Here's the problem: (This problem is from Hungerford's Algebra Chapter 5 exercise 6.7 )
Let R be a principal ideal domain and $p$ a prime in $R$ and $n$ a positive integer. Then $R/(p^{n})$ is an unitary $R/(p^{n})$-module with $(r+(p^{n}))(s+(p^{n}))=rs+(p^{n})$. I attempt to demonstrate that the $R/(p^{n})$-module $R/(p^{n})$ is injective as following steps:
(1) I notice that the ring $R/(p^{n})$ is a principal ideal ring since it is a quotient ring of $R$ which is a PID. And by this fact, I prove that each proper ideal of $R/(p^{n})$ is generated by $p^{i}+(p^{n})$ with $0<i<n$.
(2) Now, in order to prove that module $R/(p^{n})$ is injective, it is sufficient to prove that for each ideal $I=(p^{i}+(p^{n}))$ of the ring $R/(p^{n})$ the module homomorphism $f: I\rightarrow R/(p^{n})$ can be extended to a homomorphism $R/(p^{n})\rightarrow R/(p^{n})$.
The module $R/(p^{n})$ is injective? how to extend this module homomorphism $f: I\rightarrow R/(p^{n})$? Anyway, any and all help is appreciated. Thanks.
All modules involved are cyclic (generated by one element), so it is sufficient to define maps on generators and extend them by linearity.
Let $f: (p^i)/(p^n) \longrightarrow R/(p^n)$ be $R/(p^n)$-linear, with $1 \leq i \leq n$. Note that (the following is $\mod{(p^i)}$) $$0=f(0) = f(p^n)=f(p^{n-i}p^i)=p^{n-i}f(p^i)$$ this means that $f(p^i) = ap^i$ for some $a \in R$.
So you can extend $f$ simply by defining $f(1)=a$ and extend by linearity.