6
$\begingroup$

I want to show that every $\mathbb{Z}/6\mathbb{Z}$-module is a direct sum of projective modules.

As abelian group, $\mathbb{Z}/6\mathbb{Z}\cong\mathbb{Z}/2\mathbb{Z}\oplus \mathbb{Z}/3\mathbb{Z}$, but is it the direct sum of $\mathbb{Z}/2\mathbb{Z}$ and $\mathbb{Z}/3\mathbb{Z}$ as a $\mathbb{Z}/6\mathbb{Z}$-module?

Also, I know that free modules are projective, but does a module free over $\mathbb{Z}/2\mathbb{Z}$ implies that it is a free module over $\mathbb{Z}/6\mathbb{Z}$?

And is it really true that every $\mathbb{Z}/6\mathbb{Z}$-module is a direct sum of free modules?

Also, I don't know how to prove the statement for $\mathbb{Z}/4\mathbb{Z}$.

$\endgroup$
  • 2
    $\begingroup$ More generally, you can prove (e.g. using the Artin-Wedderburn theorem) that $\mathbb{Z}/n$ is semisimple (equivalently, that every $\mathbb{Z}/n$-module is projective and injective) iff $n$ is squarefree. $\endgroup$ – Qiaochu Yuan Jan 23 '16 at 4:45
  • $\begingroup$ There is no reference to injective modules into the question. $\endgroup$ – user26857 Jan 23 '16 at 8:58
4
$\begingroup$

Note that $2\mathbb{Z}/6\mathbb{Z}$ and $3\mathbb{Z}/6\mathbb{Z}$ are ideals of the ring $\mathbb{Z}/6\mathbb{Z}$; they are evidently minimal ideals, their intersection is trivial and their sum is the whole ring. So $\mathbb{Z}/6\mathbb{Z}$ is a direct sum of simple submodules, hence semisimple (and artinian). Therefore every module is projective and injective.

It is false that every module over $\mathbb{Z}/6\mathbb{Z}$ is free (or a direct sum of free modules, which is the same), because the ring has two non isomorphic simple modules.

The ring $\mathbb{Z}/4\mathbb{Z}$ has a single minimal ideal (which is proper), namely $2\mathbb{Z}/4\mathbb{Z}$, so it is not semisimple. This ideal, as a module, is neither injective (it is not a direct summand) nor projective because the obvious map $\mathbb{Z}/4\mathbb{Z}\to 2\mathbb{Z}/4\mathbb{Z}$ doesn't split.

$\endgroup$
  • $\begingroup$ The first line of the last paragraph is slightly imprecise: any simple module would be a counterexample as written. But saying "single nontrivial ideal" instead would best support that the ideal isn't a summand. Regards $\endgroup$ – rschwieb Jan 23 '16 at 11:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy