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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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  • $\begingroup$ 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. $\endgroup$ Commented Jan 9, 2010 at 21:45
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    $\begingroup$ @Ilya: Your "note" is somewhat confusing (I claim that every non-negative function is bounded below!). I think you are missing something like "on $\mathbb{Q} \times \mathbb{Q}$". $\endgroup$ Commented Jan 9, 2010 at 21:47
  • $\begingroup$ 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. $\endgroup$ Commented Jan 9, 2010 at 21:52
  • $\begingroup$ @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 :) $\endgroup$ Commented Jan 9, 2010 at 23:57
  • $\begingroup$ 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. $\endgroup$ Commented Jan 10, 2010 at 1:21

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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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    $\begingroup$ In fact, the wikipedia page that I linked to has a counterexample! $\endgroup$ Commented Jan 9, 2010 at 21:45
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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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