For any representation $M$ over the integers, $pM$ is a submodule for any prime $p$. Thus, if $M$ is irreducible, then for any prime, either $pM=0$ or $pM=M$. For a finitely generated $\mathbb{Z}$-module $M\neq pM$ for at least one $p$. If $pM=qM=0$ for two distinct primes, then $M=(p,q)M=0$. Thus any non-zero irreducible has a unique $p$ such that $pM=0$, and for all other $qM=M$. That is, $M$ is an irreducible representation over $\mathbb{F}_p$.

These are easily determined for a cyclic group of order $n$. There's one for each $n$th root of unity $\zeta$ in $\bar{\mathbb{F}}_p$; let $q=p^\ell$ be minimal such that $\zeta\in \mathbb{F}_q\subset \bar{\mathbb{F}}_p$, take the obvious 1-dimensional representation over $\mathbb{F}_q$ with a generator acting by $\zeta$, and restrict scalars to $\mathbb{F}_p$.

For any representation $M$ over the integers, $pM$ is a submodule for any prime $p$. Thus, if $M$ is irreducible, then for any prime, either $pM=0$ or $pM=M$. For a finitely generated $\mathbb{Z}$-module $M\neq pM$ for at least one $p$. If $pM=qM=0$ for two distinct primes, then $M=(p,q)M=0$. Thus any non-zero irreducible has a unique $p$ such that $pM=0$, and for all others, we have $qM=M$. That is, $M$ is an irreducible representation over $\mathbb{F}_p$.

These are easily determined for a cyclic group of order $n$. There's one for each $n$th root of unity $\zeta$ in $\bar{\mathbb{F}}_p$; let $q=p^\ell$ be minimal such that $\zeta\in \mathbb{F}_q\subset \bar{\mathbb{F}}_p$, take the obvious 1-dimensional representation over $\mathbb{F}_q$ with a generator acting by $\zeta$, and restrict scalars to $\mathbb{F}_p$.