Let \(X\) be the Del Pezzo surface of degree 4 in \(\PP^4\) obtained by blowing up 5 points of \(\PP^2\) (4.7).
Show that \(X\) contains 16 lines.
Show that \(X\) is a complete intersection of two quadric hypersurfaces in \(\PP^4\) (the converse follows from (4.7.1)).
Solution
Let \[ \pi:X\longrightarrow\PP^2 \] be the blowup of the five points \(P_1,\ldots,P_5\) in general position. Write \(L\) for the pullback of a line and \(E_i\) for the exceptional curves. The anticanonical embedding of (4.7) is given by \[ H=-K_X=3L-E_1-\cdots-E_5, \qquad H^2=4. \]
1. The five exceptional curves \(E_i\) are lines in the anticanonicalembedding.
The intersection form on the blowup is \[ L^2=1, \qquad L\cdot E_i=0, \qquad E_i\cdot E_j=0\ (i\ne j), \qquad E_i^2=-1. \] Hence \[ H\cdot E_i=1. \] Each \(E_i\cong\PP^1\), and the very ample divisor \(H\) embeds it as a curve of degree one in \(\PP^4\). Thus every \(E_i\) is a line on \(X\).
2. For every pair \(1\le i<j\le5\), the strict transform of the line through\(P_i\) and \(P_j\) is a line on \(X\).
Because the five points are in general position, the line through \(P_i,P_j\) contains none of the other three points. Its strict transform \(L_{ij}\) has class \[ L-E_i-E_j. \] Therefore \[ H\cdot L_{ij} =(3L-E_1-\cdots-E_5)\cdot(L-E_i-E_j) =3-1-1 =1. \] The curve \(L_{ij}\) is isomorphic to \(\PP^1\), so the anticanonical embedding maps it to a line. There are \[ \binom52=10 \] such curves.
3. The strict transform of the conic through \(P_1,\ldots,P_5\) is a lineon \(X\).
Five points in general position determine a unique conic \(Q\). It is irreducible: if it were the union of two lines, one of those lines would contain at least three of the five points, contrary to general position. Thus \(Q\) is a nonsingular conic and its strict transform \(\widetilde Q\) has class \[ 2L-E_1-\cdots-E_5. \] Consequently \[ H\cdot\widetilde Q=6-5=1. \] Since \(\widetilde Q\cong\PP^1\), it is mapped to a line by the anticanonical embedding.
4. The surface \(X\) contains 16 distinct lines.
Steps <1>1–<1>3 give respectively \[ 5, \qquad 10, \qquad 1 \] lines. Their divisor classes are respectively \[ E_i, \qquad L-E_i-E_j, \qquad 2L-E_1-\cdots-E_5, \] so they are distinct. Hence \(X\) contains \[ \boxed{5+10+1=16} \] lines. This proves part (a).
5. One has\[ \boxed{h^0(X,\OO_X(2H))=13.} \]
The surface \(X\) is rational, so \[ \chi(\OO_X)=1. \] Riemann–Roch on a nonsingular projective surface gives, for \(D=2H=-2K_X\), \[ \begin{aligned} \chi(\OO_X(2H)) &=\chi(\OO_X)+\frac12(2H)\cdot(2H-K_X)\\ &=1+\frac12(2H)\cdot3H\\ &=1+3H^2\\ &=13. \end{aligned} \]
By Serre duality, \[ H^2(X,\OO_X(2H)) \cong H^0(X,\OO_X(K_X-2H))^* =H^0(X,\OO_X(-3H))^* =0, \] because \(H\) is ample. Also \[ H^1(X,\OO_X(2H)) \cong H^1(X,\OO_X(-3H))^* =0 \] by Exercise V.4.12, applied to the ample divisor \(3H\). Thus \(h^0(X,\OO_X(2H))=\chi(\OO_X(2H))=13\).
6. At least two linearly independent quadric hypersurfaces of \(\PP^4\)contain \(X\).
The anticanonical embedding satisfies \[ \OO_X(1)=\OO_X(H), \] so restriction of quadratic forms gives a linear map \[ H^0(\PP^4,\OO_{\PP^4}(2)) \longrightarrow H^0(X,\OO_X(2H)). \] The source has dimension \[ \binom{4+2}{2}=15, \] whereas step 5 gives dimension \(13\) for the target. Hence the kernel has dimension at least two. Choose linearly independent quadrics \[ Q_1,Q_2 \] from this kernel. Then \[ X\subseteq Q_1\cap Q_2. \]
7. The quadrics \(Q_1\) and \(Q_2\) have no common hypersurface component.
If two independent quadrics in \(\PP^4\) have a common hypersurface component, that component is a hyperplane: their equations have a common linear factor. Thus their intersection is the union of that hyperplane and a codimension-two linear space. Since the irreducible surface \(X\) is contained in this union, it would be contained in one of those two linear subspaces. This is impossible: the complete anticanonical linear system embeds \(X\) nondegenerately in \(\PP^4\). Hence \(Q_1,Q_2\) form a regular sequence and their intersection is a pure surface complete intersection.
8. One has\[ \boxed{X=Q_1\cap Q_2} \]scheme-theoretically.
By step 7, the complete intersection \[ Y=Q_1\cap Q_2 \] is a pure two-dimensional scheme of degree \[ \deg Y=2\cdot2=4. \] On the other hand, the degree of the anticanonically embedded surface is \[ \deg X=H^2=4. \] Step 6 gives \(X\subseteq Y\). Since \(Y\) is pure of the same dimension as the irreducible reduced surface \(X\), the component of \(Y\) supported on \(X\) already contributes at least \(\deg X=4\) to the degree of \(Y\). Equality of the degrees leaves neither another two-dimensional component nor a nonreduced multiplicity along \(X\). Thus \(Y=X\) as schemes.
Therefore \(X\) is the complete intersection of two quadric hypersurfaces in \(\PP^4\), proving part (b).
9. Q.E.D.
Step 4 proves part (a), and steps <1>5–<1>8 prove part (b).