Past exam of the introduction to schemes course of the University of Oxford / 2021 / 2

Constructing the subsheaf. On an integral scheme, every nonempty open subset is an irreducible topological space and hence a connected space. A locally constant function from such an open subset to $F$ is constant. Thus the constant sheaf $\underline{F}$ has sections $F$ on every nonempty open subset, and the structure sheaf embeds in it by restriction to the generic point. Empty members of the given open cover may be discarded.

For every open subset $W\subseteq X$, defineThis is a subsheaf of the sheaf of modules $\underline{F}$: its condition is local, and compatible sections glue in $\underline{F}$, with the regularity conditions gluing in ${\mathcal{O}}_X$. On $U_i\cap U_j$, the hypothesis says that $f_j/f_i$ is a section of the sheaf of units of the structure sheaf. Therefore multiplication by it preserves ${\mathcal{O}}_X$, andIt follows that $\mathcal{L}{|}_{U_i}=f_i^{-1}{\mathcal{O}}_{U_i}$: the condition with index $i$ implies every other condition on an overlap, and the converse is part of the definition. Multiplication by $1/f_i$ is an isomorphism ${\mathcal{O}}_{U_i}\to \mathcal{L}{|}_{U_i}$, with inverse multiplication by $f_i$. Hence $\mathcal{L}$ is an invertible sheaf, or equivalently a line bundle, andThese local descriptions also give uniqueness as a subsheaf of $\underline{F}$. This construction is the divisor line bundle for the local rational equations $f_i$.

The tensor product. Refine the two open covers to $W_{ij}=U_i\cap V_j$. The ratiosare sections of the sheaf of units of the structure sheaf on the corresponding overlaps, so the products give a line bundle $\mathcal{M}\subseteq \underline{F}$ by the construction just proved. Multiplication in $F$ defines a sheaf morphismOn each $W_{ij}$ it sends the local generator $(1/f_i)\otimes (1/g_j)$ to the local generator $1/(f_i{g}_j)$. It is thus an isomorphism on an open cover, and hence an isomorphism globally. In particular,

The sign for projective twists. Write homogeneous coordinates $[x_0:\cdots :x_n]$ on projective space ${\mathbb{P}}_{\mathbb{Z}}^n$, and let $U_i=D_{+}(x_i)$ be its standard affine open cover. For any integer $m$, take the nonzero elements of the function fieldTheir ratios are $f_i/f_j=(x_j/x_i{)}^m$, which are sections of the sheaf of units of the structure sheaf on $U_i\cap U_j$, including when $m<0$. The twisting sheaf on projective space $\mathcal{O}(m)$ has local generator $x_i^m$ on $U_i$. Divide a rational section of $\mathcal{O}(m)$ by $x_0^m$ to embed it in the constant sheaf of the function field; the local generator becomes $(x_i/x_0{)}^m=1/f_i$. This identifies the constructed line bundle with $\mathcal{O}(m)$ and fixes the sign explicitly.

For $n\ge 1$, $\mathrm{Pic}({\mathbb{P}}_{\mathbb{Z}}^n)\cong \mathbb{Z}$, with generator $[\mathcal{O}(1)]$; the displayed data gives all its powers, with $\mathcal{O}(m)\otimes \mathcal{O}(r)\cong \mathcal{O}(m+r)$ by the product rule above. If $n=0$, projective space is $\mathrm{Spec}\mathbb{Z}$, its Picard group is zero, and the same formula gives only the trivial line bundle.

Discard empty members of the open cover: every Čech cochain group factor supported on an empty intersection is the zero abelian group, so this changes no Čech cohomology. Every remaining member is the one-point space $X$; the index set is nonempty even if the same open subset is listed repeatedly. Choose a well-order with least index $\alpha$. Use the ordered Čech cochain complex permitted in the question:All restriction maps are identities. For $q\ge 1$ define a cochain homotopy $h_q:C^q\to C^{q-1}$ byWe verify $\delta h_q+h_{q+1}\delta ={\mathrm{id}}_{C^q}$ for every $q\ge 1$. Fix $I=(i_0<\cdots <i_q)$. If $i_0=\alpha$, then $(h_{q+1}\delta s{)}_I=0$. In $(\delta h_{q}s{)}_I$, the term deleting $i_0$ is $s_I$ and every other term is zero because its retained first index is $\alpha$. Thus the identity holds on $I$.

If $i_0>\alpha$, thenAdding proves the identity in this case also. Each formula uses only finitely many terms on each component, so it remains valid for an infinite open cover and arbitrary abelian group $A$.

If $s\in C^q$ is a Čech cocycle, then $\delta s=0$, and the identity gives $s=\delta (h_{q}s)$: every Čech cocycle in positive degree is a Čech coboundary. Therefore, using the permitted comparison with ordered Čech cohomology,The identity is asserted in positive degrees; degree zero has $H^0\cong A$, so an unaugmented contraction to zero in every degree would be incorrect.

In an irreducible topological space, any two nonempty open subsets intersect. Let $W\subseteq X$ be a nonempty open subset. Two nonempty subsets open in $W$ are also open in $X$, and hence intersect; thus $W$ is itself an irreducible topological space. If $W$ were disconnected, a separation into two disjoint nonempty open subsets would contradict this intersection property. Hence $W$ is a connected space. The empty open subset is connected as well under the usual convention.

Discard empty members of $\mathcal{U}$ as in (b). Every finite intersection $U_{i_0}\cap \cdots \cap U_{i_q}$ is nonempty by repeated use of the intersection property and is an open subset, hence a connected space. A locally constant function on this intersection is constant. It follows that the constant sheaf hasand every restriction map between these nonempty intersections is the identity on $A$. The whole Čech cochain complex is therefore the same as for the indexed open cover of a point having the same index set. The cochain homotopy constructed in (b) applies component by component and provesThe argument concerns the constant sheaf of locally constant functions; arbitrary coefficient sheaves on an irreducible topological space need not have the same Čech cohomology.