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

BolkowN

BolkowN

Answered

2021-02-05

Let R sube K be a field extension of degree 2, and prove that KC. Prove that there is no field extension R sube K of degree 3.

Answer & Explanation

SchepperJ

SchepperJ

Expert

2021-02-06Added 96 answers

Let K be any quafratic extension of R. So, K=R(α), as deg kR=2, the elements 1,α,α2 are linearly depend over R. Thus E a,b,cR, with α2+bα+c=0, by quadratic formula. α=b+b24ac2a, with d=b24aac<0 if b24ac>0, alpha in R, contraditing, deg(KR)=2 Set, β=2α+bd

Then beta in K and β2=1 Using this element , we construct a field isomorphism between K and C, the field of complex numbers.

Now, K=R(α)=R(β), with β2=1

So, the map ϕ:K=R(β)C, with ϕ(x+yβ)=x+iy, x, y in R is the required field isomorphism between K and C Next , we showthat there is no field extension K of degree 3 over R. If K=R(α) and deg(KR)=3, then alpha satisfies an equation o the form aα3+bα2+cα+d=0, with a,b,c,dR, and f(x)=ax3+bx2+cx+d is irreducible over R Completing the proof by arriving at a contradiction. Complex root of ax3+bx2+cx+d=0 with a,b,c,dR, occur in pairs. Hence, ax3+bx2+cx+d=0 has a real root, which contradicts f(x) being irreducible over R.

This proves that there is no filed K with deg(KR)=3

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