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Order the index set of the open cover $\mathcal{U}=(U_a)$. The Čech cochain groups areand their differential is the alternating sum of restrictions:Terms obtained by deleting two indices cancel in pairs, so ${\delta }^2=0$. The Čech cohomology is ${\check{H}}^p=\ker \delta /\mathrm{im}\delta$, with zero incoming differential in degree zero. Its zeroth group is the group of global sections by sheaf gluing.

The acyclic cover theorem applies if every nonempty finite intersection is acyclic for $\mathcal{F}$. For a quasi-coherent sheaf on a variety it is sufficient that every such intersection is affine; on a separated variety, any affine open cover has this property. On ${\mathbb{P}}^n$, take the $n+1$ standard charts $U_i=D_{+}(X_i)$. Every intersection is a principal open in an affine chart, hence affine. The Čech complex has no terms above degree $n$, provingThis is the cohomological dimension bound from an affine cover and requires only quasi-coherence, not finite generation.

For the remaining projective calculations assume $n\ge 1$. There is a necessary zero-dimensional exception to the negative-twist assertion: ${\mathbb{P}}^0$ is a point, every twist is trivial there, and $H^0({\mathbb{P}}^0,\mathcal{O}(m))\cong k$ even for $m<0$.

Put $S=k[X_0,\dots ,X_n]$ with its usual grading. The twisting sheaf on projective space is the sheaf associated with the shifted graded module $S(m)$; on an intersection $U_I={\bigcap }_{i\in I}{U}_i$ its sections areOn $U_i$ a generator is $e_i=X_i^m$, with transition $e_j=(X_j/X_i{)}^m{e}_i$. These are regular units on overlaps and satisfy the cocycle relation, so they glue an invertible sheaf for every integer $m$, including negative $m$.

A global section is a compatible family of these homogeneous fractions, hence a single element of degree $m$ in ${\bigcap }_i{S}_{X_i}\subseteq \mathrm{Frac}S$. For $n\ge 1$, $X_0$ and $X_1$ are relatively prime in the unique factorization domain $S$, so $S_{X_0}\cap S_{X_1}=S$; therefore the full intersection is $S$. It follows thatFor $m\ge 0$ its dimension is $\left(\genfrac{}{}{0ex}{}{n+m}{n}\right)$. The proof uses all-chart compatibility; regularity on one chart alone would permit poles on its complement.

If some $d_i=0$, the ideal $(X_i^{d_i})$ contains $1$, so the requested containment is immediate. Otherwise every $d_i\ge 1$. A monomial $X_0^{a_0}\cdots X_n^{a_n}$ of degree ${\sum }_i{d}_i$ must have $a_i\ge d_i$ for some $i$: if not, its total degree would be at most ${\sum }_i(d_i-1)<{\sum }_i{d}_i$. Thus every degree-${\sum }_i{d}_i$ monomial, and hence every homogeneous polynomial of that degree, belongs to $(X_0^{d_0},\dots ,X_n^{d_n})$. This is monomial containment in an ideal of coordinate powers.

For $n>0$, a top-degree Čech cochain for $\mathcal{O}$ is a degree-zero Laurent polynomial on the full intersection. Write it with a common denominator asThe containment just proved gives $P={\sum }_i{Q}_i{X}_i^{d_i}$, with homogeneous $Q_i$ of degree ${\sum }_j{d}_j-d_i$. ConsequentlyThe $i$th term is regular on the intersection omitting $U_i$, and has degree zero. Give it the sign $(-1{)}^i$ in that component of the preceding Čech cochain. Its coboundary is the original fraction. Every top cochain is thus a coboundary, andThis is top Čech cohomology from missing-denominator monomials.

