Question

If $\mathit{u}\mathbf{\in }\mathit{K}$is transcendental over $\mathit{F}$prove that all element of $\mathit{F}\left(u\right)$except those in localid="1657955205153" $\mathit{F}$are transcendental over localid="1657955211327" $\mathit{F}$.

Short answer

Element of$F\left(u\right)$ is transcendental except the element of$F$.

Step-by-step solution

Definition of extension field.

Let $K$an extension field of $F$. Then $K$is said to be algebraic extension of $F$if every element of $K$ is algebraic over$F$ and it is not algebraic then it is transcendental.

Showing that F(u) is transcendental.

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

Consider the contrapositive statement. That is,

If $v\in F\left(u\right)$is algebraic the localid="1657954734721" $v$ is inlocalid="1657954739810" $F$. If the contraposition statement hold then the claim also holds.

Since localid="1657954664126" $F\left(u\right)$is isomorphic to $F\left(x\right)$by an isomorphism which is identity oflocalid="1657954745892" $F$.

Let localid="1657954780074" $F\left(u\right)=F\left(x\right)$

If localid="1657954775929" $v=0$

Then the result holds. So assume $v\ne 0$. If $v=\frac{f\left(x\right)}{g\left(x\right)}\in F\left(x\right)$with is algebraic localid="1657955030376" $f\left(x\right),g\left(x\right)\ne 0\mathrm{and}\left(\right(f\left(x\right),g\left(x\right))=1)$over localid="1657954787775" $K$.

Then let localid="1657954829744" $p\left(x\right)=a_0+a_{1}x+........+a_n{x}_n\in F\left(x\right)$be its minimal polynomial.

Since localid="1657954816725" $p\left(x\right)$is irreducible it follow that localid="1657954809448" $a_0$ and localid="1657954801632" $a_n$are both nonzero.