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Sep 5, 2011 at 12:40 comment added rita I was just trying to point out that all arguments seem to boil down to showing that $h^1({\mathcal O})>0$ (equivalently, $h^0(\Omega^1)>0$). For instance, to use the fundamental group you need to know that it is birational invariant for surfaces, which is true but not easier to prove than the birational invariance of $h^1({\mathcal O})$. Indeed,$h^1({\mathcal O})$ is one half of the first Betti number, hence a rougher invariant than $\pi_1$.
Sep 5, 2011 at 11:58 comment added Qing Liu To prove that $\mathbb P^1\times E$ is not rational, one can also use the fact that its $\pi_1$ is obviously non-trivial.
Sep 5, 2011 at 11:51 comment added Qing Liu To show that $E$ is not rational, one can aslo compute directly $H^0(E, O_E(\infty))$ which is equal to $k$, and would be of dimension $2$ over $k$ if $E$ was rational.
Sep 4, 2011 at 16:24 comment added rita Still, you need to know that $E$ is not rational... The only way I know to show that is by computing the genus.
Sep 2, 2011 at 14:12 comment added Hugo Chapdelaine Thanks a lot Qing for the very slick argument! It is completely self contained and the key result that you use is that there is no non-constant rational map going form $\mathbb{P}^1\rightarrow E$ which as you pointed is a consequence of Luroth's theorem. Cool!
Sep 2, 2011 at 14:10 vote accept Hugo Chapdelaine
Sep 2, 2011 at 12:42 comment added rita This is not so much different as it looks from the proof I gave. $h^1({\mathcal O})$ is the dimension of the Albanese variety, that for the non rational surface is precisely $E$. Now your argument shows precisely that the Albanese variety of a unirational variety is $0$.
Sep 2, 2011 at 12:28 history edited Qing Liu CC BY-SA 3.0
added 188 characters in body; added 1 characters in body
Sep 2, 2011 at 11:18 history edited Qing Liu CC BY-SA 3.0
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Sep 2, 2011 at 10:56 history answered Qing Liu CC BY-SA 3.0