I was reading a section on Projective modules, that asked us to prove that a projective module is a direct summand of a free module. I then began thinking whether every module should be a direct summand of a free module.

From what I understand, a module $M$ is a summand of a free module $F$ if $M\oplus M'=F$ for some module $M'$. I think of a module $M$ as the free module $F$ generated by its generators, which is then quotiented by the relations between those generators. Those relations form a submodule $N$. Hence, $M=F/N$. In other words, $F=M\oplus N$. Am I understanding this correctly?


2 Answers 2


As you say, projective modules are the modules that are summands of free modules. As the existence of the phrase "projective module" suggests, not all modules are projective. For an example, over $R=\Bbb Z$, the module $\Bbb Z/2\Bbb Z$ is not projective. It is a nonzero torsion module, but free modules are torsion-free and so all their submodules are torsion-free.

As you point out, each module satisfies $M\cong F/N$ where $F$ is free and $N$ is a submodule. This does not imply $F\cong M\oplus N$. Take $F=\Bbb Z$ and $N=2\Bbb Z\cong\Bbb Z$. Then $\Bbb Z\not\cong\Bbb Z\oplus\Bbb Z/2\Bbb Z$.


In vector spaces it is true that $W_2 = V/W_1$ means $V=W_1\oplus W_2$. This fails in modules and that makes their study totally different. For example any ideal $I\subset R$ of a ring is a module over $R$. Take even integers as ideal over the ring of integers. But this ring is not a direct sum of modules over itself.

  • $\begingroup$ Why is $\Bbb{Z}$ not equal to $\Bbb{Z}/2\Bbb{Z}\oplus 2\Bbb{Z}$? $\endgroup$
    – user67803
    Commented Apr 15, 2018 at 5:07
  • 2
    $\begingroup$ In the former $a.m=0$ means either $a$ or $m$ is zero. In the lattr take specific element $(1,0)$ in the direct sum, but $2.(1,0)=0$. So these two modules are not isomorphic. (That is one of them has a torsion element but the other is not, so they are not isomorphic modules). $\endgroup$ Commented Apr 15, 2018 at 7:21

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