## Projective modules; definitions & basic facts (2)

Posted: October 28, 2011 in Elementary Algebra; Problems & Solutions, Rings and Modules
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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.