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Given the polynomials $p_1(x_1, \ldots, x_n), \ldots,p_k(x_1, \ldots, x_n)$, let $\Lambda$ be the set of $x \in R^n$ where all of these polynomials are nonnegative. Does there exist a polynomial $q(x_1, \ldots, x_n)$ such that $\Lambda$ is the set of points where $q(x_1, \ldots, x_n)$ is nonnegative, i.e., $\Lambda = \{ x ~|~ q(x) \geq 0\}$?

robinson
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    Let $n=2$, let $p_1=x_1,p_2=x_2$. Is there a polynomial that is non-negative precisely on the (closed) first quadrant? – Gerry Myerson Sep 08 '11 at 01:29

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Not necessarily.

Consider the special case $p_1(x,y)=x$, $p_2(x,y)=y$. Then $q(x,y)$ should be nonnegative exactly when $x\ge 0$ and $y\ge 0$. The positive x axis is on the boundary of this set, so by continuity $q(x,0)=0$ for $x\ge 0$.

Now consider the function $x\mapsto q(x,0)$. This is a polynomial in one variable (it arises by dropping all terms with positive exponent in $y$). We've concluded that it must be identically zero for $x\ge 0$. But that must mean that it is the zero polynomial, because a nonzero polynomial in one variable cannot have more zeroes than its degree.

So in particular, for example, $q(-1,0)=0$. But that contradicts the assumption that $q(x,y)\ge0$ only for $x\ge 0$.

hmakholm left over Monica
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  • The fact that $q(x,0) = 0$ does not in itself imply that the entire $x$ axis is in the boundary of ${(x,y): q(x,y) \ge 0}$: maybe $q(x,y) < 0$ on both sides of $(x,0)$ if $x < 0$. To rule this out, you have to look at some partial derivatives. Let $k$ be the least positive integer such that $\dfrac{\partial^k q}{\partial y^k}(x,0) \ne 0$ for almost all $x$. Since $p(x,y)$ changes sign as you cross the $x$ axis for $x > 0$, $k$ must be odd. But since $p(x,y)$ does not change sign as you cross the $x$ axis for $x < 0$, $k$ must be even, contradiction! – Robert Israel Sep 08 '11 at 01:37
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    I don't need the entire x axis to be on the boundary specifically. Just plainly that if, say, $q(-1,0)=0$, then ipso facto $(-1,0) \in {(x,y)\mid q(x,y)\ge 0}$, contradicting the assumption that this was the case only for the closed first quadrant. – hmakholm left over Monica Sep 08 '11 at 01:41
  • Thank you very much for your answer. However, I wonder if you could explain the proof to me at my level (I am someone who does not know any algebra). You seem to be using the fact that if $a(x)$ and $b(x)$ are polynomials with infinitely many zeros in common, one must divide the other. Might I trouble you for a pointer to the proof? – robinson Sep 08 '11 at 01:42
  • Ah, of course. Silly of me. – Robert Israel Sep 08 '11 at 01:44
  • @robinson, the infinitely-many-common-zeroes property holds only if the smaller polynomial is irreducible, that is, if it is not the product of two nonconstant polynomials. (For example $x^2y$ and $xy^2$ have infinitely many common zeroes, but neither is a factor of the other, and neither is irreducible). Such things are studied in algebraic geometry. I'll see if I can think up an elementary proof, but don't hold your breath. – hmakholm left over Monica Sep 08 '11 at 01:50
  • I have replaced the last half of my argument with a more elementary one. (It used to appeal to general facts about divisibility of multivariate polynomials). – hmakholm left over Monica Sep 08 '11 at 02:01
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    My comment would be relevant if you were talking about ${(x,y): q(x,y) > 0}$. – Robert Israel Sep 08 '11 at 04:17