Let \(k\) be a field, let \(X=\PP_k^r\), and let \(Y\) be a closed subscheme of dimension \(q\ge1\), which is a complete intersection (II, Ex. 8.4). Prove the following assertions.
For all \(n\in\ZZ\), the natural map \[ H^0(X, \mco_X(n)) \to H^0(Y, \mco_Y(n)) \] is surjective. This gives a generalization and another proof of (II, Ex. 8.4c), where we assumed \(Y\) was normal.
The scheme \(Y\) is connected.
One has \(H^i(Y,\mco_Y(n))=0\) for \(0<i<q\) and all \(n\in\ZZ\).
One has \(p_a(Y)=\dim_k H^q(Y,\mco_Y)\).
Hint
Use exact sequences and induction on the codimension, starting from \(Y=X\), which is covered by Theorem III.5.1.
Solution
Put \(c=r-q\) and \(S=k[x_0,\ldots,x_r]\). The global complete-intersection hypothesis means that the saturated homogeneous ideal of \(Y\) is generated by \(c\) homogeneous forms \(F_1,\ldots,F_c\) of positive degrees \(d_1,\ldots,d_c\). For \(0\le j\le c\), put \(Y_j=\operatorname{Proj}(S/(F_1,\ldots,F_j))\), so \(Y_0=X\) and \(Y_c=Y\). All twists on these schemes are the restrictions of the standard twists on \(\PP_k^r\).
1. The forms \(F_1,\ldots,F_c\) are a regular sequence, and \(\dim Y_j=r-j\) for every \(j\).
Let \(\mathfrak m=(x_0,\ldots,x_r)\). The dimension of the homogeneous coordinate ring of the nonempty \(Y\) is \(q+1\); localization at the vertex \(\mathfrak m\) has that same dimension [@Har10a, Chapter I, §§2 and 7]. The regular local ring \(S_{\mathfrak m}\) is Cohen–Macaulay of dimension \(r+1\). Since quotienting by the \(c\) forms drops its dimension by exactly \(c\), the dimension criterion for a regular sequence makes them a regular sequence in \(S_{\mathfrak m}\). That criterion also gives dimension \(r+1-j\) for every prefix quotient at the vertex [@Har10a, Theorem II.8.21A].
The sequence is regular in \(S\) as well. At each stage the kernel testing injectivity is a graded module. A nonzero homogeneous element of such a kernel cannot become zero after localization at \(\mathfrak m\): multiplication by a polynomial outside \(\mathfrak m\) has a nonzero constant term, which gives a nonzero component in the original degree of that element. Thus every prefix is globally regular. Its homogeneous cone has dimension \(r+1-j\), and its Proj has dimension one less. These Proj schemes are nonempty, since \(r+1-j\ge q+1\ge2\) and a cone supported only at the vertex would have dimension zero. Hence \(\dim Y_j=r-j\). The argument is algebraic over \(k\) and does not require algebraic closedness. For \(c=0\) the assertions refer to the empty regular sequence and \(Y=X\).
2. For every \(j\) and \(n\in\ZZ\), the restriction map \(H^0(X,\OO_X(n))\to H^0(Y_j,\OO_{Y_j}(n))\) is surjective, and\[ H^i(Y_j,\OO_{Y_j}(n))=0\qquad(0<i<r-j). \]
Induct on \(j\). For \(j=0\), the restriction is the identity and the vanishing is the projective-space calculation [@Har10a, Theorem III.5.1]. For \(j\ge1\), regularity in step 1 gives the exact sequence on \(Y_{j-1}\) \[ 0\longrightarrow\OO_{Y_{j-1}}(n-d_j) \xrightarrow{F_j}\OO_{Y_{j-1}}(n) \longrightarrow\OO_{Y_j}(n)\longrightarrow0, \] where the last sheaf is extended by the closed inclusion into \(Y_{j-1}\). Sheafification of the graded regular-sequence quotient gives this sequence, and twisting by an invertible sheaf preserves exactness. Cohomology of the last term equals its cohomology on \(Y_j\) [@Har10a, Lemma III.2.10].
Write \(s=r-j=\dim Y_j\), so \(s\ge q\ge1\) and \(\dim Y_{j-1}=s+1\ge2\). The induction hypothesis gives \(H^1(Y_{j-1},\OO(n-d_j))=0\). Thus the long exact sequence makes restriction from \(Y_{j-1}\) onto \(H^0(Y_j,\OO(n))\) surjective. Composing with the preceding restriction from \(X\) proves the new surjectivity.
For \(0<i<s\), the same long exact sequence has the segment \[ H^i(Y_{j-1},\OO(n))\longrightarrow H^i(Y_j,\OO(n)) \longrightarrow H^{i+1}(Y_{j-1},\OO(n-d_j)). \] Both outer groups vanish by induction, since \(0<i<i+1\le s<s+1\). The middle group therefore vanishes. This holds for every integer \(n\), including negative twists, and completes the induction. At \(j=c\) it proves (a) and (c).
3. The constants map \(k\to H^0(Y,\OO_Y)\) is an isomorphism, and \(Y\) is connected.
Step 2 at \(n=0\) gives a surjection \[ k=H^0(\PP_k^r,\OO)\longrightarrow H^0(Y,\OO_Y). \] This is the constants map. It is injective: \(Y\) is nonempty, and a nonzero element of the field maps to a unit in every nonzero local ring of \(Y\). Thus \(H^0(Y,\OO_Y)=k\). A decomposition of \(Y\) into two nonempty open-and-closed subsets would yield the nontrivial idempotent global section equal to \(1\) on one subset and \(0\) on the other. A field has no such idempotent, so \(Y\) is connected, proving (b). No reducedness is used in this argument.
4. The arithmetic genus is \(\boxed{p_a(Y)=\dim_kH^q(Y,\OO_Y)}\).
Step 3 gives \(h^0(Y,\OO_Y)=1\), and step 2 gives \(h^i(Y,\OO_Y)=0\) for \(0<i<q\). Cohomology also vanishes for \(i>q\) by Grothendieck vanishing, since \(\dim Y=q\) [@Har10a, Theorem III.2.7]. Coherent cohomology is finite-dimensional because \(Y\) is projective [@Har10a, Theorem III.5.2]. Therefore \[ \chi(Y,\OO_Y)=1+(-1)^q h^q(Y,\OO_Y). \] The definition \(p_a(Y)=(-1)^q(\chi(Y,\OO_Y)-1)\) now gives the claimed value, proving (d).
5. Q.E.D.
Step 2 proves (a) and (c), step 3 proves (b), and step 4 proves (d), with the regular-sequence foundation in step 1.