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Let R be a ring. If a direct sum of R-modules is projective, can we imply that the direct sum is also a free module over R ?

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I assume you're excluding trivial summands; otherwise the answer would trivially be "no". But it's also "no" with non-trivial summands. Take for instance $\mathbb{Z}$ as a module over $\mathbb{Z}\oplus\mathbb{Z}$, with multiplication defined by $(a,b)\cdot c= a\cdot c$. This is a projective module, but its direct sum with itself is not a free module (which you can see e. g. by noting that many elements of $\mathbb{Z}\oplus\mathbb{Z}$ leave this module invariant whereas only the identity element would leave a free module invariant).

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