Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 139 / 3

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 an effective Cartier divisor $F$, the divisor restriction exact sequence givesThe scheme $F$ has dimension at most $n-1$, so the permitted scheme version of part (i) bounds $h^0(F,{\mathcal{O}}_F(kD))$ by $C{k}^{n-1}$. For the infinitely many $k$ in the hypothesis,once $k$ is sufficiently large. Thus infinitely many such $k$ have a nonzero section of $kD-F$. This dimension-drop argument is the section subtraction lemma for big divisors.

If “effective divisor” is interpreted as an effective Weil divisor on a normal variety, use its coherent divisor ideal instead. The quotient by that ideal is supported in dimension at most $n-1$, so the polynomial bound for sections of a fixed divisor gives the same conclusion. For $F=0$ the assertion is immediate.

Fix any ample divisor $A$. Choose $q\gg 0$ so that both $qA$ and $(q-1)A$ have nonzero global sections, and choose $F\in |qA|$. Condition (1) and the section subtraction lemma for big divisors yield an integer $j>0$ with $jD-qA\sim E_0\ge 0$. Adding an effective member of $|(q-1)A|$ gives $jD\sim A+E$. Thus (1) implies the stronger condition (2').

If $jD\sim A+E$ with $A$ ample and $E$ effective, multiplication by the section of $qE$ injects $H^0(X,qA)$ into $H^0(X,qjD)$. The positive leading coefficient of the ample Hilbert polynomial supplies condition (1) along the infinite sequence $m=qj$. Condition (2) implies (3). Conversely, if $jD\equiv A+E$, put $P=jD-E$. The Cartier divisor $P$ is numerically equivalent to $A$, hence ample by Kleiman's criterion; the actual equality $jD=P+E$ gives (2).

We have proved (1)$\Rightarrow$(2')$\Rightarrow$(2)$\Rightarrow$(1), and (3)$\Rightarrow$(2)$\Rightarrow$(3). Also (2') implies (3'), while (3') implies (3). Hence all five conditions are equivalent. They characterize a big divisor; the ample-plus-effective formulation is Kodaira's lemma. The inconsistent use of $n$ and $j$ in the printed multiplier is resolved by using $j$ throughout.

Choose a very ample divisor $A$. By the stronger form of Kodaira's lemma, $jD\sim A+E$ for some effective divisor $E$. The subsystem $s_E{H}^0(X,A)\subseteq H^0(X,jD)$ defines the embedding given by $A$ on $X\setminus \mathrm{Supp}E$. Ratios of its sections generate the function field $k(X)$. The map from the complete complete linear system of a divisor contains these ratios, so it induces the same function field and is birational onto its image.

The converse is true; smoothness is unnecessary. If $|jD|$ gives a birational map, choose $n$ algebraically independent ratios $s_1/s_0,\dots ,s_n/s_0$ among a generating set of its section ratios. For every $q$, the sectionsare linearly independent by algebraic independence. Therefore $h^0(X,qjD)\ge \left(\genfrac{}{}{0ex}{}{q+n}{n}\right)$, giving condition (1) of part (iii). This is the birational linear system criterion for bigness.

Condition (a) is the definition of a big real divisor. Fix an ample Cartier divisor $A$. For each big Cartier divisor $D_i$, part (iii) gives $j_i{D}_i\sim A+E_i$ with $E_i$ effective. ConsequentlyThe first coefficient is positive, proving (b).

For the converse, suppose $D{\sim }_{\mathbb{R}}tA+E$, with $t>0$ and $E$ effective. Apply rational approximation of an ample-plus-effective real divisor: in a finite-dimensional space generated by Cartier divisors and the finitely many principal divisors occurring in this relation, express $D$ as a positive convex combination of rational divisors $B_{\nu }$, each satisfying $B_{\nu }{\sim }_{\mathbb{Q}}{t}_{\nu }A+E_{\nu }$ with $t_{\nu }>0$ and $E_{\nu }$ effective. After clearing denominators, part (iii) shows that each $B_{\nu }$ is a positive rational multiple of a big Cartier divisor. This gives the actual equality required by (a), rather than merely a numerical or linear equivalence.

The approximation lemma keeps the finitely many effectivity inequalities and linear-equivalence equations simultaneously; see its proof for the rational-face argument and the reduction of nonnormal varieties by finite normalization. Therefore (a) and (b) are equivalent.

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