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The ample cone consists of real numerical divisor classes represented by ample real divisors. It is an open convex cone, and Kleiman's criterion identifies it with the interior of the nef cone.

A big real divisor is an actual positive real combination of big Cartier divisors. Equivalently it is real linearly equivalent, or numerically equivalent, to an ample real divisor plus an effective real divisor. The rational approximation of an ample-plus-effective real divisor and Kodaira's lemma connect these formulations. Fujino's notes on big real divisors give the definition also for nonnormal varieties; bigness under finite normalization relates it to section growth.

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 $0<ϵ\ll 1$ 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.

First work on an integral component. Write the big real divisor as $D{\sim }_{\mathbb{R}}B+E$ with $B$ ample real divisor and $E$ effective real Cartier, using Kodaira's lemma. Let $E_1,\dots ,E_k$ be the finitely many integral components of its support. For an integral projective curve $C$ not contained in this support, restriction of each effective Cartier summand to $C$ is effective, soThereforeThis proves negative curves of a big real divisor lie in finitely many divisors.

Now let $A$ be the given ample divisor. Openness of the ample cone gives a $\delta >0$ such that $B-ϵA$ is ample for $0<ϵ\le \delta$. For each of the finitely many $E_i$, the assumption that $D{|}_{E_i}$ is ample similarly gives a ${\delta }_i>0$ such that $(D-ϵA){|}_{E_i}$ is ample for $0<ϵ\le {\delta }_i$. Choose a single positive $ϵ$ smaller than all these bounds.

If $C$ is contained in some $E_i$, its intersection with $D-ϵA$ is positive by that restriction. OtherwiseIn particular $D-ϵA$ is nef:Only the finitely many exceptional support components are needed for the restriction test; no uniform bound over all subvarieties was assumed.

For a reducible projective scheme, use componentwise bigness on a projective scheme and repeat this argument on each reduced irreducible component. Collect their exceptional supports and take the minimum of all the finitely many positive bounds. Codimension one here is measured in the relevant irreducible component. Every integral curve lies in a component, so the same conclusion holds on $X$. Nilpotent structure does not affect these curve intersection numbers.

An ample real divisor is a finite positive real combination of ample Cartier divisors:Equivalently its numerical class lies in the ample cone. This is a numerical condition even when the coefficients are irrational; it does not mean that some integer multiple of $D$ must be an integral divisor.

On an integral projective variety, a big real divisor is a finite positive real combination of big Cartier divisors. Equivalently, by the real form of Kodaira's lemma,For an integral Cartier divisor, bigness means maximal section-growth order $h^0(X,mD)\ge c{m}^{\dim X}$ along sufficiently divisible positive $m$, or Iitaka dimension $\dim X$. The real linear equivalence of divisors formulation permits finite positive combinations of effective Cartier divisors, whose supports are codimension one. The definitions and the ample-plus-effective formulation on integral varieties are discussed in Fujino's notes on big real divisors.

For the paper's assertions on a general projective scheme, use componentwise bigness on a projective scheme: require bigness on every reduced irreducible component. All arguments below can then be carried out on those finitely many integral components; ampleness is also detected there. On a reducible scheme, merely asking for maximal total section growth on one component is insufficient. For example $X={\mathbb{P}}^2⨿{\mathbb{P}}^2$, with $D$ restricting to $\mathcal{O}(1)$ on the first component and $\mathcal{O}(-1)$ on the second, has quadratic total section growth, but negative intersection with every line in the second component. No finite collection of codimension-one subvarieties can contain all those lines. Thus that weaker meaning would make part (ii) false. In dimension zero the positivity statements are vacuous and every line bundle is ample; the compatible bigness convention also regards it as big.

Condition (b) immediately implies (c): a positive real multiple $tA$ of an ample divisor is an ample real divisor, and real linear equivalence of divisors implies numerical equivalence of divisors.

More explicitly, the real numerical class of $D$ is the sum of a class in the ample cone and an effective real divisor class. This separates strict positivity from the possibly degenerate effective part; it does not claim that the effective part itself is ample.

Suppose $D\equiv A^{\prime }+E^{\prime }$ as in (c), and set $P=D-E^{\prime }$. Its numerical class is ample, so $P$ is an ample real divisor. Fix an ample Cartier divisor $A$. The ample cone is open by Kleiman's criterion, so choose $\epsilon >0$ with $P-\epsilon A$ still ample. Express this ample real divisor as a positive real combination $\sum b_i{B}_i$ of ample Cartier divisors. Some positive multiple of each $B_i$ has an effective representative; dividing by that multiple gives $B_i{\sim }_{\mathbb{Q}}{G}_i$ with $G_i$ effective. Hencewhich is (b). The equivalence of numerical ampleness with a positive combination of ample Cartier representatives follows by rational approximation inside the open ample cone; the rational approximation of an ample-plus-effective real divisor proof also accounts for principal-divisor directions.

Finally, if $D^{\prime }\equiv D$, the same expression $D^{\prime }\equiv A^{\prime }+E^{\prime }$ satisfies (c). By the equivalences just proved, $D^{\prime }$ satisfies all three conditions exactly when $D$ does. Thus bigness is invariant under numerical equivalence, for real as well as Cartier divisors.

The space of real numerical divisor classes is $N^1(X{)}_{\mathbb{R}}$. Write any big class as $[D]=[A]+[E]$, with $A$ an ample real divisor and $E$ effective. For a sufficiently small perturbation $\delta \in N^1(X{)}_{\mathbb{R}}$, the class $[A]+\delta$ remains in the open ample cone, so $[D]+\delta$ remains big by part (v). Thus $\mathrm{Big}(X)$ is open. Positive scaling and addition preserve the ample-plus-effective expression, so it is a convex cone.

For $n\ge 1$, fix a very ample divisor $H$. The linear functional $\ell ([D])=D\cdot H^{n-1}$ is strictly positive on every big class: the ample part contributes positively and the effective part nonnegatively. No nonzero linear subspace can be contained in this cone, since it would contain both $v$ and $-v$. In particular the zero class is not big in positive dimension. When $n=0$, $N^1(X{)}_{\mathbb{R}}=0$, so there is no positive-dimensional subspace to consider. Therefore