Let \(X=\AA_k^1\) be the affine line over an infinite field \(k\). Let \(P, Q\) be distinct closed points of \(X\), and let \(U=X-\ts{P, Q}\). Show that \(H^1(X, \ZZ_U) \neq 0\).
More generally, for \(n\ge1\), let \(Y \subseteq X=\AA_k^n\) be the union of \(n+1\) hyperplanes in suitably general position, and let \(U=X-Y\). Show that \(H^n(X, \ZZ_U) \neq 0\). Thus the result of (2.7) is the best possible.
Solution
All sheaves and cohomology are for the Zariski topology. For an open inclusion \(j:U\hookrightarrow X\), the notation \(\ZZ_U\) on \(X\) means the extension by zero \(j_!\underline{\ZZ}_U\) of the constant sheaf, not the constant sheaf on \(X\) or the direct image \(j_*\underline{\ZZ}_U\). For a closed subset \(D\subseteq X\), write \(\underline{\ZZ}_D\) for its constant sheaf pushed forward to \(X\); its stalk is \(\ZZ\) at points of \(D\) and zero elsewhere.
1. A constant sheaf on an irreducible space is flasque, and its direct image under a closed inclusion remains flasque.
Every nonempty open subset of an irreducible space is irreducible and therefore connected. A section of the constant sheaf \(\underline{\ZZ}\) is a locally constant integer-valued function, so on such an open it is a single constant. The section group is consequently \(\ZZ\) on a nonempty open and zero on the empty open. Restriction maps are identities or maps onto zero, and are surjective. This is flasqueness. For a closed inclusion \(i:D\hookrightarrow X\), the sections of \(i_*\underline{\ZZ}\) on an open \(V\) are the sections on \(V\cap D\). Thus their restrictions are also surjective when \(D\) is irreducible. Finite direct sums of these sheaves are flasque as well, since their restrictions are finite direct sums of surjections.
2. In (a), \(\boxed{H^1(X,\ZZ_U)\cong\ZZ}\).
The constant-sheaf restriction maps give an exact sequence \[ 0\longrightarrow\ZZ_U\longrightarrow\underline{\ZZ}_X \longrightarrow\underline{\ZZ}_{\{P\}}\oplus\underline{\ZZ}_{\{Q\}} \longrightarrow0. \] Indeed, outside \(P,Q\) the first map is an isomorphism on stalks; at either omitted point the second map is an isomorphism from \(\ZZ\) onto its one nonzero skyscraper stalk. This proves exactness at every point. The two terms after \(\ZZ_U\) are flasque by step 1, since \(X\) is irreducible and each closed point is an irreducible space. They form a flasque resolution, which computes sheaf cohomology [@Har10a, Proposition III.2.5 and Remark III.2.5.1]. On global sections its differential is the diagonal map \[ \ZZ\longrightarrow\ZZ\oplus\ZZ,\qquad a\longmapsto(a,a). \] Its cokernel is \(\ZZ\) via \((b,c)\mapsto c-b\). This proves (a), whether or not the two closed points are \(k\)-rational.
3. In (b), choose affine hyperplanes \(H_0,\ldots,H_n\) so that each intersection of \(s\le n\) of them is an affine space of dimension \(n-s\), and their total intersection is empty.Such a choice exists.
For example, in coordinates \(x_1,\ldots,x_n\), take \[ H_i=V(x_i)\quad(1\le i\le n),\qquad H_0=V(x_1+\cdots+x_n-1). \] An intersection involving only coordinate hyperplanes sets the corresponding coordinates to zero and has the stated dimension. If an intersection of at most \(n\) hyperplanes involves \(H_0\), at least one coordinate is not prescribed to be zero. The equation of \(H_0\) solves uniquely for that coordinate in terms of the remaining ones, again giving the stated affine space. All \(n+1\) equations would imply \(0=1\), so their intersection is empty. These are the intersection properties of suitably general affine hyperplanes, and the rest of the proof applies to any hyperplanes with these properties.
4. For \(H_J=\bigcap_{j\in J}H_j\), there is a flasque resolution\[ 0\longrightarrow\ZZ_U\longrightarrow C^0\longrightarrow C^1 \longrightarrow\cdots\longrightarrow C^n\longrightarrow0, \qquad C^0=\underline{\ZZ}_X,\quad C^p=\bigoplus_{\#J=p}\underline{\ZZ}_{H_J}\ (1\le p\le n). \]
Order the hyperplanes by their indices. The differential is alternating restriction: for \(J=\{j_0<\cdots<j_p\}\), its component is \[ (dc)_J=\sum_{a=0}^p(-1)^a c_{J\setminus\{j_a\}}|_{H_J}. \] In degree zero, the empty-index component is the constant sheaf on \(X\). Every pair of omitted indices occurs with opposite signs in the next differential, so \(d^2=0\).
3.The stalk of \(\ZZ_U\) is zero, and the remaining stalk complex has one copy of \(\ZZ\) for every subset of \(T\), including the empty subset, with the displayed alternating maps.This is the augmented cochain complex of a simplex and is exact.For an explicit contraction, choose \(v\in T\) and extend cochains to alternating functions on ordered tuples, setting them to zero on tuples with a repeated index.The operator \(h\) which sends a cochain \(c\) to \(hc(j_1,\ldots,j_{p-1})=c(v,j_1,\ldots,j_{p-1})\) satisfies \(dh+hd=\id\); expanding both alternating sums cancels all terms except \(c\).The same formula in the augmented degree evaluates a one-index cochain at \(v\).Thus every stalk complex is exact.
3.Step 1 makes every \(C^p\) flasque, proving the resolution assertion.
5. In (b), \(\boxed{H^n(X,\ZZ_U)\cong\ZZ}\).
4.Each proper intersection contributes exactly one copy of \(\ZZ\).Consequently the resulting complex is\[ \ZZ\longrightarrow\ZZ^{\binom{n+1}{1}} \longrightarrow\cdots\longrightarrow\ZZ^{\binom{n+1}{n}}, \]in degrees \(0,\ldots,n\), with alternating subset maps.Adjoin one further copy of \(\ZZ\) in degree \(n+1\), corresponding to the full index set, and define\[ \varepsilon(c)=\sum_{i=0}^n(-1)^i c_{\{0,\ldots,n\}\setminus\{i\}}. \]The extended complex is the augmented simplex complex on \(n+1\) vertices.The contraction in step 4 proves that it is exact.In particular, \(\ker\varepsilon\) is the image of the preceding differential, and \(\varepsilon\) is surjective; one component with coefficient \(1\) maps to \(1\) or \(-1\).The original complex has no outgoing differential in degree \(n\), because the total hyperplane intersection is empty.Its degree-\(n\) cohomology is therefore\[ \ZZ^{\binom{n+1}{n}}/\im d^{n-1} \cong\ZZ^{\binom{n+1}{n}}/\ker\varepsilon \cong\ZZ. \]The flasque-resolution computation identifies this with \(H^n(X,\ZZ_U)\), proving (b).Since \(\dim\AA_k^n=n\), this also shows that the vanishing in degrees greater than \(n\) in [@Har10a, Theorem III.2.7] cannot be extended to degree \(n\) for all abelian sheaves.
6. Q.E.D.
Step 2 proves (a), and steps <1>3–<1>5 prove (b) and the sharpness assertion.