I have some confusion regarding Dummit and foote Book Algebra, page number $554$
It is written that
$$x^n-1 = \prod_{\substack{\zeta \in \mu_n}} \left(x -\zeta \right)= \prod_{\substack{d|n}} \prod_{\substack{\zeta \in \mu_d} \zeta \text{ is primitive}} \left(x -\zeta \right)$$
My question : Im not getting How these two product came ? I means $\prod_{\substack{d\mid n}} \prod_{\substack{\zeta \in \mu_d}}$ in $x^n-1$ ?
My thinking :I know that here $\zeta = e^{\frac{2i\pi k}{n}}$ and $\mu_n = \mathbb{Z}_n$.
$\{1,2,…,(n-1)\} \in \mathbb{Z_n}$
$$x^n-1 = \prod_{\substack{\zeta \in\mathbb{Z}_n }} \left(x -e^{\frac{2i\pi k}{n}} \right)$$
$$x^n-1 = \prod_{\substack{(n-1)! }} \left(x -e^{\frac{2i\pi k}{n}} \right)$$
since $1\times 2\times\cdots\times n-1=(n-1)!$
Best Answer
You understand that $x^n-1= \prod (x- \zeta)$, where $\zeta$ varies over all of the $n$th roots of unity, right? And any $n$th root of unity is a primitive $d$th root of unit of unity for some unique $d$, where $d \mid n$.
So all we're doing is collecting the roots that are primitive $d$th roots of unity for the same $d$, and then taking the product over all $d$ such that $d \mid n$. This product will pick up all of the $n$th roots of unity and each specific root will occur exactly once. The multiplicative group of the primitive $d$th roots of unity is isomorphic to the multiplicative group $U_d$ of units of the ring $\Bbb Z / d \Bbb Z$.