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.

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 nonrational 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



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. 

