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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.

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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 $B=\phi(b)$.

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Thanks, I should have seen that. – yunone Jun 22 '12 at 21:19

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