Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 134 / 3

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.

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.

Take $V=C$ for each integral projective curve in $X$. The assumed positivity gives $D\cdot C>0$, hence certainly $D\cdot C\ge 0$. By the definition of a nef divisor,This is only the curve test. It does not already prove ampleness: positivity against each individual curve can fail to be uniform on limiting classes in the closed cone of curves. The hypotheses in higher dimensions, used in the next sections, are what rule out this failure.

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.

Part (b) supplies an ample $A$ with $D-A$ big. Therefore $D=(D-A)+A$ is big as well, since adding an ample class to a big class preserves the ample-plus-effective decomposition. The permitted induction hypothesis makes $D$ ample on every proper positive-dimensional closed subscheme of lower dimension, in particular on every codimension-one integral subvariety. Zero-dimensional restrictions are ample automatically.

Apply part (ii) to this big $D$ and the ample $A$ chosen in (b). The result isFor a reducible scheme the same argument is componentwise. If a positive-dimensional irreducible component itself is a strict closed subscheme of $X$, the permitted induction assumption already makes $D$ ample on that component; otherwise its codimension-one subvarieties are lower-dimensional and the same proof applies. Only one finite minimum of bounds is required.

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.