Sketched Solutions to Exercises

4.3. (i) Since HomA(A (I),M) ∼= M (I) for a set I and ⊕ is exact, it is clear that HomA(A (I),−) is exact, i.e. A(I) is projective. (ii) “Only if” part: Let A(P ) = ⊕ p∈P Ap be the free A-module indexed by P and ψ : A (P ) → p be the A-homomorphism such that ψ(1p) = p. Then ψ is surjective. If P is projective, then ψ has a section, i.e. an A-homomorphism φ : P → A(P ) such that ψ ◦ φ = idP . This shows P is a direct summand of A(P ). “If” part: If P ⊕K is free, then by (i) P ⊕K is projective, hence HomA(P ⊕K,−) = HomA(P,−)⊕ HomA(K,−) is exact. This implies HomA(P,−) is exact, hence P is projective.