An alternative ring is an algebraic structure where all the field axioms are true except for the commutativity and associativity of multiplication, but it is alternative, i.e. for all a,b $a(ba)=(ab)a$ and $(aa)b=a(ab)$. If I prove in such a structure that for all a,b $ab=ba$ holds, does it follow that $a(bc)=(ab)c$?


I know that every alternative ring is associative or a Cayley-Dickson ring. So it is enough to decide, wether it is possible that a Cayley-Dickson ring is commutative but not associative.



Are commutative alternative rings associative?
No. There are commutative alternative rings that are not associative. An example due to Kaplansky is a commutative alternative algebra over the $\mathbb F\_3$ with basis $\lbrace x,y,z,u,v,w\rbrace$ and relations $xy=u, yz=v, xv=w, uz=-w$ (the other products are zero).




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It turns out you are interested in commutative alternative division rings. These must be associative here are the steps of the proof (due to R.H. Bruck):


* Prove that for any three elements $x,y,z$ we have $$(x(yz)x=(xy)(zx)$$ $$x(y(zy))=((xy)z)y$$
* Using these identities show that the associator $(x^3,y,z)=0$, where $(a,b,c)=a(bc)-(ab)c$
* Assume that for some three elements $a,b,c$ we have $(ab)c=t(a(bc))$, using the previous identity show that $a^3b^3c^3=t^3a^3b^3c^3$
* Show that $3(a,b,c)=0$. Thus so far we have $3(t-1)=0$ and $t^3=1$. This implies $(t-1)^3=0$, and because there are no zero-divisors we must have $t=1$.


