# 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!

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.
• Why does the fact that $\mu_a (x^n - 1) = p(x)(x^n - 1)$ show it is a homomorphism on $R_n$? – the man Feb 10 at 14:39
• @theman You need to show that for the equivalence class $f(x) +\langle x^n-1\rangle$ we have that $\mu_a$ applied to this is $f(x^a) +\langle x^n-1\rangle$. Thus for an element $g(x) (x^n-1)$ of the ideal, we need that $g(x^a) (x^{an} - 1)$ is in the ideal. This is true if and only if it's true for $g(x) =1$. – Matt Samuel Feb 10 at 14:55
• For the kernel, I forgot to say that $gcd(a,n) = 1$. I'm trying to find for which $f(x)$, $f(x^a) \in \langle x^n - 1 \rangle$ right? I'm not sure where to continue from here? – the man Feb 10 at 15:13