Quadric surface as a $\mathbb{F}_n$ surface

Question

The minimal models for rational projective smooth surfaces are $\mathbb{P}^2$ or the surfaces $\mathbb{F}_n$ for $n\neq 1$, where $$\mathbb{F_n}=\mathbb{P}_{\mathbb{P}^1}(\mathcal{O}_{\mathbb{P}^1}\oplus\mathcal{O}_{\mathbb{P}^1}(n)).$$ The right member of the equality is the projective bundle associated to the rank 2 vector bundle $\mathcal{O}_{\mathbb{P}^1}\oplus\mathcal{O}_{\mathbb{P}^1}(n)$ on $\mathbb{P}^1$.

The smooth quadric $\mathit{Q}\subset\mathbb{P}^3$ is a rational minimal surface since it does not contain exceptional curves.

My question is: am i right if i say that $\mathit{Q}\cong\mathbb{F}_0$ (birationally isomorphic)?

Answer 1

You are more than right:
not only is the quadric $Q$ birationally isomorphic to $F_0$ but it is actually isomorphic to $F_0$.

Indeed $F_0$ is clearly isomorphic to $\mathbb{P}^1\times \mathbb{P}^1$ and it is a basic result in classical geometry that every smooth quadric in $\mathbb{P}^3$ is isomorphic to $\mathbb{P}^1\times \mathbb{P}^1$.

Answer 2

Yes you are right. See Example V.2.11.1 in Hartshorne: the ruled surface $X = C \times \mathbb{P}^1$ corresponds to the normalized locally free sheaf $\mathcal{E} = \mathcal{O}_C \oplus \mathcal{O}_C$ on $C$. In your case, take $C=\mathbb{P}^1$ so $X = Q$ and we get exactly what you have your question.