Ample line bundle

Remarks

If $X$ is of finite type over a Noetherian ring $A$, then $\mathcal{L}$ is ample if and only if $\mathcal{L}^{\otimes m}$ is very ample over $\operatorname{Spec} A$ for some $m \ge 1$. Here very ample means $\mathcal{L}^{\otimes m} \cong i^*\mathcal{O}(1)$ for an immersion $i \colon X \to \mathbb{P}^N_A$.

Example: on $X = \mathbb{P}^n_k$ the twisting sheaf $\mathcal{O}(1)$ is ample. By a theorem of Serre, every coherent $\mathcal{F}$ on $\mathbb{P}^n_k$ has $\mathcal{F}(n)$ globally generated for $n$ large.

Non-example: on $\mathbb{P}^1_k$, $\mathcal{L} = \mathcal{O}(-1)$ is not ample. Take $\mathcal{F} = \mathcal{O}_X$: then $\mathcal{L}^{\otimes n} = \mathcal{O}(-n)$ has no nonzero global sections for $n \ge 1$, so it is not globally generated.

Ampleness is preserved by positive tensor powers, and $\mathcal{L}$ is ample if and only if $\mathcal{L}^{\otimes m}$ is ample for one $m \ge 1$.