Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 16 (Algebraic Geometry)

A locally ringed space is a topological space $X$ equipped with a sheaf of commutative rings ${\mathcal{O}}_X$ whose stalk ${\mathcal{O}}_{X,x}$ at every point is a local ring. A morphism of locally ringed spaces $(f,f^{\#}):(Y,{\mathcal{O}}_Y)\to (X,{\mathcal{O}}_X)$ consists of a continuous map $f:Y\to X$ and a morphism of sheaves of rings $f^{\#}:{\mathcal{O}}_X\to f_{\ast }{\mathcal{O}}_Y$, with every induced stalk map ${\mathcal{O}}_{X,f(y)}\to {\mathcal{O}}_{Y,y}$ a local homomorphism: the inverse image of the target maximal ideal is the source maximal ideal. Here $f_{\ast }{\mathcal{O}}_Y$ is the direct image sheaf, with sections ${\mathcal{O}}_Y(f^{-1}U)$ on $U$.

Let $\varphi :A\to B$ be a ring homomorphism. Define the map between the spectra of rings byThe inverse image is a prime ideal, and $f^{-1}(D(a))=D(\varphi (a))$. Thus $f$ is a continuous map, because the principal open subsets form a basis. On these principal open subsets, define the structure sheaf map using the localizationsThese maps commute with restrictions and consequently give a morphism of sheaves. At $\mathfrak{q}$, putting $\mathfrak{p}={\varphi }^{-1}\mathfrak{q}$, the stalk map is $A_{\mathfrak{p}}\to B_{\mathfrak{q}}$. An element $b/s$ belongs to its inverse image of $\mathfrak{q}{B}_{\mathfrak{q}}$ exactly when $b\in \mathfrak{p}$, so this is a local homomorphism. We have constructed a morphism of locally ringed spaces.

Conversely, take a morphism of locally ringed spaces $f:\mathrm{Spec}B\to \mathrm{Spec}A$. Its map on global sections gives a ring homomorphismLocality of the stalk map implies, for every $a\in A$,Hence the point map must be $\mathfrak{q}\mapsto {\varphi }^{-1}\mathfrak{q}$. On $D(a)$, its structure sheaf map must send $b/1$ to $\varphi (b)/1$ and the inverse of $a$ to the inverse of $\varphi (a)$. Thus it is precisely the localization map constructed above. Agreement on this basis proves agreement of the morphisms of sheaves. In particular, the inducing ring homomorphism exists and is unique.

If $\varphi$ is surjective, write $B=A/I$. The correspondence between prime ideals of $A/I$ and prime ideals of $A$ containing $I$ identifies $\mathrm{Spec}B$ homeomorphically with the closed subset $V(I)$: the principal open subsets correspond on both sides. Moreover, the stalk maps are the surjections $A_{\mathfrak{p}}\to (A/I{)}_{\mathfrak{p}/I}$, so ${\mathcal{O}}_X\to f_{\ast }{\mathcal{O}}_Y$ is surjective.

For the converse, it is important not to assume that global sections preserve an arbitrary surjection of sheaves. Here the affine module-sheaf equivalence supplies the extra information: on every $D(a)$,Thus $f_{\ast }{\mathcal{O}}_Y$ is the quasi-coherent sheaf associated with the $A$-module $B$, and the cokernel sheaf of ${\mathcal{O}}_X\to f_{\ast }{\mathcal{O}}_Y$ is associated with the module $C=\mathrm{coker}(A\to B)$, since localization of a module is exact. Surjectivity of the morphism of sheaves gives $C_{\mathfrak{p}}=0$ for every prime ideal $\mathfrak{p}$. This implies $C=0$: if $0\ne c\in C$, its proper annihilator lies in a maximal ideal $\mathfrak{m}$, and then $c/1\ne 0$ in $C_{\mathfrak{m}}$. Therefore $A\to B$ is surjective. We have proved the closed immersion criterion for affine schemes:

A proper morphism of schemes is a separated morphism of finite type which is a universally closed morphism: after every base change, the resulting map of underlying topological spaces takes closed subsets to closed subsets.

