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  

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