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Let f(x) = x3 + ax2 + bx + c ∈ $\mathbb{Q}$[x]. Let K be the splitting field of f(x). I want to construct a tower:
$\mathbb{Q}$ ⊂ K1 ⊂ ... ⊂ Kr = K

Each Ki = Ki-1(α) where either α2 ∈ Ki or α3 ∈ Ki.

I figured my first step should be to extend by a primitive nth root of unity first, and then work from there.

James Norton
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    I would check out the chapter on Galois theory in Artin. He goes through the details of the cubic as well as solvability by radicals. – jgon Apr 29 '15 at 01:19
  • You could also check Dummit & Foote's chapter on solvability by radicals, it is also very detailed. Artin's book suggested by jgon is great too. – Patrick Da Silva Apr 29 '15 at 01:20
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    Actually now that I think about it, Artin proves Cardano's formula, which seems to be (basically) what you want. – jgon Apr 29 '15 at 01:22
  • Agree with jgon. The quantities $u$ and $v$ in Cardano's method (see here for an on-site explanation) have cubes that are contained in a quadratic extension. That's essentially the tower you need (throw the thid roots of unity to the mix). – Jyrki Lahtonen Apr 29 '15 at 14:30
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    Reading more carefully: You need the third roots of unity to get that root tower. But the splitting field may not need them! Works as prescribed only over $\Bbb{Q}(\sqrt{-3})$. – Jyrki Lahtonen Apr 29 '15 at 14:32
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    For example the splitting field of the polynomial $x^3-3x+1$ over $\Bbb{Q}$ is the cubic real field $K=\Bbb{Q}(\cos(2\pi/9))$ which clearly cannot be achieved as a root tower extension. OTOH $K(e^{2\pi i/3})=\Bbb{Q}(e^{2\pi i/9})$ clearly is a root tower extension. – Jyrki Lahtonen Apr 29 '15 at 15:06

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