# Qube root of 2, zero of second order polynomial

1. Mar 4, 2012

### jostpuur

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

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

2. Mar 4, 2012

### micromass

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

3. Mar 4, 2012

### jostpuur

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.

4. Mar 4, 2012

5. Mar 4, 2012

### jostpuur

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 $X^2+bX+a$ 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.

6. Mar 4, 2012

### Norwegian

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.

7. Mar 4, 2012

### Office_Shredder

Staff Emeritus

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

8. Mar 4, 2012

### jostpuur

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.