Question

If $\mathit{u}\mathbf{,}\mathit{v}\mathbf{\in }\mathit{K}$ verify that $\mathit{F}\mathbf{\left(}\mathit{u}\mathbf{\right)}\mathbf{\left(}\mathit{v}\mathbf{\right)}\mathbf{=}\mathit{F}\mathbf{\left(}\mathit{v}\mathbf{\right)}\mathbf{\left(}\mathit{u}\mathbf{\right)}$ .

Short answer

$F\left(u\right)\left(v\right)=F\left(v\right)\left(u\right)$

Step-by-step solution

Definition of simple extension.

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

Step 2:Showing that F(u)(v)=F(v)(u) .

Let the subset $\left\{v_1,v_2,.......v_n\right\}$of localid="1657967093088" $V$ is linearly independent over localid="1657967088660" $F$and with $c_i\in F$.

Let $\left\{w-v_1,w-v_2,.........w-v_n\right\}$is linearly independent over localid="1657967099208" $F$. This implies

Simplify further,

localid="1657967125392" ${\alpha }_1(w-v_1)+{\alpha }_2\left(w-v_2\right)+.....+{\alpha }_n\left(w-v_n\right)=0$

Put the values of in the above equation

localid="1657967171416" $\left({\alpha }_1+{\alpha }_2+......{\alpha }_n\right)\left(c_1{v}_1+c_2{v}_2+....c_n{v}_n\right)-({\alpha }_1{v}_1+{\alpha }_2{v}_2+.......{\alpha }_n{v}_n)=0$

This gives

$\left(\right({\alpha }_1+{\alpha }_2+....{\alpha }_n)c_1-{\alpha }_1)v_1+\left(\right({\alpha }_1+{\alpha }_2+....{\alpha }_n)c_2-{\alpha }_2)v_2+...+\left(\right({\alpha }_1+{\alpha }_2+....{\alpha }_1+{\alpha }_2+....{\alpha }_n)c_n-{\alpha }_n)v_n=0$

Since $\left\{v_1,v_2,.....v_n\right\}$ of is linearly independent over .

Therefore $\left(\right({\alpha }_1+{\alpha }_2+...{\alpha }_n)c_i-{\alpha }_i)=0$ for all $i=1,2,3.....n$

This implies $\left(\right({\alpha }_1+{\alpha }_2+...{\alpha }_n\left)c_i\right)={\alpha }_i$ for all $i=1,2,3.....n$

$$\begin{gathered} \sum _{i-1}^n\left(\right({\alpha }_1+{\alpha }_2.....{\alpha }_n\left)c_i\right)=\sum _{i-1}^n{\alpha }_i \\ \sum _{i-1}^n{\alpha }_i\sum _{i-1}^n{c}_i=\sum _{i-1}^n{\alpha }_i \end{gathered}$$

Simplify further

$$\begin{gathered} \sum _{i-1}^n\underset{i-1}{\overset{n}{{\alpha }_i\sum c_i-}}\sum _{i-1}^n{\alpha }_i=0 \\ \sum _{i-1}^n{\alpha }_i\sum _{i-1}^n{c}_i-1_F=0 \end{gathered}$$

This gives

$\sum _{i-1}^n{\alpha }_i=0or\sum _{i-1}^n{c}_i=1_F$

Now the set $\left\{w-v_1,w-v_2,....w-v_n\right\}$ is linearly independent if and only if all the${\alpha }_i=0$ . So,

$\sum _{i-1}^n{c}_i\ne 1_F$

Or

$c_1+c_2+c_3+........c_n\ne 1_F$

Hence proved.