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)?

  • $\begingroup$ 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 $\mathbb P^1\times\mathbb P^1 = \mathbb F_0$. $\endgroup$ – Andrew Feb 1 '14 at 19:55
  • $\begingroup$ Yes it's exactly the image of Segre embedding. $\endgroup$ – idioteca Feb 1 '14 at 19:57

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 $\mathbb{P}^1\times \mathbb{P}^1$.

  • $\begingroup$ 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? $\endgroup$ – idioteca Feb 2 '14 at 10:47
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    $\begingroup$ 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!) $\endgroup$ – Georges Elencwajg Feb 2 '14 at 11:24
  • $\begingroup$ 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? $\endgroup$ – idioteca Feb 2 '14 at 14:49
  • $\begingroup$ I took the definition from Beauville's book Complex algebraic surfaces, the first one at page 25 when he starts speaking about ruled surfaces, the second one at page 40 when he starts speaking about rational surfaces. $\endgroup$ – idioteca Feb 2 '14 at 16:05
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    $\begingroup$ Yes!${}{}{}{}{}$ $\endgroup$ – 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.


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