Primitive rings; definition & examples

Posted: December 17, 2009 in Noncommutative Ring Theory Notes, Primitive Rings
Tags: , , ,

Let R be a ring and let M be a left R-module. Recall the following definitions:

1) M is called faithful if rM=(0) implies r=0 for any r \in R. In other words, M is called faithful if \text{ann}_RM = \{r \in R : \ rM=(0) \}=(0).
2) M is called simple if (0) and M are the only left R submodules of M.

Faithful and simple right R-modules are defined analogously.

Definition. A ring R is called left (resp., right) primitive if there exists a left (resp., right) R-module M which is both faithful and simple.

Remark. We will show later that left and right primitivity are not equivalent.  From now on, I will only consider left primitive rings. If a statement is true for left but not for right, I will mention that.

Example 1. If R is a ring and M is a left simple R-module, then R_1=R/\text{ann}_RM is a left primitive ring.  This is clear because M would be a faithful simple left R_1-module.

Example 2. Every simple ring is left primitive. That’s because we can choose a maximal left ideal \mathbf{m} of R and then M=R/\mathbf{m} would be a faithful simple left R-module. The reason that M is faithful is that \text{ann}_R M is a two-sided ideal contained in \mathbf{m} and therefore \text{ann}_R M = (0), because R is simple. One special case of this example is M_n(D), the ring of n \times n matrices with entries from a division ring D. If V is an infinite dimensional vector space over a field F, then \text{End}_F V is an example of a left primitive ring which is not simple [see Example 4 and this post].

Example 3. If R is a left primitive ring and 0 \neq e \in R an idempotent, then R_1=eRe is left primitive: let M be a faithful simple R-module. The claim is that M_1=eM is a faithful simple left R_1-module. Note that M_1 \neq (0) because e \neq 0 and M is faithful. Clearly M_1 is a left R_1-module.  To see why it’s faithful, let r_1 = ere \in R_1 with r_1M_1=(0). Then (0)=ere^2M=ereM=r_1M. So r_1=0, because M is faithful. To prove that M_1 is a simple R_1-module let 0 \neq x_1 \in M_1. We need to show that R_1x_1=M_1. Well, since x_1 = ex, for some x \in M, we have ex_1=ex=x_1. Thus R_1x_1=eRex_1=eRx_1=eM=M_1.

Example 4. Let D be a division ring and let M be a right vector space over D. Then R=\text{End}_D M is a left primitive ring. Here is why: M is a left R-module if we define fx = f(x), for all f \in R and x \in M. It is clear that M is faithful as a left R-module. To see why it is simple, let x,y \in M with x \neq 0. Let B=\{x_i: \ i \in I \} be a basis for M over D such that x=x_k \in B, for some k \in I. Define f \in R by f(\sum_{i \in I}x_id_i) = yd_k. Then f(x)=f(x_k)=y. So, we’ve proved that Rx = M, which shows that M is a simple R-module. If \dim_D M = n < \infty, then R \cong M_n(D), which we already showed its primitivity in Example 2. Note that if M was a “left” vector space over D with \dim_D M = n < \infty, then R would be isomorphic to the ring M_n(D^{op}) rather than the ring M_n(D).

  1. Budi Santoso says:


    I think, it will be more fine if it’s given a literature that taken from…

Leave a Reply

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

You are commenting using your 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