If $G$ is cyclic of order n, $\mathbb{Q}[G]\cong \oplus_{d|n}\mathbb{Q[\xi_d]}$

Question

I am trying to find a proof of this result.

If $G$ is cyclic of order n, $\mathbb{Q}[G]\cong \oplus_{d|n}\mathbb{Q[\xi_d]}$.

I think the proof will involve the use of characters of finite abelian groups (which extend to homomorphisms of $\mathbb{Q}$-algebras between $\mathbb{Q}[G]$ and $\mathbb{Q}[\xi_m]$).

I am just looking for a good reference where I could find this result or useful information about characters of finite abelian groups which would help me prove this.

Thanks in advance.

Answer 1

Let $G=C_n$ be a cyclic group, $g$ a generator. Let $f:\mathbb{Q}[x]\to \mathbb{Q}[G]$, $\sum a_kx^k\mapsto \sum a_kg^k$. This is a surjective ring homomorphism, and its kernel is the ideal $(x^n-1)$. So,

$$\mathbb{Q}[G]\cong \frac{\mathbb{Q}[x]}{(x^n-1)}$$

This polynomial factors over $\mathbb{Q}$ as

$$x^n-1=\prod_{d|n}\Phi_d(x)$$

where $\Phi_d(x)$ is the $d^{th}$ cyclotomic polynomial. By the Chinese remainder theorem, these two facts imply

$$\mathbb{Q}[G]\cong \prod_{d|n}\frac{\mathbb{Q}[x]}{\Phi_d(x)}$$

and then finally, $\frac{\mathbb{Q}[x]}{\Phi_d(x)}=\mathbb{Q}[\xi_d]$ is the $d^{th}$ cyclotomic field from the question.