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So How does one prove (rigorously) that $$ Frac(\mathbb{C}[x,y,t]/(y^2-x^3-t)) \not\simeq Frac(\mathbb{C}[t][x,y]/(y^2-x^3-1))? $$ So here $Frac$ denotes the fraction field of an integral domain.

Note that this gives an example of (a non-trivial) isotrivial family over $\mathbf{C}^{\times}$.

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up vote 5 down vote accepted

There is a slightly different proof which works over any field $k$ (of characteristic different from 2 and 3). The first field is just $k(x,y)$. As Rita indicated, if it is isomorphic to the second field as $k$-extension, than there exists a birational map $f: \mathbb P^2\to \mathbb P^1\times E$. This birational map induces a birational map over the algebraic closure of $k$, so we can suppose $k$ is algebraically closed. Composing with the projection to $E$, we get a dominant rational map $g: \mathbb P^2\to E$. By two arbitrary points where $g$ is defined, it passes a line $L$. As $E$ is not rational, by Lüroth $g|_L$ is constant. So $g$ is constant, contradiction. More quickly, one can say that the existence of $g$ implies that $E$ is unirational and this is impossible.

EDIT: the rational map $\mathbb P^2\to E$ is not birational ! but dominant.

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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 $0$. – rita Sep 2 '11 at 12:42
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! – Hugo Chapdelaine Sep 2 '11 at 14:12
Still, you need to know that $E$ is not rational... The only way I know to show that is by computing the genus. – rita Sep 4 '11 at 16:24
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. – Qing Liu Sep 5 '11 at 11:51
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. – Qing Liu Sep 5 '11 at 11:58

The second field is the function field of $X_2:=E\times {\mathbb P}^1$, where $E$ is a smooth elliptic curve; the first one is the function field of $X_1:=\{zy^2-x^3-tz^2=0\}\subset {\mathbb P}^3$. The surface $X_1$ is rational, as one can see by projecting onto ${\mathbb P}^2$ from the point $P$ given by $x=y=z=0$, which is a double point of $X_1$. The surface $X_2$ is not rational, since it has $h^1({\mathcal O}_{X_2})=1$. So $X_1$ and $X_2$ are not birational, and the two fields are not isomorphic (as extensions of $\mathbb C$).

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So I guess the first $X_2$ should read as $X_1$. Is it completely obvious that $h^1(\mathcal{O}_X)$ is a birational invariant? After all $\mathcal{O}_X$ is the sheaf of regular functions on $X$ which a priori could change under birational maps. – Hugo Chapdelaine Sep 2 '11 at 1:18
@Georges: I've edited and fixed the typos (I hope). Thanks! @Hugo: $h^1({\mathcal O})$ is a birational invariant because by Hodge theory it is equal to $h^0(\Omega^1) – rita Sep 2 '11 at 6:27
@Hugo (continued): there is also a famous rationality criterion by Castelnuovo, that says that a surface $S$ is rational iff $h^1({\mathcal O})=h^0(2K_S)=0$. – rita Sep 2 '11 at 8:36
I'm quite happy with your proof but I think there should be a more elementary proof which is self contained. You see the whole point of this question is to come up with a birational invariant and if we assume from the outset that $h^1(\mathcal{O})$ is a birational invariant then it (almost) kills the problem. – Hugo Chapdelaine Sep 2 '11 at 12:10
The last statement is a well known fact, have a look at Beauville's book on surfaces, the chapter on birational maps of surfaces. – rita Sep 2 '11 at 12:37

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