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

jostpuur Mar4-12 10:31 AM

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]

micromass Mar4-12 10:44 AM

Re: qube root of 2, zero of second order polynomial
 
The polynomial [itex]X^2+aX+c[/itex] will have to divide [itex]X^3-2[/itex] in that case.

jostpuur Mar4-12 11:33 AM

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.

micromass Mar4-12 12:00 PM

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

Quote by jostpuur (Post 3797948)
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...ield_theory%29

jostpuur Mar4-12 12:59 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 [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.

Norwegian Mar4-12 04:03 PM

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

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

Office_Shredder Mar4-12 04:12 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)[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 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 04:14 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)[itex]\epsilon[/itex]Q[x] has a root [itex]\alpha[/itex], then the minimal polynomial of [itex]\alpha[/itex] divides g.

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.


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