Question

If $\mathbf{u}\mathbf{\in }\mathbf{K}$and$\mathbf{c}\mathbf{\in }\mathbf{F}$, prove that$\mathbf{F}\mathbf{(}\mathbf{u}\mathbf{+}\mathbf{c}\mathbf{)}\mathbf{=}\mathbf{F}\mathbf{\left(}\mathbf{u}\mathbf{\right)}\mathbf{=}\mathbf{F}\mathbf{\left(}\mathbf{cu}\mathbf{\right)}\mathbf{.}$.

Short answer

It is proved that$\mathrm{F}(\mathrm{u}+\mathrm{c})=\mathrm{F}\left(\mathrm{u}\right)=\mathrm{F}\left(\mathrm{cu}\right)$ .

Step-by-step solution

Describe the concept of field theory

In field theory, a simple extension is s field extension that is generated by the adjunction of a single element. Every finite field is a simple extension of the prime field of the same characteristics.

Prove that F(u+c)=F(u)=F(cu)

Let be an extension field of , then it is required to show for$u\in K$ and$c\in F$ that

$F(u+c)=F\left(u\right)=F\left(cu\right)$

Let,

$$\begin{gathered} {\wedge }_{u+c}:=\left\{L\subset K:L\hspace{0.33em}isa\hspace{0.33em}fieldandu+c\in L,\hspace{0.33em}F\subset L\right\} \\ {\wedge }_{u+c}:=\left\{L\subset K:L\hspace{0.33em}isa\hspace{0.33em}fieldanduc\in L,\hspace{0.33em}F\subset L\right\} \end{gathered}$$

And,

${\wedge }_u:=\left\{L\subset K:Lisa\hspace{0.33em}fieldand\hspace{0.33em}u\in L,\hspace{0.33em}F\subset L\right\}$

Then by definition it is known that

$$\begin{gathered} F(u+c)=\underset{L\in {\wedge }_{uc}}{\cup }L \\ F\left(u\right)=\underset{L\in {\wedge }_{uc}}{\cup }L \end{gathered}$$

And,

role="math" localid="1659159041634" $F\left(uc\right)=\underset{L\in {\wedge }_{uc}}{\cup }L$

Now note that $u\in F(u+c)$. Indeed,$u+c\in F(u+c),$since $F\subset F(u+c),c\in F,$and , it follows that

$u=u+c-c\in F(u+c)$

Therefore, by definition,

$F\left(u\right)\subset F(u+c)$

On the other hand, note that $u+c\in F\left(u\right)$since $u\in F\left(u\right),c\in F$, and $F\subset F\left(u\right).$Consequently,

$F(u+c)\subset F\left(u\right)$

Since $F(u+c)$is the smallest subfield containing both $u+ce,$and F . Therefore,

$F(u+c)=F\left(u\right)$

Now note that $u\in F\left(uc\right).$Indeed, since $uc\in F\left(uc\right),F\subset F\left(uc\right)$, and $c\in F$, it follows that

$u=uc{c}^{-1}\in F\left(uc\right)$

Thus by definition,

$F\left(u\right)\subset F\left(uc\right)$

On the other hand, note that$uc\in F\left(u\right)$since $u\in F\left(u\right),c\in F$, and $F\subset F\left(u\right).$

Consequently,

$F\left(uc\right)\subset F\left(u\right)$

Since,$F\left(uc\right)$is the smallest subfield containing both $\left(uc\right)$and F. Thus,

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

Therefore, it verifies that

$F(u+c)=F\left(u\right)=F\left(cu\right)$

Hence, it is proved that $F(u+c)=F\left(u\right)=F\left(cu\right).$.