I would like to determine the factors of $X^N - 1$ over different integer rings, such as $$ \mathbb{Z}[X]\ \mbox{and}\ \mathbb{Z}_p[X]\quad \mbox{for}\quad prime\,\, p $$ In particular, I am interested when $$ \mathbb{Z}_p[X]/(X^N - 1)\ \mbox{is a field or how it decomposes into fields.} $$
Thanks for replies or pointers to references !.
Over $\mathbb{Z}$ the answer is that $x^n - 1$ decomposes into a product of cyclotomic polynomials
$$x^n - 1 = \prod_{d \mid n} \Phi_d(x)$$
where $\Phi_d(x)$ is the monic polynomial whose roots are exactly the primitive $d^{th}$ root of unity, and hence has degree $\varphi(d)$, the totient function. which With this definition the above factorization is just a way to organize the fact that every $n^{th}$ root of unity is a primitive $d^{th}$ root of unity for a unique divisor $d$ of $n$, but the surprising fact is that the cyclotomic polynomials have integer coefficients and are irreducible over $\mathbb{Z}$. The integer coefficients are not so hard to see but the irreducibility, while plausible, requires nontrivial work to show; see this question for a proof.
Over a finite field $\mathbb{F}_p$ the above factorization still holds but the cyclotomic polynomials don't usually remain irreducible. Instead what happens is the following. In general, if $f(x) \in \mathbb{F}_p[x]$ is any polynomial, we can consider its splitting field, which is some finite extension $\mathbb{F}_{p^k}$ of $\mathbb{F}_p$. This is a Galois extension with Galois group generated by the Frobenius map $x \mapsto x^p$, so it follows that the irreducible decomposition of $f(x)$ over $\mathbb{F}_p$ is given by the orbits of the Frobenius map acting on its roots.
By this I mean that if $\alpha \in \mathbb{F}_{p^k}$ is a root of $f(x)$ and applying the Frobenius map produces elements $\alpha, \alpha^p, \alpha^{p^2}, \dots $ until we get to $\alpha^{p^{n-1}}$ and then repeat, then $\prod_{i=0}^{n-1} (x - \alpha^{p^i})$ is an irreducible factor of $f(x)$, and every irreducible factor arises in this way.
So, to apply this argument to $x^n - 1$ we need to understand how Frobenius acts on its roots in the splitting field over $\mathbb{F}_p$. We would like to say that these roots are just the $n^{th}$ roots of unity, and they are, but there aren't necessarily $n$ of them in general. Specifically, writing $n = p^k m$ where $\gcd(p, m) = 1$, we have
$$x^n - 1 = x^{p^k m} - 1 \equiv (x^m - 1)^{p^k} \bmod p$$
so in general the roots are the $m^{th}$ roots of unity, repeated $p^k$ times. So we reduce to discussing $x^m - 1$ where $\gcd(m, q) = 1$.
A root of $x^m - 1$ is a primitive $d^{th}$ root of unity $\zeta_d$ for some divisor $d \mid m$ (these correspond to the roots of $\Phi_d(x)$, as above). The Frobenius map acts via $\zeta_d \mapsto \zeta_d^p$, so as we keep applying Frobenius we get the elements $\zeta_d^{p^i}$. The first time we get $\zeta_d$ back occurs for the smallest value of $i$ such that
$$p^i \equiv 1 \bmod d$$
which is the multiplicative order $\text{ord}_d(p)$ of $p \bmod d$. Since this is true for any primitive $d^{th}$ root of unity, the conclusion is that $\Phi_d(x) \bmod p$ breaks up into a product of irreducible polynomials (whose roots are orbits as above) all of which have the same degree $\text{ord}_d(p)$.
So the decomposition of $\mathbb{F}_p[x]/(x^m - 1)$ is the following (if $p \mid n$ then we don't get a product of fields because there are nilpotents so I will ignore this case). We get one copy of the finite field $\mathbb{F}_{p^i}$ for every distinct irreducible factor of $x^m - 1$ of degree $i$, by the Chinese remainder theorem. So we get
$$\boxed{ \mathbb{F}_p[x]/(x^m - 1) \cong \prod_{d \mid m} \mathbb{F}_{p^{\text{ord}_d(p)}}^{\frac{\varphi(d)}{\text{ord}_d(p)}} }.$$
In particular it is never a field unless $m = 1$ because $x^m - 1$ is always divisible by $x - 1$.
