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Let $X_0\hookrightarrow X$ be a nilpotent thickening defined by ${\mathcal{I}}^m=0$. If $X$ is an affine scheme, its closed subscheme $X_0$ is an affine scheme. Conversely, suppose $X_0$ is an affine scheme. The underlying topological spaces are identical, so $X$ is a quasi-compact scheme. For any quasi-coherent sheaf $\mathcal{F}$, each subquotient ${\mathcal{I}}^j\mathcal{F}/{\mathcal{I}}^{j+1}\mathcal{F}$ is naturally a quasi-coherent sheaf on $X_0$. By sheaf cohomology under a closed inclusion and vanishing of quasi-coherent cohomology on an affine scheme, its first sheaf cohomology on $X$ vanishes. The long exact sequence in sheaf cohomology and descending mathematical induction through this finite filtration prove $H^1(X,\mathcal{F})=0$. The Serre criterion for affineness now makes $X$ an affine scheme.

Let $i:X_{\mathrm{red}}\hookrightarrow X$ be the reduced induced subscheme, with structure sheaf ${\mathcal{O}}_{X_{\mathrm{red}}}={\mathcal{O}}_X/\mathcal{N}$. Its underlying map of topological spaces is a homeomorphism, since every prime ideal contains the nilradical.

Put $\mathcal{G}=\mathcal{F}/\mathcal{N}\mathcal{F}$. Multiplication by $\mathcal{N}$ annihilates $\mathcal{G}$, so its action by ${\mathcal{O}}_X$ factors through ${\mathcal{O}}_X/\mathcal{N}$. Thus $\mathcal{G}$ is naturally a sheaf of modules over ${\mathcal{O}}_{X_{\mathrm{red}}}$. On an affine open subscheme $\mathrm{Spec}A$, with $\mathcal{F}=\tilde{M}$, it corresponds to $M/\mathrm{Nil}(A)M$ over $A/\mathrm{Nil}(A)$, so it remains a quasi-coherent sheaf.

The sheaf cohomology of a sheaf of modules is the sheaf cohomology of its underlying sheaf of abelian groups. Under the above homeomorphism, that underlying sheaf is identical on the two spaces: the global sections functor and its flasque resolutions are identical. Equivalently, use sheaf cohomology under a closed inclusion with $i_{\ast }\mathcal{G}$. Therefore there are canonical isomorphisms, in every degree,In particular, this gives the requested equality in degree one. Equality of the underlying topological spaces alone would not compare arbitrary coefficient sheaves; the factorization of the action explains why the indicated coefficients are the same.

Let $S=k[X_0,\dots ,X_n]$ with its usual grading, and put $S(m{)}_d=S_{m+d}$. Define the twisting sheaf on projective space byMore explicitly, on $U_i=D_{+}(X_i)$ its sections are the homogeneous degree-$m$ part of $S_{X_i}$. They are a free rank-one module over $(S_{X_i}{)}_0=k[X_j/X_i:j\ne i]$, with generator $e_i=X_i^m$. These generators make sense for every integer $m$, since $X_i$ is invertible on this chart. Their transition rule isThe ratios are regular units on overlaps and satisfy the cocycle identity, so they glue to an invertible sheaf. In particular $\mathcal{O}(m)\otimes \mathcal{O}(r)\cong \mathcal{O}(m+r)$.

For $n\ge 1$, a global section is a compatible homogeneous degree-$m$ rational expression lying in every $S_{X_i}$. Inside the fraction field of $S$,To justify this, write a fraction in lowest terms in the unique factorization domain $S$. Membership in $S_{X_i}$ forces every irreducible divisor of its denominator to be associated with $X_i$. Since there are at least two nonassociated variables, the denominator must be a unit. ThereforeThe degree-$m$ monomials are indexed by nonnegative integers $a_0,\dots ,a_n$ with sum $m$. The stars and bars count givesHere and below, an all-integer binomial shorthand must use the counting convention $\left(\genfrac{}{}{0ex}{}{a}{n}\right)=0$ when $a<n$, including negative $a$. The polynomial extension of binomial coefficients to negative upper arguments does not give the stated dimensions. In dimension zero, ${\mathbb{P}}^0$ is one point and every ${\mathcal{O}}_{{\mathbb{P}}^0}(m)$ is trivial, so its space of global sections is $k$ for every $m$; this case is treated separately.

