Given a polynomial over many variables $x_1...x_n$with integer coefficients, I'd like to prove that there are nonnegative coordinates $x_1=a_1,...,x_n=a_n$ of a point at which the polynomial doesn't vanish.

I know how to prove that easily for a single-variable polynomial (find $x$ large enough for the highest-degree term to dominate all others). And I know how to prove that a many-variabled real nonzero polynomial does not vanish, by repeatedly applying the proof for a single variable (the one using a finite number of roots, not the one in the previous sentence) and reducing the number of variables; e.g. as described here. But I don't see how I can ensure the coordinates are nonnegative with this proof. I suspect the claim is true for real polynomials, though I only need it in the case when the coefficients are integers.

I think I have an idea for the proof, but it seems tedious, and I wonder if there's a better simpler approach.

Here's the outline of my idea. First, we pick a constant $C$ large enough so that if we assign $x_i=C$ for all variables, then every term with highest degree (sum of all its variables' degrees) dominates, by absolute value, the sum of all terms with lesser degrees, even if each is taken with the maximal available coefficient in the entire polynomial. Then by restricting to values above $C$, we're ensuring that terms of lesser degrees can never bring the value to $0$ if it wasn't $0$ without them, so now we can pretend they do not exist. Now only terms of equal degree remain; choose some $x_1$ that varies in degree among them, and change its assignment from $C$ to $C^2$. The value of the polynomial will be as if we let $x_1=C$ remain, but doubled its degree in all the terms, and that value, we assume, is zero. In that new polynomial, some terms now have lesser degree than others, and we can find a new uniform assignment $x_i=C_1$, greater than C, at which these terms with lesser degrees do not matter, and then, by applying induction over the number of terms, find some assignment $x_i=a_i$ where the new polynomial does not vanish. Then changing $x_1$'s assignment to its square and leaving others intact makes the previous polynomial not vanish.

  • 2
    $\begingroup$ Maybe I missed something but the set $\{ x \in \mathbb{R}^n \mid x_i < 0 \}$ is open in $\mathbb{R}^n$. So your polynomial cannot vanish identically on it unless it is the zero polynomial. $\endgroup$
    – halbaroth
    Nov 1 '20 at 17:13
  • $\begingroup$ Why can't it vanish on an open set? Are you invoking that the polynomial is an analytic function and so must vanish everywhere if it vanishes on an open set, or some other reason? $\endgroup$ Nov 1 '20 at 17:45
  • 1
    $\begingroup$ You don't need sophisticated arguments to prove it. See math.stackexchange.com/questions/36970/…. $\endgroup$
    – halbaroth
    Nov 1 '20 at 17:49

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