# Decomposing left regular representation of cyclic group over $\mathbb{Q}$

Let $G$ be a cyclic group of order $p$, where $p$ is prime. Let $V = \mathbb{Q}(G)$ be the rational group ring of $G$. How do you explicitly decompose $V$ as a direct sum of irreducible representations? If we were working over $\mathbb{C}$, then I know that the irreducible representations of $G$ are just $1$-dimensional representations where the generator of $G$ acts by a root of unity (though even there I don't know an explicit decomposition of the left regular representation). But over $\mathbb{Q}$, I don't even know the irreducible representations.

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$$\mathbb{Q}[\mathbb{Z}/n] = \mathbb{Q}[X]/(X^n - 1) = \mathbb{Q}[X]/\displaystyle\prod_{d|n} \Phi_d = \bigoplus_{d|n} \mbox{ } \mathbb{Q}[X]/\Phi_d = \bigoplus_{d|n} \mbox{ } \mathbb{Q}(\zeta_d).$$
The key point here being: the rational regular representation of $\mathbb Z/n\mathbb Z$ is isomorphic to $\mathbb Q[X]/(X^n-1)$ with the generator of the group acting as multiplication by $X$. And there is an implicit statement that the subrepresentations determined by the cyclotomic polynomials are irreducible, which may noy be obvious to all. –  Mariano Suárez-Alvarez Oct 1 '11 at 3:10
Great, so for $n$ a prime number this is the direct sum of a copy of the trivial representation and one other representation. Where can I read about this sort of thing (for instance, the fact that the subrepresentations determined by cyclotomic polynomials are irreducible)? –  Felix Oct 1 '11 at 3:23
The $\mathbb{Q}[\mathbb{Z}/n]$ action on $\mathbb{Q}(\zeta_d)$ is given by $\alpha \cdot x = \pi(\alpha)x$ where $\pi: \mathbb{Q}[\mathbb{Z}/n] \rightarrow \mathbb{Q}(\zeta_d)$ is the quotient map. It follows that the $\mathbb{Q}[\mathbb{Z}/n]$-submodules of $\mathbb{Q}(\zeta_d)$ are exactly the $\mathbb{Q}(\zeta_d)$-submodules of $\mathbb{Q}(\zeta_d)$ i.e., the ideals of $\mathbb{Q}(\zeta_d).$ As $\mathbb{Q}(\zeta_d)$ is a field, $\mathbb{Q}(\zeta_d)$ contains only two such modules: $0$ and $\mathbb{Q}(\zeta_d)$. Irreducibility follows. –  jspecter Oct 1 '11 at 3:40