Question

Let F be a subfield of R and $\mathbf{k}\mathbf{\in }\mathbf{F}$. Show that $\mathbf{F}\left(\sqrt{k}\right)\mathbf{=}\left\{a+b\overline{)\sqrt{k}}a,b\in F\right\}$ is a subfield of C that contains F. If K>0 , show that F is a subfield of R.

Short answer

Since $k>0$, and $\sqrt{k\in \mathrm{R}}$, and R is closed under addition and multiplication. This implies that given field F is a subfield of R.

Step-by-step solution

Showing F is a subfield of R R  

Let $a,b\in F$, this implies that $a,b\in$R as F is a subfield of R.

If k > 0 , then $\sqrt{k}\in \mathrm{R}$.

Showing F(k)={a+bka,b∈F} is a subfield of R

For any $a,b\in \mathrm{R}$, $a,b\sqrt{k}\in \mathrm{R}$ since R is closed under addition and multiplication.

This implies that $\mathrm{F}\left(\sqrt{\mathrm{k}}\right)=\{\mathrm{a}+\mathrm{b}\overline{)\sqrt{\mathrm{k}}}\mathrm{a},\mathrm{b}\in \mathrm{F}\}$is a subfield of R.