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?

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    $\begingroup$ I think the issue here is that there are various definitions of unirational. One is that there is a dominant rational map P^n --> X for some n (that's equivalent to your statement about the function field). But one can show without too much difficulty that it's equivalent to ask for a dominant rational map P^d --> X. where d is the dimension of X. Then the claim about function fields follows. $\endgroup$ – user5117 Apr 26 '12 at 21:51
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    $\begingroup$ Another small comment: your defining equation seems to be missing a variable. $\endgroup$ – user5117 Apr 26 '12 at 22:10
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    $\begingroup$ @aglearner, "I wonder what is the minimal possible degree". The minimal possible degree is 2. The degree cannot be 1, or else the cubic threefold $X$ would be rational (contradicting Clemens-Griffiths). On the other hand, if you fix a general line $L$, then it is easy to prove rationality of the parameter space $B$ for pairs $(p,H)$ of a point $p\in L$ and a $2$-plane $L\subset H$ such that $H$ is tangent to $X$ at $p$. Let $C$ be the parameter space for triples $(p,H,q)$ where $(p,H)$ is as above, and where $q$ lies on the residual conic to $L$ inside the plane cubic $H\cap X$.(contd.) $\endgroup$ – Jason Starr Apr 27 '12 at 11:35
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    $\begingroup$ (contd.) There is a projection $\pi:C\to B$, $\pi(p,H,q) = (p,H)$, which is a conic bundle. And there is a section $P:B\to C$, $P(p,H) = (p,H,p)$. Thus $C$ is birational to $B\times \PP^1$. On the other hand, it is easy to see that the projection $\rho:C\to X$, $\rho(p,H,q) = q$ is generically finite of degree $2$, just the number of intersection points of the residual conic with the line $L$. $\endgroup$ – Jason Starr Apr 27 '12 at 11:38
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    $\begingroup$ Jason, huge thanks, great explanation! If you would transfer these two comments into an answer, I would be very happy to accept it. To confirm, for you $L$ is a generic line on $X$. By the way, I was always confused by the expression "conic bundle". Does this mean a bundle whose fibers are $\mathbb P^1$, but some of them may be degenerate? $\endgroup$ – aglearner Apr 27 '12 at 12:49

(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.

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    $\begingroup$ Artie, thanks, since this answers completely to my first question, I will accept this answer. It is also cool that Jason Starr explained in his comments that every cubic three-fold admits a double cover that is a rational 3-fold ... $\endgroup$ – aglearner Apr 29 '12 at 13:19
  • $\begingroup$ You're welcome! $\endgroup$ – user5117 Apr 30 '12 at 13:33

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