Try Then is a root of
Let E = Q(α,ω), α = 2^{1/3}, ω = e^{2}^{π}^{i}^{/3}, (Q is the field of rationals)
Is E a simple extension of Q?
My attempt:
Let β = α + ω, find a monic irreducible polynomial, p in Q[x] with root β.
p should have degree 6, implying E = Q(β)
β^{3} = (α + ω)^{3}
= α^{3} + 3α^{2}ω + 3αω^{2} + ω^{3}
= 2 + 3α^{2}ω + 3αω^{2} + 1
= 3αω(α + ω) + 3
= 3αωβ + 3
β^{3} 3 = 3αωβ
(β^{3} 3)^{3} = (3αωβ)^{3}
(β^{3})^{3} + 3(β^{3}) ^{2}(-3) + 3(β^{3}) (-3)^{2} + (-3)^{3} = 3^{3}α^{3}ω^{3}β^{3}
β^{9} - 9β^{6} + 27β^{3} 27 = 9(2)(1)β^{3}
= 18 β^{3}
β^{9} - 9β^{6} + 9β^{3} 27 = 0
However, if I define p(x) = x^{9} 9x^{6} + 9x^{3} 27 = 0, my calculator tells me p(2^{1/3} + e^{2}^{π}^{i}^{/3}) ≠ 0.
Also I cant see how to reduce p such that p = fg with f,g in Q[x], deg(f) = 6, and f(β) = 0.
i have a different idea, prove Q(α+ω) = Q(α,ω). obviously Q(α,ω) contains Q(α+ω). so if we can show α,ω are in Q(α+ω), we're done. in fact all we NEED to do is show ω is in Q(α+ω), because then: α = α+ω - ω is also in Q(α+ω).
note that: 0 = α^{3} - 2 = (α+ω - ω)^{3} - 2 = (α+ω)^{3} - 3(α+ω)^{2}ω + 3(α+ω)ω^{2} - ω^{3} - 2
= (α+ω)^{3} - 3(α+ω)^{2}ω + 3(α+ω)(-ω-1) - 3 (since ω^{2}+ω+1 = 0)
= (α+ω)^{3} - 3(α+ω)^{2}ω - 3(α+ω)ω - 3(α+ω) - 3, so that:
ω(3(α+ω)^{2}+3(α+ω)) = (α+ω)^{3} - 3(α+ω) - 3, thus:
ω = [(α+ω)^{3} - 3(α+ω) - 3]/(3(α+ω)^{2}+3(α+ω)), which is in Q(α+ω).
by the way, there is an error in your p(x), it should be:
p(x) = x^{9}-9x^{6}-27x^{3}-27 (unless i've also made an error. could be.)