We showed in the previous section that every dense subring of the ring of linear transformations of a vector space over a division ring is left primitive. Now, we’d like to prove the converse: every left primitive ring is a dense subring of the ring of linear transformations of some vector space over some division ring. We will assume that R is a ring, M is a simple left R module and D=\text{End}_R(M). As usual, M is considered as a right vector space over D.

Remark. If \varphi \in \text{Hom}_R (M^k, M), then there exist d_i \in D such that for all x_i \in M we have \varphi (x_1, \cdots , x_k) = \sum_{i=1}^k x_i d_i.

Proof. For any 1 \leq i \leq k let \mu_i : M \longrightarrow M^k be the i-th injection map, i.e. \mu_i(x)=(0, \cdots , x, \cdots , 0), where x is the i-th coordinate. Clearly d_i=\varphi \mu_i \in D. Now

\varphi (x_1, \cdots , x_k)=\sum_{i=1}^k \varphi \mu_i(x_i)=\sum_{i=1}^k x_i d_i.

Density Theorem. (Chevalley – Jacobson) Let R be a left primitive ring, M a faithful simple left R module and D=\text{End}_R(M). Then R is a dense subring of S=\text{End}_D(M).

Proof. We have already showed that R is a subring of S. So we need to prove that for any 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 j. The proof is by induction over n: if n = 1, then, since M is simple and x_1 \neq 0, we have Rx_1=M and thus there exists f \in R such that fx_1=f(x_1)=y_1. Assuming that the result is true for n-1, we will have R(x_1, \cdots , x_{n-1})=M^{n-1}  (density condition!). We now prove a claim:

Claim.  There exists f \in R such that f(x_n) \neq 0 and f(x_1) = \cdots = f(x_{n-1})=0.

Proof of the claim. Suppose to the contrary that the claim is not true. Then f(x_1)= \cdots f(x_{n-1})=0 will imply that f(x_n)=0, for any f \in R. Define \varphi : M^{n-1} \longrightarrow M by \varphi (f(x_1), \cdots , f(x_{n-1}))=f(x_n), \ f \in R. See that \varphi is well-defined, that is \varphi \in \text{Hom}_R (M^{n-1},M). Therefore, if we put f=1_R, then by the above remark there must exist d_i \in D such that x_n = \varphi (x_1, \cdots , x_{n-1}) = \sum_{i=1}^{n-1} x_i d_i, which contradicts D-linear independence of x_1, \cdots , x_n. \ \Box

So, using the above claim,  for any 1 \leq i \leq n, we can choose f _i \in R such that f_i(x_i) \neq 0 and f_i (x_j)=0, for all j \neq i. Thus, since M is simple, Rf_i(x_i)=M, for all 1 \leq i \leq n. Hence there exist g_i \in R such that g_if_i(x_i)=y_i. Let f = \sum_{i=1}^n g_i f_i. Then for any 1 \leq j \leq n

f(x_j)=\sum_{i=1}^n g_if_i(x_j)=g_jf_j(x_j)=y_j. \ \Box

The Structure Theorem For Primitive Rings. Let R be a left primitive ring, M a faithful simple left R module and D=\text{End}_R(M).
1) If \dim_D M = n < \infty, then R \cong M_n(D).
2) If \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. It follows from the density theorem and remarks 3 and 4 in the previous post. \Box

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Comments
  1. Budi Santoso says:

    Thank you for your article. 🙂

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