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Let $\mathcal{U}=(U_0,\dots ,U_m)$ be a finite affine open cover of the projective scheme $X$. For a coherent sheaf $\mathcal{F}$, form the Čech cochain complexThe two ways to omit any pair of indices have opposite signs, giving ${\delta }^2=0$. Its cohomology groups areA projective scheme over a field is a separated scheme. Finite intersections of its affine opens are affine, and a coherent sheaf is quasi-coherent. Higher sheaf cohomology of a quasi-coherent sheaf on an affine scheme vanishes. Thus this is an acyclic cover and the acyclic cover theorem identifies the displayed Čech cohomology with $H^p(X,\mathcal{F})$. In particular its degree-zero kernel glues compatible local sections to $\Gamma (X,\mathcal{F})$. A cover by $m+1$ opens also gives vanishing for $p>m$.

For the cohomology of twists on projective space, take $S=k[X_0,\dots ,X_r]$ with the usual grading and $U_i=D_{+}(X_i)$. On an intersection indexed by the nonempty set $J$,This follows directly from the construction of the twisting sheaf on projective space: sections of the twist are the degree-$n$ homogeneous fractions in the corresponding localization. The Čech differential preserves each Laurent monomial $X^a=X_0^{a_0}\cdots X_r^{a_r}$, where $a_i\in \mathbb{Z}$ and ${\sum }_i{a}_i=n$. Such a monomial appears precisely in summands withThe complex therefore decomposes as the direct sum of finite-dimensional cochain complexes indexed by these exponent vectors. Their differentials have only the alternating signs of the simplex incidence maps.

If $N(a)=\varnothing$, this is the ordinary unaugmented simplex cochain complex: its degree-zero kernel consists of the common value on every vertex and has dimension one, while all higher cohomology vanishes. If $\varnothing \ne N(a)\ne \{0,\dots ,r\}$, choose $v\notin N(a)$. A cochain homotopy inserting $v$ into the ordered index list, with the sign of that insertion, contracts this subcomplex. Insertion or deletion of $v$ preserves the condition $N(a)\subseteq J$; in $\delta h+h\delta$, terms inserting and deleting different vertices cancel in pairs, while the term inserting then deleting $v$ is the identity. Thus this entire monomial subcomplex is acyclic. Finally, if all $a_i<0$, the monomial occurs only in the full intersection, in degree $r$, and contributes one copy of $k$ there.

For $r\ge 1$, these three cases give the complete answer:In the first case counting nonnegative exponent vectors gives $h^0=\left(\genfrac{}{}{0ex}{}{n+r}{r}\right)$. In the top case write $a_i=-1-b_i$ with $b_i\ge 0$; then $\sum b_i=-n-r-1$, givingThis is the Laurent-monomial description of top cohomology on projective space. All intermediate degrees vanish for every twist, and the top degree vanishes when $n\ge -r$. If $r=0$, then ${\mathbf{P}}_k^0=\mathrm{Spec}k$ and every twist is trivial, so $H^0({\mathbf{P}}_k^0,\mathcal{O}(n))=k$ for every integer $n$, with all higher groups zero.

Use homogeneous coordinates $X_0,\dots ,X_n$. Multiplication by these degree-one sections defines the sheaf morphismOn $U_i=D_{+}(X_i)$, write $t_j^{(i)}=X_j/X_i$, with $t_i^{(i)}=1$, and let $E_j^{(i)}$ be the standard vector in component $j$ using the local line bundle frame $X_i^{-1}$ of $\mathcal{O}(-1)$. Then $ϵ(E_j^{(i)})=t_j^{(i)}$, so $ϵ$ is surjective since $ϵ(E_i^{(i)})=1$. Its kernel is a locally free sheaf with basisIndeed any vector in the kernel is uniquely a sum of these vectors, by solving for its $i$-th component.

The Kähler differential sheaf ${\Omega }_{{\mathbf{P}}^n/k}^1$ has basis $d{t}_j^{(i)}$ on $U_i$, since this chart is a polynomial affine scheme. Define the local map $d{t}_j^{(i)}\mapsto F_j^{(i)}$. To check that these isomorphisms glue, on $U_i\cap U_{\ell }$ put $u=t_i^{(\ell )}$ and $v=t_j^{(\ell )}$. ThenThe twisting sheaf on projective space frame changes by $E_j^{(i)}=u^{-1}{E}_j^{(\ell )}$, henceFor $j=\ell$, use $t_{\ell }^{(\ell )}=1$, $d{t}_{\ell }^{(\ell )}=0$ and $F_{\ell }^{(\ell )}=0$; the same formula applies. Thus the differential and kernel frames have identical transition matrices. We obtain the cotangent form of the Euler sequence:This construction works over any field, including positive characteristic.