The valuative criterion for properness, in the Noetherian setting used here, says that a finite type morphism $f:X\to Y$ is a proper morphism if and only if the following existence and uniqueness condition holds. For every valuation ring $V$, its fraction field $K$, and compatible morphisms of schemes $a:\mathrm{Spec}K\to X$ and $b:\mathrm{Spec}V\to Y$ with $f\circ a=b{|}_{\mathrm{Spec}K}$, there is exactly one morphism of schemes $h:\mathrm{Spec}V\to X$ such that $h{|}_{\mathrm{Spec}K}=a$ and $f\circ h=b$. All valuation rings are allowed; no discreteness assumption is needed in this formulation. The following arguments also use that each of the three defining conditions for a proper morphism is preserved by base change.

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

The composite of two finite type morphisms is of finite type: on suitable affine open subsets, take the algebra generators of the first algebra together with the images of the algebra generators of the second. To check separatedness, factor the diagonal morphism of the composite asThe first arrow is a closed immersion, and the second is the base change of the closed immersion ${\Delta }_g:Y\to Y{\times }_{Z}Y$. Hence their composite is a closed immersion.

Finally, after any base change $Z^{\prime }\to Z$, the composite is $X^{\prime }\stackrel{f^{\prime }}{\to }{Y}^{\prime }\stackrel{g^{\prime }}{\to }{Z}^{\prime }$. Both arrows take closed subsets to closed subsets, so their composite does as well. This is universal closedness. All three conditions hold, and therefore

Use the graph morphism of schemesIt is a closed immersion: the square with ${\Delta }_g:Y\to Y{\times }_{Z}Y$ and the map $X{\times }_{Z}Y\to Y{\times }_{Z}Y$, $(x,y)\mapsto (f(x),y)$, is a fibre product of schemes square. The separatedness of $g$ makes ${\Delta }_g$ a closed immersion, and base change preserves a closed immersion.

The projection $\pi :X{\times }_{Z}Y\to Y$ is the base change of the proper morphism $gf:X\to Z$, so it is a proper morphism. The closed immersion ${\Gamma }_f$ is a finite morphism, hence a proper morphism by part (i). Since $f=\pi {\Gamma }_f$, part (ii) provesThis argument supplies finite type for $f$ as well; it need not be separately assumed.

We first justify surjectivity is preserved by base change for morphisms of schemes. Given $y^{\prime }$ over $y$ and a point $x$ over $y$, the fiber after the residue-field extension contains a point becauseIndeed, tensoring a nonzero vector space over a field with a field extension remains nonzero; a nonzero commutative ring has a prime ideal, giving the required point. Thus every base change of $f$ is surjective.

Now make an arbitrary base change $Z^{\prime }\to Z$. For a closed subset $C\subseteq Y^{\prime }=Y{\times }_Z{Z}^{\prime }$, its inverse image under $f^{\prime }:X^{\prime }\to Y^{\prime }$ is closed. The proper morphism $(gf{)}^{\prime }:X^{\prime }\to Z^{\prime }$ therefore has closed image on that inverse image. Surjectivity of $f^{\prime }$ givesso $g^{\prime }(C)$ is closed. This proves that $g$ is a universally closed morphism. Together with the given separatedness and finite type, it proves

Write $f$ as the ring homomorphism $A\to B$. By the permitted assumption, $B$ is a finitely generated $A$-algebra. We will prove that every element of $B$ is integral over $A$, including any nilpotent contribution. If $B=0$, there is nothing to prove.

