I am trying to understand the example mentioned in this question, but I can't follow the argument Shafarevich is using, so I want to compute an explicit example to understand what's going on.
Let $X$ be the surface $x_0^m-x_1^m+x_2^m-x_3^m$ and $L$ the line $x_0=x_1,x_2=x_3$ on $X$. I think $E$ would then be the intersection of $X$ with the plane $x_0-x_1+x_2-x_3=0$, but I don't know what $C$ would be. If anyone can help me finish working through this example, or point me in the direction of an easier solution (either in general or for a specific value of $m$) that would be greatly appreciated.
Here is the solution I ended up with (with $m=2$):
Let $H_1=Z(x_0-x_3), H_2=Z(x_0-x_1)$ be two planes in $\mathbb{P}^3$. Then $X\cap H_1=M_0\cup M_1\cup M_2$ and $X\cap H_2= L_0\cup L_1\cup L_2$ where $M_i=Z(x_0-x_3, x_1-\alpha^i x_2), L_i=Z(x_0-x_1, x_2-\alpha^i x_3)$ and $\alpha$ is a primitive 3rd root of unity. Note that $L=L_0$. As $L_0\cap M_0=\{(1:1:1:1)\}$ and $L_0\cap M_1, L_0\cap M_2=\varnothing$, we have $L_0\cdot(M_0+M_1+M_2)=1$. We also have that $L_0\cap L_1, L_0\cap L_2=\{(1:1:0:0)\}$ so $L_0\cdot L_1 = L_0\cdot L_2 = 1$. Finally, as $$\text{div }\frac{x_0-x_3}{x_0-x_1}=M_0+M_1+M_2-(L_0+L_1+L_2),$$ i.e. $M_0+M_1+M_2\sim L_0+L_1+L_2$, we conclude that $$1 = L_0\cdot(M_0+M_1+M_2) = L_0\cdot(L_0+L_1+L_2)=L_0^2+2,$$ and therefore $L^2=L_0^2=-1$.