I have to prove that every $\mathbb{Z}/6\mathbb{Z}$-module is projective.

I've found already this question Prove that every $\mathbb{Z}/6\mathbb{Z}$-module is projective and injective. Find a $\mathbb{Z}/4\mathbb{Z}$-module that is neither. but I haven't defined what does it mean to be Artinian or semisimple: the proof should be based just on the equivalent definitions of projective modules (exactness of the covariant Hom functor, being a direct summand of a free module, lifting property and split sequences ending in the module).
My idea is to use the fact that $\mathbb{Z}/6\mathbb{Z} \cong \mathbb{Z}/2\mathbb{Z} \bigoplus \mathbb{Z}/3\mathbb{Z}$, these two submodules are projective and they are fields, hence all $\mathbb{Z}/2\mathbb{Z}$ and $\mathbb{Z}/3\mathbb{Z}$-modules are free. But I don't know if taken any $\mathbb{Z}/6\mathbb{Z}$-module I can "decompose" it in a sum of a $\mathbb{Z}/2\mathbb{Z}$-module and a $\mathbb{Z}/3\mathbb{Z}$-module.


Let $M$ be a $\Bbb{Z}/6\Bbb{Z}$ module. We'll show that $M=2M\oplus 3M$.

If $m\in 2M\cap 3M$, then $m=2m'=3m''$ for some $m',m''\in M$, and this gives $$0=6m'=3m=9m''=3m''=m,$$ so $2M\cap 3M=0$.

On the other hand, $m=3m-2m$, so $2M+3M=M$.

Thus $M=2M\oplus 3M$. Note that $2M$ is a $\newcommand\ZZ{\Bbb{Z}}\ZZ/3\ZZ$ module and $3M$ is a $\ZZ/2\ZZ$ module.

Thus the claim you wanted to prove is true.

Then $2M$ is a direct sum of copies of $\ZZ/3\ZZ$ and $3M$ is a direct sum of copies of $\ZZ/2\ZZ$, and as you've noted $\ZZ/3\ZZ$ and $\ZZ/2\ZZ$ are both projective. Hence $M$ is a direct sum of projective modules and thus projective.


Another style of proof is: it’s semisimple so all short exact sequences split, in particular all modules are injective and projective.

  • 1
    $\begingroup$ “the proof should be based just on the equivalent definitions of projective modules” $\endgroup$
    – egreg
    Jan 1 '19 at 23:26
  • $\begingroup$ @egrer Wouldn't using "P is projective iff all SES ending in P split" be fair? $\endgroup$
    – Pedro Tamaroff
    Jan 1 '19 at 23:39
  • $\begingroup$ Ordinarily this is the answer I would have given, but the OP says that they don't know what semisimple is, well technically they say that they haven't defined it. Also it is essentially the answer given in the linked question. $\endgroup$
    – jgon
    Jan 2 '19 at 3:07

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