## Solutions of xf(x) = 1 in a field

Posted: December 13, 2010 in Elementary Algebra; Problems & Solutions, Groups and Fields
Tags: , ,

This is a special case of a nice result due to David Hilbert:

Problem. Let $F$ be a field and $f \in Aut(F)$ with $o(f)=2.$ Solve the equation $xf(x)=1$ in $F.$

Solution. The claim is that $S=\{a^{-1}f(a): \ 0 \neq a \in F \}$ is the solution set of the equation. If $x = a^{-1}f(a) \in S,$ then, since $o(f)=2,$ we have $xf(x)=a^{-1}f(a)f(a^{-1})a=1.$ Conversely, suppose that $xf(x)=1.$ If $x= -1,$ then, since $o(f)= 2 > 1,$ there exists some $b \in F$ such that $f(b) \neq b.$ Let $a=f(b)-b.$ Then $f(a)=b-f(b)=-a$ and hence $x=-1=a^{-1}f(a) \in S.$ So we may assume that $x \neq -1.$ Let $a=(x+1)^{-1}.$ Then

$xf(x+1)=x(f(x)+1)=xf(x)+x=x+1$

and thus

$x=(x+1)f((x+1)^{-1})=a^{-1}f(a) \in S. \ \Box$