Understanding Bresar's Example 1.10 on Simple Matrix Rings: Can Anyone Help?

[Math Amateur]

I am reading Matej Bresar's book, "Introduction to Noncommutative Algebra" and am currently focussed on Chapter 1: Finite Dimensional Division Algebras ... ...

I need help with some aspects of Bresar's Example 1.10 on a simple matrix ring over a division ring ...

Example 1.10, including some preamble, reads as follows:https://www.physicsforums.com/attachments/6238
In the above text from Bresar we read the following:

" ... and hence also \(\displaystyle (d a_{jk}^{-1} ) E_{ii} \cdot a_{jk} E_{il} = d E_{il}\) for every \(\displaystyle d \in D\). Consequently, \(\displaystyle I = M_n(D)\). ... ... "My questions are as follows:Question 1I am assuming that \(\displaystyle (d a_{jk}^{-1} ) E_{ii} \cdot a_{jk} E_{il} = d E_{il}\) because you can take the "scalars" out of the multiplication and multiply them as in

\(\displaystyle c_1 (a_{ij} ) \cdot c_2 (b_{ij} ) = c_1 c_2 (a_{ij} ) \cdot (b_{ij} )\)Is that correct?(Note: why we are messing with multiplications by scalars in a problem on rings, I don't know ... we seem to be treating the ring \(\displaystyle M_n (D)\) as an algebra over \(\displaystyle D\) ... )
Question 2


Bresar seems to be assuming that \(\displaystyle d E_{il}\) for all \(\displaystyle 1 \le i, l \le n\) and for every \(\displaystyle d \in D\) implies that \(\displaystyle I = M_n (D)\) ...

But ... ... why exactly is this true ...My thoughts ... maybe it is true because the \(\displaystyle E_{il}\) generate the ring \(\displaystyle M_n (D)\) ... or to put it another way ... any element in \(\displaystyle I\) or \(\displaystyle M_n (D)\) can be written uniquely in the form \(\displaystyle \sum_{i, j = 1}^n d_{ij} E_{ij} \) ... Help with these questions will be appreciated ...

Peter=====================================================So that readers of the above post can appreciate the relevant context of the post, I am providing the introduction to Section 1.3 Simple Rings ... as follows:View attachment 6239
View attachment 6240

 

[Opalg]

Question 1I am assuming that \(\displaystyle (d a_{jk}^{-1} ) E_{ii} \cdot a_{jk} E_{il} = d E_{il}\) because you can take the "scalars" out of the multiplication and multiply them as in

\(\displaystyle c_1 (a_{ij} ) \cdot c_2 (b_{ij} ) = c_1 c_2 (a_{ij} ) \cdot (b_{ij} )\)Is that correct?(Note: why we are messing with multiplications by scalars in a problem on rings, I don't know ... we seem to be treating the ring \(\displaystyle M_n (D)\) as an algebra over \(\displaystyle D\) ... )
Question 2


Bresar seems to be assuming that \(\displaystyle d E_{il}\) for all \(\displaystyle 1 \le i, l \le n\) and for every \(\displaystyle d \in D\) implies that \(\displaystyle I = M_n (D)\) ...

But ... ... why exactly is this true ...My thoughts ... maybe it is true because the \(\displaystyle E_{il}\) generate the ring \(\displaystyle M_n (D)\) ... or to put it another way ... any element in \(\displaystyle I\) or \(\displaystyle M_n (D)\) can be written uniquely in the form \(\displaystyle \sum_{i, j = 1}^n d_{ij} E_{ij} \) ...

You are correct on both counts.

Question 1: $D$ is a ring, but $M_n (D)$ is an algebra, whose ring of scalars is $D$. In a product like $(d a_{jk}^{-1} ) E_{ii} \cdot a_{jk} E_{il}$, you can push the scalars past the matrices to get $$(d a_{jk}^{-1} ) E_{ii} \cdot a_{jk} E_{il} = (d a_{jk}^{-1}a_{jk} ) E_{ii} \cdot E_{il} = dE_{il}.$$

Question 2: Yes, that is exactly the reasoning here.