Associative rings and algebras Division rings and central simple algebras Proposition

Hua's identity

Statement

Let $D$ be a division ring, and let $a, b \in D$ with $a \ne 0$ , $b \ne 0$ and $ab \ne 1$ . Then $a^{-1} + (b^{-1} - a)^{-1}$ is invertible, and $a - \left(a^{-1} + (b^{-1} - a)^{-1}\right)^{-1} = aba.$

Remarks

The element $b^{-1} - a$ is nonzero because $ab \ne 1$ . The proof is a direct factorization: $a^{-1} + (b^{-1} - a)^{-1} = a^{-1}\left((b^{-1} - a) + a\right)(b^{-1} - a)^{-1} = a^{-1} b^{-1} (b^{-1} - a)^{-1}$ . Its inverse is $(b^{-1} - a) b a = a - aba$ , and subtracting it from $a$ gives $aba$ . For example, in $\Q$ with $a = 2$ and $b = 3$ : $1/2 + (1/3 - 2)^{-1} = 1/2 - 3/5 = -1/10$ , and $2 - (-10) = 12 = 2 \cdot 3 \cdot 2$ . The hypothesis $ab \ne 1$ is needed, since for $b = a^{-1}$ the term $(b^{-1} - a)^{-1}$ is not defined.

Going deeper

For a student

Hua used the identity to show that a map $\sigma$ between division rings that is additive, sends $1$ to $1$ and satisfies $\sigma(x^{-1}) = \sigma(x)^{-1}$ preserves $aba$ . From this one deduces Hua's theorem: such a map is a ring homomorphism or an antihomomorphism. The identity also underlies the use of the product $aba$ in Jordan algebras, where $aba$ plays the role of the quadratic operator.

Reading path

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Group concept
Ring definition
Unit of a ring definition
Division ring definition
Hua's identity statement

Tag	`AWS7XTZ49C`
URI	`m2://AWS7XTZ49C`
Kind	statement, proposition
Writing	W0 stub
MSC	16K40
Status	This copy of the site has no build data. Run `m2 status AWS7XTZ49C`.
About	Division ring
Uses	Division ring
Same as	Wikidata Q15709387 Wikipedia Hua's identity