projection of a quadric surface Consider the quadric surface $X = \{ xy = zw \} \subset \mathbb{P}^3$ and pick a point $x \in X$. I think it is true that if we think of $\mathbb{P}^2$ as the space of lines through $x$ in $\mathbb{P}^3$, then the morphism $X \setminus \{ x \} \to \mathbb{P}^2$ which sends $y \mapsto \overline{xy}$ represents a birational map $X \to \mathbb{P}^2$. But I do not understand the geometry of $X$ well enough to prove this. Certainly this morphism fails to be injective along the two obvious lines in $X$ through $x$, but how do I see that the map is an isomorphism elsewhere? I would like to avoid computing in coordinates if at all possible.
 A: The basic idea is that the "inverse map" is given by sending $\ell \in \mathbb{P}^2$ (identify the points in $\mathbb{P}^2$ with lines through $x$) to the point $y$ where $X \cap \ell = \{x, y\}$.  (We are using that $X$ has degree 2, which means that it intersects a general line in 2 points.)  It's pretty clear that this map is inverse to the one you described, wherever things are well-defined.  It also should be clear that the sets of points in $X$ and $\mathbb{P}^2$ where the maps are not well-defined are proper closed subsets.  Lastly, you need to check that these are actually morphisms where they are defined, and I'm afraid this step, by definition, requires a certain amount of coordinate computation to verify.
A side remark to help you understand the geometry of $X$ is that you can view it as the isomorphic image of $\mathbb{P}^1 \times \mathbb{P}^1$ under the Segre map $((a:b),(c:d))\mapsto (ac:bd:ad:bc)$ to $\mathbb{P}^3$.  I chose a weird ordering for the products of the variables so that the equation $xy = zw$ would be satisfied, assuming you order your coordinates on $\mathbb{P}^3$ "alphabetically" as $(x:y:z:w)$.  In particular, this shows that through every point of $X$ are exactly 2 lines in $\mathbb{P}^3$ contained in $X$.  Indeed, the two copies of $\mathbb{P}^1$ give two separate rulings on $X$.  Shafarevich's book has a nice discussion of this surface; you might also look at Igor Dolgachev's notes on Classical Algebraic Geometry.
A: As a concrete example of this problem, consider the projection of $X$ through the point $[0:0:0:1]$ onto the plane $w=0$. Show this is a birational map (consider the Segre embedding above, where is it not defined?) and agrees with your construction above. I know this uses coordinates, but it is actually not too messy.
Edit:
Let me elaborate a little bit. If you consider the affine patch where $w=1$ we get the equation $z=xy.$ The point mentioned above maps to $(0,0,0).$ This only contains the two families of lines mentioned in the previous post, so away from these two lines at the origin (namely the $x,y$ axes), the map to $\mathbb{P}^2$ is an isomorpism. Namely, send a point in $X$ to the the approriate line, and send a line to the associate intersection with $X.$
This picture helped me:
http://bit.ly/KHkhme
A: Actually, the rational projection $p:X\dashrightarrow \mathbb P^2$ factors through the diagram
$\require{AMScd}$
\begin{CD}
\text{Bl}_xX @>{\phi}>> \mathbb P^2\\
@V{\sigma}VV \\
X 
\end{CD}
where $\sigma$ blows up the quadric surface at $p$, and $\phi$ is the projection map.
Since every line $l\subseteq\mathbb P^3$ passing $x$ either intersect $X$ at two distinct points (where the projection $p$ is regular), or $l$ has higher order of contact with $X$ (where the projection $p$ is not defined), which means $l$ is contained in the tangent space $T_xX$ due to $\text{deg}(X)=2$, so $l$ intersects $X$ only at $x$ or $l\subseteq X$ belongs to one of the two rulings.
When blowup at $x$, the exceptional divisor distributes each point to each direction in $\mathbb P(T_xX)$ so the projection $p$ factors through $\phi$ which becomes a morphism. What $\phi$ does is to contract the two lines $l_1, l_2$ on the tangent plane $T_pX$ to the two points $u,v\in \mathbb P^2$. In other words,  $\phi$ identifies $\text{Bl}_xX$ with the blowup $\text{Bl}_{u,v}\mathbb P^2$, i.e., a del Pezzo surface of degree 7.
