# The variety $x_1 + x_2 + .. + x_n = 0$, $x_1 x_2 .. x_n = 1$ for n > 4

For $n = 3$ the variety $x_1 + x_2 + .. + x_n = 0$, $x_1 x_2 .. x_n = 1$ is an elliptic curve, and for $n = 4$ it is rational [edit: or so I thought, before seeing the other replies].

What can be said, for example regarding rationality, for larger $n$ (for values of $x_i$ where the variety has no components of the same form with smaller $n$)? Is this variety of a standard type, such as toric or Calabi-Yau?

One might assume that it remains rational for $n > 4$; but, given that the degree of the product increases, that might well not be so.

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Are you sure the $n=4$ case isn't K3? – S. Carnahan Jan 14 '12 at 7:57
A quick coordinate change yields $x_1\cdots x_{n-1}(x_1 + \cdots x_{n-1}) = -1$, which seems to be Calabi-Yau in general. – S. Carnahan Jan 14 '12 at 8:01
The case $n=4$, and the generalization $x_1 x_2 x_3 x_4 + a = 0$ for arbitrary nonzero $a$, was already studied by Euler. I've given some talks about this (see math.harvard.edu/~elkies/euler_11c.pdf for the most recent version) and should be writing it up soon. Yes, it will be a Calabi-Yau hypersurface in general. – Noam D. Elkies Jan 14 '12 at 19:37
@Noam, thanks for the link! I hope you do write it up soon :) – Mariano Suárez-Alvarez Jan 15 '12 at 0:35
Very interesting Noam, and useful as an extended worked example, in what can otherwise seem a bewilderingly abstract topic! One reason I am interested in this variety is that for $N = 4$, with the extra condition $x_1 x_2 = a^2$ with rational $a$, it is birationally equivalent to a rational box with two face diagonals and body diagonal rational. – John R Ramsden Jan 15 '12 at 8:31

## 2 Answers

(expanded comment)

For $n = 3$ in $X Y (X + Y) + 1 = 0$ letting $Y = t X$ gives $t (t + 1) = - 1 / X^3 = x^3$ say. Then letting $y = 2 t + 1$ gives $4 x^3 + 1 = y^2$, which is a non-degenerate elliptic curve.

For $n = 4$, in $X + Y + Z + T = 0, X Y Z T = 1$ Letting $X Y, Z T = -u, -1/u$ resp satisfies the second and in the first gives $X Z (X + Z) = X/u + u Z$. Then letting $t = X/Z$ this becomes $t u (t + 1) Z^2 = t + u^2$, in which letting $t = u v, w = 1/Z$ gives $u v (u v + 1) = (u + v) w^2$.

I thought I had a parametrization of the latter; but having found a slip in my algebra, now I'm not so sure. Maybe it is K3.

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According to Magma web calculator n=4 is not rational. Let me know if you need the code. – joro Jan 14 '12 at 10:09
Wow, yes the code would be very useful - I have plenty of other surfaces that need checking! Is there a link to a web page the code can be downloaded from? My email address, in case you need it, is jhnrmsdn@yahoo.co.uk – John R Ramsden Jan 14 '12 at 13:27
DOH! It occurred to me that the code you mentioned must be the program given in your reply. I thought you were talking about Magma web calculator itself! – John R Ramsden Jan 14 '12 at 13:31
@John the code is in the answer. The web interface to magma is the only URL. Paste the code in the web interface or in magma. – joro Jan 14 '12 at 13:58

According to Magma web calculator $n=4$ is not rational.

After eliminating $x_4$ the surface is $x_1x_2x_3(x_1+x_2+x_3)+1=0$

The program that can be checked in the web calculator:

Q := RationalField();
P<x1,x2,x3> := PolynomialRing(Q, 3);
F<s,t> := RationalFunctionField(Q, 2);
p := x1*x2*x3*(x1+x2+x3)+1;
Solve(p, F);


Don't know how to check for larger $n$.

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