A non-associative algebra $A$ is alternative if

1. $[\ a,a,b\ ]=0$ , and
2. $[\ b,a,a\ ]=0$ ,

Remarks

• Let $A$ be alternative and suppose $\operatorname{char}(A)\neq2$ . From the fact that $[\ a+b,a+b,c\ ]=0$ , we can deduce that the associator $[\ , , ]$ is anti-commutative, when one of the three coordinates is held fixed. That is, for any $a,b,c\in A$ ,
  1. $[\ a,b,c\ ]=-[\ b,a,c\ ]$
  2. $[\ a,b,c\ ]=-[\ a,c,b\ ]$
  3. $[\ a,b,c\ ]=-[\ c,b,a\ ]$
  Put more succinctly, $$[\ a_1,a_2,a_3\ ]=\operatorname{sgn}(\pi)[\ a_{\pi(1)},a_{\pi(2)},a_{\pi(3)}\ ],$$ where $\pi\in S_3$ , the symmetric group on three letters, and $\operatorname{sgn}(\pi)$ is the sign of $\pi$ .
• An alternative algebra is a flexible algebra, provided that the algebra is not Boolean (characteristic $\neq2$ ). To see this, replace $c$ in the first anti-commutative identities above with $a$ and the result follows.
• Artin's Theorem: If a non-associative algebra $A$ is not Boolean, then $A$ is alternative iff every subalgebra of $A$ generated by two elements is associative. The proof is clear from the above discussion.
• A commutative alternative algebra $A$ is a Jordan algebra. This is true since $a^2(ba)=a^2(ab)=(ab)a^2=((ab)a)a=(a(ab))a=(a^2b)a$ shows that the Jordan identity is satisfied.
• Alternativity can be defined for a general ring $R$ : it is a ring such that for any $a,b\in R$ , $(aa)b=a(ab)$ and $(ab)b=a(bb)$ .