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Show that if $n=rs$ where $n,r,s>1$ are positive integers, then the $ \mathbb{Z}_n $-module $r \mathbb{Z}_n$ is projective but it is not free if $(r,s)=1$.

Any ideas or help how to prove this will be appreciated.

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  • $\begingroup$ What does "n=r,s" mean? $\endgroup$ – WillO May 29 '14 at 22:59
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$$0\to r\mathbb Z_n\to \mathbb Z_n \to s\mathbb Z_n\to 0$$ is a exact sequence with $f:r\mathbb Z_n \to \mathbb Z_n$ is identity map and $g:\mathbb Z_n\to s\mathbb Z_n$ is defined by $g(x)=sx$, then $f$ is one to one and $g$ is onto.

Thus, $\mathbb Z_n=r\mathbb Z_n\oplus s\mathbb Z_n$ as an $\mathbb Z_n$-module, $\mathbb Z_n$ is a free $\mathbb Z_n$-module which shows $r\mathbb Z_n$ and $s\mathbb Z_n$ are projective $\mathbb Z_n$-modules.

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