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

**Definition**. Let be a ring with unity and let be an -module. The annihilator of in is defined by We say that is **faithful** if

**Remark**. Every free -module is faithful. To see this, let be a free -module and let If is an element of a basis of then implies that and so is faithful.

**Example 1**. Let be some rings with unity and put Clearly is an -module. Let Then and thus, by Corollary 1 in this post, is a projective -module. But and hence, by the above remark, is not free. A similar argument shows that each is a projective but not free -module. As an example, we know from the Chinese remainder theorem that if is the prime factorization of then Thus each is a projective but not free – module.

**Example 2**. Let be a field and the ring of matrices with entries in Let

Clearly is an -module and as -modules. Thus is a projective -module. Suppose that is free and is a basis for Then, since we have But, by the definition of we also have So which is absurd. Hence is not free. You can extend this argument to where is any integer. Then each column space of will be a projective but not a free -module.

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

**Example 3**. Let be a commutative ring and let be (finitely generated) projective -modules. Then is a (finitely generated) projective -module.

*Proof*. So there exist -modules and free -modules such that So where (resp. ) is finite if (resp. ) is finitely generated. See that where

**Example 4**. Let be a commutative ring and let be a finitely generated projective -module. Then is a finitely generated projective -module.

*Proof*. First note that if is an -module, then is an -module too because we can define for all and Now, since is a finitely generated projective -module, there exist an -module and a free -module such that Note that

as -modules, and so is a (finitely generated) free -module. On the other hand

by Theorem 1. Now identify with and let be the inclusion map. Define the map by

for all Clearly are -module homomorphisms and is the identity map. Thus, by Lemma 3 in this post, Therefore is projective and since is finitely generated, is finitely generated too because

**Theorem**. Let be a commutative ring and let be finitely generated projective -modules. Then, as -modules, we have

*Proof*. A nice challenge for the reader!

[…] was reading this webpage https://ysharifi.wordpress.com/2011/10/28/examples-of-projective-modules/, and there was something that confused […]

Hi: I have a comment about example 2. The solution appears to rely on the reasoning of , but it isn’t clear why one could reason with “dimensions” this way. There are rings such that as 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 dimensions that the rank is unique. Good post overall!