Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 113 (Algebraic Geometry)

On the principal open subscheme $D(a)\subseteq \mathrm{Spec}A$, the induced morphism of sheaves is the ring homomorphismIf $\alpha$ is injective, this map is injective by exactness of localization: it is the localization of the injective $A$-module map $A\to B$. Explicitly, if the displayed fraction vanishes, some power of $\alpha (a)$ annihilates $\alpha (b)$; injectivity gives $a^{m}b=0$, so the original fraction also vanishes. The principal opens form an open basis, so the kernel sheaf is zero on a basis and hence zero everywhere. Conversely, an injective morphism of sheaves induces an injection on global sections. On the whole affine scheme these sections are $A\to B$, exactly $\alpha$. Thus

Take the dual numbers $A=k[\epsilon ]/({\epsilon }^2)$ over any field $k$, let $B=k$, and let $\alpha$ send $\epsilon$ to zero. Its kernel is the nonzero ideal $(\epsilon )$, but both spectra of rings have exactly one point: every prime ideal contains the nilpotent element $\epsilon$, and the only prime of $A$ is $(\epsilon )$. Consequently the induced map is a homeomorphism for the Zariski topology, although it is not an isomorphism of schemes:More generally, quotienting by a nilpotent ideal gives a nilpotent thickening with unchanged underlying topology. The structure sheaf retains information that the topology alone cannot recover.

The rings $A,B$ are nonzero integral domains. Let $\eta$ be the generic point of $Y$, corresponding to $(0)\subseteq A$. Surjectivity provides a prime ideal $\mathfrak{q}\subseteq B$ with ${\alpha }^{-1}(\mathfrak{q})=0$. Since $\ker \alpha \subseteq {\alpha }^{-1}(\mathfrak{q})$, the ring homomorphism $\alpha$ is injective.

Put $K=\mathrm{Frac}A$ and $S=A\setminus \{0\}$. The scheme-theoretic fibre at $\eta$ isEvery element of $\alpha (S)$ is nonzero, so the localization $S^{-1}B$ is a nonzero integral domain. Its zero ideal is prime, ensuring that its spectrum is nonempty; an affine spectrum of an integral domain is an integral scheme. This proves the integrality of the generic fibre of an affine dominant morphism. In fact the proof only needs injectivity of $\alpha$, rather than surjectivity at every point.

Use the ring homomorphism $k[t]\to k[x]$ given by $t\mapsto x^2$. The source scheme $X=\mathrm{Spec}k[x]$ is an integral scheme. At the rational point $y=(t)$, its residue field is $k$, and its scheme-theoretic fibre isThe class of $x$ is nonzero but has square zero, so this is not a reduced scheme. This example works over every field. An integral total space therefore need not have reduced fibres.

For an affine open subscheme $U=\mathrm{Spec}R$, quasi-coherence gives $\mathcal{I}{|}_U=\tilde{J}$ for an ideal $J\subseteq R$. Define $Z_U=\mathrm{Spec}(R/J)$ with its usual closed immersion into $U$. On a principal open subscheme $D(a)$, its restriction is $\mathrm{Spec}(R_a/J_a)$, because localization commutes with taking this quotient. These local constructions therefore agree on overlaps and glue.

The resulting underlying closed set is $Z=\{x:{\mathcal{I}}_x\ne {\mathcal{O}}_{X,x}\}$, locally $V(J)$. With inclusion $i:Z\hookrightarrow X$, its structure sheaf isAt a point corresponding to $\mathfrak{p}\supseteq J$, its stalk is $R_{\mathfrak{p}}/J_{\mathfrak{p}}$, the local ring of $\mathrm{Spec}(R/J)$. Thus the glued ringed space is a scheme and $i$ is a closed immersion, with ideal sheaf of a closed subscheme exactly $\mathcal{I}$. This is the construction of a closed subscheme from a quasi-coherent ideal. If $\mathcal{I}={\mathcal{O}}_X$, the construction gives the empty scheme; if $\mathcal{I}=0$, it gives $X$.

