If \(C\) is a curve of genus \(g\), show that the diagonal \(\Delta \subseteq C \times C\) has self-intersection \(\Delta^2=2-2 g\). (Use the definition of \(\Omega_{C / k}\) in (II, §8).)
Let \(l=C \times \mathrm{pt}\) and \(m=\mathrm{pt} \times C\). If \(g \geqslant 1\), show that \(l, m\), and \(\Delta\) are linearly independent in \(\Num(C \times C)\). Thus \(\Num(C \times C)\) has rank \(\geqslant 3\), and in particular, \[ \Pic(C \times C) \neq p_1^* \Pic C \oplus p_2^* \Pic C .\] Cf. (III, Ex. 12.6), (V, Ex. 4.10).
Solution
Let \(X=C\times C\), let \(p_1,p_2:X\to C\) be the two projections, and let \(\delta:C\hookrightarrow X\) be the diagonal embedding with image \(\Delta\).
1. The conormal sheaf of the diagonal is canonically\[ \mci_\Delta/\mci_\Delta^2\cong\Omega_{C/k}. \]
For an affine open \(U=\Spec B\subseteq C\), the diagonal in \(U\times U=\Spec(B\otimes_kB)\) has ideal \[ J=\ker\qty{B\otimes_kB\longrightarrow B}, \] where the map is multiplication. The standard description of Kähler differentials identifies \[ J/J^2\cong\Omega_{B/k}, \qquad 1\otimes b-b\otimes1\longmapsto db. \] These affine identifications are natural under localization, so they glue to the displayed isomorphism on \(C\); see D-4GCH6.
2. The normal bundle of the diagonal is\[ \mcn_{\Delta/X}\cong\qty{\Omega_{C/k}}^\vee, \]and hence\[ \boxed{\Delta^2=2-2g}. \]
By definition of the normal sheaf D-MODCONORM, step 1 gives \[ \mcn_{\Delta/X} = \sheafhom_{\OO_C}(\mci_\Delta/\mci_\Delta^2,\OO_C) \cong \qty{\Omega_{C/k}}^\vee. \] Since \(C\) is nonsingular, \(\Omega_{C/k}\) is the canonical line bundle and has degree \(2g-2\). Therefore \[ \deg\mcn_{\Delta/X}=-(2g-2)=2-2g. \] For a smooth curve on a smooth surface, its self-intersection is the degree of its normal bundle, equivalently \[ \Delta^2 = \deg\OO_X(\Delta)|_\Delta = \deg\mcn_{\Delta/X}. \] This proves part (a).
3. With\[ l=C\times\{Q\}, \qquad m=\{P\}\times C, \]for fixed points \(P,Q\in C\), the intersection numbers are\[ l^2=m^2=0, \qquad l\cdot m=l\cdot\Delta=m\cdot\Delta=1. \]
Each of \(l\) and \(m\) is a fibre of one of the two projections. Distinct fibres of the same projection are disjoint, and each fibre has trivial normal bundle, so \[ l^2=m^2=0. \] The curves \(l\) and \(m\) meet transversely at the single point \((P,Q)\), hence \(l\cdot m=1\). The diagonal meets \(l\) at \((Q,Q)\) and \(m\) at \((P,P)\), again transversely in one point, so \[ l\cdot\Delta=m\cdot\Delta=1. \]
4. If \(g\geq1\), the classes of \(l,m,\Delta\) are linearly independent in\(\Num(C\times C)\).
Suppose \[ a[l]+b[m]+c[\Delta]=0 \] in \(\Num(C\times C)\) for integers \(a,b,c\). Intersecting with \(l\), then with \(m\), and then with \(\Delta\), and using steps <1>2–<1>3, gives \[ b+c=0, \qquad a+c=0, \qquad a+b+c(2-2g)=0. \] The first two equations give \[ a=b=-c. \] Substitution into the third gives \[ -2gc=0. \] Since \(g\geq1\), it follows that \(c=0\), and hence \(a=b=0\). Thus the three numerical classes are linearly independent.
5. Consequently\[ \boxed{\operatorname{rank}\Num(C\times C)\geq3} \]and\[ \boxed{\Pic(C\times C)\neq p_1^*\Pic C\oplus p_2^*\Pic C}. \]
The rank bound follows immediately from step 4.
For any divisor \(D\) on \(C\), the numerical class of \(p_1^*D\) is \((\deg D)[m]\), and the numerical class of \(p_2^*D\) is \((\deg D)[l]\). Hence the image in \(\Num(C\times C)\) of \[ p_1^*\Pic C+p_2^*\Pic C \] is contained in the subgroup generated by \([l]\) and \([m]\). Step 4 shows that \([\Delta]\) is not in that subgroup. Therefore the divisor class \(\OO_X(\Delta)\) cannot lie in \(p_1^*\Pic C+p_2^*\Pic C\), so the asserted equality of Picard groups is impossible.
6. Q.E.D.
Steps <1>1–<1>2 prove part (a), and steps <1>3–<1>5 prove every assertion in part (b).