Let $G$ be a cyclic finite group of order $n$. I tried to determine the structure of the group ring $\mathbb Q[G]$ over the rationals $\mathbb Q$, what I got for even $n$ is $$ \mathbb Q[G] = A \oplus B \oplus C $$ where \begin{align*} A & = \left\{ \sum_{i=0}^{n-1} a_i g^i : a_0 = a_1 = \ldots = a_{n-1} \right\} \\ B & = \left\{ \sum_{i=0}^{n-1} a_i g^i : a_1 = a_3 = \ldots = a_{n-1}, a_2 = a_4 = \ldots = a_n, ~ a_i + a_{i+1} = 0, i = 1,3,5,\ldots, n\right\} \\ C & = \left\{ \sum_{i=0}^{n-1} a_i g^i : \sum_{i=0}^{n-1} a_i = 0 \mbox{ and the condition for $B$ fails } \right\} \end{align*} and we have $B \oplus C = \{ \sum_{i=0}^{n-1} a_i g^i : \sum_{i=0}^{n-1} a_i = 0 \}$. For odd $n$ we have $$ \mathbb Q[G] = A \oplus B $$ with \begin{align*} A & = \left\{ \sum_{i=0}^{n-1} a_i g^i : a_0 = a_1 = \ldots = a_{n-1} \right\} \\ B & = \left\{ \sum_{i=0}^{n-1} a_i g^i : \sum_{i=0}^{n-1} a_i = 0 \right\}. \end{align*}
How to show that this decomposition could not be refined further, i.e. the modules are simple?
As $A$ is in both cases $1$-dimensional as a $\mathbb Q$-vector spaces, it is already irreducible, similar $B$ in case one is one dimensional, as it could be generated as a $\mathbb Q$-vector space by the vector $(1,-1,1,-1,\ldots, 1,-1)^T$. But how to show that $C$ is an irreducible $\mathbb Q[G]$-module? If we look at the group ring over $\mathbb C$, as this field is algebraically closed and $G$ is abelian, it could be decomposed into one-dimensional simple $\mathbb C[G]$-modules, so that this does not work for $C$ above must be related to the case that $\mathbb Q$ is not algebraically closed, but I do not know how to make an argument out of this observation?
Your claim is not true. The correct decomposition for a finite cyclic group of order $n$ is
$$\mathbb{Q}[G] \cong \mathbb{Q}[x]/(x^n - 1) \cong \prod_{d \mid n} \mathbb{Q}[x]/\Phi_d(x)$$
by the Chinese remainder theorem, where $\Phi_d(x)$ denotes the $d^{th}$ cyclotomic polynomial, and another name for $\mathbb{Q}[x]/\Phi_d(x)$ is the cyclotomic field $\mathbb{Q}(\zeta_d)$.
The first even $n$ for which there are more than $3$ factors in this decomposition is $n = 6$, which has divisors $1, 2, 3, 6$, reflecting the factorization
$$x^6 - 1 = (x - 1)(x^2 + x + 1)(x + 1)(x^2 - x + 1).$$