Lemma 1.24 proof on left/right multiplication maps in Bresar

[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 the proof of Lemma 1.24 ...

Lemma 1.24 reads as follows:

My questions regarding the proof of Lemma 1.24 are as follows ... ...Question 1

In the above proof by Bresar, we read:

" ... ... Since $A$ is simple, the ideal generated by $b_n$ is equal to $A$.

That is $\sum_{ j = 1 }^m w_j b_n z_j = 1$ for some $w_j , z_J \in A$. ... ... "My question is ... ... how/why does the fact that the ideal generated by $b_n$ being equal to $A$ ...

imply that ... $\sum_{ j = 1 }^m w_j b_n z_j = 1$ for some $w_j , z_J \in A$ ...?

Question 2In the above proof by Bresar, we read:" ... $0 = \sum_{ j = 1 }^m R_{ z_j } \ ( \sum_{ i = 1 }^n L_{ a_i } R_{ b_i } ) \ R_{ w_j }$$= \sum_{ i = 1 }^n L_{ a_i } \ ( \sum_{ j = 1 }^m R_{ w_j b_i z_j } )$$= \sum_{ i = 1 }^n L_{ a_i } R_{ c_i }$... ... "
My questions are

(a) can someone help me to understand how$\sum_{ j = 1 }^m R_{ z_j } \ ( \sum_{ i = 1 }^n L_{ a_i } R_{ b_i } ) \ R_{ w_j }$$= \sum_{ i = 1 }^n L_{ a_i } \ ( \sum_{ j = 1 }^m R_{ w_j b_i z_j } )$
(b) can someone help me to understand how$\sum_{ i = 1 }^n L_{ a_i } \ ( \sum_{ j = 1 }^m R_{ w_j b_i z_j } )$$= \sum_{ i = 1 }^n L_{ a_i } R_{ c_i }$

Help will be appreciated ...

Peter

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*** NOTE ***So that readers of the above post will be able to understand the context and notation of the post ... I am providing Bresar's first two pages on Multiplication Algebras ... ... as follows:

 

[fresh_42]

Q1:
What does it mean to be an ideal $I$ of $A$?
For a (two-sided!) ideal, it has to hold, that $A\cdot I \subseteq I$ and $I\cdot A \subseteq I$. Since $b_n \in I$, we need to have all left and right multiples to also be in $I$. So all elements of the form $w_jb_nz_j$ are together with $b_n$ also elements of $I$.
Furthermore an ideal is closed under addition, so all summations of elements of $I$ are again in $I$, esp. any sum $\sum_{j=1}^{m} w_jb_nz_j$. This is simultaneously the most general form of any element of $I = <b_n> = A\cdot I \cdot A$.

Q2:
We have $\sum_{i=1}^{n} L_{a_i}R_{b_i} = 0 \; (^*) \;$ by assumption.
Then let us define $\sum_{j=1}^{m} w_jb_iz_j =: c_i \; (^{**}) \;$, simply as an abbreviation for these sums $c_1, \ldots , c_n$.
Because all sums are finite, we won't have to bother any order of summation.
At last let us assume we have an arbitrary element $x \in A$.

Now calculate $\left( \sum_{j=1}^{m} R_{z_j} \left( \sum_{i=1}^{n} L_{a_i}R_{b_i} \right) R_{w_j} \right)(x)$ by using $(^*)$ and $(^{**})$.

 

[Math Amateur]

Thanks again for your assistance, fresh_42 ...

Most helpful ...

Peter