Talk:Division 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.

-- user:clausen


 * 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 $$(ax)*y=\overline{(ax)y}=\overline{a(xy)}=\overline{a}\cdot\overline{xy}=\overline{a}(x*y)$$.

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


 * There is no slip in your reasoning. $$\overline{a}=a$$ since $$a$$ 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 $$a*b=\overline{ab}$$ is not isomorphic to the complex numbers. For example, $$a*(b*c)=\overline{a}bc$$, whereas $$(a*b)*c=ab\overline{c}$$, so the new algebra is not associative (and thus not isomorphic to C). 130.238.58.63 (talk) 16:05, 8 July 2008 (UTC)

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)

Topology ref
The attribution to milnor and kervaire is false. Adams proved it first using the Adam's spectral sequence. Then Adams and Atiyah proved it using K-theory anbd Adam's operations (stating the proof could fit on a postcard). 78.50.95.19 (talk) 11:12, 10 September 2022 (UTC)


 * @78.50.95.19 I should have added that this can be linked to the appropriate part of the hopf invariant one article. I also would like a reference that Hopf reduced the dimension to a power of two. I know a proof using steenrod operations bit I am not sure if there was an earlier proof or if hopf was working after steenrod work. 78.50.95.19 (talk) 11:14, 10 September 2022 (UTC)