The Kähler differential sheaf has rank $n$. Taking determinant line bundles in this locally split short exact sequence of sheaves givesThe left side is the tensor product of $n+1$ copies of $\mathcal{O}(-1)$, and $\det (\mathcal{O})=\mathcal{O}$. ConsequentlyIt is the canonical bundle of projective space. The determinant identity can also be seen directly by taking the wedge of the kernel basis followed by a lift of the quotient basis; changing that lift by a kernel vector leaves the wedge unchanged.

Let $X$ be a separated Noetherian scheme over a field $k$ and $\mathcal{F}$ a coherent sheaf. Coherent cohomology studies the groups $H^q(X,\mathcal{F})$: global sections in degree zero and the obstructions to gluing local sections in higher degrees. Coherence means finite generation locally, and on a Noetherian scheme it is preserved by kernels, cokernels and extensions. This makes coherent sheaves suitable for exact-sequence calculations.

For an ordered open cover $\mathcal{U}=(U_i)$, the Čech cochain groups arewith differentialEach double omission occurs twice with opposite signs, so ${\delta }^2=0$. The Čech cohomology of the cover is ${\check{H}}^q=\ker \delta /\mathrm{im}\delta$, and taking the direct limit over refinements defines cover-independent Čech cohomology. In degree zero the sheaf gluing axiom identifies it with $\Gamma (X,\mathcal{F})$.

The comparison with derived-functor sheaf cohomology is particularly effective algebraically. Positive-degree cohomology of a quasi-coherent sheaf on an affine scheme vanishes. A separated scheme has affine intersections of affine opens, so a finite affine cover is acyclic for $\mathcal{F}$. Its Čech cochain complex consequently computes $H^q(X,\mathcal{F})$ directly. For a short exact sequence of sheaves, sections over each affine intersection are exact on quasi-coherent sheaves. The resulting short exact sequence of Čech complexes gives the long exact sequence in sheaf cohomology. In particular, the connecting map measures the failure of a global section of the quotient to lift globally.

Several central projective results make the theory finite and computable. On a projective variety over $k$, every coherent cohomology group is finite-dimensional. The Grothendieck vanishing theorem gives $H^q(X,\mathcal{F})=0$ for $q>\dim X$. If ${\mathcal{O}}_X(1)$ is ample, Serre vanishing gives $H^{q>0}(X,\mathcal{F}(m))=0$ for sufficiently large $m$, and these large twists are globally generated sheaves. On projective space a coherent sheaf has a finite twisting resolution of a coherent sheaf on projective space: finite sums of $\mathcal{O}(a)$ resolve it. The explicit line-bundle cohomology below and the long exact sequences prove finiteness and positive-twist vanishing. For a closed embedding $i:X\hookrightarrow {\mathbb{P}}^n$, $i_{\ast }\mathcal{F}$ is coherent and $H^q(X,\mathcal{F})=H^q({\mathbb{P}}^n,i_{\ast }\mathcal{F})$, reducing these results to projective space. A sufficiently positive twist of a finite presentation by sums of $\mathcal{O}(-a)$ also proves global generation.

The Euler characteristic of a coherent sheafis therefore defined, and the long exact sequence makes it additive. With a fixed projective embedding, $\chi (X,\mathcal{F}(m))$ is its Hilbert polynomial, and for sufficiently large $m$ it equals $h^0(X,\mathcal{F}(m))$. On a smooth projective variety of dimension $r$, Serre duality gives, for a vector bundle $E$,For an arbitrary coherent sheaf the correct counterpart is $H^q(X,\mathcal{F}{)}^{\ast }\cong {\mathrm{Ext}}_X^{r-q}(\mathcal{F},{\omega }_X)$; replacing this Ext group indiscriminately by the cohomology of the ordinary sheaf dual would lose local extension information.

