Let \(Q\) be the nonsingular split quadric surface \(xy=zw\) in \(X=\PP_k^3\) over a field \(k\). We consider nonempty effective Cartier divisors \(Y\) on \(Q\), equivalently nonempty locally principal closed subschemes of codimension one. Since \(\Pic Q\cong\ZZ\oplus\ZZ\), we can speak of the type \((a,b)\) of \(Y\) (II, 6.16 and II, 6.6.1).
Let us denote the invertible sheaf \(\mcl(Y)\) by \(\mco_Q(a, b)\). Thus for any \(n \in \ZZ\), \(\mco_Q(n)=\mco_Q(n, n)\).
Use the special cases \((q,0)\) and \((0,q)\), with \(q>0\), represented by disjoint unions of \(q\) ruling lines after extending the field if necessary, to show:
If \(\abs{a-b}\le1\), then \(H^1(Q,\mco_Q(a,b))=0\).
If \(a,b<0\), then \(H^1(Q,\mco_Q(a,b))=0\).
If \(a\le-2\), then \(H^1(Q,\mco_Q(a,0))\ne0\).
Use these results to prove the following assertions.
If \(Y\) has type \((a,b)\) with \(a,b>0\), then \(Y\) is connected.
If \(k\) is algebraically closed, then for any \(a,b>0\) there is an irreducible nonsingular curve of type \((a,b)\). Use (II, 7.6.2) and (II, 8.18).
An irreducible nonsingular curve of type \((a,b)\), with \(a,b>0\), is projectively normal in \(\PP_k^3\) if and only if \(\abs{a-b}\le1\). In particular, over an algebraically closed field this gives nonsingular curves that are not projectively normal, including the rational quartic of type \((1,3)\) (I, Ex. 3.18).
For any nonempty effective Cartier divisor \(Y\) of type \((a,b)\) on \(Q\), show that \[ p_a(Y)=ab-a-b+1. \]
(The field in the ruling-line model)
Over a finite field with \(s\) elements, each ruling has only \(s+1\) \(k\)-rational fibres, so arbitrarily many distinct lines isomorphic over \(k\) to \(\PP_k^1\) cannot be chosen in that ruling. The split-quadric cohomology formulas hold over every field; the field-extension comparison in step <1>1 makes the disjoint-line model a valid computation of their numerical values in all cases.
Hint
For (c), calculate the Hilbert polynomials of the relevant sheaves and compare with the special case \((q,0)\) represented by \(q\) disjoint ruling lines. See (V, 1.5.2) for another method.
Solution
Use the Segre identification \(Q\cong\PP_k^1\times_k\PP_k^1\), whose coordinates can be taken as \(x=s_0t_0\), \(y=s_1t_1\), \(z=s_0t_1\), \(w=s_1t_0\). Then \[ \OO_Q(a,b)=p_1^*\OO_{\PP^1}(a)\otimes p_2^*\OO_{\PP^1}(b), \qquad\OO_Q(1)=\OO_Q(1,1) \] by the Segre twisting formula [@Har10a, Exercise II.5.11]. Let \(v(t)=\max(t+1,0)\) and \(w(t)=\max(-t-1,0)\) for integers \(t\).
1. For all integers \(a,b\), the cohomology dimensions and Euler characteristic are\[ \begin{aligned} h^0(Q,\OO_Q(a,b))&=v(a)v(b),\\ h^1(Q,\OO_Q(a,b))&=w(a)v(b)+v(a)w(b),\\ h^2(Q,\OO_Q(a,b))&=w(a)w(b),\\ \chi(\OO_Q(a,b))&=(a+1)(b+1). \end{aligned} \]
On \(\PP_k^1\), the cohomology of \(\OO(t)\) has dimensions \(v(t)\) in degree zero and \(w(t)\) in degree one, and vanishes in higher degrees [@Har10a, Theorem III.5.1]. Apply the Künneth formula over a field to the two projective lines and these invertible sheaves. It gives \[ H^j(Q,\OO_Q(a,b))\cong \bigoplus_{p+q=j}H^p(\PP^1,\OO(a))\otimes_k H^q(\PP^1,\OO(b)). \] Taking dimensions yields the first three formulas. The alternating sum is \((v(a)-w(a))(v(b)-w(b))\), and \(v(t)-w(t)=t+1\) for every integer \(t\), giving the fourth formula.
The ruling-line calculation in the hint is explicit in these terms. Choose \(q\) distinct rational points of the first factor and let \(D\) be their inverse image, a divisor of type \((q,0)\) consisting of \(q\) disjoint projective lines. The sequence \[ 0\to\OO_Q(-q,0)\to\OO_Q\to\OO_D\to0 \] and \(H^1(Q,\OO_Q)=0\) give \[ H^1(Q,\OO_Q(-q,0))\cong k^q/k(1,\ldots,1), \qquad h^1=q-1, \] since \(H^0(Q,\OO_Q(-q,0))=0\). Also \(\chi(\OO_D)=q\) and \(\chi(\OO_Q(-q,0))=1-q\). The other ruling gives the symmetric calculation. When \(k\) has too few rational points for this choice, make it over an infinite extension field. The cohomology base-extension comparison proved in P-AGH352HILBPOLY, step 1, preserves these dimensions, so the numerical formulas remain valid over the original field.
