Question

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

Short answer

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

Step-by-step solution

Definition of monic polynomial.

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

Step 2: Showing that  [K(u):K]≤[F(u):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 andu is algebraic over K.

Therefore uis 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 Kand $g\left(u\right)=0$.

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

$degp\left(x\right)\le degg\left(x\right)$

Now,

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

Therefore