Question

If $\left[K:F\right]$is finite and$\mathbf{u}$ is algebraic over$\mathbf{K}$ prove that $\left[\mathbf{K}\left(u\right)\mathbf{:}\mathbf{K}\right]\le \left[\mathbf{F}\left(u\right)\mathbf{:}\mathbf{F}\right]$.

Short answer

It is proved that$\left[K\left(u\right):K\right]\le \left[F\left(u\right):F\right]$.

Step-by-step solution

Definition of monic polynomial. 

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

Showing that  Ku:K≤Fu:F

Since$\left[K:F\right]$is finite and$u$is algebraic over$K$.

This implies$\raisebox{1ex}{$K$}\!\left/ \!\raisebox{-1ex}{$F$}\right.$is algebraic extension and$u$is algebraic over$K$.

Therefore$u$is algebraic over$F$. Thus$\left[F\left(u\right):F\right]$is finite.

Suppose that$g\left(x\right)$is minimal polynomial of$u$over$F$. And$p\left(x\right)$is minimal polynomial of$u$over$K$

Since$F\subseteq K$

Therefore$g\left(x\right)$can be considered as a polynomial over$K$and$g\left(u\right)=0$.

So,$p\left(x\right)$must divide $g\left(x\right)$ over $K$

$\mathrm{deg}p\left(x\right)\le \mathrm{deg}g\left(x\right)$

Now,

$\begin{array}{l}\left[K\left(u\right):K\right]=\mathrm{deg}p\left(x\right)\\ \left[K\left(u\right):F\right]=\mathrm{deg}g\left(x\right)\end{array}$

Therefore

$\left[K\left(u\right):K\right]\le \left[F\left(u\right):F\right]$.