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Does there exist $v=(a,b,c)\in\mathbb{Q^3}$ with none of $v$'s terms being zero s.t. $ a+b\sqrt[3]2+c\sqrt[3]4=0$ ?

And I was doing undergraduate algebra 2 homework when I encountered it in my head. At first It seemed like it can be proved there can be no such $v$ like how $\sqrt{2}$, or $\sqrt{2}+\sqrt{3}$ are proved to be irrational, but this case wasn't easy like those. Or maybe I was too hasty.

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Do you know field theory ? – Belgi Sep 19 '12 at 10:03
Very little. I studied group theory in algebra 1 and now I'm in algebra 2. I have currently learned Euler,Fermat's theorems and what the field of fractions is. – YD55 Sep 19 '12 at 10:44

I'll try to show some more elementary approach, not using field theory. Suppose there is such a $v$. We have $c \ne 0$ (as $\sqrt[3]2$ is irrational). Rewriting the equation, we find $\alpha, \beta \in \mathbb Q$ with \[ \sqrt[3]4 = \alpha + \beta \sqrt[3]2 \] and $\alpha, \beta \ne 0$ (as $\sqrt[3]2, \sqrt[3]4 \not\in \mathbb Q$). Taking the third power, we get \[ 4 = \alpha^3 + 3\alpha^2\beta \sqrt[3]2 + 3\alpha\beta^2\sqrt[3]4 + 2\beta^3 \] so, as $\alpha\beta^2 \ne 0$, \[ \sqrt[3]4 = \frac{4 - \alpha^3 - 3\alpha^2\beta\sqrt[3]2 - 2\beta^3}{3\alpha\beta^2} \] Which gives \[ \alpha + \beta \sqrt[3]2 = \frac{4-\alpha^3 - 2\beta^3}{3\alpha\beta^2} - \frac\alpha\beta \sqrt[3]2 \] As $\sqrt[3]2$ is irrational, we must have \[ \alpha = \frac{4-\alpha^3 - 2\beta^3}{3\alpha\beta^2}, \beta = -\frac\alpha\beta \] So $\beta^2 = -\alpha$, giving \[ -3\alpha^3 = 3\alpha^2\beta^2 = 4-\alpha^3 - 2\beta^3 \iff 2(\beta^3 - \alpha^3) = 4 \iff \beta^3 - \alpha^3 = 2 \iff \beta^3 + \beta^6 = 2 \] But $x^6 + x^3 - 2$ has no rational zeros, as $\pm 1, \pm 2$ are the only possibilities. Contradiction.

So, there is no such $v$.

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you can also put $t=x^3$ and solve directly to see no rational roots – Belgi Sep 19 '12 at 10:48

Suppose there exists such a v.

Then $a = - b\sqrt[3]2 - c\sqrt[3]4 $

You can try to show that the left hand side is irrational. Which would show there is no such v. If there really is no such v.

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Did that work out? Because I tried that way too but it didn't for me. – YD55 Sep 19 '12 at 9:45

The question can be rephrased as:

Is the triple $\ 1,\sqrt[3]2,\sqrt[3]4\ $ linearly independent over $\mathbb Q$?

And the answer is no (i.e., there is no such $v$).

Denote $\alpha:=\sqrt[3]2$. Then $\alpha^3=2$. The main point is, that the polynomial $x^3-2$ (which defines $\alpha$ as its root) is irreducible over $\mathbb Q$: cannot be written as a proper product of polynomials of degree 1 and 2. (This can be shown directly..)

In other words, the field extension $\mathbb Q(\alpha)$ of $\mathbb Q$, is --as $\alpha$ is the root of the irreducible $x^3-2$-- per definition, is isomorphic to the quotient $K:=\mathbb Q[x]/(x^3-2)$ of polynomial ring $\mathbb Q[x]$. That is, the elements of $K$ are the polynomials, but $x^3-2 = 0$ is assumed (as the only rule) in $K$.

And, similarly as $\mathbb Q[x]$ has $1,x,x^2,x^3,x^3,\ldots$ as basis, $K$ has $1,x,x^2$ as a (standard) basis over $\mathbb Q$. ($x^3$ and above powers can be rephrased by $1,x,x^2$, using the rule: for example $x^3=2,\ x^4=2x,$ etc.)

The correspondence between $K$ and $\mathbb Q(\alpha)$ is simply given by $x\mapsto\alpha$.

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I think you meant that the answer is yes, in the third line above...or else you meant to write dependent on your second line. – DonAntonio Sep 19 '12 at 10:13

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