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

Advertisements
Comments
  1. 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!

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s