The isomorphism of $\mathbb{Z}_{p}[x] /\left(x^{n}-1\right)$ using Hensel's Lemma

Question

I an trying to prove the following.Let p:prime, $n\in\mathbb{N}$ with $(n,p)=1$ ,and $x^{n}-1=f_{1}(x) \cdot \ldots \cdot f_{r}(x) \quad\left(f_{1}(x), \dots, f_{r}(x) \in \mathbb{Z}_{p}[x]\right.$ are irreducible)

then,

$\mathbb{Z}_{p}[x] /\left(x^{n}-1\right) \simeq \mathbb{Z}_{p}[x] /\left(f_{1}(x)\right) \oplus \ldots \oplus \mathbb{Z}_{p}[x] /\left(f_{r}(x)\right)$.

i can use the above isomorphism when $\mathbb{Z}_{p}$ is replaced by $\mathbb{F}_{p}$ and Hensel's Lemma.

Does anyone know how to prove this? or if you know some website that explain this claim,please let me know. Thank you very much in advance.

Answer 1

• $F(x) = x^n-1$, factorize in irreducibles $F(x) = \prod_{j=1}^J f_j(x)^{e_j}\in \Bbb{Q}_p[x]$ where the $f_j$ are monic, since the roots of $F$ have $p$-adic valuation $\ge 0$ the $f_j \in \Bbb{Z}_p[x]$.

  For $p \nmid n$ then $ \gcd(F,F') = 1$ so $e_j = 1$.

  Letting $g_j(x) \equiv 1 \bmod f_j(x), \equiv 0 \bmod f_i(x), (i\ne j)$ you have an isomorphism $\Bbb{Z}_p[x]/(F(x)) \to \prod_{j=1}^J \Bbb{Z}_p[x]/(f_i(x))$ sending $u(x) = u(x)\sum_{j=1}^J g_j(x) \to (u(x)g_1(x),\ldots, u(x)g_J(x))$.

• The splitting field $\Bbb{Q}_p(\zeta_n)/\Bbb{Q}_p$ of $F \in \Bbb{Z}_p[x]$ is Galois with Galois group $G$, let $O_K$ be its ring of integers with unique maximal ideal $(\pi)$.

  $G$ sends $O_K$ and $(\pi)$ to themselves, it is also well-defined on the residue field $O_K/(\pi) \cong \Bbb{F}_{p^d}$, which is he splitting field of $F\in \Bbb{F}_p[x]$ where $d$ is the order of $p \bmod n$.

  For each $j$ pick a root $\zeta_n^{a_j}$ of $f_j$, say $a_1= 1$. Then $f_j(x) = \prod_{\beta \in G \zeta_n^{a_j}} (x-\beta)$.

  If one of the reduction $f_j \in \Bbb{F}_p[x]$ is not irreducible then $[K:\Bbb{Q}_p] > [\Bbb{F}_{p^d}:\Bbb{F}_p]$, which means there is one non-trivial $\sigma \in G$ which is trivial on $O_K/(\pi)$, and hence the reduction $f_1 \in \Bbb{F}_p[x]$ has a double root.

  We know it isn't the case since $f_1\in \Bbb{F}_p[x]$ divides $F(x) \in \Bbb{F}_p[x]$ and $\gcd(F,F') = 1 \in \Bbb{F}_p[x]$.

  Whence $|G| = [K:\Bbb{Q}_p] = [\Bbb{F}_{p^d}:\Bbb{F}_p]=d$, the elements of $G$ send $\zeta_n \to \zeta_n^{p^m}$ for $ m \in 0 \ldots d-1$, and the factorization of $F \in \Bbb{Z}_p[x]$ corresponds to the partition of the roots of unity under action of $p$th powers.