## A silly technical question on Albert algebras

Apologies in advance for this spectacularly uninteresting question, but it has just come up in my work. (Okay, not in a truly important way, but I am trying to gauge the scope of a certain construction.)

Let $K$ be a field and $A$ be a division Albert algebra over $K$, i.e., a certain kind of $27$-dimensional commutative Jordan algebra over $K$.

Is it true that for all nonzero elements $x,y \in A$, one has $(xy) y^{-1} = x$?

Note that this is true in any finite-dimensional division algebra in which each subalgebra generated by two elements is associative, in particular in any composition algebra. But so far as I know (and by the way I know nothing about Albert algebras!), this "2-associativity" property does not hold here.

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I suggested to his daughter Nancy that they could be called "Albergras." I don't believe she ever responded to that. Her biography of her father is still available, hope the link works  iuniverse.com/bookstore/…  – Will Jagy Dec 4 2010 at 18:24
@Thanks, Will. I would like to read that book when I get the chance. I think A.A. Albert (who would have a cubic algebra named after him, come to think of it!) is one of the most eminent mathematicians that many math people have never heard of, so I'd like to have more information about him to spread. By the way this is actually related to Euclidean (cubic!) forms. I have been working hard on the topic and have some calculations about quaternion orders to share with you in the near future... – Pete L. Clark Dec 4 2010 at 18:47
(However, if I may level with you: I think your suggested name is terrible.) – Pete L. Clark Dec 4 2010 at 18:49
I don't think Nancy liked the name either. The one factual matter she did not have time to correct in her book is that Oscar Zariski's parents (page 48) actually stayed in Kobrin (now in Belarus) when Oscar left. The Mac Tutor Zariski entry is still mistaken on this point. Short review of two books by Lance Small,  ams.org/notices/200511/rev-small.pdf  – Will Jagy Dec 11 2010 at 19:30

No. In an associative algebra $(xy+yx)y^{-1} + y^{-1}(xy+yx)= 2x+ yxy^{-1} + y^{-1}xy \neq 4x$ unless $x$ and $y$ commute. Hence, it does not hold even in special Jordan algebras.
To put some detail, the equality would mean $[[x,y],y]=0$, which does not happen for general Hermitian matrices, for instance. Also, it is worth noting that the inverse $y^{-1}$ in the associative algebra remains the inverse once it is viewed as a Jordan algebra. – Denis Serre Dec 3 2010 at 9:34