exact sequence with infinite element

Question

How to prove:

Define exact sequence with infinite element from $R$-modules and $R$-homomorphism ,then exact sequence with infinite element from $\Bbb Z$-modules and $\Bbb Z$-homomorphism : $\require{AMScd}$ \begin{CD} ... @>>> \Bbb Z_4 @>>> \Bbb Z_4 @>>> \Bbb Z_4 @>>>...\\ \end{CD} exist.

is this solution true? $\require{AMScd}$ \begin{CD} ... @>i>> \Bbb Z_4 @>O>> \Bbb Z_4 @>i>> \Bbb Z_4 @>O>>...\\ \end{CD} for $R$ -module is true?

i.e: for $\Bbb Z_4$-modules (infinite $\Bbb Z_4$-module) : $Im O=\{0\}=\ker i, Im i=\Bbb Z_4=\ker O$ so this function to be found.

for $R$-modules (infinite $R$-module) : $Im O=\{0\}=\ker i, Im i=R=\ker O$

question is : is this solution true? (the question is simple)

$i$ function is $R(or\Bbb Z_4)$-identity homomorphism : ($i: R(or\Bbb Z_4) \rightarrow R(orZ_4) , i(x)=x , \forall x \in R(or \Bbb Z_4) )$

$O$ is $R(or\Bbb Z_4)$- homomorphism: (($O: R(or\Bbb Z_4) \rightarrow R(orZ_4) , i(x)=0_R(or 0_{\Bbb Z_4}) , \forall x \in R(or \Bbb Z_4) )$

Answer 1

Your solution is right, but you could ask instead: Is there a possibility not to use the zero homomorphism $f(x)=0$?

Still it works, choose $f:\mathbb{Z}_4 \to \mathbb{Z}_4, f(x)=2x$. Both $Im(f)$ and $ker(f)$ are equal to $\{0,2\}$ and this way

\begin{CD}\cdots @>{f}>>\mathbb{Z}_4 @>{f}>> \mathbb{Z}_4 @>{f} >> \mathbb{Z}_4 @>{f}>> \cdots \\\end{CD}

is an infinite exact sequence.

In General, the construction of $f(x)=n*x$ works for $\mathbb{Z}_{n^2}$, in this case $\mathbb{Z}_{2^2}$.