# Thread: Algebra, Problems For Fun (26)

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

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

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

2. Finally I got it!

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

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

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

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

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

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

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