2. All three assertions in (a) hold.
The expression for \(h^1\) in step 1 is nonzero exactly when one index is at most \(-2\) and the other is at least zero. This follows because \(w(t)>0\) exactly for \(t\le-2\) and \(v(t)>0\) exactly for \(t\ge0\). Such indices differ by at least two, proving (a)(i). When both indices are negative, both \(v\) terms vanish, proving (a)(ii). For \(a\le-2\) and \(b=0\), the formula is \(h^1=-a-1>0\), proving (a)(iii).
3. If \(a,b>0\), every divisor \(Y\) in (b)(i) is connected.
The defining effective Cartier divisor gives \[ 0\longrightarrow\OO_Q(-a,-b)\longrightarrow\OO_Q\longrightarrow\OO_Y\longrightarrow0. \] Step 1 gives \(H^0(Q,\OO_Q(-a,-b))=0\), and (a)(ii) gives \(H^1(Q,\OO_Q(-a,-b))=0\). The long exact sequence therefore identifies \(H^0(Y,\OO_Y)\) with \(H^0(Q,\OO_Q)=k\). A disconnected scheme has a nontrivial idempotent global function, obtained by taking values zero and one on two nonempty open-and-closed pieces. No such idempotent exists in \(k\), so \(Y\) is connected, even if it is nonreduced.
4. Over an algebraically closed field, nonsingular irreducible divisors of every type \((a,b)\) with \(a,b>0\) exist.
The product of the degree-\(a\) and degree-\(b\) Veronese embeddings of the two projective lines, followed by a Segre embedding, is a closed immersion whose pullback of \(\OO(1)\) is \(\OO_Q(a,b)\). Thus this sheaf is very ample, as in [@Har10a, Example II.7.6.2]. Bertini’s hyperplane theorem for this embedding gives a nonempty nonsingular divisor \(Y\) of the required type [@Har10a, Theorem II.8.18]. Step 3 makes it connected. Its regular local rings are domains, so distinct irreducible components cannot meet; the finitely many components would consequently be open and closed. Connectedness forces only one component, and nonsingularity gives reducedness. Hence \(Y\) is irreducible and nonsingular, proving (b)(ii).
5. For the curve in (b)(iii), projective normality holds exactly when \(\abs{a-b}\le1\).
The quadric is a positive-dimensional complete intersection, so P-AGH355COMPINT makes \(H^0(\PP^3,\OO(n))\to H^0(Q,\OO_Q(n,n))\) surjective for all \(n\). For each \(n\ge0\), twisting the divisor sequence gives \[ 0\to\OO_Q(n-a,n-b)\to\OO_Q(n,n)\to\OO_Y(n)\to0. \] Part (a)(i) makes \(H^1(Q,\OO_Q(n,n))=0\). Its long exact sequence thus identifies the cokernel of restriction from \(Q\) to \(Y\) with \(H^1(Q,\OO_Q(n-a,n-b))\). The first restriction, from projective space to \(Q\), is already surjective. Hence restriction from projective space to \(Y\) is surjective in every degree \(n\ge0\) exactly when all these \(H^1\) groups vanish. The curve is integral and normal by the irreducible nonsingular hypothesis, so this is precisely projective normality by P-AGH2514PROJNORM, part (d).
If \(\abs{a-b}\le1\), part (a)(i) gives all the required vanishings. If, for example, \(a\ge b+2\), take \(n=b>0\). Then \((n-a,n-b)=(b-a,0)\) with \(b-a\le-2\), and part (a)(iii) makes the group nonzero. Interchanging the factors treats \(b\ge a+2\). This proves both implications in (b)(iii).
6. For every divisor in (c),\[ \boxed{P_Y(n)=(a+b)(n+1)-ab,\qquad p_a(Y)=ab-a-b+1.} \]
Twist the effective Cartier divisor sequence by \(\OO_Q(n,n)\) and use Euler additivity. Step 1 gives, for every integer \(n\), \[ \chi(\OO_Y(n))=(n+1)^2-(n-a+1)(n-b+1) =(a+b)(n+1)-ab. \] By P-AGH352HILBPOLY, this is the Hilbert polynomial for the induced embedding of the curve in \(\PP^3\). The curve has dimension one, so \(p_a(Y)=1-\chi(\OO_Y)=ab-a-b+1\). The same computation gives \(\deg Y=a+b\) from the leading coefficient. For \(q\) disjoint ruling lines it gives \(P_Y(n)=q(n+1)\) and \(p_a(Y)=1-q\), agreeing with the hint’s Euler calculation. This proof does not require \(Y\) to be integral or nonsingular.
7. A nonsingular irreducible curve of type \((1,3)\) over an algebraically closed field is a rational quartic and is not projectively normal.
Step 6 gives degree four. Consider its projection to the second projective line. A fibre of this projection on \(Q\) is a line on which \(\OO_Q(1,3)\) has degree one. No such line is a component of this integral curve of positive type in both coordinates, so its equation restricts to a nonzero linear form on the fibre. Thus each fibre of the restricted projection is a zero-dimensional scheme of length one. The projection is proper and quasi-finite and hence finite [@Har10a, Exercise III.11.2]; it is nonconstant and has degree one. A finite birational morphism onto the normal curve \(\PP^1\) is an isomorphism, by integral closedness on its affine charts. Hence the curve is isomorphic to \(\PP^1\) and is rational. Step 5 excludes projective normality because the two type indices differ by two.
8. Q.E.D.
Steps <1>1–<1>2 prove (a), steps <1>3–<1>5 prove (b), step 6 proves (c), and step 7 verifies the stated rational-quartic example.