Zariski openness of Newton non-degenerate polynomials

Suppose you are given a convex polyhedron $\Delta$ in $\mathbb{R}^n$ (i.e. a convex hull of finitely many points in $\mathbb{Z}^n$) and consider a finite dimensional vector space $V$ over $\mathbb{C}$ defined by $$V=\left\{f(x)=\sum_{\alpha\in\mathbb{Z}^n}a_\alpha x^\alpha\in\mathbb{C}[x_1^{\pm},\cdots,x_n^\pm] \mid a_\alpha=0 \text{ if } \alpha\notin\Delta \right\}.$$ Suppose further that $\dim \Delta\leq n-1.$ We say $f\in V$ is Newton non-degenerate if $$\left\{x\in(\mathbb{C}^{*})^n\mid\frac{\partial f}{\partial x_1}(x)=\cdots=\frac{\partial f}{\partial x_n}(x)=0\right\}=\emptyset.$$ My question is

Is the set $$\{f\in V|f\text{ is Newton non-degenerate}\}$$ Zariski open in $V$?.

It is widely believed that this result was first affirmatively proved by Kouchnirenko (Theorem 6.1., Invent. Math., 1976; link), but it seems he just proved there is a Zariski open subset of $V$ consisting only of Newton non-degenerate polynomials. It might be possible that my understanding of French is not good enough, but at least he does not prove the Zariski openness of the set $\{f\in V\mid f\text{ is Newton non-degenerate}\}$.

Could you please give me another reference stating that the set $\{f\in V|f\text{ is Newton non-degenerate}\}$ is Zariski open? Or could you suggest the outline of the proof?

Thank you very much in advance

The set described in the question is not Zariski open. Let $n=2$ and $\Delta = \mathrm{ConvexHull}((2,0), (0,2))$. So we are considereding polynomials $a x^2+bxy+cy^2$. Then your set is $$\{ b^2 \neq 4 ac \} \cup \{ (t,0,0),\ (0,0,t) : t \neq 0 \}.$$ This contains the Zariski open set $b^2 \neq 4ac$, but is not Zariski open.
You might wonder if this could be fixed by restricting to the open set where the $a_{\alpha}$ are nonzero (for $\alpha \in \Delta$), but no. Take $n=3$ and $\Delta = \mathrm{ConvexHull}((2,0,0),(0,2,0), (0,0,2))$. So we are considering quadratic polynomials $(x_1, x_2, x_3) A (x_1,x_2,x_3)^T$ with $A$ some symmetric matrix. The resulting conic is smooth iff $\det A \neq 0$. It is smooth when restricted to $x_1 x_2 x_3 \neq 0$ both on the open set $\det A \neq 0$, but it is also zero if the conic is a pair of lines crossing at a point where one of the coordinates vanishes.
• Thank you very much! You are right, and thanks to your example, I could notice that my definition of Newton non-degeneracy was wrong. I would like to modify the definition as follows. Under the same setting, $f(x)$ is said to be Newton non-degenerate if for any face $\Delta^\prime <\Delta,$ the $\Delta^\prime$ principal part $$f_{\Delta^\prime}(x)=\sum_{\alpha\in\Delta^\prime}a_\alpha x^\alpha$$ satisfies $$\{x\in(\mathbb{C}^*)^n|\frac{\partial f_{\Delta^\prime}}{\partial x_1}(x)=\cdots =\frac{\partial f_{\Delta^\prime}}{\partial x_n}(x)=0\}=\phi.$$ Isn't it still Zariski open? – Templeman Feb 19 '17 at 18:31