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Assume $D$ is ample. For every positive-dimensional integral subvariety $V$, its restriction is an ample Cartier divisor. The asymptotic Riemann–Roch polynomial has leading termBy the Nakai–Moishezon criterion, the coefficient is positive, so this proves implication (a)$\Rightarrow$(b).

For (a)$\Rightarrow$(c), choose $m>0$ such that ${\mathcal{O}}_V(mD)$ is very ample. Fix a closed point $p\in V$. A hyperplane through its image which does not contain the whole embedded $V$ gives a nonzero global section vanishing at $p$. Such a hyperplane exists since $\dim V>0$. Its vanishing is therefore nonempty but not all of $V$. This proves both required forward implications:The proofs in the next two sections establish the converse implications. Reduction to integral components is legitimate by ampleness on reduced components; in dimension zero every line bundle on a projective scheme is ample and there is no positive-dimensional subvariety to test.

We prove (b)$\Rightarrow$(a) by induction on dimension, using the independently proved (c)$\Rightarrow$(a) argument in the next section. Work on an integral projective variety $V$ of dimension $d>0$. The hypothesis is inherited by all its integral closed subvarieties. Induction therefore makes $D$ ample on every strictly lower-dimensional closed reduced subvariety, and hence on every proper closed subscheme of dimension less than $d$, by ampleness on reduced components.

Choose an effective Cartier divisor $H$ which is a very ample hyperplane section of $V$. Then ${\mathcal{O}}_H(D)$ is ample. We need a vanishing statement uniform in extra positive $H$-twists:Here is a justification using Castelnuovo–Mumford regularity. Embed $H$ by ${\mathcal{O}}_H(H)$. For each of the finitely many positive cohomology degrees $j$, Serre vanishing for the ample bundle ${\mathcal{O}}_H(D)$ makes $H^j(H,{\mathcal{O}}_H(mD-jH))$ vanish for $m\ge M$. Thus the pushed-forward sheaf ${\mathcal{O}}_H(mD)$ is zero-regular. Persistence of regularity makes it $t+j$-regular for every $t\ge 0$, giving exactly the displayed vanishing. This is the uniform Serre vanishing for two ample twists lemma. It does not assume that $D$ is ample on $V$.

Apply the divisor restriction exact sequence to $H$ with twists $mD+tH$. For $i\ge 2$, both neighbouring cohomology groups on $H$ vanish, soFor a fixed $m\ge M$, sufficiently large $t$ kills the right-hand cohomology by Serre vanishing for $H$ on $V$. The higher cohomology vanishing from an ample hyperplane restriction argument gives $H^i(V,{\mathcal{O}}_V(mD))=0$ for all $i\ge 2$. ConsequentlyThis step is essential: divergence of a polynomial does not by itself prove that its top-degree coefficient is positive.

For $m$ large enough, $h^0(V,mD)>1$. Evaluation at a closed point has a one-dimensional target, so its kernel contains a nonzero section vanishing there. We have obtained condition (c) on $V$. Lower-dimensional subvarieties already have the same property by induction, so the next section's implication (c)$\Rightarrow$(a) applies. ThusFor $d=1$, higher groups with $i\ge 2$ vanish automatically and the same evaluation argument starts the induction. The regularity facts used above are stated in the Stacks Project, regularity lemmas.

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.

The Nakai–Moishezon criterion says that a Cartier divisor $D$ on a projective scheme is ample exactly whenIn particular the test includes each positive-dimensional irreducible component. Here and below, “proper” means proper over the ground field, not necessarily a strict subset of $X$. Interpreting it as a strict subset would make the later criteria false even for an integral projective curve, whose strict closed subvarieties have dimension zero.

The intersection product of Cartier divisors with cycles depends only on their numerical equivalence of divisors classes. One way to see this is to intersect all but one factor first, obtaining a one-cycle; replacing the remaining factor by a numerically equivalent divisor does not change its pairing with that cycle. Multilinearity then handles replacement of every factor. Consequently $D\equiv D^{\prime }$ makes all the displayed numbers equal. ThusThe criterion applies to the integral reduced subvarieties of a possibly nonreduced scheme; ampleness on reduced components explains why the nilpotent structure does not change this condition.

Choose $0<ϵ\ll 1$ as in (c). Thenis the sum of a nef divisor and an ample real divisor. The nef-plus-ample ampleness lemma givesFor clarity, this last lemma follows from Kleiman's criterion and the convex cone property: if $x$ is an interior point of the nef cone and $y$ lies in that cone, translating a small neighbourhood of $x$ by $y$ stays in the cone. Thus $x+y$ remains in its interior, which is the ample cone on a projective scheme.

The complete argument proves the real Nakai–Moishezon criterion rather than assuming it: curve positivity gives nefness, rational approximation and a proved section-count inequality give bigness, induction and the finite-support argument give a uniform ample subtraction, and the nef-plus-ample lemma concludes ampleness. The zero-dimensional case is automatic, and ampleness on reduced components handles reducibility and nilpotents.