let R be the ring of linear transformations under V where $\dim_{D} (V) = \infty $ show that $R$ is left primitive but it is not simple

Question

I have this question that I'm trying to solve: let $R$ be the ring of linear transformations under $V$ where $\dim_{D} (V) = \infty$; show that $R$ is left primitive but it is not simple. Can anyone help me solve this?

Answer 1

For $R$ to be left primitive it must have a faithful simple left $R$-module. In this case, $V$ is a faithful simple left $R$-module, since if $v\ne 0$, then there exists a linear transformation sending $v$ to $w$ for any $w\in V$. Thus any nonzero $R$-submodule of $V$ is $V$. Moreover it is faithful since by definition of $R$, $R\to \operatorname{End}(V)$ is just the identity map.

On the other hand, $R$ is not simple, since the finite rank transformations are a nontrivial two sided ideal in $R$. This follows from the facts $\newcommand\rk{\operatorname{rank}}$ $$\rk(A+B)\le \rk(A)+\rk(B),$$ and $$\rk(AB)\le \min\{\rk(A),\rk(B)\}.$$ Thus sums of finite rank transformations have finite rank, and products of finite rank transformations with any transformation have finite rank.