Let $X_n= \begin{bmatrix} x_1&x_2&\cdots&x_n\\ x_n&x_1&\cdots&x_{n-1}\\ \vdots&\vdots&\ddots&\vdots\\ x_2&x_3&\cdots&x_1\\ \end{bmatrix} $ be a circulant matrix. It is well known that over $\mathbb{C}$ we have the factorization $\displaystyle\det X_n=\prod_{\zeta^n=1}\left(\sum_{j=0}^{n-1}\zeta^jx_j\right)$. I am interested in how $\det X_n$ factorizes over $ \mathbb{Z}$. For example, we have
$\det X_2=x_1^2-x_2^2=(x_1-x_2)(x_1+x_2)$.
$\det X_3=x_1^3+x_2^3+x_3^3-3x_1x_2x_3=(x_1+x_2+x_3)(x_1^2+x_2^2+x_3^2-x_1x_2-x_1x_3-x_2x_3)$
$\det X_4=(x_1+x_2+x_3+x_4)(x_1-x_2+x_3-x_4)(x_1^2+x_2^2+x_3^2+x_4^2-2x_1x_3-2x_2x_4)$
Conjecture. The number of irreducible factors should be precisely the number of divisors of $n$.
Motivation. The circulant determinant is related to the representation theory of the cyclic group $C_n$ of order $n$, and we have a decomposition of the group ring $\mathbb{Q}[C_n]=\mathbb{Q}[t]/(t^n-1)=\oplus_{d\mid n} \mathbb{Q}[t]/\Phi_d(t) = \oplus_{d\mid n} \mathbb{Q}(e^{2\pi i/d})$.
Observation. We can group together the complex terms using isomorphisms of $\mathbb{Q}(e^{2\pi i/n})$ that by Galois theory should give integral terms.
Question. Is the above conjecture true? If so, what is a good reference with a proof?
(I am unfamiliar with representation theory over $\mathbb{Q}$, and could not find online whether the conjecture is true).