To obtain the negative-twist bound by induction on dimension, let $H\cong {\mathbb{P}}^{n-1}$ be a hyperplane, with inclusion $j$. Its equation gives the hyperplane exact sequence for twisting sheavesThe associated long exact sequence in sheaf cohomology and sheaf cohomology under a closed inclusion giveFor $n=1$, restriction $H^0({\mathbb{P}}^1,\mathcal{O})\to H^0(H,{\mathcal{O}}_H)$ is the surjection $k\to k$. Exactness and the already proved $H^1({\mathbb{P}}^1,\mathcal{O})=0$ give $H^1({\mathbb{P}}^1,\mathcal{O}(-1))=0$. The displayed surjections then give the same vanishing for every $m\ge 0$. This handles the point hyperplane without making a false negative-twist claim on ${\mathbb{P}}^0$.

For $n\ge 2$, the induction hypothesis in dimension $n-1$ gives $H^{n-1}(H,{\mathcal{O}}_H(m))=0$ whenever $m\ge -(n-1)$. Therefore consecutive top-degree groups are isomorphic for $m\ge 1-n$. Starting with $H^n({\mathbb{P}}^n,\mathcal{O})=0$ and stepping down through $m=0,-1,\dots ,1-n$ reaches $H^n({\mathbb{P}}^n,\mathcal{O}(-n))=0$. Stepping upward proves all positive twists too. We concludeThis is top-twist vanishing by hyperplane induction.

Finally, on $U_0$ write $x_a=X_a/X_0$ and ${\omega }_0=d{x}_1\wedge \cdots \wedge d{x}_n$. On $U_i$, let ${\omega }_i$ be the wedge of $d(X_a/X_i)$ for $a\ne i$, taken in increasing index order. For $i>0$, differentiating $X_0/X_i=x_i^{-1}$ and $X_a/X_i=x_a/x_i$ givesIn the wedge, all terms involving a second copy of $d{x}_i$ disappear; the surviving powers are $x_i^{-2}$ from $d(x_i^{-1})$ and $x_i^{-1}$ from the other $n-1$ differentials. Moving $d{x}_i$ into its original position produces the stated sign. Rescale each local generator by ${\eta }_i=(-1{)}^i{\omega }_i$. Then ${\eta }_j=(X_j/X_i{)}^{-n-1}{\eta }_i$ on every overlap, exactly the transition of the twist with $m=-n-1$. This canonical-form transition on projective space proves the canonical bundle of projective space:For line bundles the Serre duality pairing becomesIn particular $H^n(\mathcal{O}(-r))$ is dual to $H^0(\mathcal{O}(r-n-1))$, which is zero for $r\le n$, precisely the independently obtained bound. The case $r=0$ recovers $H^n(\mathcal{O})=0$, and $H^n(\mathcal{O}(-n-1))\cong k$ matches $H^0(\mathcal{O})\cong k$. Thus the calculated transition functions, global sections and top-degree vanishing agree with Serre duality.

Choose the rational point $x=[1:0:\cdots :0]$ and the coherent sheaf $\mathcal{F}=i_{\ast }k$ supported there. By skyscraper sheaf cohomology, $H^0(X,\mathcal{F})=k$.

Its dual of a sheaf ${\mathcal{F}}^{\vee }=\mathcal{H}om(\mathcal{F},{\mathcal{O}}_X)$ is zero. Away from $x$ its source is zero. At $x$, a homomorphism $k\to {\mathcal{O}}_{X,x}$ must send one to an element annihilated by the maximal ideal. Since $n>0$, that ideal contains a nonzero coordinate parameter; the local ring is an integral domain, so the image must vanish. This proves that every local homomorphism vanishes. ConsequentlyThere is no contradiction with Serre duality: the ordinary sheaf dual suffices for locally free sheaves, but general coherent sheaves require an Ext functor. This example illustrates failure of ordinary sheaf-dual Serre duality for a skyscraper sheaf.

Let $C\in \mathrm{Neg}(K_X)$. The arithmetic adjunction formula on a smooth surface givesWe first exclude $C^2\ge 0$, following the PDF's hint and the finiteness proved in (a).

