I have to find the minimal polynomial of $\sqrt 2 + \sqrt[3] 2$ over $\mathbb Q$. The suggested way of doing it is to prove that $\mathbb Q[\sqrt 2 + \sqrt[3] 2]=\mathbb Q[\sqrt 2,\sqrt[3] 2]$ first.
I can prove that. It's enough to prove that $\sqrt 2,\sqrt[3] 2\in \mathbb Q[\sqrt 2 + \sqrt[3] 2]$ and that $\sqrt 2 + \sqrt[3] 2\in \mathbb Q[\sqrt 2,\sqrt[3] 2]$. The latter is obvious and $\sqrt[3] 2\in \mathbb Q[\sqrt 2 + \sqrt[3] 2]$ will follow from $\sqrt 2\in \mathbb Q[\sqrt 2 + \sqrt[3] 2].$ It remains to prove $\sqrt 2\in \mathbb Q[\sqrt 2 + \sqrt[3] 2].$
For $\alpha=\sqrt 2 + \sqrt[3] 2,$ we have
$ \begin{eqnarray} &\alpha^0&=1,\\ &\alpha^1&=\sqrt 2 + \sqrt[3] 2,\\ &\alpha^2&=2+\sqrt[3]4+\sqrt 2\sqrt[3]2\\ &\alpha^3&=2+2\sqrt 2+6\sqrt[3]2+3\sqrt 2\sqrt[3]4\\ &\alpha^4&=4+4\sqrt 2+12\sqrt[3]4+8\sqrt 2\sqrt[3]2\\ &\alpha^5&=32+4\sqrt 2+20\sqrt[3]2+4\sqrt 2\sqrt[3]2 \end{eqnarray} $
If I can express $\sqrt 2$ as a linear combination of $\{\alpha^0,\alpha^1,\cdots,\alpha^5\}$, I'm done. I don't know how to find out whether $\{1,\sqrt 2,\sqrt[3]2,\sqrt[3]4,\sqrt 2\sqrt[3]2,\sqrt 2\sqrt[3]4\}$ are linearly independent over $\mathbb Q$ but I think I don't have to care. I can use Gaussian elimination anyway. I write the matrix
$\left[ \begin{array}{rrrrrr|r} 1 & 0 & 2 & 2 & 1 & 8 & 0\\ 0 & 1 & 0 & 2 & 1 & 1 & 1\\ 0 & 1 & 0 & 6 & 0 & 5 & 0\\ 0 & 0 & 1 & 0 & 3 & 0 & 0\\ 0 & 0 & 1 & 0 & 2 & 1 & 0\\ 0 & 0 & 0 & 3 & 0 & 5 & 0 \end{array}\right]$
and by performing row operations I obtain a row echelon form, which proves that I'll obtain the reduced row echelon form if I want, which will give me the desired coefficients $a_0,\cdots,a_5$ such that
$\sqrt 2=\sum_{i=0}^5a_i\alpha^i.$
I don't have to find the coefficients -- I'm just satisfied with their existence.
If this is correct, I have proven that
$\mathbb Q[\sqrt 2 + \sqrt[3] 2]=\mathbb Q[\sqrt 2,\sqrt[3] 2].$
But I don't know how to proceed further. Of course, I can find the minimal polynomials of $\sqrt 2$ and $\sqrt[3] 2$ over $\mathbb Q$ but I don't know how to find the minimal polynomial of $\sqrt 2$ over $\mathbb Q[\sqrt[3]2]$ or the minimal polynomial of $\sqrt [3]2$ over $\mathbb Q[\sqrt 2].$
What I did is this. I wrote the equation
$x=\sqrt 2+\sqrt [3]2$
and by squaring and cubing obtained
$W(x):=x^6-6x^4-4x^3+12x^2-24x-4=0.$
I belive this is the required polynomial but I would need to prove that it's irreducible over $\mathbb Q.$ Eisenstein's criterion doesn't work, which means I'm lost.
Could you please help me with this? I would like to know three things.
1) Is my proof of $\mathbb Q[\sqrt 2 + \sqrt[3] 2]=\mathbb Q[\sqrt 2,\sqrt[3] 2]$ correct?
2) How can I use this fact? I believe it's supposed to make the solution easier to find.
3) How can I tell whether $W(x)$ is reducible over $\mathbb Q$ or not?
EDIT: 1) has been answered by André Nicolas in a comment. All of this answers 3) but I would like to ask whether we could know that the polynomial is irreducible without knowing that $\sqrt 2+\sqrt[3]2$ is its zero. Also, it would be to know whether
4) there is a theorem that would give me $\mathbb Q[\sqrt 2 + \sqrt[3] 2]=\mathbb Q[\sqrt 2,\sqrt[3] 2]$ without the need of performing the Gaussian elimination?
For these numbers it was possible to do, but a bit higher dimension and it would turn out impossible without a computer.