# Can properties of a polynomial over $\mathbb{Q}$ be carried over to properties over $\mathbb{R}$?

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.

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If you have a vector $\vec{x}$ with integer coordinates such that $f(\vec{x})=0$, then that equality holds whether you view $f$ as over $\mathbb{Z}$ or over $\mathbb{R}$. So if $f$'s integral roots lie on axes over $\mathbb{Z}$, then $f$'s integral roots lie on axes over $\mathbb{R}$ too. Since you are specifying integer coordinates in the roots that you are considering, and integer coefficients of $f$, there are no complications from extending to $\mathbb{R}$. If you dropped either of these integer constraints, you would have trouble. – alex.jordan Mar 29 '12 at 20:33
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. – Chris Eagle Mar 29 '12 at 20:37

## 2 Answers

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

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OP is asking a question about integral roots (roots where every coordinate equals $0$). So are you sure this is a counterexample? – alex.jordan Mar 29 '12 at 20:37
The OP wants to know about the property "$f(\vec x) = 0 \implies$ some coordinate of $\vec x$ is 0". This polynomial satisfies that property over $\mathbb{Z}$ but not over $\mathbb{R}$. – Chris Eagle Mar 29 '12 at 20:39
Read the OP's second paragraph: "its integral zeroes..." The only integral zero of this example is $(0,0)$. – alex.jordan Mar 29 '12 at 20:40
@alex.jordan I actually meant what Chris think I meant. I will modify my question if this was not clear. – Ofir Mar 29 '12 at 20:46
@ChrisEagle - that's a nice conterexample, thank you. I will look over my special cases and see what they have in common that your example didn't have and try and refine my question. – Ofir Mar 29 '12 at 20:46

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 $(2, \sqrt{6}, 1)$.

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A nice algebraic example (and motivation). Check out my new requirement of f (in the edit). – Ofir Mar 29 '12 at 21:23