Showing this is an automorphism Let $R_n = F_q[x]/\langle x^n - 1 \rangle$, where $F_q[x]$ is a finite field. 
Consider $\mu_a$ which acts on $R_n$ like so; $f(x) \mu_a \equiv f(x^a) \bmod (x^n - 1)$ for $f(x) \in R_n$.
Is this an automorphism?
I'm not too sure where to start here. I know polynomial substitution is a homomorphism, so can I instantly deduce that this is a homomorphism? 
If I show it's injective, then I can deduce that it is surjective since $R_n$ is finite, but I'm not sure where to start with injectivity?
Any help would be appreciated!
 A: The lifted function $x\mapsto x^a$ on $F_q[x] $ is indeed a homomorphism by the universal property of polynomial rings, however this does not automatically imply that it is a homomorphism on $R_n$. You need to show that $\mu_a(x^n-1)=p(x)(x^n-1)$ for some $p$. This is true because
$$x^{an} - 1=(x^n-1)(x^{n(a-1)}+x^{n(a-2)}+\cdots+x^n+1)$$
Thus it is a homomorphism on $R_n$. To show it is an automorphism, you then need to prove the kernel is trivial. 
Now, $f(x)$ is a multiple of $x^n-1$ if and only if for all $n$th roots of unity $\zeta$ we have $f(\zeta)=0$. If $f(x^a)$ is a multiple of $x^n-1$, then $f(\zeta^a) = 0$ for all $n$th roots of unity. Since $a$ is relatively prime to $n$, every $n$th root of unity occurs as $\zeta^a$ for some other $n$th root of unity. Thus $f(\zeta)=0$ for all $n$th roots of unity $\zeta$, hence $f(x)$ is a multiple of $x^n-1$. Thus the kernel is trivial.
Note that this must be taken over all $n$th roots of unity, including ones that are not contained in $F_q$. We are essentially working in the algebraic closure.
