# Thread: Algebra, Problems For Fun (26)

1. ## Algebra, Problems For Fun (26)

Let $F$ be a field and $\sigma \in \text{Aut}(F)$ with $o(\sigma)=2.$ Prove that $A=\{a^{-1}\sigma(a): \ 0 \neq a \in F \}$ is the solution set of the equation $x\sigma(x)=1.$

Remark: In general, for any integer $n \geq 1$ and any $\sigma \in \text{Aut}(F)$ of order $n,$ the solution set of the equation $x\sigma(x)\sigma^2(x) \cdots \sigma^{n-1}(x)=1$ is still $A.$ (David Hilbert)

2. Finally I got it!

Let $S$ be the solution set. Clearly $A \subset S$; we want to show $A \supset S$.

Let $x \in S$; suppose $x \neq -1$.
Clearly then $y=(1+x)^{-1}\sigma(1+x) \in A$; but we have

$y(1+x)=\sigma(1+x) = 1+\sigma(x)$
$y(1+x)x=x+x\sigma(x) = x+1$

hence $yx=1$; i.e. $y=x^{-1} \in A$, and clearly $A$ is closed under taking inverses so $x \in A$.

To show $-1 \in A$ (clearly $-1 \in S$), take any $y \in F$ such that $z=y-\sigma(y) \neq 0$; then

$\sigma(z) = \sigma(y)-y = -z$

hence $z^{-1}\sigma(z) = -1 \in A$.