Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 113 (Algebraic Geometry)

Take $X=\mathbb{R}$, $U=(0,1)$, and let $\mathcal{F}$ be the skyscraper sheaf with value a nonzero abelian group $A$ at $p=3/4$. Cover $V=(-1,1)$ by $V_1=(-1,2/3)$ and $V_2=(1/2,1)$. The presheaf gives $(j^p\mathcal{F})(V)=0=(j^p\mathcal{F})(V_1)$ and $(j^p\mathcal{F})(V_2)=A$. A nonzero section on $V_2$ and the zero section on $V_1$ agree on the overlap, whose skyscraper sections vanish, but cannot be glued on $V$. Thus $j^p\mathcal{F}$ need not be a sheaf.

Sheafification preserves stalks. If $P\in U$, neighborhoods contained in $U$ are cofinal, so $(j^p\mathcal{F}{)}_P={\mathcal{F}}_P$. If $P\notin U$, no neighborhood of $P$ lies in $U$, so every term defining the presheaf stalk is zero. Therefore

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On the defining presheaf, send a section over $V\subseteq U$ by the identityand use the unique zero map when $V⊈U$. These maps commute with restrictions and therefore sheafify to the natural counit

After restricting $j^p\mathcal{F}$ back to $U$, every open set lies in $U$, so one recovers the original sheaf $\mathcal{F}$. Equivalently, the natural map $\mathcal{F}\to j^{-1}{j}_{!}\mathcal{F}$ is an isomorphism on every stalk and hence an isomorphism of sheaves.

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Use the counit from part b for the first map and the restriction map $\mathcal{G}\to i_{\ast }{i}^{-1}\mathcal{G}$ for the second. Exactness can be checked on stalks. At $P\in U$ the sequence iswhereas at $P\in Z$ it isThusis a short exact sequence of sheaves.

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Let $X=\mathrm{Spec}k[x]$ and $U=D(x)$. A global section of $j_{!}{\mathcal{O}}_U$ is a regular function on the integral scheme $U$ whose support is closed in $X$ and contained in $U$. Every nonzero regular function on $U$ has support dense in $X$, soIf this sheaf were quasi-coherent, then on the affine scheme $X$ it would be the sheaf associated with this zero module and hence would vanish. Its stalks at points of $U$ are instead ${\mathcal{O}}_{U,P}\ne 0$ by part a. This contradiction proves that extension by zero need not preserve quasi-coherence.

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The Proj construction has underlying setIts closed sets are $V_{+}(I)=\{\mathfrak{p}:I\subseteq \mathfrak{p}\}$ for homogeneous ideals $I$, and its standard opens are $D_{+}(f)$. The structure sheaf is characterized byfor homogeneous $f$ of positive degree.

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A closed immersion identifies its source homeomorphically with a closed subset and induces a surjection from the target structure sheaf to the pushed-forward source structure sheaf.

The quotient map $S\to S/I$ identifies $\mathrm{Proj}(S/I)$ with $V_{+}(I)\subseteq \mathrm{Proj}S$. On every standard open $D_{+}(f)$ it induces the surjectionwith kernel $(I_f{)}_0$. These affine-local maps glue, proving that the natural morphism $\mathrm{Proj}(S/I)\to \mathrm{Proj}S$ is a closed immersion.

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It suffices to compare the localized homogeneous ideals on every $D_{+}(f)$. If $a\in I$ is homogeneous, choose $N$ so large that $\mathrm{deg}(f^{N}a)\ge n_0$. Then $f^{N}a\in I^{\prime }$, and in $S_f$,Thus $I_f\subseteq I_f^{\prime }$, while the reverse inclusion follows from $I^{\prime }\subseteq I$. Hence $I_f=I_f^{\prime }$, so $(I_f{)}_0=(I_f^{\prime }{)}_0$ and the quotient structure sheaves agree on all standard opens. The two quotients define the same closed subscheme of $\mathrm{Proj}S$.