If $C^2=0$, adjunction and $K_X\cdot C<0$ force $p_a(C)=0$ and $K_X\cdot C=-2$. An integral projective curve of arithmetic genus zero is a smooth rational curve: normalization and the nonnegative singularity-length correction show both normalization genus and singularity correction vanish. Part (iii) makes $C$ semiample. Choose a basepoint-free $|mC|$. Over the infinite algebraically closed field, a general section avoids containing any of the finitely many curves of $\mathrm{Neg}(K_X)$ as a component. Its effective divisor $G$ then has $K_X\cdot G\ge 0$, but $K_X\cdot G=m{K}_X\cdot C<0$, a contradiction.

If $C^2>0$, Riemann–Roch theorem for algebraic surfaces and Serre duality show $h^0(X,mC)$ is unbounded. Indeed $h^2(X,mC)=h^0(X,K_X-mC)=0$ for $m\gg 0$, because the latter divisor has negative intersection with a fixed ample divisor. HenceThere is therefore some $m$ with $h^0(X,mC)>h^0(X,(m-1)C)$, giving a section whose divisor does not contain $C$. The canonical section of $mC$ does not vanish identically along any other curve. A general linear combination of these two sections consequently contains none of the finite set $\mathrm{Neg}(K_X)$, again contradicting its negative canonical intersection. Thus $C^2<0$.

Now put $a=-C^2\ge 1$ and $b=-K_X\cdot C\ge 1$. Adjunction givesIt follows that $a=b=1$ and $p_a(C)=0$. HenceThe general-section argument only avoids finitely many proper linear subspaces, so it works over any algebraically closed field, including positive characteristic.

The resolution principle for sheaf cohomology states that an exact sequencewith $H^q(X,{\mathcal{A}}^j)=0$ for every $q>0$ computes $H^i(X,\mathcal{F})$ as the cohomology of $\Gamma (X,{\mathcal{A}}^{\bullet })$. In particular one may use a flasque resolution.

An affine morphism $\varphi :X\to Y$ is one for which the inverse image of every affine open subset of $Y$ is affine. We use vanishing of quasi-coherent cohomology on an affine scheme: a quasi-coherent sheaf has zero higher sheaf cohomology on an affine variety. Take a flasque resolution $\mathcal{F}\to {\mathcal{I}}^{\bullet }$. Each direct image sheaf ${\varphi }_{\ast }{\mathcal{I}}^j$ is flasque, because restriction on $V\subseteq U\subseteq Y$ is restriction on ${\varphi }^{-1}V\subseteq {\varphi }^{-1}U$. On an affine open $U\subseteq Y$, the augmented complex of sections isIts positive-degree cohomology vanishes by the affine theorem, and its degree-zero kernel is $\Gamma ({\varphi }^{-1}U,\mathcal{F})$. Checking exactness on this affine basis shows that ${\varphi }_{\ast }{\mathcal{I}}^{\bullet }$ is a flasque resolution of ${\varphi }_{\ast }\mathcal{F}$. The global-section complexes are identical, proving cohomology under an affine morphism:

For projective space, put $S=k[x_0,\dots ,x_n]$ with its usual grading and $U_i=\{x_i\ne 0\}$. Construct the twisting sheaf on projective space by setting its sections on $U_i$ to be the degree-$m$ part $(S_{x_i}{)}_m$, viewed as a module over $(S_{x_i}{)}_0$. It is free of rank one with generator $x_i^m$, even when $m$ is negative. On overlaps the generators differ by the unit $(x_i/x_j{)}^m$, so these modules glue to a line bundle $\mathcal{O}(m)$.

