Alternative algebras in characteristic 2, especially scalar extension

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?

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
$[x,x] = \sum_{i=1}^n x_i^2 [e_i,e_i] + \sum_{i \neq j} x_i x_j [e_i,e_j]$
$= 0 + \sum_{i < j} x_i x_j ([e_i,e_j] + [e_j,e_i]) = 0$,
• Two remarks about this argument: (1) This proof doesn't require $F$ and $K$ to be fields. We could just as well have $F$ be any commutative ring and $K$ be any commutative $F$-algebra. The proof would need only a minor modification: The $F$-module $A$ would not necessarily have an $F$-basis anymore; but this doesn't hurt the argument, since all we need is that any $x\in A_K$ may be written (not necessarily uniquely) in the form $x=\sum_i x_i e_i$ for some $x_i \in K$ and some $e_i\in A$; and this is a trivial consequence of the definition of $A_K$ as the tensor product $A\otimes_F K$. – darij grinberg Jan 17 '12 at 18:59
• (2) The same argument can be used to prove another fact in algebra, which is very often swept under the rug (it looks as if it would follow from abstract nonsense, functoriality, linearity, etc., but it does not): If $F$ is a commutative ring, $K$ is a commutative $F$-algebra, $A$ is an $F$-module, and $n$ is a nonnegative integer, then $\wedge^n_K \left(A\otimes_F K\right)$ is canonically isomorphic to $\left(\wedge^n_F A\right)\otimes_F K$ (where, for every $F$-algebra $T$, the notation $\wedge^n_T$ means the $n$-th exterior power of a $T$-module). – darij grinberg Jan 17 '12 at 19:02