TITLE: blowing curves down
QUESTION [10 upvotes]: If I blow up $\mathbb{P}^2$ in $2$ points I get $3$ exceptional divisors - two are disjoint and one intersects both of them. What object do I get if I blow down this intersecting curve? It's a minimal surface with Picard number $2$ and no $(-1)$-curves, what is that? $\mathbb{P}^1\times\mathbb{P}^1$?

REPLY [2 votes]: Up to automorphisms, the map $\mathbb{P}^2\to \mathbb{P}^1\times \mathbb{P}^1$ is in fact given by $(x:y:z)\mapsto ((x:y)(x:z))$, one easily sees that it blows-up $(0:0:1)$ and $(0:1:0)$ and that it contracts the line of equation $x=0$ onto the point $((0:1),(0:1))$. The inverse is $((a:b),(c:d))\to (ac:bc:ad)$, and it blows-up $((0:1),(0:1))$ and contracts the two lines $a=0$ and $c=0$ (passing through the point $((0:1),(0:1))$) onto $(0:0:1)$ and $(0:1:0)$.
One funny other comment is that if you take the real sphere $S$ of dimension $2$ and take the stereographic projection, it gives a birational map $S\to \mathbb{R}^2$ which is not defined at the north pole (the point of the projection). Seeing $S$ in $\mathbb{P}^3$ as a a smooth quadric and putting $\mathbb{R}^2$ into the projective plane, we obtain a birational map $S\to \mathbb{P}^2$ which blows-up the point $q$ (north pole) and contract the two imaginary lines being the intersection of the quadric with the tangent plane at $q$.
Over the reals, $S$ is not isomorphic to $\mathbb{P}^1\times \mathbb{P}^1$ (look at the topology of the real part) but extending to $\mathbb{C}$ it is, and the stereographic projection is exactly the inverse of the map you wanted (up to change of coordinates).