"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)
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)
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 .
Can anyone patch up my slip in reasoning? Robinh 19:25, 7 August 2005 (UTC)
- There is no slip in your reasoning. since 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)
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 is at least two, and let be non-proportional vectors. Denote by the linear operators of left multiplication with respectively. Being a linear operator on a vector space over an algebraically closed field, has an eigenvalue, say , and a corresponding non-zero eigenvector . Thus , which implies . Since is non-zero and 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). 184.108.40.206 (talk) 16:05, 8 July 2008 (UTC)
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 220.127.116.11 (talk) 13:29, 2 March 2014 (UTC)