Question

Prove that a left R-module P is projective if and only if it is direct
summand of a free R-module.​

Answer 1

Answer:

If F is a free R-module and P ⊆ F is a submodule then P need not be free even if P is a direct summand of F. Take e.g. R = Z/6Z. Notice that Z/2Z and Z/3Z are Z/6Z-modules and we have an isomorphism of Z/6Z-modules: Z/6Z ∼= Z/2Z⊕Z/3Z Thus Z/2Z and Z/3Z are non-free modules isomorphic to direct summands of the free module Z/6Z.

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