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It is well-known the Hurwitz theorem showing that every unital composition algebra is of dimension 1, 2, 4 or 8 over the base field.

Is there an analogue in the non unital case? ¿Do non unital composition algebras of arbitrary dimension exist?

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I think a big question is how you want to axiomatize these structures, as two axiomatizations which are equivalent in the presence of a unit might become drastically different when you remove from both the assumption of a unit. Particularly, the question is whether you want a notion of conjugation still to exist (in standard axiomatizations, as for example in the book by John H. Conway and Derek Smith, conjugation is defined by $v \mapsto 2\langle v, e \rangle - v$). –  Todd Trimble Sep 10 '12 at 14:38
    
Argh, $v \mapsto 2\langle v, e \rangle e - v$, where $e$ is the unit. –  Todd Trimble Sep 10 '12 at 14:39
    
No, I would not need a conjugation. My definition for composition algebra is (cf.Zhevlakov et al. for instance): An algebra over a field F with quadratic form n(x) is called a composition algebra if: -n(xy)=n(x)n(y). -The form n(x) is strictly nondegenerate. - A has identity. In particular I am interested in removing the 2 las conditions so I just have a quadratic form n(x) [f(x,y)=n(x+y)-n(x)-n(y) bilinear] admitting composition. –  Antonio Oller Sep 10 '12 at 14:56
    
Sorry not to have replied sooner. If the quadratic form can be degenerate, what stops us from having both the algebra multiplication and $n$ be identically zero? –  Todd Trimble Sep 15 '12 at 17:05
    
And what if we keep non-degeneracy but remove the unit? –  Antonio Oller Sep 27 '12 at 18:58

1 Answer 1

If $A$ is a finite-dimensional algebra over a field, with a strictly non-degenerate multiplicative quadratic form, then the dimension of $A$ is 1, 2, 4 or 8. This result can be found in "The Book of Involutions" by Knus, Merkurjev, Rost and Tignol (Amer. Math. Soc. Coll. Publ. 44, Providence, 1998), Corollary (33.28). The argument for this is due to Kaplansky (Proc. Amer. Math. Soc 4, 1953), and consists of defining a new multiplication on $A$ which admits a unit element.

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Kaplansky's paper is freely available here: ams.org/journals/proc/1953-004-06/S0002-9939-1953-0059895-7. The simple argument is at the top of page 957 (the 2nd page). –  YCor Oct 27 at 9:22
    
Thank you, @YCor, for linking to the paper. –  Seidon Alsaody Oct 27 at 21:40

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