Let \(X=\AA_k^2=\Spec k[x, y]\), and let \(U=X-\ts{(0,0)}\). Using a suitable cover of \(U\) by open affine subsets, show that \(H^1(U, \mco_U)\) is isomorphic to the \(k\)-vector space spanned by \(\ts{x^i y^j \st i, j<0}\). In particular, it is infinite-dimensional.
Using (3.5), this provides another proof that \(U\) is not affine — cf. (I, Ex. 3.6).
Solution
Put \(A=k[x,y]\), \(U_x=D(x)\) and \(U_y=D(y)\), viewed as open subsets of \(X=\Spec A\). Their union is \(U\), since \(V(x,y)\) consists of the origin.
1. The affine cover \((U_x,U_y)\) identifies the requested group with\[ H^1(U,\OO_U)\cong A_{xy}/(A_x+A_y). \]
The two opens and their intersection are affine, with rings \[ A_x=k[x,x^{-1},y],\qquad A_y=k[x,y,y^{-1}],\qquad A_{xy}=k[x,x^{-1},y,y^{-1}]. \] The scheme \(U\) is noetherian and separated, and \(\OO_U\) is quasi-coherent. The affine-cover Čech comparison therefore computes sheaf cohomology by the complex \[ A_x\oplus A_y\xrightarrow{(a,b)\mapsto b-a}A_{xy} \] in degrees zero and one [@Har10a, Theorem III.4.5]. There is no term in degree two for this two-member cover. The image of the differential is the additive subgroup \(A_x+A_y\), giving the displayed quotient. All these maps are \(k\)-linear.
2. The classes of the Laurent monomials with both exponents negative form a basis, giving\[ \boxed{H^1(U,\OO_U)\cong\bigoplus_{i<0,\ j<0}k\,x^iy^j.} \]
1.In particular, each element is a finite linear combination of the displayed basis monomials, with no additional linear relations.
3. The group is infinite-dimensional over \(k\), and \(U\) is not affine.
For \(m\ge1\), the classes \(x^{-m}y^{-1}\) are distinct members of the basis in step 2 and are linearly independent. Hence the cohomology group is infinite-dimensional and nonzero. If \(U\) were affine, the higher cohomology of its quasi-coherent structure sheaf would vanish by T-COHAFF [@Har10a, Theorem III.3.5]. The nonzero group in degree one contradicts that vanishing, proving nonaffineness.
4. Q.E.D.
Steps <1>1–<1>2 give the requested cohomology computation and basis, and step 3 proves both consequences.