For an effective Cartier divisor and any Cartier divisor , multiplication by a defining section of gives the displayed short exact sequence of sheaves. Its long exact sequence in sheaf cohomology compares sections and cohomology on with their restrictions to .
Let be a very ample effective Cartier divisor on a projective scheme , and suppose is ample. Then for and sufficiently large . By uniform Serre vanishing for two ample twists, cohomology of vanishes uniformly for . The restriction sequence identifies the higher groups of and for . For fixed , large kills them by Serre vanishing for ; descend to . No vanishing is claimed.
We prove (b)(a) by induction on dimension, using the independently proved (c)(a) argument in the next section. Work on an integral projective variety of dimension . The hypothesis is inherited by all its integral closed subvarieties. Induction therefore makes ample on every strictly lower-dimensional closed reduced subvariety, and hence on every proper closed subscheme of dimension less than , by ampleness on reduced components.
Choose an effective Cartier divisor which is a very ample hyperplane section of . Then is ample. We need a vanishing statement uniform in extra positive -twists:
Here is a justification using Castelnuovo–Mumford regularity. Embed by . For each of the finitely many positive cohomology degrees , Serre vanishing for the ample bundle makes vanish for . Thus the pushed-forward sheaf is zero-regular. Persistence of regularity makes it -regular for every , giving exactly the displayed vanishing. This is the uniform Serre vanishing for two ample twists lemma. It does not assume that is ample on .
Apply the divisor restriction exact sequence to with twists . For , both neighbouring cohomology groups on vanish, so
For a fixed , sufficiently large kills the right-hand cohomology by Serre vanishing for on . The higher cohomology vanishing from an ample hyperplane restriction argument gives for all . Consequently
This step is essential: divergence of a polynomial does not by itself prove that its top-degree coefficient is positive.
For large enough, . 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 . Lower-dimensional subvarieties already have the same property by induction, so the next section's implication (c)(a) applies. Thus
For , higher groups with vanish automatically and the same evaluation argument starts the induction. The regularity facts used above are stated in the Stacks Project, regularity lemmas.
We prove (c)(a) by induction on dimension. It suffices to work on an integral projective variety . By induction, is ample on every lower-dimensional integral subvariety, and hence on every lower-dimensional closed subscheme by ampleness on reduced components.
Apply (c) to itself. The nonzero section of has a nonempty zero divisor . Because is integral, this is an effective Cartier divisor and . Its support has dimension less than , so , and therefore , is ample. Part (ii) makes semiample, hence some positive multiple of is basepoint-free.
Let be the resulting Kodaira map, with . No fibre can have positive dimension: such a projective fibre contains an integral projective curve , on which has degree zero. But the assumed nonzero section of some cannot vanish anywhere, since its nonempty effective divisor would have positive degree. This contradicts (c).
Thus has zero-dimensional fibres. A proper quasi-finite morphism is a finite morphism. The finite pullback of an ample line bundle is ample, so and then are ample. This proves
The fibre argument proves the semiample and curve-positive ampleness criterion. It also explains why testing only existence of a nonzero section, without requiring a zero, would be insufficient: the trivial bundle on a positive-dimensional projective variety has a nowhere-vanishing section.
Put and let be the canonical global section cutting out the effective Cartier divisor . The divisor restriction exact sequence gives
The hypothesis says is an ample line bundle. By Serre vanishing, for all sufficiently large , so the long exact sequence in sheaf cohomology makes
surjective for all such . 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 restriction
is surjective for all sufficiently large .
Choose such an for which is also a globally generated line bundle. Lift a generating collection of its global sections to . At each point of , one lift has nonzero image in the one-dimensional residue-field fibre, hence generates the stalk of by Nakayama lemma. Outside , is nowhere zero and generates . Together these sections generate it everywhere. Therefore
This proof works on an arbitrary projective scheme because the defining section of an effective Cartier divisor is a non-zero-divisor. If is empty, 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 satisfies the hypothesis on its support but has zero intersection with the exceptional curve, and therefore is not ample.
First suppose is integral. Induct on its dimension of a scheme. The previous two parts handle all and also handle when eventually vanishes. Otherwise choose with a nonzero global section of . On an integral variety it defines an effective Cartier divisor , possibly empty, and multiplication by the section gives
Thus . For , the induction hypothesis bounds the last term by ; summing on each residue class modulo gives . For , is zero-dimensional and its positive-degree sheaf cohomology vanishes, so the same recurrence is bounded.
For a general projective scheme, a nonzero section can be a zero divisor, so that argument requires an additional step. The general cohomology growth for nef twists supplies it: for every coherent sheaf with support dimension ,
Its proof uses Fujita vanishing, an ample section avoiding the associated points of , and induction on support dimension. Taking gives the required estimate for every , including nonreduced and reducible schemes; degrees above vanish.
For an effective Cartier divisor , the divisor restriction exact sequence gives
The scheme has dimension at most , so the permitted scheme version of part (i) bounds by . For the infinitely many in the hypothesis,
once is sufficiently large. Thus infinitely many such have a nonzero section of . 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 , so the polynomial bound for sections of a fixed divisor gives the same conclusion. For the assertion is immediate.
If is an effective Cartier divisor on a projective scheme and is ample, then is semiample. The divisor restriction exact sequence and Serre vanishing make surjective for large . Their finite dimensions stabilize, so restriction on global sections is eventually surjective. Lift generators on ; off , the canonical section of generates. Together these generate , including on nonreduced .
If is a big divisor and an effective Cartier divisor, infinitely many have . The divisor restriction exact sequence bounds the dimension lost upon restriction to by , using the polynomial bound for sections of a fixed divisor; this cannot exhaust the sections along the infinite growth sequence.
A Cartier divisor is ample if for every positive-dimensional integral closed subvariety some positive multiple restricts to a bundle with a nonzero section having a nonempty zero locus. Inductively the divisor is ample on all lower-dimensional subschemes. On an integral component the chosen section cuts out a nonempty effective Cartier divisor whose restriction bundle is ample. The restriction ampleness implies semiampleness for an effective divisor lemma makes the original divisor semiample. On a curve the vanishing section forces positive degree, so the semiample and curve-positive ampleness criterion proves ampleness.