Let $R$ a ring and $M$ a $R-$module. I know that $M$ is a free module if and only if $$M\cong R^{\oplus s}.$$ I know that a finitely generated module $M$ is a module s.t. there is a surjection $$R^{\oplus t}\longrightarrow M.$$

The thing, I can't describe a finitely generated module that is not a free module (I unfortunately have vector space in my mind, and I know that they are always free). Could you please give me an example of a finitely generated module that is not free ?


Take $\mathbb Z/n\mathbb Z$ as a $\mathbb Z-$module.

You will have a surjection $$\varphi:\mathbb Z\longrightarrow \mathbb Z/n\mathbb Z,$$ given by the natural homomorphism (i.e. $\varphi(n)=[n]$), but it unfortunately won't be injectif since $\ker \varphi=n\mathbb Z.$

  • $\begingroup$ k, but why there is no bijection between $\mathbb Z$ and $\mathbb Z/n\mathbb Z$ ? You just gave a surjection, not there is no bijection. $\endgroup$
    – user386627
    Dec 5 '16 at 15:59
  • 1
    $\begingroup$ Since as group they can't be isomorphic ($\mathbb Z$ is an infinite group and $\mathbb Z/n\mathbb Z$ is finite). Therefore, as module they can't be isomorphic. $\endgroup$
    – Surb
    Dec 5 '16 at 16:02

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