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A divisor is basepoint-free when its global sections have no common zero. Its complete linear system therefore defines a Kodaira map everywhere.

We prove (c)$\Rightarrow$(a) by induction on dimension. It suffices to work on an integral projective variety $V$. By induction, $D$ is ample on every lower-dimensional integral subvariety, and hence on every lower-dimensional closed subscheme by ampleness on reduced components.

Apply (c) to $V$ itself. The nonzero section of ${\mathcal{O}}_V(mD)$ has a nonempty zero divisor $E$. Because $V$ is integral, this is an effective Cartier divisor and ${\mathcal{O}}_V(E)\cong {\mathcal{O}}_V(mD)$. Its support has dimension less than $\dim V$, so $D{|}_E$, and therefore ${\mathcal{O}}_E(E)$, is ample. Part (ii) makes $E$ semiample, hence some positive multiple of $D$ is basepoint-free.

Let $f:V\to Y\subset {\mathbb{P}}^N$ be the resulting Kodaira map, with ${\mathcal{O}}_V(qD)=f^{\ast }{\mathcal{O}}_Y(1)$. No fibre can have positive dimension: such a projective fibre contains an integral projective curve $C$, on which $D$ has degree zero. But the assumed nonzero section of some ${\mathcal{O}}_C(rD)$ cannot vanish anywhere, since its nonempty effective divisor would have positive degree. This contradicts (c).

Thus $f$ has zero-dimensional fibres. A proper quasi-finite morphism is a finite morphism. The finite pullback of an ample line bundle is ample, so $qD$ and then $D$ are ample. This provesThe fibre argument proves the semiample and curve-positive ampleness criterion. It also explains why testing only existence of a nonzero section, without requiring a zero, would be insufficient: the trivial bundle on a positive-dimensional projective variety has a nowhere-vanishing section.

On ${\mathbb{F}}_2$, let $S$ be the negative section of a Hirzebruch surface, so $S^2=-2$, and let $F$ be the fiber class of a Hirzebruch surface, with $F^2=0$ and $S\cdot F=1$. The line bundle $\mathcal{O}(S+3F)$ is very ample. Indeed, the toric ampleness criterion for $aS+bF$ isand on a smooth complete toric variety every ample line bundle is very ample. These inequalities hold for $(a,b)=(1,3)$.

Explicitly, after choosing the standard lattice coordinates for ${\mathbb{F}}_2$, its lattice polytope of a toric divisor isThe monomials indexed by the lattice points of this polygon separate torus orbits and tangent directions, so their Kodaira map is a closed embedding. This directly verifies that the resolved proper toric surface is projective.

Take the invariant divisor $D_1$, whose class is $F$. Its global sections are represented in the Cox ring by $x_1$ and $x_3$. They have no common zero on $U$, so $F$ is basepoint-free. Its Kodaira map isThis is exactly the ruling ${\mathbb{F}}_k={\mathbb{P}}_{{\mathbb{P}}^1}(\mathcal{O}\oplus \mathcal{O}(k))\to {\mathbb{P}}^1$, and its fibers have divisor class $F$.

A semiample divisor on an integral projective variety is ample if it has positive degree on every integral projective curve. Its Kodaira map cannot contract a positive-dimensional fibre, since such a fibre contains a curve and the pulled-back hyperplane bundle has degree zero there. Thus the morphism is proper and a quasi-finite morphism, hence a finite morphism. The finite pullback of an ample line bundle is ample.

A Cartier divisor is semiample if some positive multiple is a basepoint-free divisor, equivalently its divisor line bundle has a globally generated positive power. Such a power defines a Kodaira map $f$ with $\mathcal{O}(mD)=f^{\ast }\mathcal{O}(1)$. Semiampleness implies nefness but need not imply ampleness: a fibre divisor of a morphism to a curve is a basic example.