Question

Let $\mathit{K}$is Galois over $\mathit{F}$show that there are only finitely many intermediate fields.

Short answer

There are only finite many intermediate fields.

Step-by-step solution

Definition of the Galois group.

The set of all $\mathit{F}$automorphism of $\mathit{K}$is an extension of role="math" localid="1657967503722" $\mathit{F}$then Galios group is define by $\mathit{G}\mathit{a}{\mathit{l}}_{\mathbf{F}}\mathit{K}$it is the Galois group.

Step-2: Showing that there are only finite many intermediate fields.

If $\mathit{K}$is an Galois group over $\mathit{F}$extension field now consider the field $\mathit{F}\left(u\right)$this field is contain every positive power of $\mathit{u}$. Now the element of $\mathit{F}\mathbf{\left(}\mathbf{u}\mathbf{\right)}$ is of the form of

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

Therefore the set is $\left\{1_F,u,u^2,.....u^{n-1}\right\}$if the set is linear independent, suppose ${\mathit{c}}_{\mathbf{0}}\mathbf{+}{\mathit{c}}_{\mathbf{1}}\mathit{u}\mathbf{+}{\mathit{c}}_{\mathbf{n}\mathbf{-}\mathbf{1}}{\mathit{u}}^{\mathbf{n}\mathbf{-}\mathbf{1}}\mathbf{=}{\mathbf{0}}_{\mathbf{f}}\mathbf{,}{\mathit{c}}_{\mathbf{i}}\mathbf{\in }\mathit{F}$and $\mathit{u}$is the root of $\mathit{p}\left(x\right)$degree n . Thus the basis of role="math" localid="1657968222219" $\mathit{F}\mathbf{\left(}\mathbf{u}\mathbf{\right)}$ is and is contain the extension fieldF .

Since the field $\mathit{F}\mathbf{\left(}\mathbf{u}\mathbf{\right)}\mathbf{\in }\mathit{F}$mean field contain a finite element so role="math" localid="1657968357889" $\mathit{\sigma }\mathbf{\in }\mathit{G}\mathit{a}{\mathit{l}}_{\mathbf{F}}\mathit{K}$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 $\mathit{F}\mathbf{\left(}\mathbf{u}\mathbf{\right)}$is contsin in the extension field $\mathit{F}$. Since field $\mathit{F}\mathbf{\left(}\mathbf{u}\mathbf{\right)}\mathbf{\in }\mathit{F}$so $\mathit{\sigma }\mathbf{\in }\mathit{G}\mathit{a}{\mathit{l}}_{\mathbf{F}\left(u\right)}\mathit{K}$.

Hence the field contain independent field over the extension field. Therefore each independent field contains his minimal polynomial of the primitive element $\mathbf{u}$over the intermediate field $\mathit{F}$.

Hence the number of independent field is the same as the number of subgroup of Galois function $\mathit{G}\mathit{a}{\mathit{l}}_{\mathbf{F}}\mathit{K}$is finite.