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 



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> Is the set $$\{f\in V|f\text{ is Newton non-degenerate}\}$$ Zariski open in $V$?.
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It is widely believed that this result was first affirmatively proved by Kouchnirenko (Theorem 6.1., Invent. Math., 1976; [link](https://eudml.org/doc/142365)), 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 



Zariski openness of Newton non-degenerate polynomials
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. 


