Consider smooth hypersurfaces in $\mathbb{P}^1\times\mathbb{P}^2$ of bidegree $(n,1)$. I managed to show explicitly that one such surface $S$ given by an equation $x^nu+y^nv=0$, where $[x:y],\ [u:v:w]$ are the homogeneous coordinates on $\mathbb{P}^1$ and $\mathbb{P}^2$ respectively, is isomorphic to the Hirzebruch surface $F_n.$ At the same time, I believe that all such surfaces are actually isomorphic to $F_n$, since they are all $\mathbb{P}^1$ bundles over $\mathbb{P}^1$ and hence are projectivizations of some $\mathcal{O}\oplus\mathcal{O}(k)$, and then $k$ must necessarily equal $k$. For a proof, see the accepted answer to this mathoverflow question: On a Hirzebruch surface.
But when I try to think of this in terms of coordinates and equations, I can't believe this is the case. For the same reason curves of high degree in $P^2$ are generally non-isomorphic.
So consider an arbitrary smooth hypersurface of bidegree $(n,1)$ given by some equation $F_u(x,y)u+F_v(x,y)v+F_w(x,y)w=0$. Is there a way to explicitly construct an isomorphism between this surface and $S$? Or are these surfaces isomorphic, but not projectively isomorphic? Or am I missing something else entirely?
Best Answer
It is not true that any such surface is isomorphic to $F_n$; in fact you can get any $F_{n-2k}$, $0 \le k \le n/2$, in this way. Indeed, as the answer to On a Hirzebruch surface shows that the surface is isomorphic to the projectivization of the rank 2 bundle $E$ on $\mathbb{P}^1$ defined by the exact sequence $$ 0 \to \mathcal{O}(-n) \stackrel{(F_u,F_v,F_w)}\longrightarrow \mathcal{O} \oplus \mathcal{O} \oplus \mathcal{O} \longrightarrow E \to 0 $$ and one can find the polynomials $(F_u,F_v,F_w)$ to get any $E \cong \mathcal{O}(k) \oplus \mathcal{O}(n-k)$: one option is to take $$ F_u = P(x,y)x^k,\qquad F_v = P(x,y)y^k,\qquad F_w = Q(x,y), $$ where $\gcd(P,Q) = 1$ and $P$ and $Q$ are polynomials of degree $n - k$ and $n$ not divisible by $x$ or $y$.
More invariantly, note that the bundle $E \cong \mathcal{O}(k) \oplus \mathcal{O}(n-k)$ is globally generated and by Bertini-type theorem a general triple of its section generates it, and the kernel of the corresponding epimorphism is isomorphic to $\mathcal{O}(-n)$.