# Let RR sube K be a field extension of degree 2, and prove that K ~= CC. Prove that there is no field extension RR sube K of degree 3.

BolkowN 2021-02-05 Answered

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.

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## Expert Answer

SchepperJ
Answered 2021-02-06 Author has 27902 answers

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 $$d=b^2-4aac<0\ if\ b^2-4ac>0$$, alpha in $$\mathbb{R}$$, contraditing, $$\deg(\frac{K}{\mathbb{R}})=2$$ 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 $$\phi: K=\mathbb{R}(\beta) \rightarrow \mathbb{C},$$ with $$\phi(x+y\beta)=x+iy,\ x,\ y$$ 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 $$\deg(\frac{K}{\mathbb{R}})=3$$, 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 $$f(x)=ax^3+bx^2+cx+d$$ is irreducible over $$\mathbb{R}$$ Completing the proof by arriving at a contradiction. Complex root of $$ax^3+bx^2+cx+d=0$$ with $$a,b,c,d \in \mathbb{R},$$ occur in pairs. Hence, $$ax^3+bx^2+cx+d=0$$ has a real root, which contradicts f(x) being irreducible over $$\mathbb{R}$$.

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

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