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Let $W$ be an integral effective Cartier divisor in an integral variety $V$ over a perfect field, and suppose $W$ is not contained in the singular locus of $V$. On a dense open subset both varieties are smooth. The Zariski tangent space of $W$ there has codimension one in that of $V$, so the differential of the local defining equation is nonzero. Thus the conormal sheaf map is injective generically. Its kernel is a subsheaf of a line bundle on the integral variety $W$ and is therefore a torsion-free sheaf; generic vanishing implies zero everywhere. Right exactness is the Conormal exact sequence for Kähler differentials.

For an effective Cartier divisor $E$ and any Cartier divisor $D$, multiplication by a defining section of $E$ gives the displayed short exact sequence of sheaves. Its long exact sequence in sheaf cohomology compares sections and cohomology on $X$ with their restrictions to $E$.

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.

A proper subvariety locally defined by one non-zero-divisor is an effective Cartier divisor. On an integral variety a nonzero local equation is automatically a non-zero-divisor. Its conormal sheaf is a line bundle, since $A/(f)\cong (f)/(f^2)$ via multiplication by $f$. Allowing the zero equation would include the whole variety and would not give this conclusion.

Let $f:{\mathbb{P}}_k^1\to \mathrm{Spec}k$ be the structure map, a projective morphism. Let $D$ be the union of the two rational points $[0:1]$ and $[1:0]$. The homogeneous polynomial $t_0{t}_1$ defines this effective Cartier divisor and gives the exact sequence of coherent sheavesThe twisting sheaf on projective space $\mathcal{O}(-2)$ has no nonzero global sections: the chart calculation of Question 3(iv) now requires $a(t)=t^{-2}b(t^{-1})$, again impossible for nonzero polynomials. We also have $\Gamma ({\mathbb{P}}^1,\mathcal{O})=k$ and $\Gamma (D,{\mathcal{O}}_D)=k\oplus k$.

The direct image sheaf under $f$ is the module of global sections on the one-point scheme. Consequently the induced sequence isIts last nonzero map has cokernel isomorphic to $k$, since $(a,b)\mapsto b-a$ has the diagonal as its kernel. Thus the induced sequence is not exact at $f_{\ast }{\mathcal{O}}_D$, proving that projective direct image need not preserve surjections.

Take the hyperplane divisor $D=(t_0=0)$. On $U_i=D_{+}(t_i)$ its local equation is $g_i=t_0/t_i$, a non-zero-divisor; on $U_0$ it is the unit one. Thus it is an effective Cartier divisor. On overlaps, $g_i/g_j=t_j/t_i$ is a unit.

The divisor line bundle ${\mathcal{O}}_X(D)$ is locally generated by $1/g_i=t_i/t_0$. Map this generator to the local generator $t_i$ of the twisting sheaf on projective space ${\mathcal{O}}_X(1)$. On overlaps both generators transform by $t_i/t_j$, so the maps glue and giveFor $n=0$ the hyperplane is empty and this is the zero Cartier divisor; the conclusion still holds.

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.

We prove (c)$\Rightarrow$(a) by induction on dimension. It suffices to work on an integral projective variety $V$. By induction, $D$ is ample on every lower-dimensional integral subvariety, and hence on every lower-dimensional closed subscheme by ampleness on reduced components.

Apply (c) to $V$ itself. The nonzero section of ${\mathcal{O}}_V(mD)$ has a nonempty zero divisor $E$. Because $V$ is integral, this is an effective Cartier divisor and ${\mathcal{O}}_V(E)\cong {\mathcal{O}}_V(mD)$. Its support has dimension less than $\dim V$, so $D{|}_E$, and therefore ${\mathcal{O}}_E(E)$, is ample. Part (ii) makes $E$ semiample, hence some positive multiple of $D$ is basepoint-free.

Let $f:V\to Y\subset {\mathbb{P}}^N$ be the resulting Kodaira map, with ${\mathcal{O}}_V(qD)=f^{\ast }{\mathcal{O}}_Y(1)$. No fibre can have positive dimension: such a projective fibre contains an integral projective curve $C$, on which $D$ has degree zero. But the assumed nonzero section of some ${\mathcal{O}}_C(rD)$ cannot vanish anywhere, since its nonempty effective divisor would have positive degree. This contradicts (c).

Thus $f$ has zero-dimensional fibres. A proper quasi-finite morphism is a finite morphism. The finite pullback of an ample line bundle is ample, so $qD$ and then $D$ are ample. This provesThe 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 $L={\mathcal{O}}_X(D)$ 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.

First suppose $X$ is integral. Induct on its dimension of a scheme. The previous two parts handle all $j>1$ and also handle $j=1$ when $h^0(X,mD)$ eventually vanishes. Otherwise choose $r>0$ with a nonzero global section of $rD$. On an integral variety it defines an effective Cartier divisor $E$, possibly empty, and multiplication by the section givesThus $h^1(X,mD)\le h^1(X,(m-r)D)+h^1(E,mD)$. For $n\ge 2$, the induction hypothesis bounds the last term by $O(m^{n-2})$; summing on each residue class modulo $r$ gives $O(m^{n-1})$. For $n=1$, $E$ 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 $\mathcal{F}$ with support dimension $d$,Its proof uses Fujita vanishing, an ample section avoiding the associated points of $\mathcal{F}$, and induction on support dimension. Taking $\mathcal{F}={\mathcal{O}}_X$ gives the required $O(m^{n-1})$ estimate for every $j\ge 1$, including nonreduced and reducible schemes; degrees above $n$ vanish.

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.

A relative effective divisor of degree $m$ on a smooth projective curve is an effective Cartier divisor in $C\times T$ that is finite and flat of degree $m$ over $T$. Its associated line bundle has degree $m$ on every fibre. The symmetric product of a curve $C^{(m)}$ represents these families and carries their universal divisor. Locally on a smooth coordinate chart, the divisor is described by a monic polynomial, whose coefficients are elementary symmetric functions of its points; this description includes collisions and works in positive characteristic.

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 $X$.

If $D$ is a big divisor and $F$ an effective Cartier divisor, infinitely many $m$ have $H^0(X,mD-F)\ne 0$. The divisor restriction exact sequence bounds the dimension lost upon restriction to $F$ by $O(m^{n-1})$, using the polynomial bound for sections of a fixed divisor; this cannot exhaust the $c{m}^n$ 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 $E$ whose restriction bundle ${\mathcal{O}}_E(E)$ 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.