3
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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
$\begingroup$

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

$\endgroup$
  • $\begingroup$ (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
  • 1
    $\begingroup$ 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
0
$\begingroup$

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».

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.