Question

Let $\mathit{\upsilon }$be an algebraic element of $\mathit{K}$whose minimal polynomial in $\mathit{F}\left[x\right]$has prime degree. If is a field such that role="math" localid="1657881622297" $\mathit{F}\mathbf{⊑}\mathit{E}\mathbf{⊑}\mathit{F}\mathbf{\left(}\mathbf{u}\mathbf{\right)}$, show thatrole="math" localid="1657881661231" $\mathit{E}\mathbf{=}\mathit{F}\mathbf{}\mathit{o}\mathit{r}\mathbf{}\mathit{E}\mathbf{=}\mathit{F}\mathbf{\left(}\mathbf{u}\mathbf{\right)}$

Short answer

$E=ForE=F\left(u\right)$

Step-by-step solution

Definition of a monic polynomial.

The monic polunomial $p\left(x\right)$over a field $F$of least degree such that $p\left(\alpha \right)=0$for an algebraic element $\alpha$over a fieldF .

Showing that E=F or E=F(u) 

Let $u$be an algebraic element of $K$whose minimal polynomial in $F\left[x\right]$has prime degree.

Since $u$is algebraic element of $K$. Let minimal polynomial of $u$is $p\left(x\right)$and $deg\left(p\left(x\right)\right)=p,p$is prime. This implies

$$\begin{gathered} \left[F\left(u\right):F\right]=degp\left(x\right) \\ =p \end{gathered}$$

If $F⊑E⊑F\left(u\right)$. Then

$\left[E:F\right]\left[E:F\left(u\right)\right]=\left[F\left(u\right):F\right]$

This implies

$\left[E:F\right]\left[E:F\left(u\right)\right]=p$

This gives

$\left[E:F\right]=1orpand\left[E:F\left(u\right)\right]=1orp$

By this either

$E=ForE=F\left(u\right)$