Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 16 / 4

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