Every root of $x^n-1$ is simple in $ \mathbb{Z}_p[x]$ Let $p$ be a prime number s.t $p$ doesn't divide $n$. Show that every roots of $x^n-\overline{1}$ is simple in $\mathbb{Z}_p$
If $\overline {a} \in \mathbb{Z}_p$ is a root of $x^n - \overline{1}$ then $a^n \equiv 1\pmod p$ and $\gcd(a,p)=1$. By Fermat's little theorem we have $a^{p-1} \equiv 1\pmod p$.
Ok, now I need to prove that if $\overline{b} \in \mathbb{Z}_p$ is root of $x^n - \overline{1}$, that is, $b^n \equiv 1 \pmod p$, then $\overline {a} = \overline{b}$, that is, $a \equiv b\pmod p$. Can you give me a way to solve that?

Hint:
If $f(a)=0$, and $a$ is not a simple root, what can you say about $f'(a)$? (The formal derivative)

Define $f(x)=x^n-1\in\Bbb Z_p[x]$, and let the group-homomorphism $$\frac{d}{dx}:\Bbb Z_p[x]\rightarrow\Bbb Z_p[x]$$ be the formal derivative.
Detecting multiplicity, and why it works

Let $f(x)\in R[x]$ be a polynomial in any polynomial ring over any commutative ring $R$. Suppose that $f(a)=0$, then $f(x)=(x-a)g(x)$. Taking the formal derivative we get $$f'(x)=g(x)+(x-a)h(x)$$ If $f'(a)=0$, we get that $g(a)=0$, and so $(x-a)\mid g(x)$, in other words $$(x-a)^2\mid f(x)$$

Applying the test

Firstly, $f(0)\neq 0$, since $0^n-1\neq 0$, so assume that there is a non-zero $a\in\Bbb Z_p$ such that $f(a)=0$. Assume also, for sake of contradiction, that $f'(a)=0$. Then $$\frac{d}{dx}(x^n-1)=nx^{n-1}\\na^{n-1}=0$$ Thus $p\mid n$ or $p\mid a$. Both are impossible, so $f'(a)\neq 0$. And we're done!

