The quadric surface $xy = zw$ is birational but not isomorphic to $\PP^2$

On \(D_+(w)\) normalize \(w=1\). The equation of \(Q\) becomes \[ z=xy. \] Hence every point of \(U\) has a unique form \[ [x:y:xy:1]. \] The maps \[ \AA^2\longrightarrow U, \qquad (a,b)\longmapsto[a:b:ab:1], \] and \[ U\longrightarrow\AA^2, \qquad [x:y:z:w]\longmapsto\left(\frac{x}{w},\frac{y}{w}\right) \] are morphisms and are inverse to one another. Thus \(U\cong\AA^2\).

The quadric \(Q\) is irreducible by P-AGH215QUADRIC, so its nonempty open subset \(U\) is dense. The standard open subset \[ V=D_+(X_2)\subseteq\PP^2 \] is likewise dense and isomorphic to \(\AA^2\) via \[ [X_0:X_1:X_2]\longmapsto \left(\frac{X_0}{X_2},\frac{X_1}{X_2}\right). \] Composing the two affine-chart isomorphisms gives an isomorphism \[ U\xrightarrow{\sim}V. \] An isomorphism between dense open subsets is a birational equivalence, so \[ Q\dashrightarrow\PP^2 \] is birational.

By P-AGH215QUADRIC, one ruling consists of the pairwise disjoint lines \[ L_{[a:b]} = \{[ac:bd:ad:bc]:[c:d]\in\PP^1\}. \] In particular, \[ L_{[1:0]}\cap L_{[0:1]}=\varnothing. \] Each is a closed irreducible curve on \(Q\).

Suppose an isomorphism \[ \alpha:Q\xrightarrow{\sim}\PP^2 \] existed. An isomorphism preserves closedness, irreducibility, intersections, and dimension. Therefore \[ \alpha(L_{[1:0]}) \quad\text{and}\quad \alpha(L_{[0:1]}) \] would be two disjoint curves in \(\PP^2\). But P-AGH37HYPMEETS proves that any two curves in \(\PP^2\) have nonempty intersection. This contradiction shows that no such isomorphism exists.

Step 2 proves birationality, and step 4 proves nonisomorphism.