non-isomorphic stably isomorphic fields Q1: What is the simplest example of two non-isomorphic fields $L$ and $K$ of characteristic $0$ such that $L(x)\simeq K(x)$ (here $x$ is an indeterminate)? 
Q2: Do we have a sufficient criterion for a general field $K$ of characteristic $0$ which guarantees that if $K(x_1,\ldots,x_n)\simeq L(x_1,\ldots, x_n)$ (here $L$ is a field and the
$x_i$'s are indeterminates) then $K\simeq L$?  
 A: An answer to Q2, generalizing Ralph's comment: "$K$ is algebraically closed" is a sufficient condition. Indeed, you can characterize $K$ inside $K(x_1,\dots,x_n)$ as the set of elements having $m$-th roots for infinitely many integers $m$. More generally, it is enough to assume that for some $m>1$, the $m$-th power map on $K$ is onto. Examples: $K$ perfect of positive characteristic, or $K=\mathbb{R}$.
A: I don't think that there are any really easy examples. In the famous paper of Beauville, Colliot-Thélène, Sansuc and Swinnerton-Dyer "Variétés stablement rationnelles non rationnelles" they construct surfaces $S$ over $\mathbb Q$ that are not rational, but such that the products $S \times \mathbb P^3$ are rational. You get an example by taking $K$ to be a purely transcendental extension of the function field of $S$ of transcendence degree $d$, and a purely transcendental extension of $\mathbb Q$ of transcendence degree $d+2$, for some $d$ between $0$ and $3$ (I don't know the correct value of $d$).
