MHB Automorphisms of Central Simple Algebras - Bresar Example 1.27 .... ....

[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 aspects of Example 1.27...

Example 1.27 reads as follows:
View attachment 6266My 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 $$M( \mathbb{C} ) = \text{End}_\mathbb{R} ( \mathbb{C} )$$ ... ?

Hopefully, someone can clarify this matter ...(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:
View attachment 6267
View attachment 6268
https://www.physicsforums.com/attachments/6269

 

[GJA]

Hi Peter,

(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 ... ...

If the ring is commutative, then the multiplication $axa^{-1}$ in Definition 1.26 will always work out to be $x$, so the only inner automorphism of a commutative ring is the identity map.

(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 ... ?

A field is a special type of commutative ring, just like a square is a special type of rectangle. The statement is more general when specified in terms of the least amount of structure necessary - in this case the commutative ring structure.

(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 $$M( \mathbb{C} ) = \text{End}_\mathbb{R} ( \mathbb{C} )$$ ... ?

$\mathbb{C}$ is not a central simple algebra over $\mathbb{R}$. Try seeing if you can determine why it's not central (this is why the author makes the comment about $A$ being central).

(b) ... can someone please explain exactly why the conjugation automorphism is an element

of $$\text{End}_\mathbb{R} ( \mathbb{C} )$$ but not of $$M( \mathbb{C} )$$ ... ?

If it were an element of $M(\mathbb{C})$ then it would need to be expressible as a a linear combination of left/right shift operators. Try looking at an example for a contradiction. For example, try looking at $z=1+i$ and seeing if you can determine a contradiction that comes from requiring

$$
1-i = \sum_{k=1}^{n}a_{k}(1+i)b_{k}
$$

Edit: I made an error in the first post because I was thinking of $a_{k}$ and $b_{k}$ as belonging to $\mathbb{R}$. The idea of looking for a counterexample in the fashion I mentioned is not as simple as I thought. Instead I will say for the time being that it's known that $z\mapsto\bar{z}$ is not a holomorphic function but any function of the form $z\mapsto\sum_{k}^{n}a_{k}zb_{k}$ is.

 

[Math Amateur]

Hi Peter,
If the ring is commutative, then the multiplication $axa^{-1}$ in Definition 1.26 will always work out to be $x$, so the only inner automorphism of a commutative ring is the identity map.
A field is a special type of commutative ring, just like a square is a special type of rectangle. The statement is more general when specified in terms of the least amount of structure necessary - in this case the commutative ring structure.
$\mathbb{C}$ is not a central simple algebra over $\mathbb{R}$. Try seeing if you can determine why it's not central (this is why the author makes the comment about $A$ being central).

If it were an element of $M(\mathbb{C})$ then it would need to be expressible as a a linear combination of left/right shift operators. Try looking at an example for a contradiction. For example, try looking at $z=1+i$ and seeing if you can determine a contradiction that comes from requiring

$$
1-i = \sum_{k=1}^{n}a_{k}(1+i)b_{k}
$$

Edit: I made an error in the first post because I was thinking of $a_{k}$ and $b_{k}$ as belonging to $\mathbb{R}$. The idea of looking for a counterexample in the fashion I mentioned is not as simple as I thought. Instead I will say for the time being that it's known that $z\mapsto\bar{z}$ is not a holomorphic function but any function of the form $z\mapsto\sum_{k}^{n}a_{k}zb_{k}$ is.

Thanks GJA ... most helpful ...

Still puzzling over the last question ...

Thanks again ...

Peter

 

[GJA]

Hi Peter,

I realized I was overlooking two critical facts to answer your last question (instead of appealing to holomorphicity) - going to chalk it up to the late hour.

1) When we write an element $f\in M(A)$ in the form $f(x)=\sum_{k}a_{k}xb_{k},$ the $a$'s and $b$'s are fixed.

2) Since $\mathbb{C}$ is commutative, any function $f\in M(\mathbb{C})$ can be expressed as

$$
f(z)=\sum_{k}a_{k}b_{k}z=\left(\sum_{k}a_{k}b_{k}\right) z = wz
$$

where $w\in\mathbb{C}$. Thus any $f(z)\in M(\mathbb{C})$ must have the form $f(z)=wz$ where $w$ is a fixed complex number. Proving the statement now can be reduced to looking at an example. Taking $z = 1$ would require $w=1$ while taking $z=i$ would require $w=-1$.