How do you prove that there does not exist numbers [itex]a,b\in\mathbb{Q}[/itex] such that [tex] 0 = a + b\sqrt[3]{2} + \sqrt[3]{2}^2 [/tex]
I think we only know that [itex]X-\sqrt[3]{2}[/itex] must divide [itex]X^2+bX+a[/itex]. [itex]X^2+bX+a[/itex] doesn't need to divide anything.
Yes it will, as it will be the minimal polynomial with [itex]\sqrt[3]{2}[/itex] as a root. http://en.wikipedia.org/wiki/Minimal_polynomial_(field_theory)
I see. I have tried to read Galois theory earlier, and now I started remembering stuff. (Although that Wikipedia-page didn't help much...) The knowledge that [itex]X^2+bX+a[/itex] must divide [itex]X^3-2[/itex] is one possible way to the proof, but actually the idea of the minimal polynomial can be used in more primitive ways too. For example, simply write [tex] X^3 - 2 = (X^2 + bX + a)(X - b) + (b^2 - a)X + ab - 2 [/tex] and proof starts to appear.
What is the logic behind this statement? From a false premiss you can of course deduce anyhing you like, but I assume that is not your intended logic here. When it comes to minimal polynomials and such, the general result is usually the other way around: if g(x)[itex]\epsilon[/itex]Q[x] has a root [itex]\alpha[/itex], then the minimal polynomial of [itex]\alpha[/itex] divides g.
If the cube root of 2 satisfies a quadratic polynomial, then x^{3}-2 is NOT the minimal polynomial (since there's a lower degree polynomial that gives us zero). So there are two possibilities 1) The minimal polynomial is degree 1 - obviously false 2) The minimal polynomial is degree 2 - in this case, the quadratic polynomial we have must be the minimal polynomial
With the anti-thesis assumption the [itex]X^2+bX+a[/itex] becomes the "new" minimal polynomial. Or at least that's one way to get to the proof. See my reponse #5 to see the essential. There are several ways to complete the proof then.