Reference and Question 7(a) of Problem Set 8
The main parts of Question 7(a) are:
Let $H$ be the line at infinity in $\mathbb{CP}^{2}$, and let $P$ and $Q$ be distinct points on $H$. Let $X$ be the blow up of $\mathbb{CP}^{2}$ at $P$ and $Q$; let $E_{1}$, $E_{2}$ be the exceptional divisors over $P$, $Q$ and let $L$ be the proper transform of $H$. What is the self-intersection number of $L$?
I know that the self-intersection number of $H$ and the exceptional divisors is $1$ and $-1$, respectively. And blowing up a point on $L$ reduces the intersection number by $1$. So, this would mean that $L$ now has the number $-1$? However, I do not know how one finds the self-intersection number of the proper transform of $L$? A non-expert explanation will be welcomed.
You are right, $L^2 = -1$.
Giving a non-expert explanation depends on what prior results you have available. Here is a reasonable guess. You may know that for a divisor $D$ "downstairs" with total transform $\tilde{D}$, the self-intersection number is preserved: $\tilde{D}^2 = D^2$. Here one intersection number is computed "upstairs" (in this case, on $X$) and the other is computed "downstairs" (in this case, on $\mathbb{P}^2$). Granting this, with $D=H$, we have $\tilde{D} = L + E_1 + E_2$. We know $H^2 = 1$. So $(L+E_1+E_2)^2 = 1$. Distributing, we get $$ L^2 + 2L.(E_1+E_2) + (E_1^2 + 2 E_1.E_2 + E_2^2) = 1. $$ We know $L.E_1 = L.E_2 = 1$ since the curves intersect transversely (normally). We know $E_1.E_2 = 0$ since the exceptional divisors are disjoint. And we know $E_1^2 = E_2^2 = -1$. So $$ L^2 + 2 \cdot 2 + (-1 + 0 -1) = 1 $$ which gives $L^2 = -1$.