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

Corollary 1. An R-module P is projective if and only if there exists an R-module K such that P \oplus K is free.

Proof. If P \oplus K is free, then P is projective by Lemma 2 in part (1). Conversely, suppose that P is projective. By Remark 2 in part (1), there exists an exact sequence 0 \longrightarrow K \overset{f_1}{\longrightarrow} F \overset{f_2}{\longrightarrow} P \longrightarrow 0, where F is free. This sequence splits by the theorem in part (1), i.e.  P \oplus K \cong F. \ \Box

Corollary 2. A finitely generated R-module P is projective if and only if there exists an R-module K such that P \oplus K is a finitely generated free R-module, i.e. P \oplus K \cong R^n for some integer n \geq 1.

Proof. One side is obvious by Corollary 1 (or Lemma 2 in part (1)). Conversely, suppose that P=\sum_{i=1}^n Rx_i is a finitely generated projective R-module. Let \{e_1, \ldots , e_n \} be a basis for R^n and define the map f_2 : R^n \longrightarrow P by f_2(\sum r_ie_i)=\sum r_i x_i. Clearly f_2 is a well-defined onto R-module homomorphism. Let K = \ker f_2. Then we have an exact sequence 0 \longrightarrow K \overset{f_1}{\longrightarrow} R^n \overset{f_2}{\longrightarrow} P \longrightarrow 0, where f_1 is the inclusion map. This sequence splits by the theorem in part (1), i.e. P \oplus K \cong R^n. \ \Box

Corollary 3. Let \{A_i : \ i \in I\} be a family of R-modules. Then A = \bigoplus_{i \in I} A_i is projective if and only if A_i is projective for all i \in I.

Proof. Suppose first that A is projective and let j \in I. By Corollary 1, there exists an R-module K and a free R-module F such that F=A \oplus K. But then F \cong A_j \oplus (\bigoplus_{i \in I, \ i \neq j} A_i \oplus K) and thus A_j is projective by Corollary 1 again. Conversely, suppose that each A_i is projective. Then for every i \in I, there exists an R-module K_i and a free R-module F_i such that F_i = A_i \oplus K_i. Let F = \bigoplus_{i \in I} F_i and K = \bigoplus_{i \in I} K_i. Then F is a free R-module and F = A \oplus K. Thus A is projective. \Box

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

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