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 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.

  • $\begingroup$ Actually I wrote "presumably for all $n \geq 3$", so the existence of a further counterexample for $n=47$ does not let me remove the "presumably". However, I see now that the author did not ask that $[K:F]$ be finite, so the $n=3$ counterexample is automatically a counterexample for all $n \geq 3$. I'll edit my answer accordingly. $\endgroup$ – Noam D. Elkies Mar 15 '13 at 4:22
  • $\begingroup$ sorry. I thought you had written "rational". I now see you had written "non-rational" ( makes a bit of a difference!). $\endgroup$ – Venkataramana Mar 15 '13 at 4:48

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.