$\operatorname{\mathcal{Jac}}\left( \mathbb{Q}[x] / (x^8-1) \right)$

Question

$\DeclareMathOperator{\Jac}{\mathcal{Jac}}$ Using the fact that $R := \mathbb{Q}[x]/(x^8-1)$ is a Jacobson ring and thus its Jacobson radical is equal to its Nilradical, I already computed that $\Jac \left( \mathbb{Q}[x] / (x^8-1) \right) = \{0\}$. But I'd like to calculate this in an elementary way, preferably using nothing beyond the very basic properties of the Jacobson radical, quotients of rings and maximal ideals. And here starts my struggle.

Answer 1

Hints:

1. $\Bbb Q[x]$ is a principal ideal domain

2. The ideals (including the maximal ideals) sitting above $(x^8-1)$ just depend on the factorization $x^8-1=(x-1)(x+1)(x^2+1)(x^4+1)$.

3. If $p$ and $q$ are coprime polynomials, then $(p)\cap (q)=(pq)$.

4. The nilradical of a commutative ring is certainly always contained in the Jacobson radical.

Answer 2

Hint: $\,{\rm max}\ M\subset K[x]/(p_1\cdots p_n)\! \iff\! M=(p_i)\,\Rightarrow\, \cap\, M_ i =\, \cap\, (p_i) = (p_1\cdots p_n) = (0)$