Assume first $n\ge 1$. These local modules all embed in the graded fraction field, and agreement on overlaps identifies global sections with ${\bigcap }_i(S_{x_i}{)}_m$. Since $S$ is a unique factorization domain and at least two distinct variables occur, a reduced fraction belonging to every $S_{x_i}$ has no nonconstant denominator. Thus the intersection is $S_m$. Counting monomials by stars and bars givesHere $h_n^i(m)={\dim }_k{H}^i({\mathbf{P}}^n,\mathcal{O}(m))$. The printed formula needs the convention that a counting binomial coefficient is zero when its upper integer is smaller than its nonnegative lower integer; the generalized polynomial convention for negative upper integers would give incorrect dimensions. For $n=0$, every twist on the single point ${\mathbf{P}}^0$ is trivial, so $h_0^0(m)=1$ for every $m$ and its higher cohomology is zero. This case supplies the base of dimension induction; the displayed top-cohomology formula below is for $n\ge 1$.

The standard $n+1$ affine opens have affine finite intersections. The acyclic cover theorem and affine vanishing identify sheaf cohomology with a Čech cochain complex ending in degree $n$. In particular $H^i({\mathbf{P}}^n,\mathcal{O}(m))=0$ for $i>n$.

Let $H=\{x_n=0\}\cong {\mathbf{P}}^{n-1}$, with inclusion $i$. Multiplication by $x_n$ gives the hyperplane exact sequence for twisting sheavesExactness is checked in the local trivializations, where a hyperplane equation is a non-zero-divisor and the quotient is its restriction to $H$. Since a closed immersion is an affine morphism, the first part identifies the cohomology of $i_{\ast }{\mathcal{O}}_H(m)$ with that on $H$.

For $n=1$, the long exact sequence in sheaf cohomology and $H\cong {\mathbf{P}}^0$ giveStarting from the given vanishing for sufficiently large $m$ and descending gives $h_1^1(m)=0$ for $m\ge -1$ and $h_1^1(m)=-m-1$ for $m\le -2$.

Now let $n\ge 2$, and assume the formulas in dimension $n-1$. Restrictionis surjective for every $m$: for $m\ge 0$ it is polynomial restriction, and for $m<0$ both groups are zero since $\dim H\ge 1$. The long exact sequence in sheaf cohomology therefore makes $H^1({\mathbf{P}}^n,\mathcal{O}(m-1))\to H^1({\mathbf{P}}^n,\mathcal{O}(m))$ injective. For $2\le j<n$, dimension induction gives $H^{j-1}(H,{\mathcal{O}}_H(m))=0$, so the same sequence makesinjective. Composing these injections up to a sufficiently large twist, where the target is zero by the given hypothesis, proves $H^j({\mathbf{P}}^n,\mathcal{O}(m))=0$ for every $m$ and $0<j<n$.

The remaining part of the long exact sequence in sheaf cohomology is consequentlywhere the last zero uses the affine-cover dimension bound on $H$. Hence $h_n^n(m-1)=h_n^n(m)+h_{n-1}^{n-1}(m)$. Using dimension induction and eventual vanishing, downward summation yieldsby telescoping Pascal's identity; for $m\ge -n$ it is zero. Thus, without invoking Serre duality in the induction,Finally, the canonical bundle of projective space is ${\omega }_{{\mathbf{P}}^n}\cong \mathcal{O}(-n-1)$, obtained by taking determinants in the dual Euler sequence. Serre duality therefore predictsIt pairs the two computed extreme-degree dimensions and pairs zero intermediate groups with zero intermediate groups, exactly as required.

A connected integral projective variety has only constant regular functions, so $h^0(X,{\mathcal{O}}_X)=1$. By Serre duality and $K_X\sim 0$, $h^2(X,{\mathcal{O}}_X)=h^0(X,{\mathcal{O}}_X(K_X))=1$. The assumed first sheaf cohomology vanishes. Therefore the Euler characteristic of a coherent sheaf isThese hypotheses characterize the K3 surface setting used below.

