Let \(k\) be a field and let \(X\) be the closed subscheme of \(\PP_k^2\) defined by a single homogeneous equation \(f(x_0,x_1,x_2)=0\) of degree \(d\ge1\). (Do not assume \(f\) is irreducible.) Assume that \((1,0,0)\) is not on \(X\). Then show that \(X\) can be covered by the two open affine subsets \(U=X \intersect \ts{x_1 \neq 0}\) and \(V=X \intersect \ts{x_2 \neq 0}\). Now calculate the Čech complex \[ \Gamma(U, \mco_X) \oplus \Gamma(V, \mco_X) \to \Gamma(U \intersect V, \mco_X) \] explicitly, and thus show that \[ \begin{aligned} \dim H^0(X, \mco_X)&=1, \\ \dim H^1(X, \mco_X)&=\frac{1}{2}(d-1)(d-2). \end{aligned} \]
(Positive degree)
The condition \(d\ge1\) is the plane-curve range of this calculation. A nonzero constant equation instead defines the empty subscheme, which also avoids \([1:0:0]\) but has zero global-section group. Thus the stated \(H^0\) formula requires the positive-degree hypothesis.
Solution
The coefficient of \(x_0^d\) in \(f\) is \(f(1,0,0)\), which is nonzero by hypothesis. Multiplying \(f\) by the inverse of this coefficient does not change \(X\), so assume that the coefficient is \(1\).
1. The two indicated affine opens cover \(X\) and have coordinate rings\[ B=k[t,a]/(f(a,1,t)),\qquad C=k[s,b]/(f(b,s,1)), \]where \(a=x_0/x_1\), \(t=x_2/x_1\) on \(U\), and \(b=x_0/x_2\), \(s=x_1/x_2\) on \(V\).
The only point of \(\PP^2\) with \(x_1=x_2=0\) is \([1:0:0]\). Since it does not belong to \(X\), the opens \(U\) and \(V\) cover \(X\). Each is a closed subscheme of a standard affine chart of projective space, with the displayed dehomogenized equation. Thus both are affine, with precisely these coordinate rings. Their intersection has coordinate ring \[ D=B[t^{-1}]=C[s^{-1}], \] and the change of coordinates is \(s=t^{-1}\) and \(b=a/t\). Homogeneity gives \(f(a,1,t)=t^d f(a/t,t^{-1},1)\), so these substitutions identify the two localized quotients.
2. As \(k\)-vector spaces, the Čech complex is the direct sum, for \(0\le i<d\), of the complexes\[ k[t]\oplus t^{-i}k[t^{-1}]\xrightarrow{(u,v)\mapsto v-u}k[t,t^{-1}], \]with the \(i\)th complex multiplied by \(a^i\).
The polynomial \(f(a,1,t)\) is monic of degree \(d\) in \(a\). Polynomial division therefore makes \(B\) a free \(k[t]\)-module with basis \(1,a,\ldots,a^{d-1}\). Existence of the remainder follows by division by a monic polynomial, and uniqueness follows because a nonzero multiple of that polynomial cannot have \(a\)-degree less than \(d\). The same argument makes \(C\) free over \(k[s]\) with basis \(1,b,\ldots,b^{d-1}\). Localizing gives \[ D=\bigoplus_{i=0}^{d-1}a^i k[t,t^{-1}]. \] The maps \(B\to D\) and \(C\to D\) are injective, since multiplication by \(t\) or \(s\) on the respective free coefficient module is injective. Under the coordinate change in step 1, their images are \[ B=\bigoplus_{i=0}^{d-1}a^i k[t],\qquad C=\bigoplus_{i=0}^{d-1}a^i t^{-i}k[t^{-1}]. \] Consequently the difference of restrictions decomposes into exactly the displayed coefficientwise maps. The scheme \(X\) is noetherian and separated, and \(\OO_X\) is quasi-coherent. Its cohomology is therefore computed by this two-affine Čech complex [@Har10a, Theorem III.4.5].
3. The kernel of the Čech differential consists precisely of the constant functions, so \(\boxed{\dim_k H^0(X,\OO_X)=1}\).
The kernel is identified with \(B\cap C\) in \(D\). In its \(a^i\) component, it is the intersection \[ k[t]\cap t^{-i}k[t^{-1}]. \] The first space has only exponents at least zero, and the second only exponents at most \(-i\). For \(i=0\) their intersection is \(k\); for every \(i>0\) it is zero. Uniqueness of Laurent coefficients therefore gives \(B\cap C=k\) in the constant \(a^0\) component. The corresponding sections are the same constant on both charts and are exactly the global constant functions.
4. The first cohomology has basis\[ \left\{[a^it^{-j}]:2\le i\le d-1,\ 1\le j\le i-1\right\}, \]and \(\boxed{\dim_k H^1(X,\OO_X)=\frac{(d-1)(d-2)}2}\).
The cokernel in component \(i\) is \[ k[t,t^{-1}]/\bigl(k[t]+t^{-i}k[t^{-1}]\bigr). \] Its surviving Laurent exponents are the integers strictly between \(-i\) and zero. There are none for \(i=0,1\), and for \(i\ge2\) they give the basis \(t^{-1},\ldots,t^{-(i-1)}\). The free decomposition of \(D\) from step 2 proves independence between different \(a^i\) components as well. Thus the displayed classes form a basis of the entire cokernel, which is \(H^1(X,\OO_X)\). Their number is \[ \sum_{i=2}^{d-1}(i-1)=\frac{(d-1)(d-2)}2 \] for \(d\ge2\), and both the basis and the formula give zero for \(d=1\). No step used irreducibility or absence of nilpotents in \(B\) or \(C\); monicity was the necessary property.
5. Q.E.D.
Step 1 proves the affine-cover assertion, step 2 calculates its Čech complex, and steps <1>3–<1>4 compute the two requested dimensions.