Problem with a semisimple ring example This is a example in: Frank W. Anderson,  Kent R. Fuller  (auth.) Rings and Categories of Modules



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*$D$ be a division ring, not a field. It's mean $D$ is non-commutative. $C_n(D)$ and $R_n(D)$ satisfy eight axioms of vector space. So what's "right" and "left" vector space above meaning?

*I can't prove that $C_n(D)$ is simple left $M_n(D)$-module and $R_n(D)$ is simple right $M_n(D)$-module.

*What's "primitive diagonal idempotents"? I know "primitive idempotents".
Thanks for regarding!
 A: 1) When working with vector spaces over division rings, side matters (as it does with noncommutative rings for that matter). Consider the quaternions $\Bbb H$ and the (right or left) vector space $\Bbb H\times\Bbb H$. The pair $(1,i)$ when multiplied on the right with $j$ is $(j,k)=x$. When multiplied on the left by $j$, it is $(j,-k)=y$. You can easily check that $x,y$ are not in the same 1-dimensonal right subspace, nor are they in the same one dimensional left subspace. So, to satisfy the vector space axioms coherently, we need to pick a side and keep the scalars on the chosen side.
2) One way to prove it is simple is to show that for any two nonzero vectors $a,b$, you can find a matrix $M$ such that $Ma=b$ for $C_n(D)$ (or $aM=b$, for $R_n(D))$. Can you see how that can be done?
3) This just is pointing out that the diagonal idempotents $E_i$ are primitive. After all, $E_iM_n(D)E_i\cong D$.
A: If $D$ is a division ring, then a left $D$-vector space is simply a left $D$-module. The same thing applies to «right objects».
