Let R sube K be a field extension of degree 2, and prove that K≅C....

Let K be any quafratic extension of $\mathbb{R}$. So, $K=\mathbb{R}(\alpha )$, as deg $\frac{k}{\mathbb{R}}=2$, the elements $1,\alpha ,{\alpha }^2$ are linearly depend over $\mathbb{R}$. Thus $\mathbb{E}a,b,c\in \mathbb{R}$, with ${\alpha }^2+b\alpha +c=0$, by quadratic formula. $\alpha =\frac{-b+-\sqrt{b^2-4ac}}{2a}$, with $d=b^2-4aac<0if{b}^2-4ac>0$, alpha in $\mathbb{R}$, contraditing, $\mathrm{deg}(\frac{K}{\mathbb{R}})=2$ Set, $\beta =\frac{2\alpha +b}{\sqrt{-d}}$

Then beta in K and ${\beta }^2=-1$ Using this element , we construct a field isomorphism between K and C, the field of complex numbers.

Now, $K=\mathbb{R}(\alpha )=\mathbb{R}(\beta )$, with ${\beta }^2=-1$

So, the map $\varphi :K=\mathbb{R}(\beta )\to \mathbb{C},$ with $\varphi (x+y\beta )=x+iy,x,y$ in $\mathbb{R}$ is the required field isomorphism between K and $\mathbb{C}$ Next , we showthat there is no field extension K of degree 3 over R. If $K=\mathbb{R}(\alpha )$ and $\mathrm{deg}(\frac{K}{\mathbb{R}})=3$, then alpha satisfies an equation o the form $a{\alpha }^3+b{\alpha }^2+c\alpha +d=0,$ with $a,b,c,d\in \mathbb{R},$ and $f(x)=a{x}^3+b{x}^2+cx+d$ is irreducible over $\mathbb{R}$ Completing the proof by arriving at a contradiction. Complex root of $a{x}^3+b{x}^2+cx+d=0$ with $a,b,c,d\in \mathbb{R},$ occur in pairs. Hence, $a{x}^3+b{x}^2+cx+d=0$ has a real root, which contradicts f(x) being irreducible over $\mathbb{R}$.

This proves that there is no filed K with $\mathrm{deg}(\frac{K}{\mathbb{R}})=3$