For a closed inclusion $i:H\hookrightarrow {\mathbb{P}}^n$, direct image sheaf formation is exact: its stalk is the original stalk at a point of $H$ and zero outside $H$. It preserves flabby sheaves, since restrictions on the target are restrictions between the corresponding open subsets of $H$. Push forward a flasque resolution of $\mathcal{F}$; exactness gives a flasque resolution of $i_{\ast }\mathcal{F}$, and its complex of global sections is the original one becauseTaking cohomology of these identical complexes proves sheaf cohomology under a closed inclusion:This argument applies even without quasi-coherence.

We derive the remaining sheaf cohomology dimensions using the given eventual vanishing, the long exact sequence in sheaf cohomology, and induction on dimension. Multiplication by the linear equation of a hyperplane $H\cong {\mathbb{P}}^{n-1}$ gives the hyperplane exact sequence for twisting sheavesExactness can be checked in each chart by quotienting its polynomial ring by the equation of $H$; multiplication is injective because that equation is a non-zero-divisor. We use the closed-inclusion cohomology identification to replace the cohomology of the final sheaf by that on $H$.

The dimension-zero base is a point, whose global-section functor is exact. Thus all its higher sheaf cohomology is zero. For $n=1$, the relevant long exact sequence isWhen $m\ge 0$, restriction of homogeneous polynomials to the hyperplane point is surjective: after choosing that point as $[1:0]$, the section $X_0^m$ restricts to a generator. Therefore the consecutive $H^1$ groups are isomorphic for $m\ge 0$. The given vanishing for large $m$ propagates down to $m=-1$. For $m<0$, both displayed $H^0$ groups are zero, and the sequence becomesStarting with $H^1(\mathcal{O}(-1))=0$, induction downward gives $h^1(\mathcal{O}(m))=-m-1$ for $m\le -2$. For degrees $i>1$, the point contributes no cohomology to the exact sequence, so consecutive $H^i$ groups are isomorphic for every $m$; the given eventual vanishing makes them all zero. Hence the complete positive-degree formula holds for ${\mathbb{P}}^1$.

Now let $n\ge 2$ and assume the formulas for $H\cong {\mathbb{P}}^{n-1}$. For every $m$, the restrictionis surjective: for $m\ge 0$ extend a polynomial in the hyperplane coordinates to the same polynomial on ${\mathbb{P}}^n$, and for $m<0$ both groups vanish because $H$ has positive dimension. Consequently the exact sequence gives an injectionFor $2\le i\le n-1$, the induction hypothesis says $H^{i-1}(H,{\mathcal{O}}_H(m))=0$, and gives the same injection in degree $i$. For each fixed $m$, composing finitely many injections reaches a sufficiently large twist, whose higher cohomology is zero by the given hypothesis. ThereforeFor $i>n$, the two neighbouring hyperplane groups $H^{i-1}(H,{\mathcal{O}}_H(m))$ and $H^i(H,{\mathcal{O}}_H(m))$ are zero by induction. Consecutive $H^i$ groups on ${\mathbb{P}}^n$ are then isomorphic, and eventual vanishing again makes them zero. This derives vanishing above the dimension too, without needing a special projective-space cohomology theorem.

For top cohomology the exact sequence now reduces toWrite $b_n(m)=h^n({\mathcal{O}}_{{\mathbb{P}}^n}(m))$. The induction hypothesis and eventual vanishing give the recurrenceConsecutive top groups agree for $m\ge -n+1$, so descending from a large twist gives $b_n(m)=0$ for $m\ge -n$. Below this range, iterating the recurrence gives a finite sum:The last equality follows by summing Pascal's identity. These short exact sequences also prove that every group whose dimension is being counted is finite-dimensional. Thus, for $n\ge 1$,With the stated counting convention, the latter is the requested single binomial formula. The base ${\mathbb{P}}^0$ has $h^0=1$ for all twists and no higher cohomology; it is not governed by the negative-twist $H^0$ rule for positive-dimensional projective space.