We now calculate the basic examples. For $n\ge 1$, use the standard cover $U_i=D_{+}(x_i)$ of ${\mathbb{P}}^n$. Sections of $\mathcal{O}(m)$ on an intersection are degree-$m$ Laurent monomials in which negative exponents are allowed only for coordinates inverted on that intersection. Decompose the Čech cochain complex by exponent vectors $(a_0,\dots ,a_n)$ with sum $m$. If no exponent is negative, the corresponding simplex complex contributes only in degree zero. If all exponents are negative, the monomial occurs only on the full intersection and contributes only in degree $n$. When the negative-index set is nonempty and proper, inserting any index outside it gives a contracting homotopy, so the monomial complex is acyclic. Hence the cohomology of twisting sheaves on projective space isIn the top group a basis consists of the Laurent monomials with every exponent at most $-1$. Writing $a_i=-1-b_i$ reduces the count to nonnegative solutions of $\sum b_i=-m-n-1$, giving the displayed binomial coefficient. In particular, all intermediate cohomology vanishes, and ${\omega }_{{\mathbb{P}}^n}=\mathcal{O}(-n-1)$ agrees with Serre duality. For ${\mathbb{P}}^0$ separately, every twist is trivial and $H^0=k$.

On the projective line this givesThus positive and negative degrees produce quite different groups, while their alternating difference is a single polynomial. The twisted cubic is isomorphic to ${\mathbb{P}}^1$ and its hyperplane bundle corresponds to ${\mathcal{O}}_{{\mathbb{P}}^1}(3)$. Consequently $h^0(X,{\mathcal{O}}_X(m))=3m+1$ for $m\ge 0$, agreeing with Q2, and $h^1(X,{\mathcal{O}}_X(m))=\max (-3m-1,0)$.

For a smooth plane curve $C$ of degree $d\ge 1$, the ideal-sheaf sequence of a projective hypersurface isThe long exact sequence in sheaf cohomology and the preceding formula give $H^0(C,{\mathcal{O}}_C)=k$ and $H^1(C,{\mathcal{O}}_C)\cong H^2({\mathbb{P}}^2,\mathcal{O}(-d))$. ThereforeFor $d=1,2$ this is zero; for a plane cubic it is one. Twisting the same exact sequence gives $\chi (C,{\mathcal{O}}_C(m))=dm+1-g$. These examples illustrate how Čech cohomology, vanishing, Serre duality and exact sequences convert local equations into global numerical invariants.

Let $\mathcal{I}\subseteq {\mathcal{O}}_X$ be the ideal sheaf of $Y$. The formal completion of a scheme along $Y$ has underlying topological space $|Y|$ and topological structure sheafThe topology is the adic topology given by the kernels of the projections to the successive infinitesimal neighbourhoods $X_n=(|Y|,{\mathcal{O}}_X/{\mathcal{I}}^{n+1})$. In the locally Noetherian scheme setting of these examples this is a formal scheme. On a Noetherian affine scheme, with $Y=V(I)\subseteq \mathrm{Spec}A$, it is the formal spectrum $\mathrm{Spf}\hat{A}$, where $\hat{A}={\underleftarrow{\lim }}_{n}A/I^{n+1}$. Thus the completion retains all orders of infinitesimal information normal to $Y$.

Here $Z=D(t)=\mathrm{Spec}k[q,t,t^{-1}]$ is an open subscheme of $X$. The relevant affine completions arewith the $q$-adic topology. The complement of $t=0$ in the first formal scheme is its formal open $D(t)$. The sections on this open are the completed localization of an adic ring:The same calculation on every smaller principal open subset identifies the structure sheaves, not just their global sections. Equivalently, every infinitesimal neighbourhood of $Z\cap (q=0)$ is precisely the restriction to $D(t)$ of the corresponding neighbourhood in $X$. By formal completion commutes with restriction to an open subscheme,The isomorphism is canonical and respects the morphisms to $\mathrm{Spf}k[[q]]$.

It would give the wrong answer to replace the formal-open structure sheaf by the ordinary localization of a ring $k[t][[q]][t^{-1}]$. That ring has a uniform bound on the negative powers of $t$ across the coefficients of any one element, whereashas no such bound. This difference between ordinary and completed localization does not obstruct the isomorphism of formal schemes; it explains why the completed localization is essential.

Let $n={\mathrm{rank}}_{\mathbb{Z}}M$. Work with commutative rings and take a quotient to mean a surjection $p:M{\otimes }_{\mathbb{Z}}R\twoheadrightarrow Q$, considered up to an isomorphism of $Q$ commuting with $p$. Equivalently, retain its kernel as a submodule of $M\otimes R$. Here a rank-$r$ locally free module is finite locally free, or equivalently a finite projective module of constant rank $r$. A ring homomorphism $R\to R^{\prime }$ sends the quotient toThe tensor product of modules is right exact and preserves finite projectivity, so this defines the functorial action. We prove scheme representability in the usual functor of points sense:naturally in $R$. The Grassmannian of locally free quotients will be constructed explicitly.

