Question

Prove that $u\in K$ is a repeated root of $f\left(x\right)\in F\left[x\right]$if and only if u is a root of both $f\left(x\right)$ and $F\left[x\right]$.

Short answer

$u\in K$ is a root of $f\left(x\right)\in F\left[x\right]$.

Step-by-step solution

Definition of field extension

A simple extension is a field extension which is generated by the adjunction of simple element. Every finite field is a simple extension of the prime field of the same characteristic.

Showing that  u∈K is a root of  fx∈Fx

Let K is an extension field of F and$u\in K$represent root of$f\left(x\right)\in F\left[x\right]$if and only if F divided by${\left(x-u\right)}^m$. Therefore,

$f\left(x\right)={\left(x-u\right)}^{m}g\left(x\right)+h\left(x\right),m⩾1,g\left(x\right)\in K\left[x\right]\left(1\right)$

And $g\left(u\right)\ne 0$and $h\left(x\right)$is a linear polynomial. Put $x=u$ in the equation 1 then:

$$\begin{gathered} f\left(u\right)={\left(u-u\right)}^{m}g\left(u\right)+h\left(u\right) \\ f\left(u\right)=h\left(u\right) \end{gathered}$$

This gives:

$f^{\text{'}}\left(u\right)=h^{\text{'}}\left(u\right)$

Now take $m=2$. Then differentiate $f\left(x\right)$ to find $f^{\text{'}}\left(x\right)$. Hence

$f={\left(x-u\right)}^{2}g+\left(x-u\right)f^{\text{'}}\left(u\right)+f\left(u\right)$

This gives that $u\in K$ represent root of $f\left(x\right)\in F\left[x\right]$if and only if u is a root of both $f\left(x\right)$ and $F\left[x\right]$.