# Automorphisms of Central Simple Algebras/Bresar Example 1.27

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• Math Amateur
In summary: This follows because the composition of two conjugation automorphisms is again a conjugation automorphism, and conjugation is invertible.This is a nice result, but it is not quite what we are looking for. We want conjugation to be an automorphism of M(C), not just an endomorphism.One way to make conjugation an automorphism is to use the fact that every automorphism of a commutative ring is an automorphism of the self-dual closure of the ring. That is, every automorphism of a commutative ring is a automorphism of the intersection of all the elements of the ring with the identity
Math Amateur
Gold Member
MHB
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 aspects of Example 1.27...

Example 1.27 (including a short preamble and Definition 1.26) reads as follows:

My questions on Example 1.27 comprise the following:
Question 1

In the above example from Bresar we read the following:

" ... ... Of course it is outer ... for the identity map is obviously the only inner automorphism of a commutative ring. ... ... "

I have two questions regarding this remark ...

(a) ... why does Bresar assert that "the identity map is obviously the only inner automorphism of a commutative ring" ... surely a commutative ring may have some units (invertible elements other than ##1## ...) and so may have some inner automorphisms ... BUT Bresar is asserting that this is not the case ... can someone please clarify this issue ...(b) ... Bresar seems to be talking about ##\mathbb{C}## ... but he is referring to "a commutative ring" in the quote above ... but why? ... ##\mathbb{C}## is a field ... ?

Can someone please clarify what Bresar means ...

Question 2

In the above example from Bresar we read the following:

" ... ... We also remark that it is an element of ##\text{End}_\mathbb{R} ( \mathbb{C} )## but not ##M( \mathbb{C} )##. ... "I have two questions on this remark ... as follows ...(a) ... ... Bresar proved in Lemma 1.25 (see previous post) that for

##M(A) = \text{End}_F (A)## ...

so ... why isn't it true that ##M( \mathbb{C} ) = \text{End}_\mathbb{R} ( \mathbb{C} )## ... ?(b) ... can someone please explain exactly why the conjugation automorphism is an element of ##\text{End}_\mathbb{R} ( \mathbb{C} )## but not of ##M( \mathbb{C} )## ... ?Help will be much appreciated ... ...

Peter=============================================================================

So that readers of this post can appreciate the notation and the context I am providing Bresar Section 1.5 ( which includes Lemmas 1.24 and 1.25) and also the start of Bresar Section 1.6 ... ... as follows:

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• Bresar - Example 1.27 ....png
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• Bresar - 1 - Section 1.5 Multiplication Algebra - PART 1 ... ....png
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• Bresar - 2 - Section 1.5 Multiplication Algebra - PART 2 ... ....png
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• Bresar - 1 - Section 1.6 ... Automorphisms of Central Simple Algebras ... PART 1 ....png
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I think I can answer question 1 easily but mind that I am a bit rusty on abstract algebra.

if the ring is commutative then any inner automorphism is the identity since by definition of the inner automorphism ##\phi(x)=axa^{-1}=aa^{-1}x=x##. The second step in the series of equalities is because the ring is commutative.

Also any field is a commutative ring (by the definition of the field if I am not mistaken) so I hope the above answer this question.

Math Amateur
It is necessary to sit with pencil (or your favorite writing implement) and paper in order to understand the answers to questions like yours, Math Amateur.

M(C) is the multiplication algebra of C, generated by left and right multiplications of an element (call it z) of C, by other elements of C. First verify that conjugation is an endomorphism of the R-algebra C. (Here it is absolutely necessary to know what the terms conjugation, R, R-algebra, and C mean.) Then verify that no matter which left and right multiplications you use to get an element of M(C), you will never obtain complex conjugation.

## 1. What is an automorphism of a central simple algebra?

An automorphism of a central simple algebra is a bijective linear transformation that preserves the algebraic structure of the central simple algebra. This means that it maps elements within the algebra to other elements within the algebra, while also preserving the operations and properties of the algebra.

## 2. What is the significance of studying automorphisms of central simple algebras?

Studying automorphisms of central simple algebras is important because it allows us to better understand the structure and properties of these algebras. It also has applications in fields such as algebraic geometry, representation theory, and number theory.

## 3. How can one determine the automorphisms of a central simple algebra?

The automorphisms of a central simple algebra can be determined by considering the structure of the algebra and its defining properties. For example, in Bresar Example 1.27, the automorphisms can be determined by considering the structure of the quaternion algebra and its basis elements.

## 4. What is Bresar Example 1.27?

Bresar Example 1.27 refers to a specific example of a central simple algebra that is used to introduce the concept of automorphisms in this context. It is a specific quaternion algebra over a field of characteristic not equal to 2 or 3.

## 5. How are automorphisms of central simple algebras related to other mathematical concepts?

Automorphisms of central simple algebras have connections to various other mathematical concepts, such as group theory, Galois theory, and algebraic number theory. For example, the automorphisms of a central simple algebra can be used to define the group of units of the algebra, which has important implications in Galois theory and algebraic number theory.

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