## Projective modules; definitions and basic facts (1)

Posted: October 28, 2011 in Elementary Algebra; Problems & Solutions, Rings and Modules
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Throughout $R$ is a ring with 1 and all modules are left $R$-modules.

Definition 1. An $R$-module $P$ is called projective if it satisfies the following condition: given $R$-modules $M, N$ and $R$-module homomorphisms $h: P \longrightarrow N$ and $f: M \longrightarrow N,$ where $f$ is onto, there exists an $R$-module homomorphism $g: P \longrightarrow M$ such that $f g = h.$

Example. Every free $R$-module $F$ is projective. To see this, let $\{y_i: \ i \in I \}$ be a basis for $F$ and suppose that $h: F \longrightarrow N$ and $f: M \longrightarrow N$ are $R$-module homomorphisms and $f$ is onto. So for every $i \in I$ there exists some $x_i \in M$ such that $f(x_i)=h(y_i).$ Now define $g: F \longrightarrow M$ by $g(\sum r_iy_i)=\sum r_i x_i.$ Since $\{y_i: \ i \in I \}$ is a basis for $F,$ the map $g$ is well-defined. Clearly $g$ is an $R$-module homomorphism and

$fg(\sum r_iy_i)=f(\sum r_ix_i)=\sum r_if(x_i)=\sum r_i h(y_i)=h(\sum r_iy_i).$

Thus $fg = h$ and so $F$ is projective.

Definition 2. Let $A_i, \ i \in \mathbb{Z},$ be $R$-modules and $f_i : A_i \longrightarrow A_{i+1}$ be $R$-module homomorphisms. The sequence $\cdots \overset{f_{n-2}} {\longrightarrow} A_{n-1} \overset{f_{n-1}}{\longrightarrow} A_n \overset{f_n}{\longrightarrow} A_{n+1} \overset{f_{n+1}} {\longrightarrow}\cdots$ is called exact if $\text{im}(f_{i-1}) = \ker f_i$ for all $i.$

Remark 1. So, by Definition 2, the sequence $0 \longrightarrow A_1 \overset{f_1}{\longrightarrow} A_2 \overset{f_2}{\longrightarrow} A_3 \longrightarrow 0$ is exact if and only if $f_1$ is one-to-one, $f_2$ is onto and $\text{im}(f_1)=\ker f_2.$

Remark 2. Recall that every $R$-module $A$ is a quotient of a free $R$-module. That means there exists a free $R$-module $F$ and an onto $R$-module homomorphism $f_2 : F \longrightarrow A.$ Let $K = \ker f_2.$ Then we have an exact sequence $0 \longrightarrow K \overset{f_1}{\longrightarrow} F \overset{f_2}{\longrightarrow} A \longrightarrow 0,$ where $f_1$ is the inclusion map.

Lemma 2. Let $A$ and $B$ be $R$-modules. If $A \oplus B$ is free, then $A$ and $B$ are projective.

Proof.  Let $F = A \oplus B.$ We only need to prove that $A$ is projective. Define the maps $\alpha : A \longrightarrow F$ and $\beta : F \longrightarrow A$ by $\alpha(x)=(x,0)$ and $\beta(x,y)=x$ for all $x \in A$ and $y \in K.$ Note that $\beta \alpha = \text{id}_A.$ We now prove that $A$ is projective. Suppose that we have $R$-module homomorphisms $h: A \longrightarrow N$ and $f: M \longrightarrow N,$ where $f$ is onto. So we have an $R$-module homomorphism $h \beta : F \longrightarrow N$ and thus, since $F$ is projective by the above example, there exists an $R$-module homomorphism $\gamma : F \longrightarrow M$ such that $f \gamma = h \beta.$ Let $g = \gamma \alpha.$ Then $fg = f \gamma \alpha = h \beta \alpha = h. \ \Box$

Lemma 3. Let $A,B$ be $R$-modules. Let $f_1: A \longrightarrow B$ and $f_2: B \longrightarrow A$ be $R$-module homomorphisms such that $f_2f_1 = \text{id}_A.$ Then $B \cong \ker f_2 \oplus A.$

Proof. First note that the condition $f_2f_1 = \text{id}_A$ implies that $f_1$ is one-to-one because if $f_1(a)=0$ for some $a \in A,$ then $a = f_2f_1(a)=0.$ So we only need to prove that $B = \ker f_2 \oplus \text{im}(f_1).$ Let $b \in \text{im}(f_1) \cap \ker f_2.$ Then $b = f_1(a),$ for some $a \in A,$ and f_2(b)=0.\$ But $f_2f_1=\text{id}_A$ and so $a = f_2f_1(a)=f_2(b)=0.$ Thus $b = f_1(a)=0$ and so $\text{im}(f_1) \cap \ker f_2 = \{0\}.$ Now let $b$ be any element of $B.$ Then $f_2f_1(f_2(b))=f_2(b),$ because $f_2(b) \in A$ and $f_2f_1=\text{id}_A.$ Hence $b - f_1(f_2(b)) \in \ker f_2$ and so $b \in \ker f_2 \oplus \text{im}(f_1). \ \Box$

Definition 3. We say that the exact sequence $0 \longrightarrow A_1 \overset{f_1}{\longrightarrow} A_2 \overset{f_2}{\longrightarrow} A_3 \longrightarrow 0$ splits if $A_2 \cong A_1 \oplus A_3.$

Theorem. An $R$-module $P$ is projective if and only if every exact sequence $0 \longrightarrow A_1 \overset{f_1}{\longrightarrow} A_2 \overset{f_2}{\longrightarrow} P \longrightarrow 0$ splits.

Proof. Suppose first that $P$ is projective and $0 \longrightarrow A_1 \overset{f_1}{\longrightarrow} A_2 \overset{f_2}{\longrightarrow} P \longrightarrow 0$ is an exact sequence of $R$-modules. So $f_2$ is onto and thus, since $P$ is projective, there exists an $R$-module homomorphism $g: P \longrightarrow A_2$ such that $f_2g = \text{id}_P.$ Therefore, by Lemma 3, we have $A_2 \cong \ker f_2 \oplus P = \text{im}(f_1) \oplus P \cong A_1 \oplus P,$ because $f_1$ is one-to-one. Conversely, suppose that every exact sequence $0 \longrightarrow A_1 \overset{f_1}{\longrightarrow} A_2 \overset{f_2}{\longrightarrow} P \longrightarrow 0$ splits. By Remark 2, there exists an exact sequence $0 \longrightarrow K \overset{f_1}{\longrightarrow} F \overset{f_2}{\longrightarrow} P \longrightarrow 0$ such that $F$ is free. This sequence, by hypothesis, splits and so $F \cong P \oplus K.$ Thus $P \oplus K$ is free and we’re now done by Lemma 2. $\Box$