TITLE: Are commutative alternative rings associative?
QUESTION [8 upvotes]: 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.

REPLY [13 votes]: 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).

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 $t=1$.