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.
Best Answer
Here is an extremely easy way of showing that $\sqrt{2} \in \Bbb{Q}(\sqrt{2} + \sqrt[3]{2})$. Write $\alpha = \sqrt{2} + \sqrt[3]{2}$. Then $(\alpha-\sqrt{2})^3 = 2$, so that
$$\alpha^3 - 3\alpha^2(\sqrt{2}) + 6\alpha -2\sqrt{2} = 2.$$
It follows that $\alpha^3+ 6\alpha - 2 = \sqrt{2}(3\alpha^2+ 2)$, so that
$$\sqrt{2} = \frac{\alpha^3 + 6\alpha - 2}{3\alpha^2 + 2}.$$
Since $\Bbb{Q}(\alpha)$ is a field, the expression on the right hand side is in here, so that $\sqrt{2}$ is in here. No using horrible row reduction to calculate anything! It follows that $\sqrt[3]{2}$ is in here. Hence $\Bbb{Q}(\alpha)$ contains the fields $\Bbb{Q}(\sqrt{2})$ and $\Bbb{Q}(\sqrt[3]{2})$ which means that $[\Bbb{Q}(\alpha) : \Bbb{Q}]$ is a multiple of 3 and 2. Since we already know that it is at most $6$, it follows now that since $2$ and $3$ are coprime that $[\Bbb{Q}(\alpha) : \Bbb{Q}] = 6$. Whatever monic polynomial of degree 6 that you find for $\alpha$ over $\Bbb{Q}$ will then be irreducible, and hence will be the minimal polynomial of $x$ over $\Bbb{Q}$.