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.
[Math] Decomposing left regular representation of cyclic group over $\mathbb{Q}$
representation-theory
Related Solutions
The question of how to classify all finite-dimensional representations of $C_n$ over an arbitrary field $F$ can be studied using the structure theorem for finitely-generated modules over a principal ideal domain, in this case $F[x]$. The structure theorem asserts that any finitely-generated module is uniquely a finite direct sum of modules of the form $F[x]/p(x)^r$ where $p \in F[x]$ is irreducible and $r$ is a non-negative integer.
If $T$ is an operator acting on $F^k$ for some $n$, then $F^k$ becomes a finitely-generated module over $F[x]$ with $x$ acting by $T$. $T$ gives a representation of the cyclic group $C_n$ if and only if $T^n = 1$, in which case the summands $F[x]/q(x)^r$ in the decomposition of $F^k$ must have the property that $q(x)^r | x^n - 1$.
If $F$ has characteristic $0$ or has characteristic $p$ and $p \nmid n$, then $x^n - 1$ is separable over $F$, hence $r \le 1$ and $F^k$ is a direct sum of irreducible representations, all of which are of the form $F[T]/q(T)$ where $q$ is an irreducible factor of $x^n - 1$ over $F$.
If $F$ has characteristic $p$ and $p | n$, then writing $n = p^s m$ where $p \nmid m$ we have $$x^n - 1 = (x^m - 1)^{p^s}$$
from which it follows that $r \le p^s$ (but now it is possible to have $r > 1$). If $r > 1$, then the corresponding representation $F[T]/q(T)^r$ is indecomposable and not irreducible, where $q$ is an irreducible factor of $x^m - 1$ over $F$. The irreducible representations occur precisely when $r = 1$. In other words,
The irreducible representations of $C_{p^s m}$, where $p \nmid m$, over a field of characteristic $p$ all factor through the quotient $C_{p^s m} \to C_m$.
One can also see this more directly as follows. If $V$ is an irreducible representation of $C_{p^s m}$ over a field of characteristic $p$ and $T : V \to V$ is the action of a generator, then $$T^{p^s m} - 1 = (T^m - 1)^{p^s} = 0.$$
Thus $T^m - 1$ is an intertwining operator which is not invertible, so by Schur's lemma it is equal to zero.
The $W_i$ are just the eigenspaces of the matrix $\rho(g)$, with eigenvalue $\omega^i$ on $W_i$. In this case, there is a neat way to explicitly write down the eigenspaces: the vector $$v_i=(1,\omega^i,\omega^{2i},\dots,\omega^{(n-1)i})$$ is easily seen to be an eigenvector for $\rho(g)$ with eigenvalue $\omega^i$. So, $W_i$ is just the span of $v_i$ for each $i$.
(In case the formula for $v_i$ seems miraculous, note that it is easy to derive: if $v=(a_0,a_1,\dots,v_{n-1})$ then $\rho(g)v=(a_1,a_2,\dots,a_0)$ so to be an eigenvector with eigenvalue $\omega^i$ we must have $a_1=\omega^ia_0$, $a_2=\omega^ia_1=\omega^{2i}a_0$, and so on.)
Best Answer
$$\mathbb{Q}[\mathbb{Z}/n] = \mathbb{Q}[X]/(X^n - 1) = \mathbb{Q}[X]/\displaystyle\prod_{d|n} \Phi_d = \bigoplus_{d|n} \mbox{ } \mathbb{Q}[X]/\Phi_d = \bigoplus_{d|n} \mbox{ } \mathbb{Q}(\zeta_d).$$