Remark. If R is a left primitive ring, then, by the density theorem and remark 4 in this post, there exists a division ring D such that  either R \cong M_n(D), for some positive integer n or for any positive integer n there exists a subring R_n of R and an onto ring homomorphism R_n \longrightarrow M_n(D).

Example 1. A finite ring R is left primitive if and only if R \cong M_n(F), for some integer n and some finite field F.

Proof. The “if” part is trivial because M_n(F) is a simple ring and hence left primitive. Now suppose that R is a finite left primitive ring. By the above remark we’ll have two possibilities.

Case 1. There exists a division ring D such that R \cong M_n(D). Then D is finite and so a field.

Case 2. There exists a division ring D such that for any positive integer n there exists a subring R_n and an onto ring homomorphism R_n \longrightarrow M_n(D). But this case is not possible because R_n is finite for all n but the cardinality of M_n(D) goes to infinity as n goes to infinity.

Example 2. A ring R has the property (*) if for every x \in R there exists an integer n > 1, depending on x, such that x^{n}=x. A left primitive ring R which has the property (*) is a division ring.

Proof. Again, the same as example 1, we’ll consider two cases.

Case 1. R \cong M_n(D), for some integer n and division ring D. But for n \geq 2 the ring M_n(D) doesn’t have the property (*) because for all m \geq 2 : \ e_{12}^m = 0 \neq e_{12}. So n = 1 and thus R \cong D.

Case 2. Suppose that there exists an onto ring homomorphism from a subring R_2 of R to M_2(D). Then R_2/I \cong M_2(D), for some ideal I of R_2. Then, since R has the property (*), the ring R_2/I must have the property (*). But we showed in Case 1 that M_2(D) does not satisfy (*). So this case is impossible.

Example 3. Let R be a left primitive ring and suppose that x(xy-yx)=(xy-yx)x, for all x,y \in R. Then R is a division ring.

Proof. The proof is the same as example 2. The only thing we’ll need is to note that for n \geq 2 the ring M_n(D) doesn’t satisfy the above property because, for example, if we let x=e_{11}, \ y=e_{12}, then x(xy-yx)=e_{12} but (xy-yx)x=0.

Example 4. Let R be a left primitive ring such that 1+r^2 is a unit for all r \in R. Then R is a division ring.

Proof. Let M be a faithful simple left R-module and D=\text{End}_R(M). So, by the above remark, we only need to prove that the dimension of M, as a vector space over D, is 1. Suppose that the dimension is at least 2 and let \{x,y \} \subset M be a D-linearly independent set. By the density theorem, there exists r \in R such that rx=y and ry=-x. But then r^2x=ry=-x and thus (1+r^2)x=0, which gives us the contradiction x=0 because 1+r^2 is a unit.

Example 5. (Wedderburn-Artin) Let R be a left primitive ring. If R is left Artinian, then R \cong M_n(D), for some division ring D and positive integer n.

Proof. Let M be a faithful simple left R-module and D=\text{End}_R(M). If M is finite dimensional as a vector space over D, we’re done by the above remark. Otherwise, choose an infinite D-linearly independent set \{x_1,x_2, \cdots \} \subset M. For any positive integer n let I_n = \text{ann}_R \{x_1, \cdots , x_n \}. Clearly I_{n+1} \subseteq I_n for all n. Now, by the density theorem, there exists r \in R such that rx_1= \cdots = rx_n=0 and rx_{n+1}=x_1 \neq 0. Thus r \in I_n but r \notin I_{n+1}, which means I_n \supset I_{n+1}. So we have I_1 \supset I_2 \supset \cdots , which is a contradiction because R is Artinian.

Advertisements

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