## Some applications of density theorem

Posted: December 19, 2009 in Noncommutative Ring Theory Notes, Primitive Rings
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Remark. If $R$ is a left primitive ring, then, by the density theorem and remark 4 in this post, there exists a division ring $D$ such that  either $R \cong M_n(D),$ for some positive integer $n$ or for any positive integer $n$ there exists a subring $R_n$ of $R$ and an onto ring homomorphism $R_n \longrightarrow M_n(D).$

Example 1. A finite ring $R$ is left primitive if and only if $R \cong M_n(F),$ for some integer $n$ and some finite field $F.$

Proof. The “if” part is trivial because $M_n(F)$ is a simple ring and hence left primitive. Now suppose that $R$ is a finite left primitive ring. By the above remark we’ll have two possibilities.

Case 1. There exists a division ring $D$ such that $R \cong M_n(D).$ Then $D$ is finite and so a field.

Case 2. There exists a division ring $D$ such that for any positive integer $n$ there exists a subring $R_n$ and an onto ring homomorphism $R_n \longrightarrow M_n(D).$ But this case is not possible because $R_n$ is finite for all $n$ but the cardinality of $M_n(D)$ goes to infinity as $n$ goes to infinity.

Example 2. A ring $R$ has the property $(*)$ if for every $x \in R$ there exists an integer $n > 1,$ depending on $x,$ such that $x^{n}=x.$ A left primitive ring $R$ which has the property $(*)$ is a division ring.

Proof. Again, the same as example 1, we’ll consider two cases.

Case 1. $R \cong M_n(D),$ for some integer $n$ and division ring $D.$ But for $n \geq 2$ the ring $M_n(D)$ doesn’t have the property $(*)$ because for all $m \geq 2 : \ e_{12}^m = 0 \neq e_{12}.$ So $n = 1$ and thus $R \cong D.$

Case 2. Suppose that there exists an onto ring homomorphism from a subring $R_2$ of $R$ to $M_2(D).$ Then $R_2/I \cong M_2(D),$ for some ideal $I$ of $R_2.$ Then, since $R$ has the property $(*),$ the ring $R_2/I$ must have the property $(*).$ But we showed in Case 1 that $M_2(D)$ does not satisfy $(*).$ So this case is impossible.

Example 3. Let $R$ be a left primitive ring and suppose that $x(xy-yx)=(xy-yx)x,$ for all $x,y \in R.$ Then $R$ is a division ring.

Proof. The proof is the same as example 2. The only thing we’ll need is to note that for $n \geq 2$ the ring $M_n(D)$ doesn’t satisfy the above property because, for example, if we let $x=e_{11}, \ y=e_{12},$ then $x(xy-yx)=e_{12}$ but $(xy-yx)x=0.$

Example 4. Let $R$ be a left primitive ring such that $1+r^2$ is a unit for all $r \in R.$ Then $R$ is a division ring.

Proof. Let $M$ be a faithful simple left $R$-module and $D=\text{End}_R(M).$ So, by the above remark, we only need to prove that the dimension of $M,$ as a vector space over $D,$ is $1.$ Suppose that the dimension is at least $2$ and let $\{x,y \} \subset M$ be a $D$-linearly independent set. By the density theorem, there exists $r \in R$ such that $rx=y$ and $ry=-x.$ But then $r^2x=ry=-x$ and thus $(1+r^2)x=0,$ which gives us the contradiction $x=0$ because $1+r^2$ is a unit.

Example 5. (Wedderburn-Artin) Let $R$ be a left primitive ring. If $R$ is left Artinian, then $R \cong M_n(D),$ for some division ring $D$ and positive integer $n.$

Proof. Let $M$ be a faithful simple left $R$-module and $D=\text{End}_R(M).$ If $M$ is finite dimensional as a vector space over $D,$ we’re done by the above remark. Otherwise, choose an infinite $D$-linearly independent set $\{x_1,x_2, \cdots \} \subset M.$ For any positive integer $n$ let $I_n = \text{ann}_R \{x_1, \cdots , x_n \}.$ Clearly $I_{n+1} \subseteq I_n$ for all $n.$ Now, by the density theorem, there exists $r \in R$ such that $rx_1= \cdots = rx_n=0$ and $rx_{n+1}=x_1 \neq 0.$ Thus $r \in I_n$ but $r \notin I_{n+1},$ which means $I_n \supset I_{n+1}.$ So we have $I_1 \supset I_2 \supset \cdots ,$ which is a contradiction because $R$ is Artinian.