# I Multiplication Maps on Algebras ... Bresar, Lemma 1.25 ...

1. Dec 5, 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 help with the proof of Lemma 1.25 ...

Lemma 1.25 reads as follows:

My questions on the proof of Lemma 1.25 are as follows:

Question 1

In the above text from Bresar we read the following:

" ... ... Therefore $[ M(A) \ : \ F ] \ge d^2 = [ \text{ End}_F (A) \ : \ F ]$ ... ... "

Can someone please explain exactly why Bresar is concluding that $[ M(A) \ : \ F ] \ge d^2$ ... ... ?

Question 2

In the above text from Bresar we read the following:

" ... ... Therefore $[ M(A) \ : \ F ] \ge d^2 = [ \text{ End}_F (A) \ : \ F ]$

and so $M(A) = [ \text{ End}_F (A) \ : \ F ]$. ... ... "

Can someone please explain exactly why $[ M(A) \ : \ F ] \ge d^2 = [ \text{ End}_F (A) \ : \ F ]$ ... ...

... implies that ... $M(A) = [ \text{ End}_F (A) \ : \ F ]$ ...

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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• ###### Bresar - 2 - Section 1.5 Multiplication Algebra - PART 2 ... ....png
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2. Dec 5, 2016

### Staff: Mentor

We know that $M(A) = \langle L_a , R_b \,\vert \, a,b \in A \rangle$ is generated by left- and right-multiplications by definition.
Lemma 1.24 guarantees us that $\{L_{u_i} \cdot R_{u_j} = L_{u_i} \circ R_{u_j}\,\vert \, 1 \leq i,j \leq d \}$ are linear independent $^{*})$. These are $d^2$ many, so the dimension of $M(A)$ can only be greater than $d^2$, because we already have found $d^2$ linear independent vectors, which can be extended to a basis.

$^{*}) \; 0 = \sum \lambda_{ij}L_{u_i}R_{u_j} = \sum L_{u_i} R_{b_i} \Longrightarrow$ (Lemma 1.24) $b_i = \sum \lambda_{ij}u_j = 0 \Longrightarrow \lambda_{ij}=0$ because $\{u_k\}$ is a basis, hence $L_{u_i}R_{u_j} = L_{u_i} \cdot R_{u_j} = L_{u_i} \circ R_{u_j}$ are linear independent.
Well, $L_a$ as well as $R_b$ are endomorphisms of $A$, i.e. $\mathbb{F}-$linear mappings $A \rightarrow A$.
Therefore $M(A) \subseteq End_\mathbb{F}(A)$ and we have a subspace $M(A)$ which has at least dimension $d^2$. On the other hand $End_\mathbb{F}(A)$ has exactly the dimension $d^2$, so there is no room left between $M(A)$ and $End_\mathbb{F}(A)$.
That $\dim End_\mathbb{F}(A)=d^2$ can fastest be seen, if we think of matrices: Since $\{u_1,\ldots, u_d\}$ is a basis of $A$, all linear functions, i.e. every element of $End_\mathbb{F}(A)$ can be written as a $(d \times d)-$matrix with respect to this basis.

3. Dec 6, 2016

### Math Amateur

Thanks fresh_42 ... most helpful in helping me to grasp the meaning of Lemma 1.25 ...

But just a clarification ... You write:

" ... ... Lemma 1.24 guarantees us that $\{L_{u_i} \cdot R_{u_j} = L_{u_i} \circ R_{u_j}\,\vert \, 1 \leq i,j \leq d \}$ are linear independent $^{*})$. "

What do you mean when you write " $L_{u_i} \cdot R_{u_j} = L_{u_i} \circ R_{u_j}$ " ...

There appear to be two "multiplications" involved, namely $\cdot$ and $\circ$ ... but what are these ...?

and, further what is the meaning and significance of the equality " $L_{u_i} \cdot R_{u_j} = L_{u_i} \circ R_{u_j}$ "

Can you help ...

Still reflecting on your post ...

Peter

4. Dec 6, 2016

### Staff: Mentor

I simply wanted to indicate, that multiplication $"\cdot"$ here is the successive application of mappings $"\circ"$, no matter how it is written even if without multiplication sign. In the end it is $(L_{u_i}R_{u_j})(x) = L_{u_i}(R_{u_j}(x))=L_{u_i}(x\cdot u_j) = u_i \cdot x \cdot u_j$.

5. Dec 6, 2016

### zinq

I am learning a bit about algebra(s) that I never knew! What exactly does the property of being "central simple" have to do with the conclusions above?

Also: now I want to understand the "Brauer group".

