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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.

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See also this question: mathoverflow.net/questions/118540/… –  Artie Prendergast-Smith Mar 15 '13 at 16:19

2 Answers 2

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.

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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. –  Noam D. Elkies Mar 15 '13 at 4:22
    
sorry. I thought you had written "rational". I now see you had written "non-rational" ( makes a bit of a difference!). –  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.

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