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I am doing an exercise from "an introduction to homological algebra by Charles A.Weibel" and I am not able to get it, but if the following comes true then I think I can solve the exercise, but I don't know whether the following is true or not.

Let $M$ and $N$ are free $R$-modules. Let $L$ be an $R$-module such that $M \cong L \oplus N$. Does it implies that $L$ is a free $R$-module.

Please help me.

eyp
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  • Equivalently, you could ask if $M/N$ is free. – Wuestenfux Sep 24 '18 at 11:37
  • @Wuestenfux: I got your point, but I am not able to figure out how this would help? – eyp Sep 24 '18 at 11:53
  • Quotient of two free modules? – Wuestenfux Sep 24 '18 at 11:55
  • @Wuestenfux: If M is of finite rank, then I can prove that L is free R module from the hint given by you, but when M is not of finite rank, how should I proceed? – eyp Sep 24 '18 at 12:17
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    @Sneh you cannot proceed. The statement is not true: direct summands of a free module need not be free. – freakish Sep 24 '18 at 12:42
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    @Wuestenfux This is not equivalent. Direct summand is not just any quotient. And it is very easy to construct a quotient of free modules that is not free, e.g. $\mathbb{Z}/2\mathbb{Z}$. – freakish Sep 24 '18 at 12:44
  • @freakish: Thank you. I got your point. I was thinking of M and N as direct sum of copies of R and then mod out. I was wrong. – eyp Sep 24 '18 at 12:58

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