$(x,yz,t) = (x,y,t)z + (x,z,t)y$ holds in every Jordan algebra.

Question

The identity

$$(x,yz,t) = (x,y,t)z + (x,z,t)y$$

holds in every Jordan algebra. Remember that a Jordan algebra satisfies $xy=yx$ and $(x^2,y,x) = 0$ for all $x,y$. Here $(a,b,c) = (ab)c - a(bc)$ is the associator.

Comments: For one hand $$(x,yz,t) = (x(yz))t - x((yz)t)$$ and on the other hand $$(x,y,t)z + (x,z,t)y = ((xy)t)z - (x(yt))z + ((xz)t)y - (x(zt))y.$$

I can't see why these expressions are the same using only the two identities above.

Answer 1

I am not sure if the equality holds if $char(K)=2$.

However if $char(K)\neq2$, Shirshov proved (for example, see this book) that any Jordan Algebra satisfies the identity $((bc)a)d+((db)a)c+((cd)a)b-((ac)b)d-((da)b)c-((cd)b)a=0$

Making the changes $a\rightarrow t,b\rightarrow x,c\rightarrow y,d\rightarrow z$

You will have $\color{blue}{((xy)t)z}+\color{green}{((zx)t)y}+((yz)t)x\color{blue}{-((ty)x)z}-\color{green}{((zt)x)y}-((yz)x)t=0$

Or simply $\color{blue}{((xy)t)z-((ty)x)z}+\color{green}{((zx)t)y-((zt)x)y}=((yz)x)t-((yz)t)x$

Using the commutativity you get the desired result $\color{blue}{(x,y,t)z}+\color{green}{(x,z,t)y}=(x,yz,t)$.