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

A semisimple ring example

  1. $D$ 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?

  2. 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.

  3. What's "primitive diagonal idempotents"? I know "primitive idempotents".

Thanks for regarding!

  • $\begingroup$ for (2), can you prove that statement when $D$ is a field? $\endgroup$ – Mariano Suárez-Álvarez Oct 20 '13 at 3:26
  • $\begingroup$ The primitive diagonal idempotents are, well, the obvious things: the idempotents which are both primitive and diagonal! $\endgroup$ – Mariano Suárez-Álvarez Oct 20 '13 at 3:26
  • $\begingroup$ Please do not post pieces of books without adding a complete reference to the book. $\endgroup$ – Mariano Suárez-Álvarez Oct 20 '13 at 3:30

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, $E_iM_n(D)E_i\cong D$.

  • $\begingroup$ (2) is OK. (1) can u explain that why can't $x,y$ coexist in the same right (left) subspace. (3) can u proof that: $M_n(D)=M_n(D)E_1 \oplus ... \oplus M_n(D)E_n$ $\endgroup$ – Rachel Oct 26 '13 at 9:05
  • 1
    $\begingroup$ Dear @Rachel : Sorry, I regret my wording for (1). What I really wanted to express is that $(1,i)\Bbb H\neq \Bbb H(1,i)$. So for example, $(j,k)\in (1,i)\Bbb H$ but it isn't in $\Bbb H(1,i)$. What trouble are you having with (3)? If you multiply $M_n(D)$ on the right by each $E_i$, you should see that you get a left ideal consisting of matrices which are nonzero on a single column for each $i$. Their direct sum is pretty obviously the whole matrix ring. Let me know if you're still stuck. $\endgroup$ – rschwieb Oct 28 '13 at 13:02
  • $\begingroup$ (1) is OK. With (3), I'm really misunderstanding about the form of $E_i$. Since, I'm still stuck that: "you should see that you get a left ideal consisting of matrices which are nonzero on a single column for each $i$" $\endgroup$ – Rachel Nov 1 '13 at 23:53
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    $\begingroup$ Dear @Rachel : OK, let's break it down : $E_i$ is an $n\times n$ matrix that is zero everywhere except on the $i$th diagonal element, where it is $1$. Take that matrix and multiply it by any $n\times n$ matrix on the left. What does the product look like? The set of such products is the left ideal generated by $E_i$ ($M_n(R)E_i$) $\endgroup$ – rschwieb Nov 3 '13 at 14:03

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».


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