Assume first $0<r<n$, and choose a basis $e_1,\dots ,e_n$ of $M$. For every ordered subset $I=\{i_1<\cdots <i_r\}\subseteq \{1,\dots ,n\}$, consider those quotients for which $p(e_{i_1}),\dots ,p(e_{i_r})$ are a basis of $Q$. This choice gives a unique identification $Q\cong R^r$, in which the quotient is represented by an $r\times n$ matrix $A_I$ whose columns indexed by $I$ are the identity matrix. Its remaining $r(n-r)$ entries are arbitrary elements of $R$. Thus this subfunctor is represented by the affine schemeThere is no further change-of-basis ambiguity: the selected quotient images already specify its basis uniquely.

For another subset $J$, write $A_I^{(J)}$ for the square submatrix consisting of the $J$ columns. The same quotient lies in the $J$ chart exactly when its determinant is a unit. Consequently the chart overlap is the principal open subsetOn that overlap the matrix normalized in the $J$ basis isThe entries are regular on $D({\delta }_{IJ})$ by the adjugate matrix. Conversely, the $I$ columns of $A_J$ form $(A_I^{(J)}{)}^{-1}$, so normalization back to $I$ recovers $A_I$. The transition is therefore an isomorphism $U_{IJ}\cong U_{JI}$.

The cocycle identity follows from the same matrix computation, not just an assertion about bases. On a triple overlap, normalization first to $J$ and then to $K$ givesThis is direct normalization from $I$ to $K$. By gluing of schemes along open subschemes, the standard affine charts of the quotient Grassmannian therefore glue to a scheme $G$ over $\mathbb{Z}$.

The universal matrices also glue a universal quotient. On $U_I$, use the surjectionOn overlaps, identify its targets by $(A_I^{(J)}{)}^{-1}$, precisely the transition matrix above. The cocycle identity glues these free sheaves into a rank-$r$ locally free sheaf $\mathcal{Q}$ and glues the maps into $M\otimes {\mathcal{O}}_G\twoheadrightarrow \mathcal{Q}$. Its kernel $\mathcal{S}$ is locally free of rank $n-r$: on $U_I$ an explicit basis isHence we have the universal exact sequence

It remains to verify that these charts capture every quotient, including a nonfree projective target. Given $p:R^n\twoheadrightarrow Q$, define $V_I\subseteq \mathrm{Spec}R$ as the locus where the selected images $p(e_i)$, $i\in I$, form a basis. This is open: locally trivialize the locally free sheaf associated with $Q$, and it is the nonvanishing locus of the corresponding maximal minor. The sets $V_I$ cover $\mathrm{Spec}R$. Indeed, at every prime $\mathfrak{p}$, the images of all $n$ vectors span the $r$-dimensional residue-field quotient, so some $r$ of them form a basis there. Their determinant is a unit in $R_{\mathfrak{p}}$, hence they form a basis over that local ring and on an open neighbourhood of $\mathfrak{p}$.

On $V_I$, trivialize $Q$ using exactly those selected images. The remaining columns give regular functions and hence a morphism $V_I\to U_I$. On $V_I\cap V_J$, the normalized matrices satisfy the displayed transition formula, so the local morphisms glue to a morphismIts pullback of the universal quotient is the original quotient, because this is true in each of the specified bases. Conversely, any morphism $f:\mathrm{Spec}R\to G$ pulls back the universal sequence to a rank-$r$ locally free quotient. On each inverse-image chart its matrix is exactly the normalized matrix used to reconstruct $f$, so the two constructions are inverse. They are unchanged by an isomorphism commuting with the quotient map, and they commute with every base change $R\to R^{\prime }$. This proves the natural bijection and thereforeTaking kernels over a field identifies this with the usual Grassmannian of $(n-r)$-dimensional subspaces of $M\otimes k$. The construction itself does not assume the target module is globally free.

For $r=0$, the only quotient is zero; for $r=n$, a surjection to a locally free module of the same rank is an isomorphism, as can be checked locally by its determinant. Both endpoint functors are represented by $\mathrm{Spec}\mathbb{Z}$. If $r>n$, no such quotient exists over a nonempty test scheme, and the representing scheme is empty. Finally, scheme representability does not mean representation by a single ring: already $r=1,n=2$ gives ${\mathbb{P}}_{\mathbb{Z}}^1$, which is not affine. This is the standard algebraic-geometric meaning of representability for the functor on rings here.

