# Talk:Division algebra

WikiProject Mathematics (Rated Start-class, Mid-importance)
This article is within the scope of WikiProject Mathematics, a collaborative effort to improve the coverage of Mathematics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.
Mathematics rating:
 Start Class
 Mid Importance
Field:  Algebra

## Associativity

"In this page, however, we will assume associativity".

huh? Fields and associative algebras are both associative (in multiplication, which is what we're interested in here, right?)

What's the purpose of this statement? Associativity was never at issue.

The statement is at the end of a paragraph which gives a different (non-associative) definition. If the statement were omitted, it would be unclear which definition is being used in the rest of the article. --Zundark 08:36 5 Jun 2003 (UTC)

Why do we have separate pages for division ring and division algebra? There isn't much difference.

Waltpohl 07:46, 24 Feb 2004 (UTC)

It seems that 'algebra' here is not always assumed associative.

Charles Matthews 09:12, 24 Feb 2004 (UTC)

Examples would be nice!

rs2 2004.03.10, 01:02 (UTC)

## Field or Ring

Shouldn't the article refer to rings ? The way I read it, all fields are divisional algebras. -- Nic Roets 13:52, 20 July 2005 (UTC)

Yes, all fields are division algebras. From what I know, division algebras are generalizations of fields. Basically, a division algebra is an object in which you can still do division, but it does not need to be commutative or associative. Oleg Alexandrov 15:30, 20 July 2005 (UTC)

Hi

I have edited Example of a non-associative algebra but I must have slipped up somewhere and have seemingly proved that it isn't an algebra at all. I got ${\displaystyle (ax)*y={\overline {(ax)y}}={\overline {a(xy)}}={\overline {a}}\cdot {\overline {xy}}={\overline {a}}(x*y)}$.

Can anyone patch up my slip in reasoning? Robinh 19:25, 7 August 2005 (UTC)

There is no slip in your reasoning. ${\displaystyle {\overline {a}}=a}$ since ${\displaystyle a}$ is real. In fact this division algebra you give is isomorphic to the usual multiplication on the complex plane, the isomorphism given by complex conjugation. perkinsrc008 12:19, 14 April 2008 (UTC)
The algebra defined from the complex numbers by ${\displaystyle a*b={\overline {ab}}}$ is not isomorphic to the complex numbers. For example, ${\displaystyle a*(b*c)={\overline {a}}bc}$, whereas ${\displaystyle (a*b)*c=ab{\overline {c}}}$, so the new algebra is not associative (and thus not isomorphic to C). 130.238.58.63 (talk) 16:05, 8 July 2008 (UTC)

## References

How come there are no references here for any proofs? Can anybody point in in the direction of why this is true:

* It is known about the dimension of a finite-dimensional division algebra A over a field K:
* dim A= 1 if K is algebraically closed,
* dim A= 1, 2, 4 or 8 if K is real closed  —Preceding unsigned comment added by Moxmalin (talk • contribs) 17:10, 20 March 2008 (UTC)

Why is the dimension of a finite-dimensional division algebra over an algebraically closed field equal to one? Suppose the dimension of such an algebra ${\displaystyle A}$ is at least two, and let ${\displaystyle a,b}$ be non-proportional vectors. Denote by ${\displaystyle L_{a},L_{b}}$ the linear operators of left multiplication with ${\displaystyle a,b}$ respectively. Being a linear operator on a vector space over an algebraically closed field, ${\displaystyle L_{b}^{-1}L_{a}}$ has an eigenvalue, say ${\displaystyle r}$, and a corresponding non-zero eigenvector ${\displaystyle x}$. Thus ${\displaystyle L_{b}^{-1}L_{a}(x)=rx}$, which implies ${\displaystyle (a-rb)x=0}$. Since ${\displaystyle x}$ is non-zero and ${\displaystyle a,b}$ linearly independent, this is impossible. QED.
As for the statement about real closed fields, it follows via a model theoretic trick from the corresponding result for the real numbers. A detailed proof for the theorem is given in the article 'In which dimensions does a division algebra over a given ground field exist' by Darpö, Dieterich and Herschend (Enseign. Math. (2) 51 (2005), no. 3-4). 130.238.58.63 (talk) 16:05, 8 July 2008 (UTC)

## Early work

This article should cite C. S. Peirce, who proved that the only dimensions for which division is defined are 1, 2, 4, and 8. See Charles Sanders Peirce (1881), "On the Algebras in which Division is Unambiguous", Addendum III in Peirce, Benjamin, "Linear Associative Algebra", American Journal of Mathematics v. 4, pp. 226-229, republished 1882 as Linear Associative Algebra with the addenda and notes by C. S. Peirce, D. Van Nostrand, New York, 133 pages, pp. 129-133. (This book can be downloaded from Google books.) — Preceding unsigned comment added by 71.183.59.192 (talk) 13:29, 2 March 2014 (UTC)

It would help to give a reference to a modern source putting Peirce's work into historical context. Deltahedron (talk) 14:27, 2 March 2014 (UTC)

## Infinite dimensional?

There is not a single example of an infinite-dimensional division algebra in the article, and the articles on Banach algebras and normed division algebras say that only finite-dimensional ones exist. There seems to be some sort of mistake.--Leon (talk) 15:50, 9 October 2016 (UTC)