In which Fields, does $x^n-x$ have a multiple zero?

Question

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.

Answer 1

You are on the right track. Note that if $f(a)=0$ then $a^{n}=a$, which means $a^{n-1}=1$.

Now then $f'(a)=na^{n-1}-1=n-1$. So, $f'(a)=0$ implies $n-1=0$ in your field, which means that the field has characteristic $p$ that divides $n-1$.

For all other fields there is no multiple root.

Answer 2

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...