An integral scheme is nonempty, reduced and irreducible. For an affine scheme $X=\mathrm{Spec}R$, reducedness is equivalent to $\sqrt{(0)}=0$: nilpotent elements vanish in every stalk, and conversely a nonzero element has nonzero localization somewhere. Irreducibility is equivalent to the nilradical being prime. Indeed, if $ab$ lies in every prime, then $V(ab)=V(a)\cup V(b)=X$; irreducibility makes one of these closed subsets all of $X$, so $a$ or $b$ lies in the nilradical. Conversely, if the nilradical is prime, its point is a generic point of the whole spectrum. Thus reducedness and irreducibility together say that zero is a prime ideal, with $R\ne 0$. Therefore

For a global idempotent $e$, its germ in any local ring is either zero or one. To see this, $e(1-e)=0$, and at least one of $e,1-e$ is a unit. Multiplication by that unit forces the other to vanish. On each affine chart,These equalities also hold globally. Hence $V(1-e)$ is a clopen set. Conversely, on a clopen subset $U$ and its complement, glue the sections one and zero. They define a global idempotent $e_U$, and the two constructions are inverse. This proves the idempotent–clopen correspondence.

Assume each connected component is open. Every clopen subset is a union of connected components, since its intersection with a connected component is clopen in that component. Each individual component is also closed, so it defines a nonzero idempotent. A decomposition into two nonempty disjoint clopen subsets corresponds exactly to a sum of two nonzero orthogonal idempotents. A union containing at least two components admits such a decomposition by separating one component from the rest. A single component admits none. Consequently the precise conclusion is

The printed definition needs both nonzero and orthogonal. Literally forbidding every sum of two nonzero idempotents gives a false statement in characteristic two. For $X=\mathrm{Spec}({\mathbb{F}}_2\times {\mathbb{F}}_2)$, the component idempotent $(1,0)$ decomposes as $(1,1)+(0,1)$; both summands are nonzero idempotents but are not orthogonal. In fact each of the three nonzero idempotents in this ring decomposes into the other two, although $X$ has two open connected components. Excluding zero is also necessary: zero corresponds to the empty subset, not to a component. With the standard primitive idempotent definition the preceding proof establishes the intended bijection in every characteristic.

The Absolute Frobenius morphism is the identity on the underlying topological space and acts on the structure sheaf by $a\mapsto a^p$. On an affine scheme it is induced by the ring homomorphism $A\to A$, $a\mapsto a^p$. This is additive in characteristic $p$, and the inverse image of a prime ideal $\mathfrak{p}$ is again $\mathfrak{p}$, since $a^p\in \mathfrak{p}$ exactly when $a\in \mathfrak{p}$. These affine maps agree under localization and hence glue.

A perfect ring of characteristic p is one whose Frobenius endomorphism is bijective. Since the Absolute Frobenius morphism fixes points, its map on the stalk at $x$ is precisely the Frobenius endomorphism of ${\mathcal{O}}_{X,x}$. A morphism of sheaves is an isomorphism exactly when it is an isomorphism on all stalks. ThereforeIn the reverse direction the inverse sheaf map, together with the identity on points, gives the inverse morphism of locally ringed spaces. This is perfection in the Frobenius sense, not the unrelated homological use of the word perfect.

For the function field assertions below, interpret the normal schemes as integral, as is implicit in writing a single field $k(X)$ or $k(Y)$. General normal schemes need not be irreducible; without this interpretation those fields are not defined globally as single fields. Set $K=k(Y)\subseteq L=k(X)$. For an affine open $V=\mathrm{Spec}A$ in $Y$, finiteness gives $f^{-1}V=\mathrm{Spec}B$, with $A\subseteq B$ finite and both rings normal domains. In fact $B$ is the integral closure of $A$ in $L$: any element of $L$ integral over $A$ is also integral over $B$ and therefore belongs to $B$. This local description yields the two forms of Frobenius factorization through a finite normal cover.

