Question

$$\begin{gathered} \mathbf{Assume}\mathbf{}\mathbf{that}\mathbf{}{\mathbf{1}}_{\mathbf{F}}\mathbf{+}{\mathbf{1}}_{\mathbf{F}}\mathbf{\ne }{\mathbf{0}}_{\mathbf{F}}\mathbf{}\mathbf{if}\mathbf{}\mathbf{u}\mathbf{\in }\mathbf{F}\mathbf{}\mathbf{,}\mathbf{let}\mathbf{}\sqrt{\mathbf{u}}\mathbf{}\mathbf{denote}\mathbf{}\mathbf{a}\mathbf{}\mathbf{root}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{u}\mathbf{}\mathbf{in}\mathbf{}\mathbf{K}\mathbf{.}\mathbf{}\mathbf{prove}\mathbf{}\mathbf{that}\mathbf{} \\ \mathbf{F}\mathbf{(}\sqrt{\mathbf{u}}\mathbf{+}\sqrt{\mathbf{v}}\mathbf{)}\mathbf{=}\mathbf{F}\mathbf{\left(}\sqrt{\mathbf{u}}\sqrt{\mathbf{v}}\mathbf{\right)} \end{gathered}$$

Short answer

$\mathrm{F}\left(\sqrt{\mathrm{u}}+\sqrt{\mathrm{v}}\right)=\mathrm{F}\left(\sqrt{\mathrm{u}}\sqrt{\mathrm{v}}\right)$

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

is every element of K is algebraic over F.

Showing that F(u+v)=F(uv)

Let $1_{\mathrm{F}}+1_{\mathrm{F}}\ne 0_{\mathrm{F}}\mathrm{and}\mathrm{K}$ is an extension field of $\mathrm{F}\mathrm{and}\mathrm{u}\in \mathrm{F}\mathrm{and}\sqrt{\mathrm{u}}$be a root of

${\mathrm{x}}^2-\mathrm{u}\mathrm{in}\mathrm{K}.$

Since $\sqrt{\mathrm{u}}\sqrt{\mathrm{v}}\in \mathrm{F}\left(\sqrt{\mathrm{u}}\sqrt{\mathrm{v}}\right)$

This implies $\sqrt{\mathrm{u}}+\sqrt{\mathrm{v}}\in \mathrm{F}\left(\sqrt{\mathrm{u}}\sqrt{\mathrm{v}}\right)\mathrm{and}\mathrm{F}\underline{\subset }\mathrm{F}\left(\sqrt{\mathrm{u}}\sqrt{\mathrm{v}}\right).\mathrm{So},\mathrm{F}\left(\sqrt{\mathrm{u}}\sqrt{\mathrm{v}}\right)\underline{\subset }\mathrm{F}\left(\sqrt{\mathrm{u}}\sqrt{\mathrm{v}}\right)$

As$\sqrt{\mathrm{u}}+\sqrt{\mathrm{v}}\in \mathrm{F}\left(\sqrt{\mathrm{u}}\sqrt{\mathrm{v}}\right)\mathrm{and}\mathrm{u},\mathrm{v}\in \mathrm{F}.\mathrm{This}\mathrm{implies}\mathrm{u},\mathrm{v}\in \mathrm{F}\left(\sqrt{\mathrm{u}}+\sqrt{\mathrm{v}}\right).$

Thus $\mathrm{u}-\mathrm{v}=\left(\sqrt{\mathrm{u}}+\sqrt{\mathrm{v}}\right)\left(\sqrt{\mathrm{u}}-\sqrt{\mathrm{v}}\right)$by this

$$\begin{gathered} \left(\sqrt{\mathrm{u}}-\sqrt{\mathrm{v}}\right)=\frac{\left(\sqrt{\mathrm{u}}+\sqrt{\mathrm{v}}\right)}{\mathrm{u}-\mathrm{v}}\in \mathrm{F}\left(\sqrt{\mathrm{u}}+\sqrt{\mathrm{v}}\right) \\ \sqrt{\mathrm{u}}=\frac{\left(\sqrt{\mathrm{u}}+\sqrt{\mathrm{v}}\right)+\left(\sqrt{\mathrm{u}}-\sqrt{\mathrm{v}}\right)}{2} \\ \sqrt{\mathrm{v}}=\left(\sqrt{\mathrm{u}}+\sqrt{\mathrm{v}}\right)-\sqrt{\mathrm{u}} \\ \sqrt{\mathrm{u}}\sqrt{\mathrm{v}}\in \mathrm{F}\left(\sqrt{\mathrm{u}}+\sqrt{\mathrm{v}}\right) \end{gathered}$$

So,

$\mathrm{F}\underline{\subset }\mathrm{F}\left(\sqrt{\mathrm{u}}+\sqrt{\mathrm{v}}\right)$

This implies $\mathrm{F}\left(\sqrt{\mathrm{u}},\sqrt{\mathrm{v}}\right)\in \mathrm{F}\left(\sqrt{\mathrm{u}}+\sqrt{\mathrm{v}}\right)$By this

$\mathrm{F}\sqrt{\mathrm{u}}+\sqrt{\mathrm{v}}=\mathrm{F}\left(\sqrt{\mathrm{u}}\sqrt{\mathrm{v}}\right).$