Question

If $\mathit{u}\mathbf{\in }\mathit{K}$and ${\mathit{u}}^{\mathbf{2}}$is algebraic over F , prove that is algebraic over F .

Short answer

It is proved u is algebraic over F .

Step-by-step solution

Definition of field extension

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

Show that u

Let K be an extension field of F and $\mathit{u}\mathbf{\in }\mathit{K}$and ${\mathit{u}}^{\mathbf{2}}$ is algebraic over F .

Since ${\mathit{u}}^{\mathbf{2}}$ is algebraic over F , there exists a polynomial $\mathit{f}\left(x\right)$over F such that$\mathit{f}\left(u^2\right)\mathbf{=}\mathbf{0}$ .

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

Replacing x by u , we get the following:

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

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

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

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

Therefore, u is algebraic over F .