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

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)?

• How do you define "the" smooth quadric? The first one I would think of is the image of the Segre embedding $\mathbb P^1\times\mathbb P^1\to \mathbb P^3$, so it's isomorphic to $\mathbb P^1\times\mathbb P^1 = \mathbb F_0$. – Andrew Feb 1 '14 at 19:55
• Yes it's exactly the image of Segre embedding. – idioteca Feb 1 '14 at 19:57

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 $\mathbb{P}^1\times \mathbb{P}^1$.
• Thank you. I needed to be sure about the birational isomorphism. Infact, $\mathbb{P}^1\times\mathbb{P}^1$ is also isomorphic to $\mathbb{P}^2$, but there is not a birational isomorphism. Is this correct? – idioteca Feb 2 '14 at 10:47
• Dear idioteca, $\mathbb{P}^1\times \mathbb{P}^1$ is not isomorphic to $\mathbb{P}^2$. Beware that every isomorphism is a birational isomorphism but the converse is false. All the surfaces involved in your question are isomorphic with each other, but only birationally isomorphic to $\mathbb P^2$ (and it is better not to mention $\mathbb P^2$ at all in your context!) – Georges Elencwajg Feb 2 '14 at 11:24
• But i have these two definitions of rational surface: 1- S is a rational surface if it is birationally equivalent to $\mathbb{P}^1\times\mathbb{P}^1$. 2- S is a rational surface if it is birationally equivalent to $\mathbb{P}^2$. They are the same right? – idioteca Feb 2 '14 at 14:49
• Yes!${}{}{}{}{}$ – Georges Elencwajg Feb 2 '14 at 17:47
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.