divisor of zeros $(s)_0$ is an effective divisor linearly equivalent to $D_0$;
(b) every effective divisor linearly equivalent to $D_0$ is $(s)_0$ for some $s\in\Gamma(X,\mathscr L)$.
Proof.
(b) If $D>0$ and $D=D_0+(f)$, then $(f)\ge -D_0$. Thus $f$ gives a global section of $\mathscr L(D_0)$ whose divisor of zeros is $D$.
Why can $(f)\ge -D_0$ conclude that $f$ gives a global section of $\mathscr L(D_0)$ whose divisor of zeros is $D$?
Best Answer
We may write $D_0$ as a Cartier divisor: $D_0 = \{(U_i, f_i)\}$, where $f_i \in K(X)^\times$ are nonzero rational functions. Then, $(f) + D_0$ being effective means exactly that $f_i*f \in \mathcal{O}(U_i)$ for all $i$.
Now, we claim that the collection of sections $f_i*f \in \mathcal{O}(U_i) = \mathscr{L}(D_0)(U_i)$ glue to a global section of $\mathscr{L}(D_0)$. Over the intersections $U_i \cap U_j$, the transition function of $\mathscr{L}(D_0)$ sends $$f_i*f \mapsto f_i*f *(f_j/f_i) = f_j*f.$$ Hence, these glue to a global section with the required property.
There are probably other ways to see this.