## Primitive rings; basic facts

Posted: December 18, 2009 in Noncommutative Ring Theory Notes, Primitive Rings
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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 $R.$