Recently my work has led me to consider octonion algebras. Not having much of a background with non-associative anything, I decided to check out a basic text on the subject, R.D. Schafer's *Introduction to Nonassociative Algebras*. 


I was reading it happily for a good while, until I got to the discussion of alternative algebras. A little background: let $A$ be an arbitrary $F$-algebra: i.e., just endowed with an $F$-bilinear product $A \times A \rightarrow A$, no further assumptions. In analogy to the more familiar commutator, it is also useful to define for any $x,y,z \in A$ the **associator** 


$[x,y,z] = (xy)z - x(yz)$,


the obvious point being that an algebra is associative iff all of its associators identically vanish. But there is a more subtle merit to this: the associator is an $F$-trilinear map from $A^3$ to $A$. From this it follows that it is entirely determined by its values on any $F$-basis $\{e\_i\}\_{i \in I}$ of $A$. And from *that* it follows that associativity can be checked on basis elements and moreover that associativity is faithfully preserved by scalar extension: clearly any trilinear map on an $F$-vector space is identically zero iff its extension to some field extension $K/F$ is identically zero.


On to alternativity: an $F$-algebra $A$ is said to be **alternating** if for all $x,y \in A$,


$[x,x,y] = [x,y,y] = 0$.


(These two identities are easily seen to imply the **flexible law** $[x,y,x] = 0$.)


Note however that these identities are not multilinear any more: e.g. the left alternator $[x,x,y]$ is quadratic in $x$ and linear in $y$. Thus both of the above consequences of multilinearity are in question: is it sufficient to check left alternativity on basis elements, and is a scalar extension of an alternative algebra necessarily alternative?


Presumably the first question has a negative answer. Compare for instance the quadratic form $q(x,y) = xy$: it vanishes on the two standard basis elements of $F^2$ yet is nondegenerate. What we need to do is *linearize*, i.e., replace the quadratic form with the associated bilinear form. In the present context, this amounts to replacing the alternating condition with the skew-symmetric condition, i.e.,: for all $x,y,z \in A$,


$[x,y,z] = -[y,x,z] = [y,z,x]$. 


The skew-symmetry condition looks much better: as a pair of equalities among trilinear maps, again it suffices to check it on basis elements and again it is faithfully preserved by scalar extension.


As is well-known, alternation implies skew-symmetry and the converse holds when $\frac{1}{2} \in F$.



> 
> But what about when $F$ has characteristic $2$?
> 
> 
> 


In this case, unfortunately (and somewhat embarrassingly) I am not even seeing why if $A$ is an alternating $F$-algebra and $K/F$ is a field extension, then $A\_K = A \otimes\_F K$ is an alternating $K$-algebra. Schafer does address this in his book: for $x \in A$, we have 
the left and right multiplication operators $L\_x, R\_x$ as elements of $\operatorname{End}\_K(A)$.  

Then equation (3.1) asserts that left and right alternating laws are equivalent to 


$L\_{x^2} = (L\_x)^2$ and $R\_{x^2} = (R\_x)^2$. 


He also says that the skew-symmetry of the associator is equivalent to (3.2), which is:


$R\_x R\_y - R\_{xy} = L\_{xy} - L\_y L\_x = L\_y R\_x - R\_x L\_y = L\_x L\_y - L\_{yx} = R\_y L\_x - L\_x R\_y = R\_{yx} - R\_y R\_x$. 


(I have no problem with these identities.) 


Then he says (on the top of p. 28) that "It follows from (3.1) and (3.2) that any scalar extension of an alternative algebra is alternative." 


Unfortunately I don't follow. Can someone help me out? 



Alternative algebras in characteristic 2, especially scalar extension
A good samaritan has delivered an answer directly to my email account. It is indeed almost obvious, as long as one ignores the bit about the multiplication operators!


Take for instance the left alternative property: for all $x,y \in A$, $[x,x,y] = 0$. As I said above, this condition is linear in $y$, so it is enough to check that for each element $e$ of a fixed $F$-basis of $A$, the bilinear form $B\_e(x,y) = [x,y,e]$ is alternating. Thus, the left alternation property itself will be preserved by base extension iff the alternating property of a bilinear map is preserved by base extension. I must have had a little mental block about this, but of course it is true: let $\{e\_i\}$ be an $F$-basis for $A$ hence a $K$-basis for $A\_K = A \otimes\_F K$. Then any $x \in A\_K$ may be uniquely written as $x = \sum\_i x\_i e\_i$ with $x\_i \in K$, and thus


$[x,x] = \sum\_{i=1}^n x\_i^2 [e\_i,e\_i] + \sum\_{i \neq j} x\_i x\_j [e\_i,e\_j]$


$= 0 + \sum\_{i < j} x\_i x\_j ([e\_i,e\_j] + [e\_j,e\_i]) = 0$,


since alternation implies skew-symmetry always. 


