I was looking in the wikipedia article about Non-associative Algebra and came across this interesting line:
Each of the properties associative, commutative, anticommutative, Jordan identity, and Jacobi identity individually imply flexible.
It was quite easy to find why the first three properties imply flexible algebra. I was wondering about the last two (Jacobi and Jordan). I went to look for the proof in Schafer but either I missed it or it was given in terms I could not understand.
could someone give a proof similar to the commutative proof?
Thanks!
EDIT: I'm looking for proofs of the following statements:
1) given Jacobi identity prove flexible:
$ \forall x,y,z: (xy)z+(yz)x+(zx)y=0 \Rightarrow \forall x,y: x(yx)=(xy)x $
2) given Jordan identity prove flexible:
$ \forall x,y: (xy)x^2=x(yx^2) \Rightarrow \forall x,y: x(yx)=(xy)x $
EDIT 2: OK, so it appears that Schafer does have a proof for the jordan identity implying flexiblity on chapter V. once I can understand it I hope I can write it here.
I've had troubles with this very question. This is an attempt to answer.
Both claims are false. You can even craft a counterexample yourself for 2) by chosing a field $F$ of $char(F)=2$. However if $char(F)\neq 2$, Schafer proved (be careful while checking the assumptions) this claim to be True, see here.
For 1) there is a counterexample too. Consider the Algebra generated by $\{e_1,e_2,e_3\}$ over a field $F$ with nonzero products given by $e_3e_1=e_2,\ e_3e_3=e_1$. Then for any $x,y,z$ we have $(xy)z=0$, interchanging we conclude that $(xy)z+(yz)x+(zx)y=0+0+0=0$ but taking $v=e_1+e_2+e_3$ we must have $e_2=e_3(ve_3)\neq(e_3v)e_3=0$ hence the Algebra can not be flexible.