About 2 weeks ago, I tried to solve the following problem.
Find the minimal polynomial of $\alpha=\sqrt{2+\sqrt[3]{3}}$ over $\mathbb{Q}$.
My attempt
First, I tried to find the polynomial with rational(integer) coefficients having $\alpha$ as a root, and $f(x)=x^6-6x^4+12x^2-11$ is a polynomial such that $f(\alpha)=0$. Unfortunately, $6$ and $12$ is not divided by $11$, so I could not use the Eisenstein's criterion.
Instead of directly showing that $f(x)$ is irreducible over $\mathbb{Q}$, I tried to show that $[\mathbb{Q}(\alpha):\mathbb{Q}]=6$. Since $[\mathbb{Q}(\alpha):\mathbb{Q}]=[\mathbb{Q}(\alpha):\mathbb{Q}(\alpha^2)][\mathbb{Q}(\alpha^2):\mathbb{Q}]$ and $\alpha^2=2+\sqrt[3]{3}$, we know that $[\mathbb{Q}(\alpha^2):\mathbb{Q}]=3$. Thus if we succeed to show that $[\mathbb{Q}(\alpha):\mathbb{Q}(\alpha^2)]=2$, then the proof is over. However, I could not do it.
I supposed that $\alpha\in\mathbb{Q}(\alpha^2)=\mathbb{Q}(\sqrt[3]{3})$, then there are $a,b,c\in \mathbb{Q}$ such that
$$
\alpha=a+b\sqrt[3]{3}+c\sqrt[3]{9}.
$$
By squaring both sides of the equation, we obtain
$$
(a^2-6bc-2)+(3c^2-2ab-1)\sqrt[3]{3}+(b^2+2ca)\sqrt[3]{9}=0
$$
and $a^2-6bc-2=3c^2-2ab-1=b^2+2ca=0$. However, I don't know how to show that this system of equations does not have a rational root and I'm stuck here.
Question: Is there a (or an alternative) way to solve the problem?
Best Answer
Here is an alternative solution. Working in $\mathbb F_3$, the minimal polynomial for $\alpha$ factorises as $(x^2+1)^3$, and $x^2+1$ is irreducible. Thus 2 divides $[\mathbb Q(\alpha):\mathbb Q]$, and you showed 3 divides this as well.