I am trying to show that $\mathbb{Z}/3\mathbb{Z}$ is a projective $\mathbb{Z}/12\mathbb{Z}$-module but not a free $\mathbb{Z}/12\mathbb{Z}$-module.

So, far I have been able to show that $4(\mathbb{Z}/12\mathbb{Z})\cong \mathbb{Z}/3\mathbb{Z}$ and $3(\mathbb{Z}/12\mathbb{Z})\cong \mathbb{Z}/4\mathbb{Z}$. Moreover, $\mathbb{Z}/3\mathbb{Z}\bigcap \mathbb{Z}/4\mathbb{Z}=\{0\}$. So that $\mathbb{Z}/12\mathbb{Z}\cong \mathbb{Z}/3\mathbb{Z}\bigoplus \mathbb{Z}/4\mathbb{Z}$.

Now, I want an easy way of seeing that $\mathbb{Z}/3\mathbb{Z}$ is not a free on $\mathbb{Z}/12\mathbb{Z}$.

Also, that $\mathbb{Z}/3\mathbb{Z}$ is a projective $\mathbb{Z}/12\mathbb{Z}$-module.

  • $\begingroup$ You already showed that its projective. Direct summands of free modules are projective. $\endgroup$ – Jyrki Lahtonen May 15 '18 at 13:25
  • $\begingroup$ Any finite free $\mathbb Z/12$-module has $12k$ elements for some $k\in\mathbb N_0$. $\endgroup$ – Pedro Tamaroff May 15 '18 at 13:27


Free modules over $R$ look like $\oplus_{i\in I}R$.

Can't you just count the elements of such a module when $I$ is a finite set?


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