$X^6 + 3X^4+3X^2-1$ is the minimal polynomial of $\sqrt{ \sqrt[3]{2}-1}$ over $\mathbb Q$

Question

It can easily be seen that $\sqrt{ \sqrt[3]{2}-1}$ is a root of $X^6 + 3X^4+3X^2-1$, which should be its minimal polynomial.

Let $a=\sqrt{ \sqrt[3]{2}-1}$.

Then $\sqrt[3]{2} = a^2+1$.

Therefore $\sqrt[3]{2} \in \mathbb Q[a]$ and $\mathbb Q[\sqrt[3]{2}] \subset \mathbb Q[a]$

The goal here is to prove that $[\mathbb Q[a]:\mathbb Q]=6$.

It is useful that $$[\mathbb Q[a]:\mathbb Q]=[\mathbb Q[\sqrt[3]{2}]:\mathbb Q]\times [\mathbb Q[a]:\mathbb Q[\sqrt[3]{2}]]$$

Ie$$[\mathbb Q[a]:\mathbb Q]=3\times [\mathbb Q[a]:\mathbb Q[\sqrt[3]{2}]]$$

Nevertheless I can't prove that $[\mathbb Q[a]:\mathbb Q[\sqrt[3]{2}]]=2$.

Furthermore usual irreducibility criteria are of no use here.

Answer 1

Hint: $X^6 + 3X^4+3X^2-1$ is irreducible over $\mathbb Q$ if, and only if, $\underbrace{(X-1)^6 + 3(X-1)^4+3(X-1)^2-1}_{\large{=X^6-6X^5+18X^4-32X^3+36X^2-24X+6}}$ is irreducible over $\mathbb Q$.

Answer 2

If $a$ didn't have degree 6, then it follows from your arguments that it would have degree 3 or 1. Thus its minimal polynomial would have degree 3 or 1 and would divide $g(X) = X^6 + 3X^4 + 3X^2 - 1$. Moreover the expression of $g(x)$ as a product of two factors would necessarily involve polynomials with integral coefficients. You can show by a direct calculation that this can't occur.