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For ample rational Cartier classes $B,C$ on an integral projective $n$-fold with $n\ge 1$, $B^n>n{B}^{n-1}\cdot C$ implies $B-C$ is big. Scale to very ample integral divisors. Choose an effective Cartier divisor $G\in |C|$. Repeated restriction gives $h^0(m(B-C))\ge h^0(mB)-m{h}^0(G,mB)$: multiply each negatively twisted restriction by a section avoiding its associated points to inject it into ${\mathcal{O}}_G(mB)$. Asymptotic Riemann–Roch and Serre vanishing give the positive leading lower bound along sufficiently divisible section indices after undoing the scaling. No complex-analytic Morse theory or characteristic-zero vanishing is used.

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.

Choose an ample Cartier divisor $H$. Put initially $A=\delta H$, $A^{\prime }=tH$ for small positive real $\delta ,t$. Since $D$ is nef, the nef-plus-ample ampleness lemma makes $D+tH$ ample. The polynomialhas $P(0,0)=D^n\cdot [X]>0$. Thus the desired strict inequality holds when $\delta ,t$ are sufficiently small and positive.

We must also arrange rationality of the two specified classes; $D$ itself need not be rational. Choose rational ample classes $B$ and $C$ sufficiently near $D+tH$ and $(\delta +t)H$, and defineThen $A^{\prime }$ is close to $tH$ and $A$ is close to $\delta H$, so both are ample real divisors by openness of the ample cone. Also $D+A^{\prime }=B$ and $A+A^{\prime }=C$ are rational and ample. Continuity preserves the strict inequality, giving

Here is the needed algebraic Morse inequality for ample divisors, with its section-count proof. Choose rational Cartier divisor representatives of $B,C$ and a common positive integer $q$ making $qB,qC$ very ample integral Cartier divisors. For the section-count argument rename these scaled representatives $B,C$; undoing this scaling restricts section indices to sufficiently divisible multiples and leaves bigness unchanged. Choose an effective Cartier divisor $G\in |C|$ by taking a defining section that avoids the associated points of ${\mathcal{O}}_X$. Repeated divisor restriction exact sequences giveBecause $C$ is very ample, for each $j$ a section of $jC$ avoiding the finitely many associated points of ${\mathcal{O}}_G$ gives an injection into ${\mathcal{O}}_G(mB)$. Thus every summand is at most $h^0(G,{\mathcal{O}}_G(mB))$. By Serre vanishing and asymptotic Riemann–Roch for the ample $B$,For $n=1$, the restriction term is the constant length of $G$, giving the same formula directly. The positive coefficient proves that $B-C$, hence $D-A$, is big. Scaling back preserves bigness, so

For a general projective scheme, enforce the same strict inequality separately on each positive-dimensional reduced irreducible component $V_j$, using its own dimension $n_j$. At $(\delta ,t)=(0,0)$ every such expression equals the positive number $D^{n_j}\cdot [V_j]$. Finitely many conditions are preserved by one sufficiently small choice and one sufficiently close rational approximation on $N^1(X{)}_{\mathbb{R}}$. The top-dimensional inequalities imply the printed inequality for $[X]$ with its positive generic multiplicities; the section proof on each component makes $D-A$ componentwise big. This avoids inferring bigness on every component from just a positive sum. The displayed inequality is used for $n\ge 1$. For $n=0$, its literal $n-1$ intersection power is undefined; handle this vacuous positivity case separately. Every line bundle is ample, all numerical classes are zero, and the componentwise bigness convention makes the conclusions automatic.

For a projective scheme $X$ of dimension of a scheme $n$ and a Cartier divisor $D$, asymptotic Riemann–Roch givesHere Euler characteristic of a coherent sheaf means $\chi ={\sum }_{i=0}^n(-1{)}^i{h}^i$, and $D^n$ is the degree of the top intersection product with the fundamental cycle of $X$, including its component multiplicities. No ampleness assumption on $D$ is needed.

By asymptotic Riemann–Roch and the preceding bounds for higher sheaf cohomology,A nef divisor satisfies $D^n\ge 0$: for a fixed ample divisor $A$, the divisors $D+\epsilon A$ are ample for positive rational $\epsilon$, and continuity of the intersection product gives the inequality as $\epsilon \to 0$. If $h^0(X,mD)=0$ eventually, the left side is nonpositive. Dividing by $m^n$ forces $D^n=0$, since $h^1\ge 0$ and $D^n\ge 0$. The same identity then gives $h^1(X,mD)=O(m^{n-1})$.

The divisor is understood to be on $X$; the printed “on $D$” is a typo. Combining asymptotic Riemann–Roch with cohomology growth for nef twists givesSince a nef divisor has nonnegative top self-intersection number, this limit is positive exactly when $D^n>0$. This is the volume of a nef divisor criterion.

For a nef divisor on an integral projective variety of dimension $n$,Indeed, combine asymptotic Riemann–Roch with cohomology growth for nef twists. Thus its normalized section-growth limit is $D^n/n!$, and it is big exactly when $D^n>0$.