Comment by jeremyscanvic
21 hours ago
> Also, from the fundamental theorem of algebra, once the Jacobian polynomial {\mathrm{det} DF} is non-zero, it must be constant.
I wouldn't have guessed this is true. I'm wondering what the proof looks like!
21 hours ago
> Also, from the fundamental theorem of algebra, once the Jacobian polynomial {\mathrm{det} DF} is non-zero, it must be constant.
I wouldn't have guessed this is true. I'm wondering what the proof looks like!
I’m fairly confident that the blog post is trying to say something like this:
Given a polynomial function from C^n to C^n, the following statements are equivalent: (a) det DF is nonzero everywhere. (b) det DF = c for some constant c != 0
The backward direction (b implies a) is trivial. The forward direction can be proven by observing that det DF is itself a polynomial function from C^n to C. If n were 1, then this would follow directly from the fundamental theorem of algebra: a non constant polynomial has degree at least 1 and hence has at least one zero. Extending this logic to higher dimension is not especially difficult.
I do find the way it’s stated in the article to be confusing.
Ah right this makes a lot of a sense - thanks!