I am reading a wonderful article of Arnaud Beauville, called *La théorie de Hodge et quelques applications*
<http://math.unice.fr/~beauvill/conf/Bordeaux2.pdf>


There is one place on page 12 that I can not understand. Beauville seem to claim the following: 


**Claim.** Denote by $K$ the field of meromorphic functions of the (complex) cubic $x\_1^3+x\_2^3+x\_3^3+x\_4^3+x\_5^3=0$. Then one can show that there exists an embedding $K\subset \mathbb C(y\_1,y\_2,y\_3)$. 


**Question.** How to show this?


I have to say, I doubt this statement (*not anymore*). Since the cubic is *unirational*, we have $K\subset \mathbb C(y\_1,...,y\_n)$ for some $n$. But it sounds strange that one can chose $n=3$. Is this a misprint or I miss something?


(for a connection to Luroth problem see <http://en.wikipedia.org/wiki/Rational_variety>)


**PS.** I guess Artie's comment explains that Beauville is 100% correct. So I would like to ask one more (non-trivial ?) question:


**Question 2** Is it known what is the minimal number $d$ such that there is a degree $d$ (dominant) morphism from a rational complex projective three-fold to the cubic? 



Discussion of Luroth's problem in an article of Beauville
(As requested, I'm turning my comment into an answer.)


The issue seems to be that there are differing definitions of unirational. One is that there exists a dominant rational map $\mathbf{P}^n \dashrightarrow X$ for *some* $n$. (That corresponds to the inclusion of function fields $K \subset \mathbb C(y\_1,...,y\_n)$.) But it's easy to show that it's equivalent to ask for a dominant rational map $\mathbf{P}^d \dashrightarrow X$, where $d$ is the dimension of X. (In your case, this gives the inclusion $K \subset \mathbb C(y\_1,y\_2,y\_3)$.)


To see the equivalence of the two, suppose $\phi: \mathbf{P}^n \dashrightarrow X$ is a dominant rational map with $n$ strictly larger than dim X. Then a general hyperplane $H$ in $\mathbf{P}^n$ will intersect the general fibre of $\phi$, so the restriction of $\phi$ to $H$ is still dominant. By induction, you're done.


