Question

Let $\mathit{E}$be an intermediate field that is normal over $\mathit{F}$and $\mathit{\sigma }\mathbf{\in }\mathit{G}\mathit{a}{\mathit{l}}_{\mathbf{F}}\mathit{K}$prove that role="math" localid="1657965963713" $\mathit{\sigma }\mathbf{\left(}\mathbf{E}\mathbf{\right)}\mathbf{=}\mathit{E}$.

Short answer

$\mathit{\sigma }\mathbf{\left(}\mathbf{E}\mathbf{\right)}\mathbf{=}\mathit{E}$

Step-by-step solution

Definition of the intermediate field.

The intermediate field is the fixed field and it is the subfield of the extension field.

Step-2: Showing that σ(E)=E .

Let $\mathit{E}$is an intermediate field of the extension field $\mathit{K}$over the group $\mathit{F}$and $\mathit{\sigma }\mathbf{\in }\mathit{G}\mathit{a}{\mathit{l}}_{\mathbf{F}}\mathit{K}$so much that $\mathit{\sigma }\left(E\right)\mathbf{=}\mathit{E}$. If the intermediate field $\mathit{E}$is the fixed and contains the number of sungroups one is $\mathit{H}$and$\mathit{c}\mathbf{,}\mathit{d}\mathbf{\in }\mathit{E}$and . So,

$$\begin{gathered} \mathit{\sigma }\left(c+d\right)\mathbf{=}\mathit{\sigma }\left(c\right)\mathbf{+}\mathit{\sigma }\left(d\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathit{c}\mathbf{+}\mathit{d} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{\sigma }\left(cd\right)\mathbf{=}\mathit{\sigma }\left(c\right)\mathit{\sigma }\left(d\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathit{c}\mathit{d} \end{gathered}$$

Since the fixed field ${\mathit{E}}_{\mathbf{H}}$is closed under the addition and multiplication property of one to one. So $\mathbf{\sigma }\left(o_K\right)\mathbf{=}{\mathbf{0}}_{\mathbf{K}}\mathbf{,}\mathbf{\sigma }\left(1_F\right)\mathbf{=}{\mathbf{1}}_{\mathbf{F}}$ for every automorphism of fixed field. Thus the fixed field is non zero it contain the finite elements. now,

$$\begin{gathered} \mathit{\sigma }\left(-E\right)\mathbf{=}\mathbf{-}\mathit{\sigma }\left(E\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{-}\mathit{E} \\ \mathit{\sigma }\left(E^{-1}\right)\mathbf{=}\mathit{\sigma }{\left(E\right)}^{\mathbf{-}\mathbf{1}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}{\mathit{E}}^{\mathbf{-}\mathbf{1}} \end{gathered}$$

Therefore the fixed field has a finite element and $\mathit{H}$is the subgroup of fixed field so $\mathit{H}\mathbf{\in }\mathit{G}\mathit{a}{\mathit{l}}_{\mathbf{F}}\mathit{K}$and $\mathit{\sigma }\mathbf{\in }\mathit{H}$then $\mathit{\sigma }\mathbf{\in }\mathit{G}\mathit{a}{\mathit{l}}_{\mathbf{F}}\mathit{K}$so $\mathit{\sigma }\left(E\right)\mathbf{=}\mathit{E}$for every $\mathit{F}\mathbf{⊑}{\mathit{E}}_{\mathbf{H}}$.