TITLE: In which Fields, does $x^n-x$ have a multiple zero?
QUESTION [4 upvotes]: In which Fields, does $x^n-x$ have a multiple zero?
Attempt: Let $f(x) = x^n-x = x(x^{n-1}-1)$ and $f'(x) = nx^{n-1}-1$
If $f(x)$ has a multiple zero, then, $f(x)$ and $f'(x)$ have a common factor. 
Let $(x-a)$ be a common factor , then $f(a)=f'(a)=0$
Also, there is no multiple zero at $0$ because of the absence of an $x^2$ term in $f(x)$.
I do have an intuition that I might be missing something very basic. However, I am not able to go ahead.
How do I move ahead? Thank you for your help.

REPLY [1 votes]: A very big hint: you've already shown that $0$ isn't a multiple root.  Now, any non-zero common factor of $f(x)$ and $f'(x)$ is also a common factor of $f'(x) = nx^{n-1}-1$ and $g(x) = x^{n-1}-1 = \frac{f(x)}{x}$; this means that it's also a factor of $f'(x)-ng(x)$.  You should find that this expression has a particularly simple form...
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