Question

Show that $\frac{{\mathbf{x}}^{\mathbf{3}}}{\mathbf{x}\mathbf{+}\mathbf{1}}\mathbf{\in }\mathit{F}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$ is transcendental over localid="1657957684452" $\mathit{F}$.

Short answer

$\frac{x^3}{x+1}\in F\left(x\right)$is transcendental.

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 localid="1657957698421" $K$is algebraic overlocalid="1657957720740" role="math" $F$ and it is not algebraic then it is transcendental.

Showing that  x3x+1∈F(x)x3x+1∈F(x) is transdental.

Let $F\left(x\right)$is the field of quotients of $F\left(x\right)$ and $F\left(u\right)\cong F\left(x\right)$and $u=\frac{x^3}{x+1}$.

If $u$ is algebraic over $F$the $F\left(u\right)$ is an algebraic extension of $F$. Now as x satisfies a cubic equation of $F\left(x\right)$.

$y^3-u(y+1)$over $F\left(u\right)$.

This implies,

$$\begin{gathered} x^3-\frac{x^3}{x+1}(x+1)=x^3-x^3 \\ =0 \end{gathered}$$

This means that $x$ is algebraic over $F\left(u\right)$. Now since $F\left(u\right)$ is an algebraic extension of$F$ and $x$ is algebraic over$F\left(u\right)$. Also,

$F⊑F\left(u\right)⊑F\left(u\right)\left[x\right]$

This implies that $x$is algebraic over $F$. This is a contradicition as $x$ is in determinant.

Therefore$\frac{x^3}{x+1}\in F\left(x\right)$ is transcendental.