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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.

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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 $t=1$.
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  • $\begingroup$ You have $yz = v$ and $yz = -w$. One of these must be a typo. $\endgroup$
    – Dustin Cartwright
    Commented Sep 4, 2011 at 0:09
  • $\begingroup$ I have one more question concerning this counterexample. If here $\mathbb F_3$ denotes the Galois field of order 3, then this is a finite alternative ring. But by the Artin-Zorn theorem every finite alternative ring is a field. What have I misunderstood? $\endgroup$
    – Sz_Z
    Commented Sep 4, 2011 at 17:59
  • $\begingroup$ @Zoltan: The Artin-Zorn theorem has the assumption that there are no zero-divisors. In my example there are quite a few zero divisors. Did you want to add that assumption? $\endgroup$
    – Gjergji Zaimi
    Commented Sep 4, 2011 at 18:17
  • $\begingroup$ Yes, I want to add that assumption. I need such an alternative ring - if there are any - that is a coordinate ring of a projective plane. $\endgroup$
    – Sz_Z
    Commented Sep 4, 2011 at 18:32

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