Question

If $\mathbf{[}\mathbf{K}\mathbf{:}\mathbf{F}\mathbf{]}$ is finite, $\mathit{\sigma }\mathbf{\in }\mathit{G}\mathit{a}{\mathit{l}}_{\mathbf{F}}\mathit{K}$, and$\mathit{u}\mathbf{\in }\mathit{K}$ is such that $\mathit{\sigma }\mathbf{\left(}\mathbf{u}\mathbf{\right)}\mathbf{=}\mathit{u}$, show that $\mathit{\sigma }\mathbf{\in }\mathit{G}\mathit{a}{\mathit{l}}_{\mathbf{F}}\mathit{K}$.

Short answer

It is proved that $\mathbf{\sigma }\mathbf{\in }{\mathbf{Gal}}_{\mathbf{F}}\mathbf{K}$.

Step-by-step solution

Definition of Galois group of K over F

The set of$\mathit{F}$ all automorphisms of role="math" localid="1659522723174" $\mathit{K}$is an extension of $\mathit{F}$then Galois group is denoted by $\mathit{G}\mathit{a}{\mathit{l}}_{\mathbf{F}}\mathit{K}$. It is the Galois group of $\mathit{K}$over $\mathit{F}$.

Showing that σ∈GalFK.

Consider the field $\mathbf{F}\mathbf{\left(}\mathbf{u}\mathbf{\right)}$. This field containing every positive power of $\mathit{u}$. Now the elements of $\mathbf{F}\mathbf{\left(}\mathbf{u}\mathbf{\right)}$is of the form of role="math" localid="1659522915882" $\mathbf{F}\mathbf{\left(}\mathbf{u}\mathbf{\right)}$is given below:

$\begin{array}{c}\mathbf{f}\mathbf{\left(}\mathbf{u}\mathbf{\right)}\mathbf{=}\mathbf{p}\mathbf{\left(}\mathbf{u}\mathbf{\right)}\mathbf{q}\mathbf{\left(}\mathbf{u}\mathbf{\right)}\mathbf{+}\mathbf{r}\mathbf{\left(}\mathbf{u}\mathbf{\right)}\\ \mathbf{=}{\mathbf{0}}_{\mathbf{F}}\mathbf{q}\mathbf{\left(}\mathbf{u}\mathbf{\right)}\mathbf{+}\mathbf{r}\mathbf{\left(}\mathbf{u}\mathbf{\right)}\\ \mathbf{=}\mathbf{r}\mathbf{\left(}\mathbf{u}\mathbf{\right)}\\ \mathbf{=}{\mathbf{b}}_{\mathbf{0}}{\mathbf{1}}_{\mathbf{F}}\mathbf{+}{\mathbf{b}}_{\mathbf{1}}\mathbf{u}\mathbf{+}\mathbf{.}\mathbf{..}\mathbf{+}{\mathbf{b}}_{\mathbf{n}\mathbf{-}\mathbf{1}}{\mathbf{u}}_{\mathbf{n}\mathbf{-}\mathbf{1}}\end{array}$

Therefore, the set is $\mathbf{\{}{\mathbf{1}}_{\mathbf{F}}\mathbf{,}\mathbf{u}\mathbf{,}{\mathbf{u}}^{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{..}{\mathbf{u}}^{\mathbf{n}\mathbf{-}\mathbf{1}}\mathbf{\}}$. Now if the set is linear independent. Suppose ${\mathit{c}}_{\mathbf{0}}\mathbf{+}{\mathit{c}}_{\mathbf{1}}\mathit{u}\mathbf{+}\mathbf{...}\mathbf{+}{\mathit{c}}_{\mathbf{n}\mathbf{-}\mathbf{1}}{\mathit{u}}^{\mathbf{n}\mathbf{-}\mathbf{1}}\mathbf{=}{\mathbf{0}}_{\mathbf{F}}$with ${\mathbf{c}}_{\mathbf{i}}\mathbf{\in }\mathbf{F}$and $\mathbf{u}$is the root of $\mathbf{p}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$degree $\mathit{n}$.

Thus, the basis of$\mathbf{F}\mathbf{\left(}\mathbf{u}\mathbf{\right)}$ is$\mathit{n}$ and$\mathbf{F}\mathbf{\left(}\mathbf{u}\mathbf{\right)}$ is contained in the extension field $\mathbf{F}$. Since, field$\mathit{F}\mathbf{\left(}\mathbf{u}\mathbf{\right)}\mathbf{\in }\mathit{F}$ means field contains finite elements. So,$\mathit{\sigma }\mathbf{\in }\mathit{G}\mathit{a}{\mathit{l}}_{\mathbf{F}}\mathit{K}$ androle="math" localid="1659523123134" $\mathit{u}\mathbf{\in }\mathit{K}$ , thenrole="math" localid="1659523137133" $\mathit{\sigma }\mathbf{\left(}\mathbf{u}\mathbf{\right)}\mathbf{=}\mathit{u}$ with$\mathit{u}\mathbf{\in }\mathit{K}$ . Therefore,$\mathit{\sigma }\mathbf{\left(}\mathbf{u}\mathbf{\right)}\mathbf{\in }\mathit{K}$ . Hence, the basis of$\mathit{F}\mathbf{\left(}\mathbf{u}\mathbf{\right)}$is $\mathit{n}$and$\mathbf{f}\mathbf{\left(}\mathbf{u}\mathbf{\right)}$is contained in the extension field $\mathit{F}$.

Since, field role="math" localid="1659523301530" $\mathit{F}\mathbf{\left(}\mathbf{u}\mathbf{\right)}\mathbf{\in }\mathit{F}$. So, role="math" localid="1659523314185" $\mathit{\sigma }\mathbf{\in }\mathit{G}\mathit{a}{\mathit{l}}_{\mathbf{F}}\mathit{K}$.