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If there exists some birational equivalence between ruled surfaces $C\times\mathbb{P}^1$ and $C'\times\mathbb{P}^1$ over a sufficiently nice field, then one can cancel the projective line and conclude that the curves $C$ and $C'$ are isomorphic. I already know how to prove this, but I wanted to attack this using the function fields, which should identify the curves.

Is there an purely algebraic argumentation that shows, that if the function fields of these surfaces are isomorphic, then so are the function fields of the curves?

More algebraically, given finitely generated extension fields $K$, $K'$ of transcendence degree 1 over an algebraically closed field $k$, such that their purely transcendental extensions $K(t)$ and $K'(t)$ of degree one are isomorphic, how do we conclude that $K$ and $K'$ are isomorphic (without passing to their curves/surfaces over $k$ and applying geometric tools)?

Addendum. I should explain my intention. First of all, the geometric statement induces an algebraical (more precisely field theoretical) statement, I wouldn't believe if there weren't this geometric interpretation (and the proof, of course). I expect that an algebraical proof (if known) is hard, but it would be nice to know if there was at least an algebraical reason why this should be true.

Ben
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  • " I already know how to prove this" Could you please briefly explain or give a reference ? – Georges Elencwajg Nov 21 '13 at 17:48
  • Dear @Georges, I'll add a reference soon. Here is a (kind of overkill, but still nice) proof. Let's assume $C$ and $C'$ are nonsingular and the field is algebraically closed of zero characteristic. The Albanese variety to a curve is a birational invariant, it preserves products, and the Albanese of $\mathbb{P}^1$ is trivial, hence $Alb(C)\cong Alb(C')$. Furthermore the Albanese is dual to the Jacobian, so the curves have the same Jacobian; by Torelli, they are isomorphic. – Ben Nov 21 '13 at 18:38
  • Ah, the use of the Albanese variety is a very ingenious and beautiful idea: I'll try to remember that products are preserved by Albanese. Thanks a lot for your comment, dear Ben. – Georges Elencwajg Nov 21 '13 at 19:25
  • Dear Ben, just for reference I gave another argument in the answer to this question: http://math.stackexchange.com/questions/140464/number-of-birational-classes-of-dimension-d-geometric-genus-0-varieties/518801#518801 but it's not really any less geometric than yours. To be honest, trying to prove such a statement by restricting oneself to algebraic methods seems to me like trying to play Bach with one hand behind your back! But of course I am sure you have a reason for your approach. –  Nov 21 '13 at 20:47
  • Dear @AsalBeagDubh, thank you for this link. Indeed, it's not less geometric, but it definitely is less high brow! But neither of our arguments provides an algebraical reasoning for the statement to be true. I'm always getting curious whenever there are two equivalent categories and a proven statement in one of them, whilst a proof in the other category seems to be out of reach. – Ben Nov 22 '13 at 14:01

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