We use the integral closure as an intersection of valuation rings in its general form: if an integral domain $R$ lies in a field $K$, then its integral closure inside $K$ is the intersection of all valuation rings of $K$ containing $R$. Here $K$ need not be the fraction field of $R$. One direction follows from valuation rings are integrally closed. For the other, if $0\ne x\in K$ is not integral, the ideal $x^{-1}R[x^{-1}]$ is proper: an expression for $1$ in that ideal, multiplied by a sufficiently large power of $x$, would be a monic equation for $x$. Choose a maximal ideal containing this ideal, and apply the valuation domination lemma to the corresponding local ring inside $K$. The resulting valuation ring $V$ contains $R$ and has $x^{-1}$ in its maximal ideal, so $x\notin V$. This proves the claimed intersection criterion. The valuation domination lemma is obtained by choosing, using Zorn's lemma, a maximal local overring dominating the given local ring; the usual alternative between adjoining $z$ and adjoining $z^{-1}$ forces this maximal overring to be a valuation ring.

The Noetherian ring $B$ has finitely many minimal prime ideals ${\mathfrak{q}}_1,\dots ,{\mathfrak{q}}_r$. SetTake any valuation ring $V$ of $K_i$ containing $R_i$. The maps $B\to D_i\to K_i$ and $A\to R_i\to V$ form a square for the valuative criterion for properness. Its lift gives $B\to V$, whose composite with $V\hookrightarrow K_i$ is the original map. Consequently the image $D_i$ is contained in $V$. The intersection criterion proves that every element of $D_i$ is integral over $R_i$.

Fix $b\in B$. For each $i$, choose a monic polynomial $P_i(T)\in A[T]$ whose value at the image of $b$ in $D_i$ is zero; coefficients over $R_i$ can be lifted to $A$. The product $P(T)={\prod }_i{P}_i(T)$ is monic and satisfiesThe nilradical of a Noetherian ring is nilpotent. Explicitly, choose finitely many generators $n_j$ with $n_j^{e_j}=0$; every product of $1+{\sum }_j(e_j-1)$ generators then contains a vanishing power. Thus $(\sqrt{(0)}{)}^N=0$ for some $N$, and $P(T{)}^N$ is a monic polynomial annihilating $b$. Hence $B$ is integral over $A$.

Finally choose algebra generators $b_1,\dots ,b_s$, with monic equations of degrees $d_1,\dots ,d_s$. Reducing powers in these equations shows that the finitely many monomialsspan $B$ as an $A$-module. Therefore $B$ is a finite $A$-module, proving

An $A$-module homomorphism $h:M\to N$ induces, by localization of a module, compatible mapson all principal open subsets $D(a)$. These glue to an ${\mathcal{O}}_X$-linear morphism of sheaves $\tilde{h}:\tilde{M}\to \tilde{N}$.

Conversely, taking global sections of an ${\mathcal{O}}_X$-linear morphism of sheaves gives an $A$-module homomorphism $h:M\to N$, since $\Gamma (X,\tilde{M})=M$ and $\Gamma (X,\tilde{N})=N$. On $D(a)$, compatibility with restriction determines the image of $m/1$, and ${\mathcal{O}}_X$-linearity determines the image of $a^{-r}m$. It must be $h(m)/a^r$. Thus the morphism of sheaves is uniquely recovered from $h$. Both constructions preserve addition and composition, proving the affine module-sheaf equivalence on morphisms:No finite-generation assumption is used.

Use the local-presentation definition of a quasi-coherent sheaf: locally it admits a presentation ${\mathcal{O}}^{(I)}\to {\mathcal{O}}^{(J)}\to \mathcal{F}\to 0$ by direct sums of copies of the structure sheaf, with no finiteness requirement on the index sets. If $M$ is an $A$-module, choose a free module presentation $A^{(I)}\to A^{(J)}\to M\to 0$. Exactness of localization of a module gives the corresponding local presentation of $\tilde{M}$. Hence a sheaf which is associated with a module on every affine open subset is a quasi-coherent sheaf.

For the converse, begin with a quasi-coherent sheaf and a local presentation on an affine open subset $V=\mathrm{Spec}R$. The free sheaves have global sections $R^{(I)}$ and $R^{(J)}$: a section has locally finite support, and quasi-compactness of $V$ makes its support finite globally. Part (i) identifies the presentation map with an $R$-module homomorphism. Exactness of localization then identifies its cokernel sheaf with ${\tilde{M}}_V$ for the module cokernel $M_V$. Thus there is an affine open cover on which $\mathcal{F}$ is associated with modules.

