Injective dimension of $\mathbb Z_n$ as a $\mathbb Z$-module 
What is the injective dimension of $\mathbb Z_n$ as a $\mathbb Z$-module? 

Can one use the well-known fact that $id(M)$ is less than or equal to $i$ iff $Ext^{i+1}(N,M)=0$ for all $N$? Thanks in advance!
 A: If you look at the wikipedia page, $\mathbb Q/\mathbb Z$ is injective. The obvious map $\mathbb Z_n\to\mathbb Q/\mathbb Z$ is an injection, and the quotient of this inclusion is isomorphic to $\mathbb Q/\mathbb Z$, so you have an injective resolution:
$$0\rightarrow\mathbb Z_n\rightarrow \mathbb Q/\mathbb Z\rightarrow \mathbb Q/\mathbb Z\to 0$$
So all you have to do is prove that the $\mathbb Z_n$ is not itself injective to get the dimension is $1$.
A: From the following exact sequence $0\to\mathbb Z\stackrel{n\cdot}\to\mathbb Z\to\mathbb Z/n\mathbb Z\to 0$ we get $\operatorname{Ext}^2(N,\mathbb Z/n\mathbb Z)=0$ for all $N$, so the injective dimension of $\mathbb Z/n\mathbb Z$ as $\mathbb Z$-module is at most $1$. However, it is not injective since it isn't divisible, so its injective dimension must be $1$. 
(I have used that the injective dimension of $\mathbb Z$ as $\mathbb Z$-module is $1$, and this follows easily from the short exact sequence $0\to\mathbb Z\to\mathbb Q\to\mathbb Q/\mathbb Z\to 0$.)  
A: In $\textrm{Mod-}\mathbb{Z}$, a module is injective if and only if it is divisible. If a module $M$ is not divisible, embed it into its injective envelope $E(M)$. Then the exact sequence
$$
0\to M\to E(M)\to E(M)/M\to0
$$
shows that the injective dimension of $M$ is at most $1$, because $E(M)/M$ is divisible, hence injective.
This can be generalized to (right) hereditary rings, that is, rings in which quotients of injective right modules are injective.
Exercise. A ring is right hereditary if and only if submodules of projective modules are projective.
