Question

If $\mathit{u}\mathbf{\in }\mathit{K}$is transcendental over $\mathit{F}$, prove that $\mathit{F}\mathbf{\left(}\mathit{u}\mathbf{\right)}\mathbf{\cong }\mathit{F}\mathbf{\left(}\mathit{x}\mathbf{\right)}$, where $\mathit{F}\mathbf{\left(}\mathit{x}\mathbf{\right)}$is the field of quotients of $\mathit{F}\mathbf{\left(}\mathit{x}\mathbf{\right)}$ .

Short answer

$F\left(U\right)\cong F\left(x\right)$

Step-by-step solution

Definition of extension field.

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

Showing that  F(u)≅F(x) .

Let $u\in K$ is transcendental over $F$And $F\left(x\right)$ is the field of quotients of$F\left(x\right)$.

Define a map $\varphi$from$F\left(x\right)$to $F\left(u\right)$. That is

Define map role="math" localid="1657949959888" $\varphi :F\left(x\right)\to F\left(x\right)asf\left(x\right)g\left(x\right)=f\left(u\right)g{\left(u\right)}^{-1}$

Inverse function of the map above exists that is

${\varphi }^{-1}:F\left(u\right)\to F\left(x\right)\mathrm{such}\mathrm{that}f\left(u\right)g{\left(u\right)}^{-1}=f\left(x\right)g{\left(x\right)}^{-1}$

such that

The function ${\varphi }^{-1}$ is well define since there every element of $F\left(u\right)$ has a unique representation $f\left(u\right)g{\left(u\right)}^{-1}$ this is because two distinct such that representation contradicts the assumption that$u\in K$ is transcendental over$F$.

But$u\in K$ is transcendental over$F$Therefore$F\left(u\right)\cong F\left(x\right)$.