4
$\begingroup$

The following question arose while trying to generalize some combinatorial statements from $\mathbb{Z}$ to $\mathbb{R}$.

Suppose I have a multivariate homogenous polynomial $f$ with coefficients in $\mathbb{Z}$, and its integral zeroes lie only on the axes, i.e. $f(\vec x) = 0 \implies$ some coordinate of $\vec x$ is 0.

I want to prove that if I look at $f$ as a polynomial over $\mathbb{R}$ then it satisfies there the same property - its zeroes (this time, real zeroes) must line on the axes. I am not sure that it is true, but I am sure that it is correct in many cases and I was told that it might have a proof that uses model theory (though I prefer an "explicit" proof). If this is incorrect, a counterexample would be nice.

Note: because $f$ is homogenous, it can be seen that its main property carries on to $\mathbb{Q}$.

EDIT: Chris found a nice counterexample. Can all the counterexamples be characterized some how? Chris also showed that a counterexample which is a sum of squares can be found.

What about the following case: $f$ is a sum of squares of product of hyperplanes, i.e. $f = \sum P_i^2$ where $P_i$ is a product of linear forms.

  • 0
    The relevant model-theoretic notion is an *elementary extension*. If a field $F$ was an elementary extension of a field $K$, then properties of this sort would carry across. $\mathbb{R}$ isn't an elementary extension of $\mathbb{Q}$ though.2012-03-29

2 Answers 2

3

This doesn't work. For example, $x^2-2y^2$.

  • 0
    @ChrisEagle - I modified my 2nd question immediately after posting because I've realized your solution can be modified (as you've shown). I am not sure if I should keep trying...2012-03-29
1

Well, the polynomial $Y^2Z = X^3-XZ^2$ is the homogenization of $y^2 = x^3-x$, an elliptic curve which has no rational points other than the trivial ones $(0,0)$ and $(1,0)$ (this isn't hard to show). So the integral (projective) points on $Y^2 = X^3-XZ^2$ are:

  • The "point at infinity" $(0,1,0)$

  • The trivial points $(0,0,1)$ and $(1,0,1)$

However, there are certainly many non-trivial real solutions that are nowhere 0, like $(2, \sqrt{6}, 1)$.

  • 0
    A nice algebraic example (and motivation). Check out my new requirement of f (in the edit).2012-03-29