Let's suppose that I have an element $e$ of order $p$ in the group of complex numbers whose elements all have order $p^n$ for some $n\in\mathbb{N}$ (henceforth called $K$), and the module generated by $(e)$ is irreducible.
How do I show that the injective hull of the module generated by $(e)$ is in fact, equal to $K$?
Hint: The abelian group of complex numbers that are $p^n$-th roots of unity for some $n$ is isomorphic to ${\mathbb Z}\left[\frac{1}{p^n}\right]/{\mathbb Z}$, with the submodule generated by the $p$-th roots of unity corresponding to ${\mathbb Z}\frac{1}{p}$. Now the strategy is exactly the same as for Injective hull of the trivial k[x]-module (can you see the common generalization of these examples?).