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Feb 19, 2017 at 18:32 vote accept Templeman
Feb 19, 2017 at 18:31 comment added Templeman 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?
Feb 19, 2017 at 18:22 comment added Templeman Thank you very much! You are right, and thanks to your example, I could notice that my definition of Newton non-degeneracy was wrong.
Feb 18, 2017 at 12:29 history answered David E Speyer CC BY-SA 3.0