Does Little Fermat imply that if $p$ is prime then $x^p=x $ in $\Bbb Z/p\Bbb Z [x] $?

Question

As in the title, I'm trying to understand the implications of Fermat's Little Theorem in $\Bbb Z/p\Bbb Z [x] $. Fermat states that if $p $ is prime then for all integer $x$, $x^p-x $ is a multiple of $p$. As far as I can see, this should mean that $x^p-x=0$ in $\Bbb Z/p\Bbb Z [x] $, but I'm not sure. Is that correct?

Answer 1

The polynomial $x^p-x$ is not the zero polynomial in $\Bbb Z/p\Bbb Z [x]$.

The polynomial function $x^p-x$ is the zero function $\Bbb Z/p\Bbb Z \to \Bbb Z/p\Bbb Z$. That's what Fermat says.

Polynomials and polynomial functions are different objects but they can be identified when the base field is infinite.