Choose an ordered affine open cover $\mathcal{U}=(U_i)$ of the separated scheme. For a quasi-coherent sheaf $\mathcal{F}$, the Čech cochain groups and differential areTerms corresponding to empty intersections are zero. Restriction commutes with restriction, so the terms in $d^2$ cancel in pairs. DefineHere an affine cover computes the Čech groups independently of the chosen affine cover. The reason is that finite intersections of affine opens in a separated scheme are affine: the intersection of two is the inverse image of the closed diagonal inside their affine product. Repeating gives the finite-intersection assertion. Quasi-coherent sheaves have no higher cohomology on these intersections. Apply an injective resolution of the sheaf and form the double complex of its Čech cochains. In one direction the higher cohomology vanishes on every affine intersection, leaving precisely the displayed Čech complex. In the other direction the augmented Čech complex of the flasque resolution terms is exact, leaving the global-section resolution computing sheaf cohomology. Thus the two cohomologies agree, and refinement maps between affine-cover complexes induce the same groups. No such affine-intersection claim is being made for arbitrary nonseparated schemes.

For an exact sequence of quasi-coherent sheaves, sections on an affine intersection give an exact sequence of modules, including surjectivity: this follows from the equivalence between quasi-coherent sheaves and modules on an affine scheme. Taking products yields a short exact sequence of Čech cochain complexes. To describe the connecting map, lift a degree-$p$ cocycle in $C^p({\mathcal{F}}^{\prime \prime })$ to a cochain in $C^p(\mathcal{F})$. Its differential maps to zero in $C^{p+1}({\mathcal{F}}^{\prime \prime })$ and therefore is a cocycle in $C^{p+1}({\mathcal{F}}^{\prime })$. Changing the lift adds a coboundary there; changing the original representative does the same. This defines the connecting homomorphism. The same lift-and-differentiate argument shows that a class is in its kernel exactly when it lifts to a cocycle in the middle complex, and verifies exactness at the other terms. Thus the resulting sequence isThe sequence must be in quasi-coherent sheaves for this affine-sections argument; that is the intended category here.

For the punctured affine plane, use $U=D(x)$ and $V=D(y)$. Their rings are $k[x^{\pm 1},y]$ and $k[x,y^{\pm 1}]$, and the intersection ring is $k[x^{\pm 1},y^{\pm 1}]$. The two-open Čech complex isConsequentlyThe denominator contains precisely the span of Laurent monomials for which at least one exponent is nonnegative. Every Laurent polynomial has a unique monomial expansion, so the remaining monomials are linearly independent in the quotient and span it. Thereforea countably infinite-dimensional vector space. The brackets here denote cohomology classes. This calculates the Čech cohomology of the punctured affine plane, rather than merely observing that some nonzero cohomology exists.

A finite morphism is a finite type morphism. It is also a separated morphism: on an affine scheme chart $\mathrm{Spec}B\to \mathrm{Spec}A$, its diagonal morphism is represented by the surjective multiplication ring homomorphism $B{\otimes }_{A}B\to B$, and is therefore a closed immersion by Question 1.

We prove the lifting condition in the valuative criterion for properness. Choose an affine open subset of $Y$ containing the image of the closed point of $\mathrm{Spec}V$. Every point of $\mathrm{Spec}V$ is a generalization of its closed point, so the whole map factors through this affine open subset. Its inverse image under the finite morphism is $\mathrm{Spec}B$, with $B$ a finitely generated $A$-module. Every $b\in B$ is integral over $A$: apply the determinant trick to multiplication by $b$ on a finite set of module generators to obtain a monic polynomial which annihilates all generators and hence annihilates $1$.

In the given valuative square, we have compatible ring homomorphisms $A\to V$ and $B\to K$. The image in $K$ of each $b\in B$ satisfies a monic polynomial over $V$. To see directly why it belongs to $V$, suppose $x\in K\setminus V$. The defining property of a valuation ring gives $x^{-1}\in V$, and it lies in the maximal ideal since it cannot be a unit. A monic equation for $x$, multiplied by $x^{-n}$, would then express $1$ as a sum of elements in that maximal ideal, a contradiction. This proves that valuation rings are integrally closed and that $B\to K$ factors through $B\to V$. The factorization is unique because $V\hookrightarrow K$ is injective. Thus the lift exists uniquely, proving

If $X$ is an affine scheme, each reduced irreducible component is a closed subscheme, hence an affine scheme. We prove the converse by explicit affine gluing along closed subschemes.

