So far we have only given a trivial example of projective modules, i.e. free modules (see the example in this post).

Definition. Let $R$ be a ring with unity and let $M$ be an $R$-module. The annihilator of $M$ in $R$ is defined by $\text{ann}_R(M)=\{r \in R: \ rx = 0, \ \forall x \in M \}.$ We say that $M$ is faithful if $\text{ann}_R(M)=(0).$

Remark. Every free $R$-module is faithful. To see this, let $M$ be a free $R$-module and let $r \in \text{ann}_R(M).$ If $x \in M$ is an element of a basis of $M,$ then $rx = 0$ implies that $r = 0$ and so $M$ is faithful.

Example 1. Let $R_1 , \ldots , R_n, \ n \geq 2,$ be some rings with unity and put $R=\bigoplus_{i=1}^n R_i.$ Clearly $R_1$ is an $R$-module. Let $K = \bigoplus_{i=2}^n R_i.$ Then $R=R_1 \oplus K$ and thus, by Corollary 1 in this post, $R_1$ is a projective $R$-module. But $(0,1, \ldots , 1) \in \text{ann}_{R}(R_1)$ and hence, by the above remark, $R_1$ is not free. A similar argument shows that each $R_i$ is a projective but not free $R$-module. As an example, we know from the Chinese remainder theorem that if $n = p_1^{k_1}p_2^{k_2} \ldots p_r^{k_r}$ is the prime factorization of $n,$ then $\displaystyle \frac{\mathbb{Z}}{n \mathbb{Z}} = \bigoplus_{i=1}^r \frac{\mathbb{Z}}{p_i^{k_i} \mathbb{Z}}.$ Thus each $\displaystyle \frac{\mathbb{Z}}{p_i^{k_i} \mathbb{Z}}$ is a projective but not free $\displaystyle \frac{\mathbb{Z}}{n \mathbb{Z}}$ – module. $\Box$

Example 2. Let $k$ be a field and $R=M_2(k),$ the ring of $2 \times 2$ matrices with entries in $k.$ Let

$P = \left \{ \begin{pmatrix} a \\ b \end{pmatrix} : \ \ a,b \in k \right \}.$

Clearly $P$ is an $R$-module and $R \cong P \oplus P,$ as $R$-modules.  Thus $P$ is a projective $R$-module. Suppose that $P$ is free and $\{x_i : \ i \in I \}$ is a basis for $P.$ Then, since $\dim_k R = 4,$ we have $\dim_k P = 4|I|.$ But, by the definition of $P,$ we also have $\dim_k P = 2.$ So $2 = 4|I|,$ which is absurd. Hence $P$ is not free. You can extend this argument to $R = M_m(k),$ where $m \geq 2$ is any integer. Then each column space of $R$ will be a projective but not a free $R$-module. $\Box$

The next two examples are important in the theory of Azumaya algebras.

Example 3. Let $R$ be a commutative ring and let $P_1, P_2$ be (finitely generated) projective $R$-modules. Then $P_1 \otimes_R P_2$ is a (finitely generated) projective $R$-module.

Proof. So there exist $R$-modules $K_i$ and free $R$-modules $F_i$ such that $P_i \oplus K_i = F_i, \ i=1,2.$ So $F_1 \cong R^m, \ F_2 = R^n,$ where $m$ (resp. $n$) is finite if $P_1$ (resp. $P_2$) is finitely generated. See that $R^{mn} \cong F_1 \otimes_R F_2 \cong (P_1 \otimes_R P_2) \oplus K,$ where $K \cong (P_1 \otimes_R K_2) \oplus (P_2 \otimes_R K_1) \oplus (K_1 \otimes_R K_2). \ \Box$

Example 4. Let $R$ be a commutative ring and let $P$ be a finitely generated projective $R$-module. Then $\text{End}_R (P)$ is a finitely generated projective $R$-module.

Proof. First note that if $A$ is an $R$-module, then $\text{End}_R(A)$ is an $R$-module too because we can define $(rf)(a)=f(ra)$ for all $r \in R, a \in A$ and $f \in \text{End}_R(A).$ Now, since $P$ is a finitely generated projective $R$-module, there exist an $R$-module $K$ and a free $R$-module $F \cong R^n$ such that $F = P \oplus K.$ Note that

$\text{End}_R(F) \cong \text{End}_R(R^n) \cong M_n(R) \cong R^{n^2},$

as $R$-modules, and so $\text{End}_R(F)$ is a (finitely generated) free $R$-module. On the other hand

$\text{End}_R(F) = \text{End}_R(P \oplus K) \cong \begin{pmatrix} \text{End}_R(P) & \text{Hom}_R(K,P) \\ \text{Hom}_R(P,K) & \text{End}_R(K) \end{pmatrix},$

by Theorem 1. Now identify $\text{End}_R(P)$ with $\begin{pmatrix} \text{End}_R(P) & 0 \\ 0 & 0 \end{pmatrix}$ and let $f_1: \text{End}_R(P) \longrightarrow \text{End}_R(F)$ be the inclusion map. Define the map $f_2 : \text{End}_R(F) \longrightarrow \text{End}_R(P)$ by

$f_2 \left (\begin{pmatrix} u & v \\ w & t \end{pmatrix} \right) = \begin{pmatrix}u & 0 \\ 0 & 0 \end{pmatrix},$

for all $u \in \text{End}_R(P), v \in \text{Hom}_R(K,P), w \in \text{Hom}_R(P,K), t \in \text{End}_R(K).$ Clearly $f_1,f_2$ are $\text{End}_R(F)$-module homomorphisms and $f_2f_1$ is the identity map. Thus, by Lemma 3 in this post, $R^{n^2} \cong \text{End}_R(F) \cong \text{End}_R(P) \oplus \ker f_2.$ Therefore $\text{End}_R(P)$ is projective and since $\text{End}_R(F)$ is finitely generated,  $\text{End}_R(P)$ is finitely generated too because $\text{End}_R(P) \cong \text{End}_R(F)/\ker f_2. \ \Box$

Theorem. Let $R$ be a commutative ring and let $P_1,P_2$ be finitely generated projective $R$-modules. Then, as $R$-modules, we have

$\text{End}_R(P_1) \otimes_R \text{End}_R(P_2) \cong \text{End}_R(P_1 \otimes_R P_2).$

Proof. A nice challenge for the reader! $\Box$

2. Hi: I have a comment about example 2. The solution appears to rely on the reasoning of $dim_k(I)=dim_k(R)dim_R(I)$, but it isn’t clear why one could reason with “dimensions” this way. There are rings such that $R\cong R^2$ as $R$ modules, and so they do not have a unique dimension/rank. However, in the case of a finite dimensional algebra like this matrix ring, it’s easy to argue via $k$ dimensions that the $R$ rank is unique. Good post overall!