It remains to show this on an arbitrary affine open subset $U=\mathrm{Spec}A$, rather than only the selected cover. Around each point of $U$, choose an affine neighborhood contained in $U$ with such a local presentation. Choose a finite cover $U={\bigcup }_{i=1}^{r}D(s_i)$ refining these neighborhoods. On each $D(s_i)$ the restriction is associated with a module. Its restriction to $D(s_i{s}_j)$ is therefore the corresponding localization, and the same is true after intersecting with $D(a)$. Put $M=\Gamma (U,\mathcal{F})$. The sheaf gluing axiom giveswhere the arrow is the difference of the two restriction maps. Localization of a module commutes with kernels and finite products. Localizing this formula at $a$ therefore gives exactly the gluing formula for $D(a)$, covered by the $D(a{s}_i)$. Consequentlycompatibly with restriction. These isomorphisms on the principal open subsets identify $\tilde{M}$ with $\mathcal{F}{|}_U$. We have proved the global-section localization for quasi-coherent sheaves and, in particular,

Work on an affine open subset $U=\mathrm{Spec}A$ and write $\mathcal{F}{|}_U=\tilde{M}$ and $\mathcal{E}{|}_U=\tilde{N}$ using part (ii). We first show that any section $e\in \Gamma (U,\mathcal{E})$ lifts to $\Gamma (U,\mathcal{G})$.

A surjection of sheaves supplies local lifts $g_i$ on a finite cover by principal open subsets $D(s_i)$. Their differencessatisfy $c_{ij}+c_{jk}=c_{ik}$ on triple intersections. We spell out why this Čech cocycle is a coboundary. Choose a common sufficiently large power $n$ so that $s_j^n{c}_{ij}$ can be represented in $M_{s_i}$ for every pair. Increasing $n$ also kills the finitely many localization errors in the cocycle identities: after extending $s_k^n{c}_{ik}$ to $M_{s_i}$ and $s_k^n{c}_{jk}$ to $M_{s_j}$, their difference in $M_{s_i{s}_j}$ is $s_k^n{c}_{ij}$. This follows because any error which vanishes after localizing at $s_k$ is killed by a power of $s_k$; there are only finitely many indices.

The $D(s_i^n)$ still cover $U$, so the ideal generated by the $s_i^n$ is $A$. Choose $a_i\in A$ with ${\sum }_i{a}_i{s}_i^n=1$ and putusing the extended representatives just chosen. On an overlap, the cocycle identity givesThus $g_i-b_i$ agree and glue to a lift of $e$. This is the degree-one argument behind the exactness of the unit-ideal localization Čech complex.

The same reasoning applies to every $D(a)\subseteq U$. Hence we have short exact sequences of modules of sections on $U$ and on $D(a)$. Localize the first sequence at $a$ and compare them. The maps on the outer modules are the isomorphisms from part (ii):An elementary diagram chase makes the middle map an isomorphism: subtract a lift from the right to reduce surjectivity to the left, and use injectivity on the left to prove injectivity. These isomorphisms identify $\mathcal{G}{|}_U$ with the quasi-coherent sheaf associated with $\Gamma (U,\mathcal{G})$. Doing this on every affine open subset provesThe proof even works without the Noetherian assumption.

Take the projective line $X={\mathbb{P}}_k^1$ and the coherent sheaves $\mathcal{F}={\mathcal{O}}_X$ and $\mathcal{G}={\mathcal{O}}_X(-1)$, where the latter is a twisting sheaf on projective space. Both are line bundles, so each stalk is a free rank-one ${\mathcal{O}}_{X,x}$-module and therefore ${\mathcal{F}}_x\simeq {\mathcal{G}}_x$ at every point.

