Let $X$ be a smooth projective variety and $D_0$ be a divisor on $X$ (over the algebraically closed field $K$). Then according to Harthshorne Proposition $7.7(b)$(Chapter $2$) every effective divisor on $X$ is of the form $(s)_0$ (i.e. zero of $s$) for some $s \in H^0(L(D_0))$.
In this context I have $2$ questions : $(i)$ What if the line bundle associated to $D_0$ does not have any non-zero global section? (Maybe this can't happen follows from something obvious which I am unable to see)
$(ii)$ Can we conclude from above that every nonzero effective divisor on $X$ always posses atleast $1$ non-zero global section?
Can someone briefly indicate the chain of argument for $(ii)$.
For $(ii)$ can we argue as follows :if $E$ is an effective divisor then it's linearly equivalent to an $(s)_0$ for some non-zero $s \in H^0(L(D))$ for some divisor $D$ on $X$. Then $L(E) \cong L((s)_0) \cong L(D)$ and hence $E$ has a section. Please correct me if this is wrong
Best Answer
Yes, your reasoning for (ii) is correct.
(i) If $h^0(X,\mathcal{O}(D_0)) = 0$, then $D_0$ cannot be effective by your cited result. Namely if $D_0$ were effective, then there exists $0 \neq s \in \Gamma(X, \mathcal{O}(D_0))$ such that $D_0 \sim (s)_0$ and in particular $\Gamma(X, \mathcal{O}(D_0)) \neq 0$.
(ii) If $D$ is effective, then it has non-trivial global sections. This is the contrapositive of (i).
Indeed, Hartshorne also notes immediately after that proposition the following statement: Let $D_0$ be an effective divisor, then $|D_0| \cong (\Gamma(X, \mathcal{O}(D_0)) - \{0 \})/k^{\times}$ where $|D_0|$ denotes the complete linear system of $D_0$. Non-trivial global sections up to $k^{\times}$ correspond to the realization of $D_0$.
In general on nice enough schemes, invertible sheaves $\mathcal{L}$ admit non-zero rational sections $s$ such that $\mathcal{L} \cong \mathcal{O}(\operatorname{div}{s})$, Vakil 14.2.E. The sheaves corresponding to effective divisors are precisely the ones which admit such non-zero global sections.