(non-trivial) isotrivial family of elliptic curves over C^{\times} 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 $\mathbf{C}^{\times}$.
 A: 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.
A: 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 $\mathbb C$).
