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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.

Given a coherent sheaf $\mathcal{F}$ and an ample Cartier divisor $A$ on a projective scheme, one integer $a_0$ ensures $H^i(X,\mathcal{F}(aA+N))=0$ for all $i>0$, all $a\ge a_0$, and every nef divisor $N$. The uniformity in $N$ is stronger than Serre vanishing. Fujino's note on Fujita vanishing states the theorem for arbitrary projective schemes over a field.

Let $H$ be a very ample effective Cartier divisor on a projective scheme $X$, and suppose $L{|}_H$ is ample. Then $H^i(X,L^m)=0$ for $i\ge 2$ and sufficiently large $m$. By uniform Serre vanishing for two ample twists, cohomology of $L^m(tH){|}_H$ vanishes uniformly for $t\ge 0$. The restriction sequence identifies the higher groups of $L^m(tH)$ and $L^m((t+1)H)$ for $i\ge 2$. For fixed $m$, large $t$ kills them by Serre vanishing for $H$; descend to $t=0$. No $H^1$ vanishing is claimed.

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.

Put $L={\mathcal{O}}_X(D)$ and let $s_D$ be the canonical global section cutting out the effective Cartier divisor $D$. The divisor restriction exact sequence givesThe hypothesis says $L{|}_D$ is an ample line bundle. By Serre vanishing, $H^1(D,L^m{|}_D)=0$ for all sufficiently large $m$, so the long exact sequence in sheaf cohomology makessurjective for all such $m$. These are finite-dimensional vector spaces. Their dimensions form a nonincreasing sequence of nonnegative integers, so the maps are isomorphisms from some point on. Exactness then implies that the restrictionis surjective for all sufficiently large $m$.

Choose such an $m$ for which $L^m{|}_D$ is also a globally generated line bundle. Lift a generating collection of its global sections to $X$. At each point of $D$, one lift has nonzero image in the one-dimensional residue-field fibre, hence generates the stalk of $L^m$ by Nakayama lemma. Outside $D$, $s_D^m$ is nowhere zero and generates $L^m$. Together these sections generate it everywhere. ThereforeThis proof works on an arbitrary projective scheme because the defining section of an effective Cartier divisor is a non-zero-divisor. If $D$ is empty, $L\cong {\mathcal{O}}_X$ and the conclusion is immediate. Semiampleness is the conclusion: the pullback of a line avoiding the centre of a blowup of a smooth algebraic surface of ${\mathbb{P}}^2$ satisfies the hypothesis on its support but has zero intersection with the exceptional curve, and therefore is not ample.

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.

Choose a very ample divisor $A$ such that $A-D$ has a nonzero global section; this is possible by taking a sufficiently high ample twist. Since $X$ is integral, multiplication by its $m$th power injects $H^0(X,mD)$ into $H^0(X,mA)$. By Serre vanishing and the Hilbert polynomial, $h^0(X,mA)=O(m^n)$. ConsequentlyThis is the polynomial bound for sections of a fixed divisor. The same proof works for a coherent sheaf of support dimension $d$, by choosing the multiplying section to avoid its associated points; the bound is then $O(m^d)$.

For a coherent sheaf $\mathcal{F}$ of support dimension $d$ on a projective scheme and a fixed Cartier divisor $D$, $h^0(X,\mathcal{F}(mD))=O(m^d)$. Choose a sufficiently high ample divisor $A$ whose section avoids the associated points of $\mathcal{F}$ and for which $D+A$ is ample. Multiplication by the $m$th power of that section injects $\mathcal{F}(mD)$ into $\mathcal{F}(m(D+A))$. The ample Hilbert polynomial, together with Serre vanishing, bounds the latter section space by $O(m^d)$.

If $E$ is an effective Cartier divisor on a projective scheme and ${\mathcal{O}}_E(E)$ is ample, then $E$ is semiample. The divisor restriction exact sequence and Serre vanishing make $H^1(X,(m-1)E)\to H^1(X,mE)$ surjective for large $m$. Their finite dimensions stabilize, so restriction on global sections is eventually surjective. Lift generators on $E$; off $E$, the canonical section of $mE$ generates. Together these generate ${\mathcal{O}}_X(mE)$, including on nonreduced $X$.

For ample line bundles $L,B$ and a coherent sheaf $\mathcal{F}$ on a projective scheme, one $M$ gives $H^i(X,\mathcal{F}\otimes L^m\otimes B^t)=0$ for every $i>0$, $m\ge M$ and $t\ge 0$. If $B$ is very ample, Serre vanishing for the finitely many fixed twists $\mathcal{F}(-iB)$ makes $\mathcal{F}\otimes L^m$ zero-regular. Persistence of Castelnuovo–Mumford regularity gives all nonnegative $B$-twists. For general ample $B$, use a very ample power and treat finitely many residues of $t$.