Left and Right Multiplication Maps on Algebras .... Bresar, Lemma 1.24

[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:View attachment 6253
My questions regarding the proof of Lemma 1.24 are as follows ... ...Question 1


In the above proof by Bresar, we read:" ... ... Since \(\displaystyle A\) is simple, the ideal generated by \(\displaystyle b_n\) is equal to \(\displaystyle A\).

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

imply that ... \(\displaystyle \sum_{ j = 1 }^m w_j b_n z_j = 1\) for some \(\displaystyle w_j , z_J \in A\) ...?Question 2In the above proof by Bresar, we read:" ... \(\displaystyle 0 = \sum_{ j = 1 }^m R_{ z_j } \ ( \sum_{ i = 1 }^n L_{ a_i } R_{ b_i } ) \ R_{ w_j }\)\(\displaystyle = \sum_{ i = 1 }^n L_{ a_i } \ ( \sum_{ j = 1 }^m R_{ w_j b_i z_j } ) \)\(\displaystyle = \sum_{ i = 1 }^n L_{ a_i } R_{ c_i } \)

... ... "

My questions are


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

Help will be appreciated ...

Peter

===========================================================*** 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:
View attachment 6254
https://www.physicsforums.com/attachments/6255

 

[jvicens]

/Thank you for your question, Peter. I am happy to help you understand the proof of Lemma 1.24 in Bresar's book.

Question 1:

The fact that the ideal generated by b_n is equal to A means that every element in A can be written as a linear combination of the elements in the set {b_n}. This is because the ideal generated by b_n is the smallest ideal in A that contains b_n, and since A is simple, the only ideals in A are {0} and A itself. Therefore, every element in A can be written as a linear combination of the elements in {b_n}.

Now, since we are trying to prove that \sum_{ j = 1 }^m w_j b_n z_j = 1 for some w_j , z_J \in A, we can use the fact that every element in A can be written as a linear combination of the elements in {b_n}. This means that we can write w_j and z_j as linear combinations of the elements in {b_n}, and therefore we can write \sum_{ j = 1 }^m w_j b_n z_j as a linear combination of the elements in {b_n}. Since the ideal generated by b_n is equal to A, this means that \sum_{ j = 1 }^m w_j b_n z_j must be equal to 1, since 1 is also an element of A.

I hope this helps to clarify how the fact that the ideal generated by b_n is equal to A implies that \sum_{ j = 1 }^m w_j b_n z_j = 1 for some w_j , z_J \in A.

Question 2:

(a) In the first step of the proof, we are using the fact that the multiplication in A is associative. This means that we can rearrange the parentheses in the expression \sum_{ j = 1 }^m R_{ z_j } \ ( \sum_{ i = 1 }^n L_{ a_i } R_{ b_i } ) \ R_{ w_j } to get \sum_{ j = 1 }^m (R_{ z_j } \sum_{ i = 1 }^n L_{ a_i } R_{ b_i }) \ R_{ w_j }. Now, we can use the fact that R_{ z_j } and R_{ w_j } are right