I'm trying to prove that $\mathbb{Z}[C_n] \cong \mathbb{Z}[X]/(X^n-1)$ where $C_n$ is the cyclic group of order $n$. Let $t$ be a generator of $C_n$. There is a unique ring homomorphism
$$\Phi:\mathbb{Z}[X] \to \mathbb{Z}[C_n]$$ sending $X$ to $t$. Obviously $(X^n-1) \subseteq \ker \Phi$. Let $f = \sum_{i=0}^m a_iX^i$ be in the kernel of $\Phi$. If $m \leq n-1$, then clearly $a_i = 0$ for all $i$, hence $f = 0$. Otherwise we have $m > n-1$ and applying $\Phi$ to $f$ we see that $$-\sum_{i=0}^{m-n}a_{i+n}t^i = \sum_{i=0}^{n-1}a_it^i.$$ From here on I tried to find a suitable polynomial g in $\mathbb{Z}[X]$ such that $g(X)(X^n-1) = f$ but failed. Can someone help me? Thanks in advance
EDIT: Here $\mathbb{Z}[G]$ denotes the group ring over $G$, i.e. the ring of all finite linear combinations $\sum_{g}a_gg$ where $a_g \in \mathbb{Z}$ and $g \in G$.