BolkowN

2021-02-05

Let $\mathbb{R}$ sube K be a field extension of degree 2, and prove that $K\cong \mathbb{C}$. Prove that there is no field extension $\mathbb{R}$ sube K of degree 3.

SchepperJ

Expert

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

So, the map $\varphi :K=\mathbb{R}\left(\beta \right)\to \mathbb{C},$ with 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 $K=\mathbb{R}\left(\alpha \right)$ and $\mathrm{deg}\left(\frac{K}{\mathbb{R}}\right)=3$, 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 $f\left(x\right)=a{x}^{3}+b{x}^{2}+cx+d$ 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 $\mathrm{deg}\left(\frac{K}{\mathbb{R}}\right)=3$

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