Number of prime ideals of a ring

Question

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.

marlu · Answer 1 · 2013-06-25T00:06:41.470

17

The prime ideals of $\mathbb Q[x]/(x^m-1)$ correspond to prime ideals in $\mathbb Q[x]$ containing $x^m-1$, by taking the preimage under the surjection $\mathbb Q[x] \to \mathbb Q[x]/(x^m-1)$. Since $\mathbb Q[x]$ is a PID, those ideals are $(f)$ for $f$ an irreducible factor of $x^m-1$. We have the factorization $$x^m-1 = \prod_{d|m} \Phi_d$$ where $\Phi_d$ is the $d$th cyclotomic polynomial, which is irreducible. So the number of irreducible factors of $x^m-1$ is $\tau(m)$, the number of positive divisors of $m$.

edited Jun 25 '13 at 00:06

answered Mar 15 '13 at 10:54

marlu

14,264

1

@City and so more generally, $x^m-1$ can be replaced with any polynomial, and the same discussion about prime ideals corresponding to prime factors of the polynomial applies. – rschwieb Mar 15 '13 at 13:10
$\sigma(m)$ usually means the sum of the positive divisors of $m$. – M Turgeon Jun 21 '13 at 18:47
Either use $\sigma_0(m)$ or $\tau(m)$. – lhf Jun 21 '13 at 18:51
Thanks, I've corrected that. – marlu Jun 25 '13 at 00:07

score 12 · Answer 2 · answered Mar 15 '13 at 13:45

12

A small addendum to marlu's answer: The Chinese Remainder Theorem implies $$\mathbb{Q}[x]/(x^m-1) \cong \prod_{d|m} \mathbb{Q}(\zeta_d),$$which is a finite direct product of fields. So the (prime) ideal structure is quite easy.

answered Mar 15 '13 at 13:45

Martin Brandenburg

181,922

score 11 · Accepted Answer · edited Jun 23 '22 at 18:52

This is obtained by stringing together several basic facts.

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.)
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.
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 $n$.

Number of prime ideals of a ring

3 Answers3