Skew-symmetry on Lie algebras over a field with characteristic not 2

Question

I´d like to see the proof (I know it could be elemental) of this fact:

Let $L$ be a Lie algebra over a field $\mathbb{F}$ with characteristic not 2. Then $[x,x]=0$ for any $x \in L$ if and only if$ [x,y]=−[y,x]$.

Thank you all

Answer 1

I suppose that this not what you really want to ask, since the equivalence of those two conditions holds by the simple fact that both conditions hold for any Lie algebra.

However, for any algebra $(A,\star)$ over a field $F$ whose characteristic is not $2$, it is true that the conditions

1. $(\forall x,y\in A):x\star x=0$;
2. $(\forall x,y\in A):x\star y=-y\star x$

are equivalent. In fact, if the second conditions holds and if $x\in A$, then $x\star x=-x\star x$, which means that $2x\star x=0$. SInce the characteistic is not $2$, it follows from this that $x\star x=0$. And if the first condition holds, then, if $x,y\in A$,\begin{align}0&=(x+y)\star(x+y)\\&=x\star x+x\star y+y\star x+y\star y\\&=x\star y+y\star x\end{align}and therefore $x\star y=-y\star x$.

Answer 2

If $[x,x]=0$ for al $x$ then $0=[x+y,x+y]=[x,x]+[x,y]+[y,x]+[y,y]=[x,y]+[y,x],$ so $[x,y]=-[y,x].$

Conversely, if $[x,y]=-[y,x]$ for all $x,y$ then $[x,x]=-[x,x],$ so $2[x,x]=0.$ Since $F$ has characteristic not 2 then $[x,x]=0.$