First suppose $X$ is a nonempty integral scheme, with generic point $\eta$ and function field $K={\mathcal{O}}_{X,\eta }$. Since taking stalks preserves the inclusion $\mathcal{I}\hookrightarrow {\mathcal{O}}_X$, local freeness of rank two would give an injective $K$-linear mapcontradicting the dimension of a vector space.

For a nonempty Noetherian scheme that is reduced, there are finitely many irreducible components. Choose one, and remove the union of the others. The resulting nonempty open subscheme is irreducible and reduced, hence integral. Restricting the proposed locally free sheaf to it gives the contradiction above. Therefore no locally free ideal of rank two exists on a nonempty reduced Noetherian scheme.

The rank bound for locally free ideals on reduced schemes in fact removes the Noetherian assumption: on an affine open where the rank is fixed, localization at a minimal prime ideal gives a field, so a free ideal has rank at most one. Nonemptiness is necessary for the literal statement: on the empty scheme the zero sheaf is vacuously locally free of every stipulated rank. The question is interpreted with this usual nonempty hypothesis.

Let $K$ be the function field of the integral scheme $X$ and choose a nonzero rational section of a line bundle $s\in {\mathcal{L}}_{\eta }$. Choose an open trivializing cover $U_i$ with generators $e_i$ for the invertible sheaf $\mathcal{L}$. On each nonempty $U_i$, write $s=a_i{e}_i$ with $a_i\in K^{\ast }$. If $e_i=u_{ij}{e}_j$ on an overlap, then $a_j=a_i{u}_{ij}$, so $a_i/a_j=u_{ij}^{-1}$ is a regular unit. Consequently the local rational functions $a_i$ define a Cartier divisor $D$.

The divisor line bundle ${\mathcal{O}}_X(D)$ restricts to $a_i^{-1}{\mathcal{O}}_{U_i}$. Define its local isomorphism from $\mathcal{L}$ by $e_i\mapsto a_i^{-1}$. On an overlap,so the maps glue. HenceThe rational section maps to the rational function $1$, which also fixes the sign convention for $D$. Choosing a different nonzero rational section changes $D$ by a principal Cartier divisor, so its linear equivalence of Cartier divisors class is unchanged. No projectivity or Noetherian hypothesis is required.

For each point $x$, choose a global section $s$ whose germ $s_x$ generates the rank-one free ${\mathcal{O}}_{X,x}$-module ${\mathcal{L}}_x$. The section $s\otimes s$ is global, and tensor products commute with taking stalks. Thereforegenerates $(\mathcal{L}{\otimes }_{{\mathcal{O}}_X}\mathcal{L}{)}_x$. Thus the tensor square is globally generated. More generally, the tensor product of globally generated line bundles is globally generated: choose a generator section for each factor at the point and tensor them. This works on every scheme, including nonreduced schemes.

We prove the vanishing by an injective resolution of sheaves and the flasque-kernel section-lifting lemma.

For a short exact sequence of sheaves $0\to \mathcal{F}\to \mathcal{G}\to \mathcal{H}\to 0$ with flasque kernel, every section of $\mathcal{H}$ on an open set lifts to $\mathcal{G}$ there. Indeed choose a maximal partial lift by the Zorn lemma, using the sheaf gluing axiom on chains. At any point outside its domain choose a local lift. The difference on the overlap is a section of $\mathcal{F}$ and extends to the new open set by flasqueness; subtract that extension from the new lift and glue. Maximality forces the lift's domain to be the whole open set. Thus taking sections preserves this short exact sequence.

Every injective sheaf is flasque: for $U\subseteq V$, the monomorphism between the extension by zero sheaves ${\mathbb{Z}}_U\hookrightarrow {\mathbb{Z}}_V$ and the defining extension property of injectivity make the restriction $\mathcal{J}(V)\to \mathcal{J}(U)$ surjective. Embed $\mathcal{F}$ into an injective sheaf $\mathcal{J}$. The quotient $\mathcal{Q}$ is flasque as well: lift a section of $\mathcal{Q}$ on a smaller open to $\mathcal{J}$, extend it in $\mathcal{J}$, and project.

