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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
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