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-   -   qube root of 2, zero of second order polynomial (http://www.physicsforums.com/showthread.php?t=583746)

 jostpuur Mar4-12 11:31 AM

qube root of 2, zero of second order polynomial

How do you prove that there does not exist numbers $a,b\in\mathbb{Q}$ such that

$$0 = a + b\sqrt[3]{2} + \sqrt[3]{2}^2$$

 micromass Mar4-12 11:44 AM

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

The polynomial $X^2+aX+c$ will have to divide $X^3-2$ in that case.

 jostpuur Mar4-12 12:33 PM

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

I think we only know that $X-\sqrt[3]{2}$ must divide $X^2+bX+a$.

$X^2+bX+a$ doesn't need to divide anything.

 micromass Mar4-12 01:00 PM

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

Quote:
 Quote by jostpuur (Post 3797948) I think we only know that $X-\sqrt[3]{2}$ must divide $X^2+bX+a$. $X^2+bX+a$ doesn't need to divide anything.
Yes it will, as it will be the minimal polynomial with $\sqrt[3]{2}$ as a root. http://en.wikipedia.org/wiki/Minimal...ield_theory%29

 jostpuur Mar4-12 01:59 PM

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 Wikipedia-page didn't help much...)

The knowledge that $X^2+bX+a$ must divide $X^3-2$ 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

$$X^3 - 2 = (X^2 + bX + a)(X - b) + (b^2 - a)X + ab - 2$$

and proof starts to appear.

 Norwegian Mar4-12 05:03 PM

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

Quote:
 Quote by micromass (Post 3797906) The polynomial $X^2+aX+c$ will have to divide $X^3-2$ in that case.
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)$\epsilon$Q[x] has a root $\alpha$, then the minimal polynomial of $\alpha$ divides g.

 Office_Shredder Mar4-12 05:12 PM

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

Quote:
 Quote by Norwegian (Post 3798326) 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)$\epsilon$Q[x] has a root $\alpha$, then the minimal polynomial of $\alpha$ divides g.

If the cube root of 2 satisfies a quadratic polynomial, then x3-2 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

 jostpuur Mar4-12 05:14 PM

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

Quote:
 Quote by Norwegian (Post 3798326) 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)$\epsilon$Q[x] has a root $\alpha$, then the minimal polynomial of $\alpha$ divides g.
With the anti-thesis assumption the $X^2+bX+a$ 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.

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