Finally calculate the canonical line bundle of a smooth variety from its transition functions. On $U_0$ write $t_j=X_j/X_0$ for $1\le j\le n$. On $U_i$, $i>0$, take the coordinates $u_j=X_j/X_i$ in increasing order of $j\ne i$. ThusIn their wedge product, $d{u}_0=-t_i^{-2}d{t}_i$ and
$d{u}_j=t_i^{-1}d{t}_j-t_j{t}_i^{-2}d{t}_i$. Every term involving a second $d{t}_i$ vanishes, so reordering the remaining differentials givesThis computation is valid in every characteristic. Multiply the ordered local generator on $U_i$ by $(-1{)}^i$. The resulting generators ${\eta }_i$ satisfy ${\eta }_i=(X_i/X_0{)}^{-n-1}{\eta }_0$ on $U_i\cap U_0$, and hence the corresponding transition rule on every overlap. They are exactly the transitions of $\mathcal{O}(-n-1)$. Therefore the canonical bundle of projective space is

For a line bundle, Serre duality statesThe intermediate groups on both sides vanish. In the two outer degrees, the top group at twist $m$ has the same dimension as the global-section group at twist $-m-n-1$:The polynomial degree on the right is nonnegative precisely when $m\le -n-1$, and its dimension is $\left(\genfrac{}{}{0ex}{}{-m-1}{n}\right)$. Interchanging $m$ and $-m-n-1$ checks the other outer degree. On ${\mathbb{P}}^0$ both sides are one-dimensional. Thus all the formulas are consistent with Serre duality, which was used only for this final consistency check, not to prove the cohomology formulas.

For the Sheaf of relative Kähler differentials on $X={\mathbb{P}}_{\mathbb{C}}^3$, the cotangent form of the Euler sequence isThe last arrow is onto because at least one homogeneous coordinate is invertible on each standard affine chart. Locally its kernel is the rank-three module generated by the differentials of the three affine coordinates, yielding the displayed Sheaf of relative Kähler differentials. The cohomology of twisting sheaves on projective space gives $H^q(X,{\mathcal{O}}_X(-1))=0$ for every $q$, while $H^0(X,{\mathcal{O}}_X)=\mathbb{C}$ and its positive-degree groups vanish. The long exact sequence in sheaf cohomology therefore gives the cotangent-sheaf cohomology of projective space in this dimension:The isomorphism for $q=1$ is the connecting map from the constant global sections of ${\mathcal{O}}_X$. The Euler characteristic of a coherent sheaf is the alternating sum of dimensions, explaining the minus sign.

For the curve, put $S=\mathbb{C}[t_0,t_1,t_2,t_3]$. It is a unique factorization domain, so irreducibility of $F$ makes $(F)$ a prime ideal. Also $F$ does not divide $G$: otherwise irreducibility of $G$ would make them associates, contrary to their distinct degrees. Thus the image of $G$ is a nonzero element of the integral domain $S/(F)$. It follows that $(F,G)$ is a regular sequence. Its Koszul complex is an exact sequence, and graded sheafification giveswhere $i:Z\hookrightarrow X$ is the closed immersion. The signs make the composite $F(-G)+GF=0$. The shifts come respectively from $5+7$, $5$, and $7$.

