Injective hull of $\mathbb{ Z}_n$

Question

What is the injective hull of $\mathbb Z_n$?

I know that in case $n=p$ is prime, the injective hull would be isomorphic to $\mathbb Z_{p^∞}$, but in general case, I have no idea. Can anyone be of some help to me? Thanks in advance!

Answer 1

I assume (as rschwieb points out) that the question concerns the category of $\mathbb{Z}$-modules.

1) $\mathbb{Z}_{p^\infty}$ is the injective hull in the case when $n=p^k$ as well. Basically because we have $\mathbb{Z}_p\subseteq \mathbb{Z}_{p^k} \subseteq \mathbb{Z}_{p^\infty} $ and the extension $\mathbb{Z}_{p} \subseteq \mathbb{Z}_{p^\infty}$ is essential, hence so is the extension $\mathbb{Z}_{p^k} \subseteq \mathbb{Z}_{p^\infty}$.

2) In the general case one can write $$\mathbb{Z}_n \simeq \mathbb{Z}_{p_1^{k_1}} \times \mathbb{Z}_{p_2^{k_2}} \times \dots \times \mathbb{Z}_{p_m^{k_m}}, $$ where $p_i$'s are primes and $k_i$'s are integers. Then we have embedding to the produt of injective hulls of the components $$\mathbb{Z}_{p_1^{k_1}} \times \mathbb{Z}_{p_2^{k_2}} \times \dots \times \mathbb{Z}_{p_m^{k_m}} \subseteq \mathbb{Z}_{p_1^\infty} \times \mathbb{Z}_{p_2^\infty} \times \dots \times \mathbb{Z}_{p_m^\infty}, $$ which can be seen to be the injective hull of the product: Basically, for each component of the group its injective hull must be a direct summand of the inj. hull of the product. Since the product is injective , it folows that this is the inj. hull.

3) I would also expect that the injective hull can be alternatively described as $\mathbb{Z}_{n^\infty}$, i.e. somthing like $\mathbb{Z}[1/n]/\mathbb{Z}$, which is probably the mlre natural description.