# I Multiplication Maps on Algebras ... Further questions ...

1. Dec 4, 2016

### 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 ...

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

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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:

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2. Dec 4, 2016

### Staff: Mentor

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: $1=1\,\cdot\,1'= 1'$.
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.
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 $0$.
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.
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.

3. Dec 4, 2016

### Math Amateur

Thanks fresh_42 ...