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.

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Comments
  1. Budi Santoso says:

    This great blog with modern mathematics, actualy in abstract algebra. I’ll follow your blog, sir. 🙂

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