The long exact sequence in sheaf cohomology now gives $H^1(X,\mathcal{F})=0$, since $\Gamma (X,\mathcal{J})\to \Gamma (X,\mathcal{Q})$ is surjective, and for $i\ge 2$ gives $H^i(X,\mathcal{F})\cong H^{i-1}(X,\mathcal{Q})$, since injectives have zero positive-degree derived functors. Repeat the construction for the flasque quotient. After finitely many such shifts any specified positive degree reduces to a vanishing first group. HenceNo separation, compactness or scheme hypothesis is used.

Write $U_0=D_{+}(t_0)=\mathrm{Spec}\mathbb{C}[x]$, where $x=t_1/t_0$, and $U_1=D_{+}(t_1)=\mathrm{Spec}\mathbb{C}[y]$, with $y=1/x$ on the overlap. Since $\mathcal{G}$ is a quasi-coherent sheaf, it is associated to a $\mathbb{C}[x]$-module $M$ on $U_0$. The preimages under the open inclusion $f$ of $U_0,U_1,U_0\cap U_1$ are $U_0,D(x),D(x)$, all affine schemes. Thus $f$ is an affine morphism and its direct image sheaf $f_{\ast }\mathcal{G}$ is quasi-coherent by direct image of a quasi-coherent sheaf under an affine morphism.

The two-chart cover is an acyclic cover by vanishing of quasi-coherent cohomology on an affine scheme, so the acyclic cover theorem identifies its Čech cohomology with sheaf cohomology. Its only potentially nontrivial positive-degree differential isThis is surjective because of the second summand. There are no normalized Čech terms in degree two or higher for a two-open cover, soThis is the acyclic direct image from an affine chart of the projective line. The pushforward need not itself be a flasque sheaf; the acyclic affine cover is what justifies this computation.

Put $R_j=S/I_j$ and $R_0=S$. The hypothesis says multiplication by $F_j$ on the graded ring $R_{j-1}$ is injective for $j=2,3$. Checking homogeneous elements suffices: an arbitrary annihilated element splits into homogeneous components, each annihilated separately because $F_j$ and $I_{j-1}$ are homogeneous. Also $F_1\ne 0$ because it has degree $d_1>0$, and $S$ is an integral domain. Thus $F_1,F_2,F_3$ is a regular sequence, and the suggested short exact sequences becomeSheafification and twisting preserve exactness. With $X_0={\mathbb{P}}_{\mathbb{C}}^4$, the resulting sequences, interpreted on the ambient projective space via the closed inclusions, areCohomology under a closed immersion identifies the displayed sheaf cohomology with that on $X_j$. For every integer $m$, cohomology of twisting sheaves on projective space gives $H^q(X_0,\mathcal{O}(m))=0$ for $1\le q\le 3$. Inducting through the long exact sequence in sheaf cohomology givesIndeed the two adjacent groups have degrees $q$ and $q+1$ on $X_{j-1}$, both in its vanishing range. This is intermediate cohomology vanishing for a projective complete intersection.

We simultaneously prove $H^0(X_j,{\mathcal{O}}_{X_j}(m))=0$ for $m<0$ and $H^0(X_j,{\mathcal{O}}_{X_j})=\mathbb{C}$. Both hold on $X_0$ by the same projective-space formula. For $j\le 3$, the preceding stage has $H^1(X_{j-1},\mathcal{O}(m-d_j))=0$. If $m<0$, both degree-zero groups on that stage vanish, so the new one vanishes. If $m=0$, the left degree-zero group vanishes because $-d_j<0$, while the middle group is $\mathbb{C}$. Therefore the natural restriction of constant functions is an isomorphism at every stage:These are isomorphisms of $\mathbb{C}$-algebras, not just vector spaces. The global regular functions on a positive-dimensional projective complete intersection are constants even if the complete intersection is singular or nonreduced; smoothness and reducedness were not assumed.