First suppose $X$ is the scheme-theoretic union of two closed subschemes which are affine schemes $Y$ and $Z$: their defining ideal sheaves have zero intersection. Their scheme-theoretic intersection $W$ is a closed subscheme of both, hence an affine scheme. WriteThe two maps $A\to C$ and $B\to C$ are surjective. Form the fiber product of ringsBoth projections are surjective. Let $I=\ker (R\to A)$ and $J=\ker (R\to B)$. ThenFor the last equality, the common image map $R\to C$ is surjective, and its kernel of a ring homomorphism consists exactly of pairs whose two coordinates map to zero. Therefore $\mathrm{Spec}R$ has closed pieces isomorphic to $Y$ and $Z$, with intersection $W$. They cover it because each prime ideal contains $I$ or $J$, as $IJ=0$.

A finite closed cover determines the topology: a subset is closed if and only if its intersections with both closed pieces are closed. The identifications of $Y,Z,W$ therefore give a homeomorphism between $X$ and $\mathrm{Spec}R$. To check the structure sheaf, on $X$ the elementary ideal-intersection sequence givesIt is exact on stalks, by ${\mathcal{I}}_Y\cap {\mathcal{I}}_Z=0$. The same sequence holds on $\mathrm{Spec}R$, using $I\cap J=0$. Under the homeomorphism, the two middle sheaves and their restriction maps identify, so their kernels identify as sheaves of rings. This identifies the structure sheaves, giving an isomorphism of schemes $X\simeq \mathrm{Spec}R$. The intersection $W$ need not be reduced.

Now a Noetherian scheme has finitely many irreducible components $Y_1,\dots ,Y_r$. Induct on $r$. For $r=1$, the reduced scheme is its single reduced irreducible component, so is an affine scheme. For $r>1$, let $Y=Y_1$ and let $Z$ be the reduced union of the remaining irreducible components. Its components are the corresponding $Y_i$, so $Z$ is an affine scheme by induction. The defining ideal sheaves of $Y$ and $Z$ have zero intersection: their intersection is the ideal sheaf of the reduced union of all components, which is zero because $X$ is reduced. The two-piece argument proves that $X$ is an affine scheme. The empty scheme is $\mathrm{Spec}0$ and causes no exception. Hence

Take the affine scheme $X={\mathbb{A}}_k^1$, its function field $K=k(t)$, and the constant sheaf $\mathcal{K}$ with value $K$. By Question 4(ii), $\mathcal{K}$ is a flasque sheaf. At the two distinct closed rational points $p_0=0$ and $p_1=1$, let $S_0,S_1$ be skyscraper sheaves with value $K$: on an open subset $V$, $S_i(V)=K$ if $p_i\in V$ and is zero otherwise. Both are flasque sheaves.

There is a natural morphism of sheaveswhich sends a constant rational function to the same element in every summand present on the open subset. At $p_i$, its stalk map is the identity $K\to K$, and at any other point the target stalk is zero. Thus $\rho$ is a surjection of sheaves, although it is not surjective on global sections. Put $\mathcal{F}=\ker \rho$.

This can even be an example of sheaves of ${\mathcal{O}}_X$-modules: let regular functions act on each copy of $K$ by multiplication after embedding them in $K$. At the closed points this action is multiplication by a rational function, not evaluation at the point. With these actions, $\rho$ is ${\mathcal{O}}_X$-linear.

The exact sequence $0\to \mathcal{F}\to \mathcal{K}\to S_0\oplus S_1\to 0$, together with flasque acyclicity, givesHence $H^0(X,\mathcal{F})=0$ and $H^1(X,\mathcal{F})=K^2/\Delta K\simeq K\ne 0$. In contrast, use the one-member affine open cover $\mathcal{U}=\{X\}$. Its alternating Čech cochain complex has no positive-degree terms, so ${\check{H}}^1(\mathcal{U},\mathcal{F})=0$. We have the explicit discrepancyThe sheaf $\mathcal{F}$ is not a quasi-coherent sheaf, as also follows from its nonzero higher sheaf cohomology on an affine scheme. Thus this example does not contradict the acyclic cover theorem for quasi-coherent sheaves on affine covers.

Put $P={\mathbb{P}}_k^2$ and $S=k[t_0,t_1,t_2]$. A homogeneous polynomial $F$ of positive degree $d$ is nonzero, and $S$ is an integral domain. Multiplication by $F$ therefore gives an exact sequence of coherent sheaveswhere $i:X\hookrightarrow P$ is the closed immersion.

