Let $\phi\colon D\to D'$ be a map of division rings, such that $\phi$ is a homomorphism of the additive groups, respects unity, and if $a\neq 0$, $\phi(a)\neq 0$, and $\phi(a)^{-1}=\phi(a^{-1})$. It's a theorem of L.K. Hua that $\phi$ is either a homomorphism of anti-homomorphism, but I'm struggling to prove it.

It's a result of Jacobson and Rickart that any Jordan homomorphism of a ring into a domain is either a homomorphism or an anti-homomorphism (Theorem 2 of this paper) so I think it's sufficient to show $\phi$ is a Jordan homomorphism, that is, $\phi(aba)=\phi(a)\phi(b)\phi(a)$ for all $a,b\in D$, as the other two properties of a Jordan homomorphism are already assumed.

I was able to derive Hua's identity that if $a,b,ab-1$ are invertible elements of a ring, then $$ ((a-b^{-1})^{-1}-a^{-1})^{-1}=aba-a. $$ Now $\phi(aba)=\phi(a)\phi(b)\phi(a)$ holds if $a,b=0$, or if $ab-1=0$, so I'm only trying to verify for the other case where $a,b,ab-1$ are all units. Applying $\phi$ to $aba$ and using the above identity, I get something bad $$ ((\phi(a)-\phi(b)^{-1})^{-1}-\phi(a)^{-1})^{-1}+\phi(a). $$ What's the correct way to tell that $\phi$ is a homomorphism or anti-homomorphism? Thanks.

up vote 1 down vote accepted

You are basically done. Just show your work more clearly, and the answer is clear.

$$ aba = ((a-b^{-1})^{-1}-a^{-1})^{-1} + a $$

$$ \phi(aba) \stackrel{1}{=} ((\phi(a)-\phi(b)^{-1})^{-1}-\phi(a)^{-1})^{-1}+\phi(a) \stackrel{2}{=} \phi(a)\phi(b)\phi(a). $$

The first equality applies $\phi$ to Hua's identity, using that it respects addition, subtraction, and inverses. The second equality uses Hua's identity with $A=\phi(a)$ and $B=\phi(b)$.

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.