TITLE: Number of prime ideals of a ring
QUESTION [15 upvotes]: Could anyone tell me how to find the number of distinct prime ideals of the ring $$\mathbb{Q}[x]/\langle x^m-1\rangle,$$ where $m$ is a positive integer say $4$, or $5$?
What result/results I need to apply to solve this problem? Thank you for your help.

REPLY [11 votes]: This is obtained by stringing together several basic facts.
1) Let $P(x) = x^n-1$.  Then $P(x) = \prod_{d \mid n} \Phi_d(x)$ is a product of cyclotomic polynomials.  The product extends over all positive integers $d$ dividing $n$.  It is known that each $\Phi_i$ is irreducible over $\mathbb{Q}$: see e.g. $\S$ 10.1.2 of these notes.  Since $\operatorname{gcd}(P(x),P'(x)) = 1$, $P(x)$ is a squarefree polynomial: it has no repeated irreducible factors.  (Or look directly at the roots of $\Phi_d$ -- they are the primitive $d$th root of unity -- to see that the $\Phi_d$'s are distinct monic polynomials.)
2) If $P(x) \in \mathbb{Q}[x]$ is a product of distinct irreducible polynomials $P_1(x) \cdots P_r(x)$, then by the Chinese Remainder Theorem, $\mathbb{Q}[x]/(P(x)) \cong \prod_{i=1}^r \mathbb{Q}[x]/(P_i(x))$.  Since each $P_i$ is irreducible, this is a product of $r$ fields.
3) The prime ideals in a finite product $\prod_{i=1}^r R_i$ of commutative rings are precisely those of the form $R_1 \times \ldots \times R_{i-1} \times \mathfrak{p}_i \times R_{i+1} \times \ldots \times R_r$, with $\mathfrak{p}_i$ a prime ideal of $R_i$.  In particular a product of $r$ fields has precisely $r$ ideals: we must take each $\mathfrak{p}_i = 0$.
Putting these together we get that the number of prime ideals in $\mathbb{Q}[x]/(x^n-1)$ is $d(n)$, the number of (positive) divisors of $n$.