Question

If $\mathit{u}\mathbf{\in }\mathit{K}$ is algebraic over F and $\mathit{c}\mathbf{\in }\mathit{F}$, prove that $\mathit{u}\mathbf{+}{\mathbf{1}}_{\mathbf{F}}$and cu are algebraic over F.

Short answer

$u+1_{\mathrm{F}}$ and cu are algebrai over F.

Step-by-step solution

Definition of field extension.

Let K an extension field of F, then K is said to be algebraic extension of Fif every element of K is algebraic over F .

Showing that u+1F and   is algebraic.

Let $u\in K$ is algebraic over F and $c\in F$. Therefore there exist a polynomial $f\left(x\right)\in F\left[x\right]$such that $f\left(u\right)=0$.

Let a polynomial $g\left(x\right)=f(x-1)\in F\left[x\right]$

Thus

$$\begin{gathered} g\left(u+1\right)=f\left(u+1-1\right) \\ =f\left(u\right) \end{gathered}$$

As $f\left(u\right)=0$. This implies $g\left(u+1\right)=0$.

This mean there exist a polynomial g(x) such that $g\left(u+1\right)=0$.

Therefore the element$u+1_F$is algebraic over F .

And consider,

$$\begin{gathered} h\left(cx\right)=f\left(c^{-1}cx\right) \\ =f\left(c^{-1}cu\right) \\ =f\left(u\right) \end{gathered}$$

And as$f\left(u\right)=0$

This means there exist a polynomial $h\left(cx\right)$such that $h\left(cu\right)$.

Therefore the element cu is algebraic over F.