A ring automorphism in cyclotomic field

Question

My friend asked me a question, see this. I've thought about that for some time, but I cannot do it, I don't want to let her wait too long, can you explain it for me? Thanks in advance!

Answer 1

(a) I think you can show it is a ring morphism. To show surjectivity, it suffices to check that $\zeta$ is in the image of $\varepsilon_n.$ Since $n$ is coprime to $p,$ by the Euclidean Algorithm there exist integers $k,l$ (we can take $k>0$) such that $kn + lp=1,$ and in particular $kn = 1 \pmod p.$ Then $\varepsilon_n ( X^k) = \zeta^{kn} = \zeta$ as desired.

Now for computing the kernel: We know $\Phi_p(X) = 1+ X + X^2 + \cdots + X^{p-1}.$ Now note that if I list out the elements of the ring $\mathbb{Z}/(p) : 0, 1, \ldots, p-1,$ and I multiply every term in the list by $n: 0, n, 2n, \ldots, (p-1)n,$ then the new list is simply a permutation of the original list (because multiplication by an invertible element is an automorphism of the group of units). Therefore $$\varepsilon_n (\Phi_p(X)) = 1+ \zeta^n + \zeta^{2n} + \cdots + \zeta^{(p-1)n}= 1+ \zeta + \zeta^2 + \cdots + \zeta^{p-1}=0.$$

This gives that $( \Phi_p(X) )\subseteq \ker \varepsilon_n.$ Now suppose $f(X)\in \mathbb{Z}[X]$ is such that $f(\zeta^n)=0.$ Note that $\Phi_p(X)$ is the minimal polynomial of $\zeta$ over $\mathbb{Q}$ (it is a famous result that the coefficients are in fact integers), and that our previous work implies that it is the minimal polynomial of $\zeta^n$ as well. The minimal polynomial of $\zeta^n$ must divide any polynomial having $\zeta^n$ as a root, so $\Phi_p(X) \mid f(X)$ as polynomials in the ring $\mathbb{Q}[X].$ Now do some work to show this is actually true in $\mathbb{Z}[X]$ as well (Hint: Gauss' Lemma), and conclude $\ker \varepsilon_n \subseteq ( \Phi_p (X) ).$

(b) From (a), The First Isomorphism Theorem gives $ \dfrac{\mathbb{Z}[X] }{ ( \Phi_p(X) )} \cong \mathbb{Z}[\zeta]$ via the map $\alpha_n: f(X) +( \Phi_p(X) ) \mapsto f(\zeta^n).$

In particular for $n=1$ we have $\dfrac{\mathbb{Z}[X] }{ ( \Phi_p(X) )} \cong \mathbb{Z}[\zeta]$ via the map $\alpha_1: f(X) +( \Phi_p(X) ) \mapsto f(\zeta).$

So then $ \alpha_1^{-1} \circ \alpha_n : \mathbb{Z}[\zeta]\to \mathbb{Z}[\zeta]$ is an isomorphism which sends $f(\zeta) \mapsto f(\zeta^n),$ which is precisely the map $\sigma_n$ in the question.

Answer 2

Hints:

Applying the first isomorphism theorem to the map $\epsilon_n$ in part (a), you get an isomorphism of $\Bbb Z[X]/(\ker(\epsilon_n))$ with $\Bbb Z[\zeta]$. Do you believe that $\Bbb Z [\zeta]\cong \Bbb Z[X]/(\Phi_p(X)) $? Combine these isomorphisms to produce the $\sigma_n$.