Fact 1. Let R be a left primitive ring and M a faithful simple left R module. By Schur’s lemma D=\text{End}_R(M) is a division ring and M can be considered as a right vector space over D in the usual way. Let S=\text{End}_D(M) and define \varphi : R \longrightarrow S by \varphi(r)(x)=rx, for all r \in R and x \in M. Then \varphi is a well-defined ring homomorphism. Also \varphi is one-to-one because M is faithful. So R can be viewed as a subring of S.

Fact 2. Every left primitive ring R is prime. To see this, suppose M is a faithful simple left R module and I,J be two non-zero ideals of R with IJ=(0).  Now JM is a submodule of M and M is simple. Therefore either JM=(0) or JM=M. If JM=(0), then we get (0) \neq J \subseteq \text{ann}_R M = (0), which is nonsense. Finally, if JM=M, then we will have (0)=(IJ)M=I(JM)=IM. Thus I \subseteq \text{ann}_R M = (0) and so I=(0), a contradiction!

Fact 3. A trivial result of Fact 2 is that the center of a left primitive ring is a commutative domain. A non-trivial fact is that every commutative domain is the center of some left primitive ring. For a proof of this see: T. Y. Lam,  A First Course in Noncommutative Ring Theory, page 195.

Fact 4. Let R be a prime ring and M a faithful left R module of finite length. Then R is left primitive. To see this, let (0)=M_0 \subset M_1 \subset \cdots \subset M_n=M be a composition series of M. Therefore M_k/M_{k-1} is a simple left R module for every 1 \leq k \leq n. We also let I_k=\text{ann}_R (M_k/M_{k-1}). Then each I_k is an ideal of R and it’s easy to see that I_1I_2 \cdots I_nM = (0). Thus I_1I_2 \cdots I_n = (0), because M is faithful. Hence I_{\ell} = (0), for some \ell, because R is prime. Therefore M_{\ell}/M_{\ell - 1} is a faithful simple left R module.

Fact 5. Every left primitive ring R is semiprimitive. This is easy to see: let M be a faithful simple left R module and J=J(R), as usual, be the Jacobson radical of R. The claim is that J=(0). So suppose that J \neq \{0\}  and choose 0 \neq x \in M. Then Rx=M, because M is simple, and so JM=Jx. Also either JM=(0), which is impossible because then J \subseteq \text{ann}_R M=(0), or JM=M. If Jx=JM=M, then ax =x, for some a \in J. Thus (1-a)x=0, which gives us the contradiction x = 0, because 1-a is invertible in R.


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