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What can be said about the polynomials $f\in\mathbb Q[x, y]$ which are nonnegative on $\mathbb R\times \mathbb R$?

Motivation: this may lead to progress in the question about polynomial onto map $\mathbb Z\times \mathbb Z\to\mathbb N$, but I post it separately as it's interesting in itself.

Note: there are examples of polynomials nonnegative on $\mathbb Z\times \mathbb Z$, but not bounded from below on $\mathbb R\times \mathbb R$, e.g. $(x^2-x)y^2$, so this doesn't apply directly.

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So, you've got some answers involving Hilbert 17, but is that really what you're looking for? If not, perhaps you could say something about what kinds of things you're looking for in answers. –  Charles Siegel Jan 9 '10 at 21:45
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@Ilya: Your "note" is somewhat confusing (I claim that every non-negative function is bounded below!). I think you are missing something like "on $\mathbb{Q} \times \mathbb{Q}$". –  Pete L. Clark Jan 9 '10 at 21:47
    
As far as an application to the original problem, I am reasonably sure that the "leading homogeneous term" of the polynomial must be a sum of squares (of polynomials, since it's homogeneous), but I haven't checked that yet either. –  Qiaochu Yuan Jan 9 '10 at 21:52
    
@Qiaochu Yuan: no, that's not true. You can always change $(x, y) \to (x, y + x^20)$, so you either prove there are no solutions, or lots of them. This is the primary difficulty I'm having about the problem :) –  Ilya Nikokoshev Jan 9 '10 at 23:57
    
I'm not sure I understand. The leading homogeneous term of (x^2 - x)(y + x^{20})^2 is x^{42}, which is a sum of squares. –  Qiaochu Yuan Jan 10 '10 at 1:21

2 Answers 2

up vote 4 down vote accepted

The following theorem of Artin -- his solution of Hilbert's 17th problem, but in a stronger form than Hilbert himself asked for -- answers the question.

Theorem (Artin, 1927): Let $F$ be a subfield of $\mathbb{R}$ that has a unique ordering, and let $f(t) = f(t_1,\ldots,t_n) \in F(t_1,\ldots,t_n)$ be a rational function such that $f(a) \geq 0$ for all $a = (a_1,\ldots,a_n) \in F^n$ for which $f$ is defined. Then $f$ is a sum of squares of rational functions with coefficients in $F$.

A proof can be found in Jacobson, Basic Algebra II, Section 11.4.

Note that the tempting strengthening -- that if $f$ is a polynomial, it is a sum of squares of polynomials -- is false, as Hilbert himself showed.

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In fact, the wikipedia page that I linked to has a counterexample! –  Charles Siegel Jan 9 '10 at 21:45

Well, that's a function $\mathbb Q\times\mathbb Q\to \mathbb Q_{\geq 0}$. However, it is the same as a function $\mathbb{R}\times\mathbb{R}\to\mathbb{R}_{\geq 0}$, by continuity, and so by Hilbert's 17th Problem, it's a sum of squares of rational functions (with real coefficients, but I'm willing to be that we can do it over $\mathbb{Q}$ as well).

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