Let ${\mathcal{I}}_Z$ be the ideal sheaf of a closed subscheme. Split this Koszul resolution intoFor all integers $d$, $H^1(X,{\mathcal{O}}_X(d))=H^2(X,{\mathcal{O}}_X(d))=0$. Also $H^1(X,{\mathcal{O}}_X)=H^2(X,{\mathcal{O}}_X)=0$. The two long exact sequences in sheaf cohomology, together with sheaf cohomology under a closed inclusion, consequently identifyBy Serre duality, $H^3(X,{\mathcal{O}}_X(-m))\cong S_{m-4}^{\ast }$ for $m\ge 4$. Counting degree-$d$ monomials in four variables gives $\dim S_d=\left(\genfrac{}{}{0ex}{}{d+3}{3}\right)$. Thus the source has dimension $\left(\genfrac{}{}{0ex}{}{11}{3}\right)=165$, and the two target spaces have dimensions $\left(\genfrac{}{}{0ex}{}{4}{3}\right)=4$ and $\left(\genfrac{}{}{0ex}{}{6}{3}\right)=20$. The rank-nullity theorem proves the requested bound:

In fact equality holds. Under the Serre duality pairings, the dual of this map isIf $Ga=Fb$, primeness of $F$ and $F\nmid G$ imply $F\mid a$. But $a$ has degree one and $F$ has degree five, forcing $a=0$, and then $b=0$. The dual map is injective, so the original map is surjective and its kernel has dimension $141$. Thus the lower bound is attained for every pair allowed in the question. No smoothness assumption is needed. The regular sequence cuts out a projective complete intersection of dimension one. Also $H^0(X,{\mathcal{I}}_Z)=H^1(X,{\mathcal{I}}_Z)=0$, by the negative twists and the intermediate cohomology vanishing in the first short exact sequence, so the second gives $H^0(Z,{\mathcal{O}}_Z)=\mathbb{C}$. This is the genus of a complete-intersection space curve: the first cohomology has dimension $141$, equal to its arithmetic genus.

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.

For the closed inclusion $f$, the direct image sheaf has stalk ${\mathcal{F}}_x$ at $x\in X$ and zero stalk at points outside $X$: outside the closed set there is an open neighbourhood disjoint from $X$. Hence $f_{\ast }$ is exact, since exact sequences of sheaves are detected on stalks.

Take a flasque resolution $\mathcal{F}\to {\mathcal{I}}^{\bullet }$. Direct image preserves flasqueness, because restrictions are the restrictions on inverse-image opens. Exactness makes $f_{\ast }{\mathcal{I}}^{\bullet }$ a flasque resolution of $f_{\ast }\mathcal{F}$. MoreoverThe complexes computing sheaf cohomology are identical. ThereforeThis is sheaf cohomology under a closed inclusion; the closedness hypothesis is used in the exactness of direct image.

The direct image of a coherent sheaf under a closed immersion puts $\mathcal{G}=f_{\ast }\mathcal{F}$ in the category of coherent sheaves on ${\mathbb{P}}_k^n$. By sheaf cohomology under a closed inclusion and the supplied compatibility of twisting with direct image,Here is a proof of the required Serre vanishing on projective space. A coherent sheaf on ${\mathbb{P}}_k^n$ is the sheaf associated with a graded module for a finite graded $k[t_0,\dots ,t_n]$-module. Equivalently, it has a presentation by finite sums of twisting sheaves. Use a finite twisting resolution of a coherent sheaf on projective space: resolve the graded module by a finite graded free resolution, using the Hilbert syzygy theorem, and sheafify; exactness of localization preserves the resolution. Its terms are finite sums of $\mathcal{O}(a)$.

Choose $d$ sufficiently large that all twists $a+d$ occurring in these finitely many terms are nonnegative. The cohomology of twisting sheaves on projective space then vanishes in every positive degree for every resolution term. In a short exact sequence $0\to {\mathcal{K}}_{j+1}\to {\mathcal{E}}_j\to {\mathcal{K}}_j\to 0$, with ${\mathcal{K}}_0=\mathcal{G}$, the long exact sequence in sheaf cohomology identifies $H^p({\mathcal{K}}_j(d))$ with $H^{p+1}({\mathcal{K}}_{j+1}(d))$ for $p>0$. Iterating to the final acyclic term provesThe standard graded-module description used here is given in Stacks Project, Section 30.15; the vanishing follows from the displayed finite-resolution argument.