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

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?

• No, it does not hold for the variable in the polynomial ring, since the statement is for an integer. – Tobias Kildetoft Jul 18 '17 at 19:19
• Polynomials and the functions they determine by evaluation should not be confused, especially over finite rings. The former family is infinite, the latter finite. – Ravi Jul 18 '17 at 19:20
• @DougM Yeah, I removed my comment and made a new one when I saw you had removed it :) – Tobias Kildetoft Jul 18 '17 at 19:21
• @Tobias I see, thank you – Richard Jul 18 '17 at 19:22
• See also this answer for some background on formal polynomials vs. polynomial functions. – Bill Dubuque Jul 18 '17 at 21:43

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.