They are nevertheless not isomorphic. On the standard charts with coordinates $t$ and $t^{-1}$, choose frames $e_0,e_1$ of $\mathcal{O}(-1)$ with $e_1=t^{-1}{e}_0$. A global section would give polynomials $a(t)$ and $b(t^{-1})$ satisfyingThe left side has only nonnegative powers of $t$, and the right side only strictly negative powers. Thus both vanish, giving $\Gamma (X,\mathcal{O}(-1))=0$. For the structure sheaf, the corresponding equality is $a(t)=b(t^{-1})$, whose solutions are precisely the constants. HenceA morphism of sheaves which is an isomorphism on all stalks would be an isomorphism; this example only asserts the existence of separate stalk isomorphisms, not that they come from one morphism of sheaves.

Recall that a flasque sheaf has surjective restriction maps on every inclusion of open subsets. We first prove the flasque-kernel section-lifting lemma: in an exact sequence $0\to \mathcal{A}\to \mathcal{B}\to \mathcal{C}\to 0$ of sheaves, with $\mathcal{A}$ a flasque sheaf, every section of $\mathcal{C}$ on any open subset $V$ lifts to a section of $\mathcal{B}$ on $V$.

Fix such a section $c$. Order its lifts on open subsets $W\subseteq V$ by extension. A chain has an upper bound by the sheaf gluing axiom, so Zorn's lemma gives a maximal lift $b_W$. If $W\ne V$, choose a point outside $W$ and a neighborhood $T\subseteq V$ on which $c$ has a local lift $b_T$. On $W\cap T$, the difference $b_T-b_W$ is a section of $\mathcal{A}$. Extend it to $T$ by flasqueness and subtract it from $b_T$. The corrected lift now agrees with $b_W$ and glues on $W\cup T$, contradicting maximality. Thus $W=V$, proving the lemma.

It follows also that if both $\mathcal{A}$ and $\mathcal{B}$ are flasque sheaves, then $\mathcal{C}$ is a flasque sheaf. Given a section of $\mathcal{C}$ on $U\subseteq V$, first lift it to $\mathcal{B}$ on $U$, extend that lift to $V$, and project to $\mathcal{C}$.

Next, an injective sheaf of modules $\mathcal{I}$ is a flasque sheaf. For $U\subseteq V$, let $j_U$ and $j_V$ denote the open inclusions into $X$. Extension by zero gives a monomorphism $j_{U!}{\mathcal{O}}_U\to j_{V!}{\mathcal{O}}_V$: on every stalk it is either $0\to {\mathcal{O}}_{X,x}$, the identity, or $0\to 0$. By the adjunction for extension by zero,Injectivity therefore makes $\Gamma (V,\mathcal{I})\to \Gamma (U,\mathcal{I})$ surjective. This proves injective module sheaves are flasque.

Take an injective resolution of sheaves of $\mathcal{F}$. Write ${\mathcal{C}}^0=\mathcal{F}$ and, successively,The initial ${\mathcal{C}}^0$ is a flasque sheaf; each ${\mathcal{I}}^r$ is a flasque sheaf; and the quotient argument proves inductively that all ${\mathcal{C}}^r$ are flasque sheaves. The section-lifting lemma makes each displayed sequence exact after global sections. Consequently the complex of global sections of the injective resolution of sheaves has zero positive cohomology. Since that complex defines sheaf cohomology,

The affine space $X={\mathbb{A}}_k^n$ is an integral scheme, hence its underlying topological space is irreducible. Every nonempty open subset of an irreducible topological space is irreducible and in particular connected. A section of the constant sheaf with value the function field $K=k(t_1,\dots ,t_n)$ is a locally constant function, so on any such open subset it is constant. Thus its sections are $K$ on each nonempty open subset and zero on the empty set. All restriction maps are either identities on $K$ or maps onto zero.

The constant sheaf is therefore a flasque sheaf. It has its natural ${\mathcal{O}}_X$-module structure because a regular function on a nonempty open subset embeds in the function field and acts by multiplication. Part (i) now givesThis is an instance of constant sheaves on irreducible spaces are flasque; no restriction on $n$ is needed.

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

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.

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.

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.