Subfield of rational function field and which is not a rational function field Let $K = k(x_{1}, x_{2},...,x_{n}), n\geq 2, k$ is a field. Is there exist a subfield of $K$ which is not a rational function field? Thanks.
 A: This is essentially the question of whether a $k$-unirational variety is
necessarily $k$-rational.  The short answer is No.
The following longer answer mostly summarizes some of the exposition at
http://en.wikipedia.org/wiki/Rational_variety; for more information
see that page and the references it gives.
The existence of a non-rational subfield $F$ of $K$ depends on $k$ and $n$.
If $k$ is algebraically closed and of characteristic zero,
then the answer is No for $n=2$ by a theorem of
Castelnuovo (and for $n=1$ by a theorem of Lüroth),
but Yes for $n=3$, and thus for all $n \geq 3$
(you did not require $F/K$ to be a finite extension).
In characteristic $p>0$ things can get much stranger:
Zariski gave examples for $n=2$ where the extension $K/F$ is inseparable;
and more recently Shioda constructed, for each $n \geq 2$
and every power $q$ of $p$,
an example where $K/F$ is inseparable and $F$ is the function field of
the Fermat hypersurface of dimension $n$ and degree $q+1$
(which is of general type once $q \geq n+3$), see Propositions
1 and 3 in

Shioda, T.: An Example of Unirational Surfaces in Characteristic $p$,
  Math. Ann. 211 (1974), 233-236.

A: A result of H.W.Lenstra (Inventiones Math., 1974):
(a clearly written paper)
For a prime number $p$ let $K=Q(x_1,x_2,\ldots, x_p)$, be a pure transcendental extension over the rational numbers.  Let $F$ be the subfield consisting of those elements of $K$ that are fixed by the cyclic permutation of the variables.
Then $F/Q$ is not purely  transcendental for $p=47$ (Swan, 1969) and for infinitely many primes $p$.
The paper contains much more, about abelian group of permutations and their invariant subfields.
