Question

If is $u\in K$transcendental over F and $0_F\ne c\in F$, prove that each of $u+1_F,cu$, and u² is transcendental over F.

Short answer

It is proved $u+1_F,cu$,and u² is transcendental over F.

Step-by-step solution

Definition of algebraic field

Let K be an extension field of F. Then, K is said to be algebraic of F if every element of Kis algebraic over F .

Showing that u+1F is transcendental over F

Let K be an extension field of F and $u\in K$ is transcendental over F and $0_F\ne c\in F$.

$u\in K$is a transcendental over $u\in K$and $0_F\ne c\in F$. Then;

$1_F\in F$.

Every element of F is algebraic over F. This implies $1_F\in F$ is algebraic over F.

If possible, each of $u+1_F,cu$ and u² is algebraic over F.

If $u+1_F$ is algebraic over F,then $u+1_F-1_F$is also algebraic over F.

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

Therefore,$u+1_F$ is transcendental over F .

Show that each u+1F,cu and u2 is transcendental over F

Secondly, if cu is algebraic over F,than $c^{-1}\left(cu\right)=u$ is algebraic over F.

This is again a contradiction as $u\in K$ is transcendental over F. So, cu is transcendental over F.

u² is algebraic over F. Therefore, there exists a polynomial $f\left(x\right)$ over F such that $f\left(u^2\right)=0$. Consider the polynomial $g\left(x\right)=f\left(x^2\right)\in F$.

Replacing x and u, we getthe following;

$g\left(u\right)=f\left(u^2\right)\in F$

$f\left(u^2\right)=0$. This implies, $g\left(u\right)=0$.

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

Therefore,u is algebraic over F. This is again a contradictionas$u\in K$ is transcendental over F.

All cases given is a contradiction. So, u² is tradicition F over .

Each of $u+1_F,cu$ and u² is transcendental over F .