Help proving a short exact sequence

Question

Show the following sequence is an exact sequence of $\mathbb Z$-modules when $n$ is a positive integer such that $n=rs$: $$ 0 \to r\mathbb{Z}_n \to \mathbb{Z}_n \to s\mathbb{Z}_n \to 0. $$

should i attack this problem using the statement tha a short sequence is exact if and only if $$ Ker(f)=Im(g) $$ which function should i consider for $f$ and $g$ ?

Answer 1

You are given the sequence $$ 0\to r\Bbb Z_n\xrightarrow{g}\Bbb Z_n\xrightarrow{f} s\Bbb Z_n\to 0\tag{1} $$ where $n=rs$, $g(rx+n\Bbb Z)=rx+n\Bbb Z$, and $f(x+n\Bbb Z)=sx+n\Bbb Z$.

Note that \begin{align*} \DeclareMathOperator{im}{im}\im g &= \{rx+n\Bbb Z:x\in\Bbb Z\} \\ &= \{x+n\Bbb Z:r\mid x\} \end{align*} Furthermore \begin{align*} \ker f &= \{x+n\Bbb Z:\exists \ell\in\Bbb Z, sx=\ell n\} \\ &= \{x+n\Bbb Z:\exists\ell\in\Bbb Z,sx=\ell rs\} \\ &= \{x+n\Bbb Z:\exists\ell\in\Bbb Z,x=\ell r\} \\ &= \{x+n\Bbb Z:r\mid x\} \end{align*} Hence $\ker f=\im g$.

To see that $g$ is injective, note that \begin{align*} \ker g &= \{rx+n\Bbb Z:\exists\ell\in\Bbb Z,rx=n\ell\} \\ &= \{rx+n\Bbb Z:\exists\ell\in\Bbb Z,rx=rs\ell\} \\ &= \{rx+n\Bbb Z:\exists\ell\in\Bbb Z,x=s\ell\} \\ &= \{rs\ell+n\Bbb Z:\ell\in\Bbb Z\} \\ &= \{n\ell+n\Bbb Z:\ell\in\Bbb Z\} \\ &\simeq 0 \end{align*}

To see that $f$ is surjective note that \begin{align*} \im f &= \{sx+n\Bbb Z:x\in\Bbb Z\} \\ &= s\Bbb Z_n \end{align*} This proves that (1) is exact.

Answer 2

Let $f:rZ_n\to Z_n$ and $g:Z_n\to sZ_n$

Take $f$ as a identity map and $g(x)=sx$ then you can easily see that $Image(f)=ker(g)$ and $g$ is onto.