Question

If$\mathit{u}\mathbf{\in }\mathit{K}$is transcendental over$\mathit{F}$and${\mathbf{0}}_{\mathbf{F}}\mathbf{\ne }\mathit{c}\mathbf{\in }\mathit{F}$, prove that each of $\mathit{u}\mathbf{+}{\mathbf{1}}_{\mathbf{F}}\mathbf{,}\mathit{c}\mathit{u}$, and ${\mathit{u}}^{\mathbf{2}}$is transcendental over .

Short answer

It is proved thateach of,$u+1_F,cu$and $u^2$is transcendental over$F$ .

Step-by-step solution

Definition of an algebraic field 

Let $\mathit{K}$be an extension field of $\mathit{F}$.$\mathit{K}$ is said to be algebraic of$\mathit{F}$ if every element of $\mathit{K}$is algebraic over$\mathit{F}$.

Show thatu+1F is transcendental over F 

Let$\mathit{K}$ be an extension field of$\mathit{F}$ , and $\mathit{u}\mathbf{\in }\mathit{K}$is transcendental over$\mathit{F}$ and ${\mathbf{0}}_{\mathbf{F}}\mathbf{\ne }\mathit{c}\mathbf{\in }\mathit{F}$.

$\mathit{u}\mathbf{\in }\mathit{K}$is a transcendental over $\mathit{u}\mathbf{\in }\mathit{K}$and${\mathbf{0}}_{\mathbf{F}}\mathbf{\ne }\mathit{c}\mathbf{\in }\mathit{F}$ . Then;

${\mathbf{1}}_{\mathbf{F}}\mathbf{\in }\mathit{F}$

Every element of$\mathit{F}$ is algebraic over $\mathit{F}$. This implies ${\mathbf{1}}_{\mathbf{F}}\mathbf{\in }\mathit{F}$is algebraic over$\mathit{F}$ .

If possible, each of $\mathit{u}\mathbf{+}{\mathbf{1}}_{\mathbf{F}}\mathbf{,}\mathit{c}\mathit{u}$, and ${\mathit{u}}^{\mathbf{2}}$is algebraic over$\mathit{F}$ .

If $\mathit{u}\mathbf{+}{\mathbf{1}}_{\mathbf{F}}$is algebraic over$\mathit{F}$ , $\mathit{u}\mathbf{+}{\mathbf{1}}_{\mathbf{F}}\mathbf{-}{\mathbf{1}}_{\mathbf{F}}$will be algebraic over$\mathit{F}$ .

This implies that$\mathit{u}$ is also algebraic over $\mathit{F}$. This is a contradiction as$\mathit{u}\mathbf{\in }\mathit{K}$ is transcendental over$\mathit{F}$ . Thus, the assumption is wrong.

Therefore$\mathit{u}\mathbf{+}{\mathbf{1}}_{\mathbf{F}}$, is transcendental over$\mathit{F}$ .

Show that each u+1F,cu, andu2 is transcendental over  F                

Secondly, if$\mathit{c}\mathit{u}$is algebraic over $\mathit{F}$, ${\mathit{c}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\mathbf{c}\mathbf{u}\mathbf{\right)}\mathbf{=}\mathit{u}$will bealgebraic over$\mathit{F}$.

This is again a contradiction as$\mathit{u}\mathbf{\in }\mathit{K}$is transcendental over $\mathit{F}$. So,$\mathit{c}\mathit{u}$is transcendental over$\mathit{F}$.

If${\mathit{u}}^{\mathbf{2}}$is algebraic over $\mathit{F}$, there will be a polynomial$\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$over$\mathit{F}$, such that $\mathit{f}\mathbf{\left(}{\mathbf{u}}^{\mathbf{2}}\mathbf{\right)}\mathbf{=}\mathbf{0}$. Consider the polynomial $\mathit{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathit{f}\mathbf{\left(}{\mathbf{x}}^{\mathbf{2}}\mathbf{\right)}\mathbf{\in }\mathit{F}$.

Replace$\mathit{x}$and $\mathit{u}$:

$\mathit{g}\mathbf{\left(}\mathbf{u}\mathbf{\right)}\mathbf{=}\mathit{f}\mathbf{\left(}{\mathbf{u}}^{\mathbf{2}}\mathbf{\right)}\mathbf{\in }\mathit{F}$

.$\mathit{f}\mathbf{\left(}{\mathbf{u}}^{\mathbf{2}}\mathbf{\right)}\mathbf{=}\mathbf{0}$This implies$\mathit{g}\mathbf{\left(}\mathbf{u}\mathbf{\right)}\mathbf{=}\mathbf{0}$.

This means that there exists a polynomial$\mathit{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$such that$\mathit{g}\mathbf{\left(}\mathbf{u}\mathbf{\right)}\mathbf{=}\mathbf{0}$.

Therefore,$\mathit{u}$is algebraic over $\mathit{F}$. This is again a contradiction as$\mathit{u}\mathbf{\in }\mathit{K}$is transcendental over$\mathit{F}$.

All cases givea contradiction. So,$\mathit{u}$is transcendental over $\mathit{F}$.

Each of $\mathit{u}\mathbf{+}{\mathbf{1}}_{\mathbf{F}}\mathbf{,}\mathit{c}\mathit{u}$, and${\mathit{u}}^{\mathbf{2}}$is transcendental over $\mathit{F}$.