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.
[Math] projection of a quadric surface
algebraic-geometryprojective-geometrysurfaces
Related Solutions
Any non-constant morphism $\Phi\colon X\to Y$ between projective smooth curves has a degree $d$ and the morphism $\Phi$ will be an isomorphism if and only if that degree is $1$.
The incredibly good news is that you can calculate that degree by just looking at the fibre $\Phi^{-1}(y)$ of $\Phi$ at just one point ( any point!) of $y\in Y$: the degree is the dimension $d=dim_k \Gamma (\Phi^{-1}(y),\mathcal O)$ .
It is not terribly difficult to explain what the right-hand side means, but this is not even necessary here: we have $d=1$ as soon as for some non-empty $V\subset Y$ the restricted morphism $\Phi^{-1}(V)\to V$ is an isomorphism.
Since this is true in your case, we are done.
Edit
A more elementary proof (maybe the one Hartshorne had in mind at the level of Chapter 1, before "degree" is introduced) would be to consider the inverse isomorphism $\psi:V\to U$, to complete it to a morphism $\bar \psi:\mathbb P^1\to \mathbb P^1 $ (just as you did for $\phi$) and realize that $\bar \psi $ is an inverse to $\bar \phi $, which is thus an isomorphism.
Any line in $\mathbb{P}^{3}$ is an intersection of two planes. In order to understand what lines are inside the quadric $xw=yz$, it is enough to understand when the hyperplane sections of $xw=yz$ in $\mathbb{P}^3$ contain a line.
Let $ax+by+cz+dw=0$ be a plane in $\mathbb{P}^3$. By Bezout's theorem, the intersection of $ax+by+cz+dw=0$ with $xw=yz$ is a curve of degree 2. If this curve is irreducible, then there is no hope of finding a line there! If this curve is a union of two lines, then we have successfully found a pair of lines.
So we need to understand what conditions need to be imposed on the coefficients $[a, b, c, d]\in\mathbb{P}^3$ such that the plane $ax+by+cz+dw=0$ intersects $xw=yz$ in a pair of lines.
Claim. $ax+by+cz+dw=0$ intersects $xw=yz$ in a pair of lines if and only if $ad=bc$.
Proof. $(\Leftarrow)$ Multiply $ax+by+cz+dw=0$ by $w$ and use $xw=yz$ to get $ayz+byw+czw+dw^2=0$. This is a plane curve in $\mathbb{P}^{2}$ (with coordinates $y, z, w$), unsurprisingly. Now, multiply both sides by $b$ to get $$ abyz+b^2yw+bczw+bdw^2=0 $$ Use the hypothesis $ad=bc$ to get $$ abyz+b^2yw + adzw + bdw^2 = 0 $$ which conveniently factors as $$ (az+bw)(by+dw)=0 $$ so we get a pair of lines.
$(\Rightarrow)$ I will leave this as an exercise. Try to factor the quadric equation, and show that such a factorization forces $ad=bc$. $\square$
So now we can answer the question "What are all the lines on the quadric surface $xw=yz$?" Well, the proof of the claim shows that the lines on $xw=yz$ are of the form $az+bw=0$ and $by+dw=0$ (viewed in $\mathbb{P}^2$ with coordinates $y,z,w$) such that $ad=bc$. Here $a, b, c, d$ come from the hyperplane $ax+by+cz+dw=0$. Now once you can fix any $[a, b]\in\mathbb{P}^{1}$, you get the line $az+bw=0$. And if you fix $[b, d]\in\mathbb{P}^{1}$, you get the line $by+dw=0$. I think these two families of lines are the desired rulings.
I am very interested in seeing a more concise and conceptual answer!
Best Answer
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.