We compute the necessary cohomology of twisting sheaves on projective space. The standard three affine open subsets $D_{+}(t_i)$ and all their intersections are affine schemes. For a quasi-coherent sheaf, vanishing of quasi-coherent cohomology on an affine scheme and the acyclic cover theorem identify its sheaf cohomology with the Čech cohomology of this cover. In degree $m$, the Čech cochain complex isDecompose it into subcomplexes indexed by Laurent monomials $t_0^{a_0}{t}_1^{a_1}{t}_2^{a_2}$ with $a_0+a_1+a_2=m$. Let $N=\{i:a_i<0\}$. Such a monomial occurs in the summand for a nonempty index set $J$ precisely when $N\subseteq J$, since only the variables in $J$ are inverted.

If $N$ is empty, the monomial contributes the cochain complex of a full two-simplex: its degree-zero cohomology is $k$ and its higher cohomology vanishes. If $|N|=1$, the dimensions in degrees $0,1,2$ are $1,2,1$; the first map has rank one and the second map has rank one with the same kernel as the first map's image. If $|N|=2$, there is one copy of $k$ in each of degrees $1$ and $2$ and the map between them is an isomorphism. These two cases are acyclic over every field, including characteristic two. If $N=\{0,1,2\}$, only the degree-two term occurs, contributing one copy of $k$. It follows thatand that $H^0(P,{\mathcal{O}}_P)=k$, $H^0(P,{\mathcal{O}}_P(-d))=0$, and $H^2(P,{\mathcal{O}}_P)=0$. Moreover the Laurent-monomial description of top cohomology on projective space gives a basis of $H^2(P,{\mathcal{O}}_P(-d))$ consisting ofFor $d\ge 3$, subtracting one from each $b_i$ counts these as $\left(\genfrac{}{}{0ex}{}{d-1}{2}\right)$. For $d=1,2$ there are no such triples, and the formula $(d-1)(d-2)/2$ is also zero.

Use the long exact sequence in sheaf cohomology of the initial sequence, and cohomology under a closed immersion to identify the terms for $i_{\ast }{\mathcal{O}}_X$. The vanishings just proved giveThereforeThis proves the cohomology of a projective plane hypersurface without imposing smoothness, irreducibility or reducedness. In particular, the second number is an arithmetic genus, not a claim about the genus of a normalization.

For a point $y\in Y$ with residue field $\kappa (y)$, the scheme-theoretic fibre is the base change of a morphism of schemesThe map $\mathrm{Spec}\kappa (y)\to Y$ is the canonical point map. This fibre product of schemes retains the structure sheaf, including any nilpotents, rather than just the set $f^{-1}(y)$. For an affine scheme map induced by $A\to B$, at $\mathfrak{p}\in \mathrm{Spec}A$ it is

Work over an affine open subscheme $V=\mathrm{Spec}A$ of $Z$. Since $g$ is a closed immersion, its inverse image is $\mathrm{Spec}(A/I)$ for some ideal $I$. Since $f$ is a closed immersion, the inverse image of this affine scheme in $X$ is $\mathrm{Spec}((A/I)/J)$ for an ideal $J\subseteq A/I$. Let $\tilde{J}$ be the inverse image of $J$ in $A$. The composite on this open is induced by the surjectionThe affine-local characterization from part (a) therefore shows that $g\circ f$ is a closed immersion. In particular, closed-subscheme ideals compose by taking the inverse image of the second quotient ideal.

Over an affine open subscheme $\mathrm{Spec}A$ of the target, a closed immersion has the form $\mathrm{Spec}(A/I)\to \mathrm{Spec}A$. Pulling it back along an affine scheme map $\mathrm{Spec}B\to \mathrm{Spec}A$ givesThe identification follows directly by tensoring the quotient presentation; it does not require $B$ to be flat over $A$. Since $B\to B/IB$ is surjective, this is again a closed immersion. These affine calculations cover every base change of a morphism of schemes.

Put $A=k[t_1,t_2]$. The punctured affine plane is covered by the principal open subschemes $D(t_1)$ and $D(t_2)$. Since $A$ is an integral domain, restriction identifies its global sections withThis intersection is $A$: if $a/t_1^r=b/t_2^s$, then $t_2^{s}a=t_1^{r}b$, and unique factorization forces $t_1^r$ to divide $a$. Therefore the inclusion $j:X\hookrightarrow \mathrm{Spec}A$ induces the identity on global regular functions.

If $X$ were an affine scheme, affine scheme reconstruction from global sections would make $j$ an isomorphism of schemes, because its global-section map is an isomorphism. But its image omits the closed point $(t_1,t_2)$. The punctured affine plane is therefore not affine. This is a useful example of global regular functions failing to distinguish a nonaffine open subscheme from its affine ambient space.