First derive the intersection constraints. Since $D$ is nef and $F,M$ are effective, $D\cdot F,D\cdot M\ge 0$; their sum is $D^2=0$, so both vanish. Since the movable divisor $M$ is nef, $M^2,M\cdot F\ge 0$, and $M\cdot D=0$ forces both to vanish. In particularFor every component $E$ of $F$, nefness and $D\cdot F=0$ imply $D\cdot E=0$. The isotropic orthogonality consequence of the Hodge index theorem gives $E^2\le 0$. If $E^2=0$, Riemann–Roch theorem for algebraic surfaces and Serre duality give $h^0(X,E)\ge 2$, since $-E$ cannot be effective. Choose $E^{\prime }\in |E|$ different from $E$. It cannot contain $E$: otherwise $E^{\prime }-E$ would be a nonzero effective numerically trivial divisor, contradicting its positive intersection with an ample divisor. Choose a member of $|M|$ avoiding $E$. Replacing one copy of $E$ in $F$ by $E^{\prime }$ now produces a member of $|D|$ with smaller multiplicity along $E$, contradicting the definition of the fixed part. Therefore

The strict statement as printed needs the qualification if $F\ne 0$. Part (a) already shows $F^2=0$ in the present situation, and the next part proves $F=0$; an unconditional $F^2<0$ would contradict that conclusion.

To prove the needed conditional statement, suppose $F\ne 0$. A fixed part has $h^0(X,F)=1$. Indeed, a different $F^{\prime }\in |F|$ has smaller multiplicity along some component of $F$; otherwise $F^{\prime }-F$ would be a nonzero effective linearly trivial divisor. Adding a member of the movable part avoiding that component contradicts fixedness.

The Hodge index theorem for algebraic surfaces, together with $D\cdot F=0$ and $D^2=0$, gives $F^2\le 0$. If equality held, Riemann–Roch theorem for algebraic surfaces would give $\chi (X,F)=2$, and Serre duality would give $h^2(X,F)=h^0(X,-F)=0$. Hence $h^0(X,F)=2+h^1(X,F)\ge 2$, a contradiction. ThereforeThis is the fixed-part elimination on a K3 surface argument.

Choose an ample divisor $A$. The Hodge index theorem for algebraic surfaces implies $D\cdot A>0$: if it were zero, $D^2=0$ would force the numerical class of $D$ to be zero. Consequently $-D$ cannot be effective, because every nonzero effective divisor has positive intersection with $A$, while $-D\cdot A<0$.

The Riemann–Roch theorem for algebraic surfaces and the preceding computation give $\chi (X,{\mathcal{O}}_X(D))=2+D^2/2=2$. By Serre duality, $h^2(X,D)=h^0(X,-D)=0$. Thus

Use the rational principal-parts resolution on an algebraic curve to represent a class by finitely many rational differential principal parts. Sum their algebraic residues of a rational differential. The sum of residues of a global rational differential is zero: trace to the projective line, where partial fractions prove the assertion. Hence the sum descends to cohomology. Pairing an invertible sheaf with its differential-valued dual gives the perfect Serre duality pairing. Since the global regular functions on an integral projective curve over an algebraically closed field are $k$, evaluation at the constant function $1$ identifies this trace canonically with an isomorphism to $k$.

For a Cartier divisor on a smooth projective surface, the displayed formula computes its Euler characteristic of a coherent sheaf. Combined with Serre duality, it turns intersection numbers into lower bounds for section spaces.

For a compact complex $n$-manifold and a holomorphic vector bundle $E$, analytic Serre duality givesThe pairing integrates the wedge product of bundle-valued Dolbeault cohomology representatives, contracting $E$ with its dual.

The tangent cone to a theta divisor at a degree-$g-1$ line bundle $L$ is cut out in $H^1({\mathcal{O}}_X)$ by the determinant of $\xi ⌣-:H^0(L)\to H^1(L)$. Under Serre duality its matrix entries are $⟨\xi ,s_i{t}_j⟩$ for bases of $H^0(L)$ and $H^0(K_X\otimes L^{-1})$. For $h^0(L)=1$, its tangent hyperplane is annihilated by the nonzero holomorphic differential form $s_1{t}_1$.