MHB Centre of an Algebra .... and Central Algebras ....

[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 some remarks of Bresar on the centre of an algebra ...

Commencing a section on Central Algebras, Bresar writes the following:https://www.physicsforums.com/attachments/6243
In the above text we read the following:

" ... The center of a unital algebra obviously contains scalar multiples of unity ... ... "Now the center of a unital algebra $$A$$ is defined as the set $$Z(A)$$ such that

$$Z(A) = \{ c \in A \ | \ cx = xc \text{ for all x } \in A \} $$Now ... clearly $$1 \in Z(A)$$ since $$1x = x1$$ for all $$x$$ ...

BUT ... why do elements like $$3$$ belong to $$Z(A)$$ ... ?

That is ... how would we demonstrate that $$3x = x3$$ for all $$x \in A$$ ... ?

Hope someone can help ...

Peter

 

[Euge]

Well, using the distributive properties and the fact that $x = x1$ for all $x$,

$$3x = x + x + x = x1 + x1 + x1 = x(1 + 1 + 1) = x3$$

 

[Opalg]

More generally, one of the axioms for an algebra $A$ over a field $F$ says that $$\lambda(xy) = (\lambda x)y = x(\lambda y)$$ for all $\lambda\in F$ and $x,y\in A.$ It follows that $$\lambda x = \lambda(x1) = x(\lambda 1) = x\lambda.$$

 

[Math Amateur]

Well, using the distributive properties and the fact that $x = x1$ for all $x$,

$$3x = x + x + x = x1 + x1 + x1 = x(1 + 1 + 1) = x3$$

Thanks Euge ...

Just wondering how you did this without using the axiom that Opalg refers to ... I think it is called "The Compatibility Axiom" ... ... namely ... ... for an algebra $A$ over a field $F$ says that $$\lambda(xy) = (\lambda x)y = x(\lambda y)$$ for all $\lambda\in F$ and $x,y\in A.$ ...

As Opalg says ... It follows that $$\lambda x = \lambda(x1) = x(\lambda 1) = x\lambda.$$

... but you don's appear to need it ... why?

Peter

 

[Math Amateur]

More generally, one of the axioms for an algebra $A$ over a field $F$ says that $$\lambda(xy) = (\lambda x)y = x(\lambda y)$$ for all $\lambda\in F$ and $x,y\in A.$ It follows that $$\lambda x = \lambda(x1) = x(\lambda 1) = x\lambda.$$

Hi Opalg ... thanks for the help ...

But ... just need a further point clarified ...

You write:

" ... ... It follows that $$\lambda x = \lambda(x1) = x(\lambda 1) = x\lambda.$$ ..."
But in this statement $$1 \in A$$ ... that is we are dealing with $$1_A$$ ... ...

and your equation

$$\lambda x = \lambda(x1) = x(\lambda 1) = x\lambda.$$

uses the equality or equivalence $$\lambda 1_A = \lambda $$Now ... although it is clear from the axioms that apply that $$\lambda 1_F = \lambda$$ ... ...... why is it the case that $$\lambda 1_A = \lambda$$ ... what axiom

underpins this statement.

Can you help?

Peter

 

[Euge]

Thanks Euge ...

Just wondering how you did this without using the axiom that Opalg refers to ... I think it is called "The Compatibility Axiom" ... ... namely ... ... for an algebra $A$ over a field $F$ says that $$\lambda(xy) = (\lambda x)y = x(\lambda y)$$ for all $\lambda\in F$ and $x,y\in A.$ ...

As Opalg says ... It follows that $$\lambda x = \lambda(x1) = x(\lambda 1) = x\lambda.$$

... but you don's appear to need it ... why?

Peter

An algebra over a field is, in particular, a vector space of the field -- so the distributive properties hold. With that, the result follows immediately.

 

[Opalg]

... why is it the case that $$\lambda 1_A = \lambda$$ ... what axiom

underpins this statement.

This is not an axiom, but a convention, as stated by Bresar just after his Definition 1.14. In the section of text that you reproduce above, he states "... we identify $F$ with $F\cdot 1$, and write $\lambda$ instead". The rest of the sentence is omitted, but presumably it says "... write $\lambda$ instead of $\lambda 1$".

 

[Math Amateur]

This is not an axiom, but a convention, as stated by Bresar just after his Definition 1.14. In the section of text that you reproduce above, he states "... we identify $F$ with $F\cdot 1$, and write $\lambda$ instead". The rest of the sentence is omitted, but presumably it says "... write $\lambda$ instead of $\lambda 1$".

Thanks to Opalg and Euge for considerable help on this issue ...

You posts were a real help ...

Peter