Prove that if p is a prime, then each element of $\Bbb{Z}_p$ is a root of $x^p-x$.

Question

Question is given in the title. I have tried picking an arbitrary element $\overline{a}$ and then using the division algorithm to try to show that my remainder is zero when dividing $x^p-x$ with $x-\overline{a}$. This gives me a remainder of $(\overline{a}^{p-1}-1)x$. How do I see that this is zero? I feel blind to it, yet I feel like my method so far is correct. Thanks in advance!

Answer 1

The statement follow by Fermat's theorem. Let $[a]\in\mathbb{Z}_p$ then we have

$$a^p \equiv a \text{ (mod } p);$$

equivalently $$ a^p-a\equiv 0 \text{ (mod } p).$$

So every elements in $\mathbb{Z}_p$ of the polynomial $q(x)=x^p-x$. Moreover in this field $q(x)$ can be decompose in simple factors

$$q(x)=x(x-[a_1])...(x-[a_{p-1}])$$ in unique way.