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Hua's theorem

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hbghlyj posted 2023-4-24 20:15 |Read mode
Hua's identity: $a, b$为除环的元素, $ ab\neq 0,1 $, 则 $ {\displaystyle a-\left(a^{-1}+\left(b^{-1}-a\right)^{-1}\right)^{-1}=aba} $.
将$ b $换作$ -b^{-1} $得到一等价形式$ {\displaystyle \left(a+ab^{-1}a\right)^{-1}+(a+b)^{-1}=a^{-1}.} $
证明
$$(a-aba)\left(a^{-1}+\left(b^{-1}-a\right)^{-1}\right)=1-ab+ab\left(b^{-1}-a\right)\left(b^{-1}-a\right)^{-1}=1.$$
虽然定理的条件是“除环”, 但只要 $ a,b,ab-1 $ 为环的单位, 该证明适用于一般的环.

该恒等式用于证明Hua's theorem: 设 $ \sigma $ 是除环之间的映射, 满足$$\sigma (a+b)=\sigma (a)+\sigma (b),\quad \sigma (1)=1,\quad \sigma (a^{-1})=\sigma (a)^{-1},$$则$ \sigma $为homomorphism或antihomomorphism. 这与fundamental theorem of projective geometry有联系.

Cohn, Paul M. (2003). Further algebra and applications, Theorem 9.1.3

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