Would anyone explain how the direct sum of projective modules are free (over a commutative ring)? What I know is an R-module P is said to be projective if to every surjective homomorphism $\alpha : B \rightarrow C $ of R-modules and to every homomorphism $\gamma: P \rightarrow C $ there exists a homomorphism $\beta: P \rightarrow B$ with $ \alpha \beta = \gamma$.
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1If you could prove that direct sums of projective modules are free, then every projective module would be free, because ${0}$ is certainly projective. – egreg Sep 29 '16 at 10:51
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A direct sum of projective modules is not necessarily free.
An example would be $F\times F\times F$ for a field $F$, in which $F\times \{0\}\times \{0\}$ and $\{0\}\times F\times \{0\}$ are projective ideals and their sum is not free.
The direct sum of projectives is again projective, though. This follows naturally from the mapping property and the definition of the direct sum.
rschwieb
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