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Choose $P\in X(k)$ and put $D=(g+1)P$. The Riemann-Roch theorem givesAs constants supply only a one-dimensional subspace, the Riemann-Roch space $L(D)$ contains a nonconstant rational function $f$. It has no poles except possibly at $P$. It must have a pole, since a global regular function on an irreducible projective variety is constant. Let its pole order at $P$ be $m$; then $1\le m\le g+1$.

On the open set where $f$ is regular, define $\varphi =[f:1]$. At its pole use the other affine chart, $\varphi =[1:1/f]$. The local ring of a smooth algebraic curve is a discrete valuation ring, and the positive valuation of $1/f$ at $P$ makes the latter expression regular there. The two expressions agree on their overlap, so they give a regular map $X\to {\mathbb{P}}^1$.

This map is nonconstant. It is proper because its source is projective and its target is separated, so its image is closed; a nonconstant image in the projective line must be the whole line. Each finite-value fibre consists of zeros of a nonzero rational function $f-a$, and the infinity fibre is the finite pole set. Hence every fibre is finite, giving a quasi-finite morphism. The proper quasi-finite criterion therefore makes $\varphi$ a finite morphism.

Finally its pullback of the infinity divisor is exactly the pole divisor $mP$. The degree of a morphism of curves equals the degree of this pullback, or equivalently the function-field extension degree $[k(X):k(f)]$. ThusThe argument includes inseparable maps: the degree here is the full field-extension degree, not just the separable degree. This proves the gonality bound from Riemann-Roch with an explicitly constructed finite regular map.

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.

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.