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A prime Weil divisor is an integral closed subscheme of codimension one. If $Y$ has generic point ${\eta }_Y$, regularity in codimension one makes the local ring ${\mathcal{O}}_{X,{\eta }_Y}$ a discrete valuation ring with fraction field $K(X)$. Its normalized discrete valuationis the order of vanishing along $Y$: writing $f=u{\pi }^m$ for a unit $u$ and uniformizer $\pi$ gives ${\nu }_Y(f)=m$. Additivity of exponents makes this a group homomorphism.

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Restriction sends a Weil divisor on $X$ to the sum of its components meeting $U$. Every prime divisor on $U$ has a codimension-one closure in the Noetherian scheme $X$, so the induced map $\mathrm{Cl}(X)\to \mathrm{Cl}(U)$ is surjective.

If the class of $D$ restricts to zero, then $D{|}_U={\mathrm{div}}_U(f)$ for some $f\in K(X{)}^{\ast }$. The divisor $D-{\mathrm{div}}_X(f)$ is supported on $Z$, hence is an integral combination of $Z_1,\dots ,Z_n$. Conversely, every such combination restricts to zero. Thereforewhere the first map sends the $i$th basis vector to $[Z_i]$, is exact.

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At the generic point of $L_0$, the functions $x_1,x_2$ are units and the equation gives $x_0=x_3^3/(x_1{x}_2)$. Taking $x_3$ as a uniformizer yieldsAt $L_1$, the functions $x_0,x_2$ are units and $x_1=x_3^3/(x_0{x}_2)$, so ${\nu }_{L_1}(x_0/x_3)=-1$. Symmetrically, ${\nu }_{L_2}(x_0/x_3)=-1$.

The complement of the three lines is $U=D_{+}(x_3)$, withThis unique factorization domain has trivial divisor class group, so part b says that $[L_0],[L_1],[L_2]$ generate $\mathrm{Cl}(X)$. The units on $U$ are scalar multiples of $u_0^a{u}_1^b$, and their boundary valuation vectors are generated byThe resulting integer matrix has Smith normal form $\mathrm{diag}(1,3,0)$. Consequently

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For an open cover $\mathcal{U}=(U_i)$, the Čech cochain complex iswith differentialIts cohomology is ${\check{H}}^p(\mathcal{U},\mathcal{F})$.

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Suppose ${\mathbb{P}}_k^n$ had a cover by $n$ affine opens. Since projective space is separated, every finite intersection in this cover is affine. The acyclic cover theorem would therefore compute the cohomology of every quasi-coherent sheaf by its Čech complex. A cover with only $n$ members has no degree-$n$ cochains, so it would implyBut top cohomology of projective space givesa contradiction. Hence no such affine cover exists.

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Choose a frame of a line bundle on each $U_i$. On $U_i\cap U_j$, the frames differ by $g_{ij}\in {\mathcal{O}}_X^{\ast }(U_i\cap U_j)$, and compatibility on triple intersections is the Čech cocycle condition $g_{ij}{g}_{jk}=g_{ik}$. Replacing the local frames by units $h_i$ changes $g_{ij}$ by the Čech coboundary $h_i^{-1}{h}_j$. Conversely, a multiplicative one-cocycle glues the trivial line bundles ${\mathcal{O}}_{U_i}$ into a line bundle. Tensor product multiplies cocycles, so

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A Cartier divisor on an integral scheme is given by an open cover $(U_i)$ and rational functions $f_i\in K(X{)}^{\ast }$ such that $f_i/f_j\in {\mathcal{O}}_X^{\ast }(U_i\cap U_j)$. Two are linearly equivalent when their quotient is represented by one global rational function. The Cartier class group is the group of Cartier divisors modulo these principal divisors.

Let ${\mathcal{K}}_X^{\ast }$ be the sheaf of nonzero rational functions. Cartier divisors are the global sections of ${\mathcal{K}}_X^{\ast }/{\mathcal{O}}_X^{\ast }$, and the exact sequencegives a long exact cohomology sequence. On an integral scheme ${\mathcal{K}}_X^{\ast }$ is flasque: every nonempty restriction map is the identity on $K(X{)}^{\ast }$. Hence $H^1(X,{\mathcal{K}}_X^{\ast })=0$, and exactness gives

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