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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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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