6. Dec 6, 2016

### Staff: Mentor

A $\mathbb{K}-$algebra $\mathcal{A}$ is central simple, if the center $\mathcal{C}(\mathcal{A})=\{c\in \mathbb{A}\,\vert \,ca=ac \,\forall \,a\in\mathcal{A}\}$ of $\mathcal{A}$ equals $\mathbb{K}$ and $\mathcal{A}$ as a ring is simple, i.e. without ideals.

The Brauer group is a long time no see. So I've read the definition again. A funny thing, that you're talking about. How does it come?

According to the theorem of Artin-Wedderburn, every central simple algebra is isomorphic to a matrix algebra $\mathbb{M}(n,\mathcal{D})$ over a division ring $\mathcal{D}$, here $\mathcal{D}=\mathbb{K}$. Now all $\mathbb{M}(n,\mathbb{K})$ are considered equivalent, i.e. $\mathbb{M}(n,\mathbb{K}) \text{ ~ } \mathbb{M}(m,\mathbb{K})$ and the elements of the Brauer group (of $\mathbb{K}$) are the equivalence classes. E.g. $[1] = [\mathbb{M}(1,\mathbb{K})]=[\mathbb{K}]$ and the inverse element is the opposite algebra $\mathcal{A}^{op}$ with the multiplication $(a,b) \mapsto ba$.

However, the really interesting question here is: Do all Scottish mathematicians (Hamilton, Wedderburn, ...) have a special relationship to strange algebras and why is it so?

7. Dec 6, 2016

### Staff: Mentor

I wasn't quite satisfied with this lapidary description of an equivalence relation here. Unfortunately the English and German Wikipedia page are a one-to-one translation. But the French has been a little bit better. Starting with a central simple algebra $\mathcal{A}$ over a field $\mathbb{K}$, we have $\mathcal{A} \otimes_{\mathbb{K}} \mathbb{L} \cong \mathbb{M}(n,\mathbb{L})$ for a finite field extension $\mathbb{L} \supseteq \mathbb{K}$

Now $\mathcal{A} \text{ ~ } \mathcal{B}$ are considered equivalent, if there are natural numbers $n,m$ and an isomorphism such that $\mathcal{A} \otimes_{\mathbb{K}} \mathbb{M}(n,\mathbb{K}) \cong \mathcal{B} \otimes_{\mathbb{K}} \mathbb{M}(m,\mathbb{K})$.
The (Abelian) Brauer group are now the equivalence classes with multiplication $\otimes$.

(At least as far as my bad French allowed me to translate it.)

8. Dec 6, 2016

### Staff: Mentor

@fresh_42 - Re: Scots & maths - try the Jack polynomial.

9. Dec 6, 2016

### Staff: Mentor

Hmmm ... I wonder whether they spoke Gaelic ...

10. Dec 6, 2016

### Math Amateur

Hi fresh_42 ...

Just a further clarification ... ...

You write:

" ... ... We know that $M(A) = \langle L_a , R_b \,\vert \, a,b \in A \rangle$ is generated by left- and right-multiplications by definition. ... ... "

Now ... if $M(A)$ is generated by $L_a$ and $R_b$ then it should contain elements like $L_a L_b L_c$ and $L_a^2 R_b^2 R_c$ ... and so on ...

BUT ... how do elements like these fit with Bresar's definition of $M(A)$ ... as follows:

$M(A) := \{ L_{a_1} R_{b_2} + \ ... \ ... \ + L_{a_1} R_{b_2} \ | \ a_i, b_i \in A, n \in \mathbb{N} \}$

... ...

... ... unless ... we treat $L_a L_b L_c = L_{abc} R_1 = L_t R_u$

where $t = abc$ and $u = 1$ ... ...and $t, u \in A$ ... ...

... and ...

we treat $L_a^2 R_b^2 R_c = L_{aa} R_{cbb} = L_r R_s$

where $r = aa$ and $s = cbb$ ...

Can you help me to clarify this issue ....

Peter

11. Dec 7, 2016

### Staff: Mentor

That's correct. In the lines ahead of Definition 1.22 Bresar mentions the rules by which $\{L_{a_1}R_{b_1}+\ldots +L_{a_n}R_{b_n}\}$ becomes an algebra. Without them, it would simply be a set of some endomorphisms.

12. Dec 7, 2016

### zinq

But also, #10 is almost immediate from the definitions of Ls and Rt:

(La Lb) x = (LaoLb) x​

= La (Lb x)​

= La (bx)​

= a(bx)​

= (ab)x​

= Lab x.
And virtually the same reasoning to show

(Rc Rd) x = Rdc x.​

(Also note that, any La and any Rb commute:

La Rb = Rb La.​

This can be proved in a similar manner.)

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