Hartshorne Exercise V.1.5
(a) If $C$ is a curve of genus $g$, show that the diagonal $\Delta\subseteq C\times C$ has self intersection $\Delta^2=2-2g.$
(b) Let $l=C\times\text{pt}$ and $m=\text{pt}\times C.$ If $g\ge 1$ show that $l,m,$ and $\Delta$ are linearly independent in $\op{Num}(C\times C)$. Thus $\op{Num}(C\times C)$ has rank $\ge 3$, and in particular, $\op{Pic}(C\times C)\ne p_1^*\op{Pic} C\oplus p_2^*\op{Pic} C.$ Cf. (III, Ex. 12.6), (IV, Ex. 4.10).
(a) Let $C$ be a curve of genus $g$, and let $X:=C\times _k C$. Let $\delta:C\ra X$ be the diagonal closed immersion, and denote $\Delta:=\delta (C)\cong C$. By definition,
$$\Omega_{C/k}=\mathcal{N}^\vee_{\Delta/X}=\mathcal{I}_\Delta/\mathcal{I}_\Delta^2=\OO(-\Delta)|_{\Delta}.$$
Taking first the dual of the above equalities and then the degree yields,
$$2-2g=-\deg (K_C)=\Delta^2.$$
(b) Recall that,
$$\op{Num}X=\op{Pic}X/\op{Pic}^nX.$$
Let $g\ge1$, but suppose for a contradiction that some nontrivial $\Z$-linear combination of $l=C\times{\text{pt}}$, $m={\text{pt}}\times C$, and $\Delta$, $al+bm+c\Delta$ with $a,b,c\in\Z$ were numerically equivalent to $0$. Then in particular,
$$(al+bm+c\Delta).\Delta=a+b+c(2-2g)=0\implies g=\frac{a+b+2c}{2}.$$
Let $r={p}\times C$ and $s=C\times {p}$ with $p\ne \text{pt}$. Then
$$(al+bm+c\Delta).r=a+c=0,\:\text{and}$$
$$(al+bm+c\Delta).s=b+c=0.$$
It follows that $g=0$, which is a contradiction. Hence, $\op{rank}\op{Num}X\ge 3$. The problems also asks us to conclude that $\op{Pic}(C\times C)\ne p_1^*\op{Pic}C\times p_2^*\op{Pic}C$ from exercises III.12.6 and IV.4.10. After reading these exercises, its clear to me that this is an immediate corollary. However I have not done these exercises, and have not even read the last several chapters on cohomology yet. Once I feel ready, I will attempt these exercises.
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