Please note the following two definitions from the 3rd edition of the book "The Linear Algebra a Beginning Graduate Student Ought to Know" by Jonathan S. Golan, we find the following text under chapter 4:
" Let $F$ be a field. An anticommutative F-algebra $(K,\bullet)$ is a Lie Algebra over $F$ if and only if it satisfies the [Jacobi identity]."
Next we are told the following about "$K^{-}$":
"Let $F$ be a field and let $(K,\bullet)$ be an associative F-algebra. Definite a new operation $\bullet$ on $K$ by setting $v \bullet w = v*w - w*v$. Then $(K,\bullet)$ is a Lie algebra over $F$, which is usually denoted by $K^{-}$. This operation in $K^{-}$ is known as the Lie product defined on the given F-algebra $K$."
So, we can take any associative F-algebra, define a new operation and thereby construct a Lie algebra "$K^{-}$". However, what if the underlying associative F-algebra we choose has an anti-commutative product (and the characteristic of $F$ is not equal to 2) s.t.:
$$ v\bullet w = v*w - w*v = v*w + v*w = 2v*w $$
Indeed, we can verify that the expression "$v*w - w*v$" satisfies the Jacobi Identity as the textbook suggests (granted the associativity of the "$*$" product) and it follows that "$v\bullet w$" satisfies the requirements for being a Lie algebra (since we can also see that it is anti-commutative). But... by our equality, this implies that the product "$2v*w$" also satisfies the requirements for being a Lie algebra, and (granted that $2 \ne 0 $) it follows that the product "$*$" from $v*w$ itself is already a Lie product!!!
This means that every anti-commutative associative F-algebra is a Lie algebra (if the underlying field has characteristic not equal to 2). Or am I mistaken?
Note:
I am aware of posts like this one and this one this one too.
I.e. that for an associative Lie algebra we can derive:
$$ 0 = u \bullet (v \bullet w) + v \bullet (w \bullet u) + w \bullet (u \bullet v) = u \bullet (v \bullet w) + v \bullet (-u \bullet w) + (-u \bullet w) \bullet v$$ $$ = u \bullet (v \bullet w) + (u \bullet v) \bullet w + -u \bullet (-v \bullet w) = 3 u \bullet v \bullet w = 0$$
Implying that for any triplet $u,v,w \in K$: $$ u \bullet v \bullet w = 0_{K} = (u \bullet v) \bullet w = u \bullet (v \bullet w) $$ I.e. (under associativity) the equivalence of the Jacobi identity with the property that the product of any vector with another vector which itself is a product of two vectors must always go to zero.
I am not asking about this equivalence. Rather I am asking if it is true that every associative anti-commutative F-algebra is a Lie algebra or is my outlined argument incorrect?
Thank you!
