Question

If K is radical extension of F prove that $\mathbf{[}\mathbf{K}\mathbf{:}\mathbf{F}\mathbf{]}\mathbf{}$is finite.

Short answer

$\mathbf{[}\mathit{K}\mathbf{:}\mathit{F}\mathbf{]}\mathbf{=}\mathit{n}\mathbf{.}$

Step-by-step solution

Definition of the splititing field.

A splititing field of a minimal polynomial with coefficients in an extension field is a smallest field extension of that field over which the polynomial splits into linear factors.

Step-2: Showing that [K:F]=n.  is finite.

If $\mathit{F}\mathbf{\left(}\mathit{u}\mathbf{\right)}\mathbf{}$ is a field containing the element $\mathit{F}\mathbf{\in }\mathbf{\left[}\mathit{u}\mathbf{\right]}$ it has every positive power of with . Now the element of the field $\mathit{F}\mathbf{\left(}\mathit{u}\mathbf{\right)}\mathbf{}$ is ${\mathbf{b}}_{\mathbf{0}}\mathbf{+}{\mathbf{b}}_{\mathbf{1}}\mathbf{u}\mathbf{+}{\mathbf{b}}_{\mathbf{2}}{\mathbf{u}}^{\mathbf{2}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}{\mathbf{b}}_{\mathbf{i}}{\mathbf{u}}^{\mathbf{i}}\mathbf{,}{\mathbf{b}}_{\mathbf{i}}\mathbf{\in }\mathbf{F}$,since $\mathit{F}\mathbf{\left(}\mathit{u}\mathbf{\right)}\mathbf{}$contain the element $\mathit{f}\mathbf{\left(}\mathit{u}\mathbf{\right)}\mathbf{}$ for every positive number localid="1657968925317" $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{\in }\mathit{F}\mathbf{\left[}\mathit{x}\mathbf{\right]}\mathbf{.}\mathbf{}$Now

If the mapping the elementu is a homorphism of rings or not.

$$\begin{gathered} \mathbf{\phi }\mathbf{:}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\to }\mathbf{F}\mathbf{\left[}\mathbf{u}\mathbf{\right]}\mathbf{} \\ \mathbf{\phi }\mathbf{\left(}\mathbf{f}\mathbf{\right(}\mathbf{x}\mathbf{\left)}\mathbf{\right)}\mathbf{=}\mathbf{f}\mathbf{\left(}\mathbf{u}\mathbf{\right)}\mathbf{} \end{gathered}$$

Therefore this is a homorphism of the rings. If degree of the polynomial $\mathit{p}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{}$ has an order n now using the division algorthim.

$$\begin{gathered} \mathbf{F}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{p}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{q}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{+}\mathbf{r}\mathbf{\left(}\mathbf{x}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{r}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}{\mathbf{b}}_{\mathbf{0}}\mathbf{+}{\mathbf{b}}_{\mathbf{1}}\mathbf{x}\mathbf{+}{\mathbf{b}}_{\mathbf{2}}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}{\mathbf{b}}_{\mathbf{n}\mathbf{-}\mathbf{1}}{\mathbf{x}}^{\mathbf{n}\mathbf{-}\mathbf{1}}\mathbf{,}{\mathbf{b}}_{\mathbf{n}\mathbf{-}\mathbf{1}}\mathbf{\in }\mathbf{F}\mathbf{\left[}\mathbf{x}\mathbf{\right]} \end{gathered}$$

Conquenctly

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

Therefore the minimal polynomial has degree $\mathbf{\le }\mathit{n}\mathbf{-}\mathbf{1}$ and must be divisible by $\mathit{p}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{}$ which has a degreen . Now for zero polynomial for each ${\mathit{c}}_{\mathbf{i}}\mathbf{=}{\mathbf{0}}_{\mathbf{F}}$ thus $\left\{1_F,u,u^2,....u^{n-1}\right\}$ is linear independent over the extension field F.

Hence the basis of $\mathit{F}\mathbf{\left(}\mathit{u}\mathbf{\right)}\mathbf{}$and $\mathit{F}\mathbf{\left(}\mathit{u}\mathbf{\right)}\mathbf{}$ is the subgroup of the splititing fieldK so $\mathbf{[}\mathit{K}\mathbf{:}\mathit{F}\mathbf{]}\mathbf{=}\mathit{n}\mathbf{}$ is finite .