## Primitive rings; definition & examples

Posted: December 17, 2009 in Noncommutative Ring Theory Notes, Primitive Rings
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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 $M_n(D).$

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Comments
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