For which field $F$ and prime $p$, the polynomial $x^p-x\in F[x]$ has a multiple root?

Question

For which field $F$ and prime $p$, the polynomial $x^p-x\in F[x]$ has a multiple root ?

If $F$ has caracteristic $p$, then $(x^p-x)'=-1$ and thus has no multiple root. So is it in any field of characteristic $q\neq p$ ?

Answer 1

We use the derivative criterion

given a root $r$ of the polynomial $f$, $r$ is a multiple root of $f$ iff it is a root also of its formal derivative $f'$.

Suppose the characteristic $q$ of the field is different from $p$. Then $$ x^{p} - x = (p x^{p-1} - 1) \frac{1}{p} x + x \cdot \frac{1 - p}{p}. $$ If $q \mid p-1$ (so that $q > 0$ and $p > 2$), then $$ \gcd(x^{p} - x, p x^{p-1} - 1) = p x^{p-1} - 1, $$ so that the roots of the derivative $p x^{p-1} - 1$ are multiple roots of $x^{p} - x$, see the examples of Dietrich Burde and Don Antonio.

If $q \nmid p - 1$, one more division will show that $$ \gcd(x^{p} - x, p x^{p-1} - 1) = 1, $$ so there are no multiple roots.

Answer 2

For example, take $p=7$. Then $$ x^7-x=x(x + 2)^3(x + 1)^3 $$ has multiple roots over $\mathbb{F}_3$.

Answer 3

What about $\;F=\Bbb F_2\;,\;\;p=3\;$ ?:

$$x^3-x=x(x^2-1)=x(x-1)^2$$