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Fix an algebraically closed field $k$ and a positive integer $d$. My question is, what is the number of birational classes of dimension $d$, projective varieties over $k$ with geometric genus 0? If it makes the question answerable over fields where we don't know resolution of singularities in higher dimensions, what if we impose smooth?

Obviously for $d=1$ there is a unique birational class. If $d=2$ I know for example that $\mathbb{P}^2$ and $\mathbb{P}^1\times\mathbb{P}^1$ are not isomorphic, but they are of course birational. Are there explicit examples of two non-birational geometric genus 0 surfaces? Is there some nice way of enumerating the birational classes?

Thanks

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1 Answer 1

(Let me give an answer to this old question to remove it from the pile.)

Any ruled surface has geometric genus $0$. On the other hand, if $C$ and $C'$ are curves of genus $\ge 1$, a birational map between surfaces $\mathbf{P}^1 \times C$ and $\mathbf{P}^1 \times C'$ induces finite morphisms $C \rightarrow C'$ and $C' \rightarrow C$. But Riemann--Hurwitz shows that's impossible unless $C$ and $C'$ are isogenous elliptic curves.

So one way to say it is that, for a fixed value $g>1$ of the irregularity $H^1(X,\mathcal{O}_X)$, there are "as many" birational classes of surfaces of geometric genus $0$ and irregularity $g$ as there are curves of genus $g$.

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" ..., a birational map between surfaces $\mathbf{P}^1 \times C$ and $\mathbf{P}^1 \times C'$ induces finite morphisms $C \rightarrow C'$ and $C' \rightarrow C$." Why is that? –  Georges Elencwajg Nov 22 '13 at 7:37
    
Dear Georges, fix a section $C\to C\times \mathbf P^1$, and a birational isomorphism from $C\times \mathbf P^1 \to D\times \mathbf P^1$. Now, start with the image of $C$ in $C\times \mathbf P^1$ and map this to the product $D\times \mathbf P^1$... –  Ari Nov 22 '13 at 8:42
    
..Assuming $g(C)$ and $g(D)$ are positive, the fibres of the first ruled surface map to the fibres of the second ruled surface and the image of $C$ is not contained in a fibre of $D\times \mathbf P^1\to D$. This implies that the composition $C\to C\times \mathbf P^1\to D\times \mathbf P^1\to D$ is non-constant. Let us show that the degree of this morphism is one. Then we are done. In fact, the degree of this morphism equals the intersection product of $C$ with a (general) fibre of $D\times \mathbf P^1$... –  Ari Nov 22 '13 at 8:43
    
...The latter equals (by the birational isomorphism above) the intersection product of $C$ with a fibre of $C\times \mathbf P^1$. Thus, it equals one. –  Ari Nov 22 '13 at 8:43
    
Dear @GeorgesElencwajg: Ari has answered on my behalf, and improved my argument! –  Asal Beag Dubh Nov 22 '13 at 23:51

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