Multiplication Maps on Algebras .... Further questions ....

[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 some further help with the statement and proof of Lemma 1.24 ...

Lemma 1.24 reads as follows:

My further questions regarding Bresar's statement and proof of Lemma 1.24 are as follows:Question 1

In the statement of Lemma 1.24 we read the following:

" ... ... Let ##A## be a central simple algebra. ... ... "I am assuming that since ##A## is central, it is unital ... that is there exists ##1_A \in A## such that ##x.1 = 1.x = 1## for all ##x \in A## ... ... is that correct ... ?
Question 2

In the proof of Lemma 1.24 we read the following:

" ... ... Suppose ##b_n \ne 0##. ... ... "I am assuming that that the assumption ##b_n \ne 0## implies that we are also assuming

that ##b_1 = b_2 = \ ... \ ... \ = b_{n-1} = 0## ... ...

Is that correct?

Question 3

In the proof of Lemma 1.24 we read the following:

" ... ...where ##c_i = \sum_{ j = 1 }^m w_j b_i z_j## ; thus ##c_n = 1## for some ##w_j, z_j \in A## ... ...

This clearly implies that ##n \gt 1##. ... ... "My question is ... why/how exactly must ##n \gt 1## ... ?

Further ... and even more puzzling ... what is the relevance to the proof of the statements that ##c_n = 1## and ##n \gt 1## ... ?

Why do we need these findings to establish that all the ##b_i = 0## ... ?

Hope someone can help ...

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:

 

[fresh_42]

Question 1

In the statement of Lemma 1.24 we read the following:

" ... ... Let ##A## be a central simple algebra. ... ... "I am assuming that since ##A## is central, it is unital ... that is there exists ##1_A \in A## such that ##x.1 = 1.x = 1## for all ##x \in A## ... ... is that correct ... ?

Central means the center ## C(A) ## of ##A## is equal to the corresponding scalar domain, i.e. a field or a division ring at least. Since the center is part of the algebra, the field ##\mathbb{F} =C(A) ## is contained in the algebra. Then ##1_\mathbb{F} \in A##. So ##1_\mathbb{F}=1_A##, because it does what a one has to do: ##1_\mathbb{F}\cdot a = a## and there cannot be two different ones: ##1=1\,\cdot\,1'= 1'##.

Question 2

In the proof of Lemma 1.24 we read the following:

" ... ... Suppose ##b_n \ne 0##. ... ... "I am assuming that that the assumption ##b_n \ne 0## implies that we are also assuming

that ##b_1 = b_2 = \ ... \ ... \ = b_{n-1} = 0## ... ...

Is that correct?

No. We only assume at least one ##b_i \neq 0## to derive a contradiction. So without loss of generality, we assume it to be ##b_n##, for otherwise we would simply change the numbering. We don't bother ## b_1, \ldots ,b_{n-1}##. They may be equal to zero or not. Only the case in which all are zero is ruled out, for then we would have nothing to show: the ##a_i## would be linear independent.

Question 3

In the proof of Lemma 1.24 we read the following:

" ... ...where ##c_i = \sum_{ j = 1 }^m w_j b_i z_j## ; thus ##c_n = 1## for some ##w_j, z_j \in A## ... ...

This clearly implies that ##n \gt 1##. ... ... "My question is ... why/how exactly must ##n \gt 1## ... ?

If ##n=1## then ##\left( \sum_{i=1}^n L_{a_i}R_{c_i} \right)(x) = (L_{a_1}R_{c_1})(x)=a_1 \cdot x \cdot c_1=a_1 \cdot x=0## for all ##x \in A##, because ##c_n=c_1=1##. Especially ##a_1\cdot 1_\mathbb{F} = a_1 = 0##, but the ##a_i## are linear independent, so they cannot equal ##0##.

Further ... and even more puzzling ... what is the relevance to the proof of the statements that ##c_n = 1## and ##n \gt 1## ... ?

It is important for the sum in ##0=\sum_{i=1}^{n-1} L_{a_i}R_{xc_i-c_i x}## being a real sum and not only trivially fulfilled.

Why do we need these findings to establish that all the ##b_i = 0## ... ?

They are used to show all ##c_i \in \mathbb{F}##. Then ##c_1a_1+\ldots c_na_n## is not only an equation in ##A## but a linear one with coefficients in ##\mathbb{F}##. Now if ##0=L_{c_1a_1+\ldots c_na_n}## then ##0=L_{c_1a_1+\ldots c_na_n}(1_\mathbb{F})=(c_1a_1+\ldots c_na_n)\cdot 1_\mathbb{F}=c_1a_1+\ldots c_na_n## and the ##a_i## are linear dependent over ##\mathbb{F}##, because at least ##c_n \neq 0## and we have our contradiction.

 

[Math Amateur]

Thanks fresh_42 ...

That post was INCREDIBLY helpful!

Peter