See part (1) here. Throughout is a ring with 1 and all modules are left -modules.

**Corollary 1**. An -module is projective if and only if there exists an -module such that is free.

*Proof*. If is free, then is projective by Lemma 2 in part (1). Conversely, suppose that is projective. By Remark 2 in part (1), there exists an exact sequence where is free. This sequence splits by the theorem in part (1), i.e.

**Corollary 2**. A finitely generated -module is projective if and only if there exists an -module such that is a finitely generated free -module, i.e. for some integer

*Proof*. One side is obvious by Corollary 1 (or Lemma 2 in part (1)). Conversely, suppose that is a finitely generated projective -module. Let be a basis for and define the map by Clearly is a well-defined onto -module homomorphism. Let Then we have an exact sequence where is the inclusion map. This sequence splits by the theorem in part (1), i.e.

**Corollary 3**. Let be a family of -modules. Then is projective if and only if is projective for all

*Proof*. Suppose first that is projective and let By Corollary 1, there exists an -module and a free -module such that But then and thus is projective by Corollary 1 again. Conversely, suppose that each is projective. Then for every there exists an -module and a free -module such that Let and Then is a free -module and Thus is projective.

In the next post I will give a few examples of projective modules.