Question

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

Short answer

It is proved that$\mathit{\sigma }\mathbf{\in }\mathit{G}\mathit{a}{\mathit{l}}_{\mathbf{F}}$.

Step-by-step solution

Definition of Galois group of K over F

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

Showing that σ∈GalFK .

Consider the field $\mathit{F}\mathbf{\left(}\mathit{u}\mathbf{\right)}$. This field containing every positive powerof u. Now the elements of F(u)is of the form of f (u)is given below:

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

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

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

Since, field $\mathit{F}\mathbf{\left(}\mathit{u}\mathbf{\right)}$$\mathbf{\in }\mathit{F}$. So, $\mathit{\sigma }\mathbf{\in }\mathit{G}\mathit{a}{\mathit{l}}_{\mathbf{F}}\mathit{K}$.