qube root of 2, zero of second order polynomial
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] 
Re: qube root of 2, zero of second order polynomial
The polynomial [itex]X^2+aX+c[/itex] will have to divide [itex]X^32[/itex] in that case.

Re: qube root of 2, zero of second order polynomial
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. 
Re: qube root of 2, zero of second order polynomial
Quote:

Re: qube root of 2, zero of second order polynomial
I see.
I have tried to read Galois theory earlier, and now I started remembering stuff. :cool: (Although that Wikipediapage didn't help much...) The knowledge that [itex]X^2+bX+a[/itex] must divide [itex]X^32[/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. 
Re: qube root of 2, zero of second order polynomial
Quote:

Re: qube root of 2, zero of second order polynomial
Quote:
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 
Re: qube root of 2, zero of second order polynomial
Quote:

All times are GMT 5. The time now is 03:35 AM. 
Powered by vBulletin Copyright ©2000  2014, Jelsoft Enterprises Ltd.
© 2014 Physics Forums