If \(H\) is an ample divisor on the surface \(X\), and if \(D\) is any divisor, show that \[ \left(D^2\right)\left(H^2\right) \leqslant(D . H)^2 . \]
Now let \(X\) be a product of two curves \(X=C \times C^{\prime}\). Let \(l=C \times \mathrm{pt}\), and \(m=\mathrm{pt} \times C^{\prime}\). For any divisor \(D\) on \(X\), let \(a=D . l\), \(b=D . m\). Then we say \(D\) has type \((a, b)\). If \(D\) has type \((a, b)\), with \(a, b \in \ZZ\), show that \[ D^2 \leqslant 2 a b , \] and equality holds if and only if \(D \equiv b l+a m\).
Hint: Show that \(H=l+m\) is ample, let \(E=l-m\), let \(D^{\prime}=\left(H^2\right)\left(E^2\right) D-\left(E^2\right)(D \cdot H) H-\left(H^2\right)(D \cdot E) E\), and apply (1.9). This inequality is due to Castelnuovo and Severi. See Grothendieck [2].
Solution
1. If \(H\) is ample, then \(H^2>0\).
This is one of the two numerical conditions in the Nakai–Moishezon criterion T-SRFNAKAI.
2. For any divisor \(D\) on \(X\), define\[ F=(H^2)D-(D\cdot H)H. \]Then \(F\cdot H=0\) and \(F^2\leq0\).
Bilinearity gives \[ F\cdot H =(H^2)(D\cdot H)-(D\cdot H)H^2 =0. \] If \(F\not\equiv0\), the Hodge index theorem T-SRFHODGE gives \(F^2<0\). If \(F\equiv0\), then \(F^2=0\). Thus in all cases \(F^2\leq0\).
3. For every divisor \(D\) and ample divisor \(H\),\[ \boxed{(D^2)(H^2)\leq(D\cdot H)^2}. \]
Expanding the square in step 2 gives \[ \begin{aligned} F^2 &=(H^2)^2D^2 -2(H^2)(D\cdot H)^2 +(D\cdot H)^2H^2\\ &=H^2\qty((H^2)D^2-(D\cdot H)^2). \end{aligned} \] By steps <1>1–<1>2, the left side is nonpositive and \(H^2\) is positive. Dividing by \(H^2\) yields the stated inequality. This proves part (a).
4. On \(C\times C'\), the ruling classes satisfy\[ l^2=m^2=0, \qquad l\cdot m=1, \]and\[ H=l+m \]is ample.
Distinct fibres of either projection are disjoint and numerically equivalent, so each fibre has self-intersection zero. A fibre of the first projection and a fibre of the second meet transversally in one point, hence \(l\cdot m=1\). Thus \[ H^2=(l+m)^2=2>0. \]
Let \(\Gamma\subseteq C\times C'\) be an irreducible curve. The two projections cannot both be constant on \(\Gamma\). Their degrees on the normalization of \(\Gamma\) are nonnegative, and at least one is positive. These degrees are respectively the intersection numbers \(m\cdot\Gamma\) and \(l\cdot\Gamma\). Therefore \[ H\cdot\Gamma=l\cdot\Gamma+m\cdot\Gamma>0. \] The Nakai–Moishezon criterion T-SRFNAKAI now shows that \(H\) is ample.
5. If \(D\) has type \((a,b)\), put\[ F=D-bl-am. \]Then\[ F\cdot l=F\cdot m=F\cdot H=0 \]and\[ F^2=D^2-2ab. \]
By definition, \[ D\cdot l=a, \qquad D\cdot m=b. \] Using step 4, \[ \begin{aligned} F\cdot l &=a-b(l^2)-a(m\cdot l)=0,\\ F\cdot m &=b-b(l\cdot m)-a(m^2)=0. \end{aligned} \] Hence \(F\cdot H=0\). Also \[ \begin{aligned} F^2 &=D^2-2D\cdot(bl+am)+(bl+am)^2\\ &=D^2-2(ba+ab)+2ab\\ &=D^2-2ab. \end{aligned} \]
6. Every divisor \(D\) of type \((a,b)\) satisfies\[ \boxed{D^2\leq2ab}. \]
By steps <1>4–<1>5, \(H\) is ample and \(F\cdot H=0\). The Hodge index theorem therefore gives \(F^2\leq0\). Step 5 identifies this inequality with \[ D^2-2ab\leq0. \]
7. Equality holds precisely when\[ \boxed{D\equiv bl+am}. \]
If \(D^2=2ab\), then step 5 gives \(F^2=0\). Since \(F\cdot H=0\) and \(H\) is ample, the strict form of the Hodge index theorem T-SRFHODGE forces \(F\equiv0\). Thus \[ D\equiv bl+am. \]
Conversely, if \(D\equiv bl+am\), then numerical equivalence preserves self-intersection and step 4 gives \[ D^2=(bl+am)^2=2ab. \]
8. Q.E.D.
Steps <1>1–<1>3 prove part (a), and steps <1>4–<1>7 prove part (b), including the equality criterion.