Projective modules; definitions and basic facts (1)

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

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


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