## Dense subrings

Posted: December 18, 2009 in Noncommutative Ring Theory Notes, Primitive Rings
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Throughout $D$ is a division ring, $M$ a right vector space over $D,$ and  $R$ is a subring of $S=\text{End}_D(M).$ Clearly $M$ can be viewed as a left $R$ module by defining $fx=f(x),$ for all $f \in R$ and $x \in M$.

Definition. $R$ is said to be a dense subring of $S$ if for every $D$-linearly independent set $\{x_1, \cdots, x_n \} \subset M$ and any set $\{y_1, \cdots, y_n \} \subset M,$ there exists $f \in R$ such that $f(x_j)=y_j,$ for all $1 \leq j \leq n.$

Remark 1. If $R$ is a dense subring of $S,$ then $R$ is left primitive.

Proof. Well, $M$ is clearly a faithful left $R$ module. To see why it is simple, let $0 \neq x \in M$ and $y \in M.$ By the density condition, there exists $f \in R$ such that $fx=f(x)=y.$ Thus $Rx=M.$

Remark 2. We proved in example 4 in here that $S$ itself is left primitive. In the above remark we showed that any dense subring of $S$ is also left primitive.

Remark 3. If $R$ is a dense subring of $S$ and $\dim_D M = n < \infty,$ then $R=S \cong M_n(D).$

Proof. Let $\{x_1, \cdots , x_n \}$ be a basis for $M,$ as a vector space over $D,$ and let $g \in S.$ Then, since $R$ is dense, there exists $f \in R$ such that $f(x_j)=g(x_j),$ for all $j.$. But then $g= f \in R$ and hence $S=R.$

Remark 4. If $R$ is a dense subring of $S$ and $\dim_D M = \infty,$ then for any positive integer $n$, there exists a subring $R_n$ of $R$ and an onto ring homomorphism $\varphi : R_n \longrightarrow M_n(D).$

Proof. Let $\{x_1, \cdots , x_n \}$ be a $D$-linearly independent subset of $M$ and let $N=\sum_{j=1}^n x_j D,$ i.e. the $D$-vector subspace of $M$ spanned by $x_1, \cdots , x_n.$ Now define $R_n = \{f \in R: \ f(N) \subseteq N \}.$ Clearly $R_n$ is a subring of $R$. Finally we define the map $\varphi : R_n \longrightarrow \text{End}_D(N) \cong M_n(D)$ by $\varphi(f)(x)=f(x),$ for all $f \in R_n$ and $x \in N.$ It’s easy to see that $\varphi$ is a well-defined ring homomorphism. To prove that $\varphi$ is onto, let $g \in \text{End}_D(N).$ Then, by the density condition, there exists $f \in R$ such that $f(x_j)=g(x_j),$ for all $j.$ Thus $f \in R_n,$ because $g \in \text{End}_D(N),$ and clearly $\varphi(f)=g.$