## Jacobson density theorem

Posted: December 19, 2009 in Noncommutative Ring Theory Notes, Primitive Rings
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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. $\Box$