Let $\alpha = \sqrt[3]{4+\sqrt{5}}$. I would like to prove that $\left[ \mathbb{Q} \left( \alpha \right ) : \mathbb{Q} \right] = 6$. We have $\alpha^3 = 4 + \sqrt{5}$, and so $(\alpha^3 – 4)^2 = 5$, hence $\alpha$ is a root of the polynomial $f(x)=x^6 – 8 x^3 + 11$.
I tried to prove with various approaches that $f(x)$ is irreducible over $\mathbb{Q}$ without success, so I devised the following strategy.
Since $x^2 – 5$ is irreducible over $\mathbb{Q}$, we have $\left[ \mathbb{Q} \left( \sqrt{5} \right ) : \mathbb{Q} \right] = 2$. Now from $\alpha^3 = 4 + \sqrt{5}$ we get $\sqrt{5} \in \mathbb{Q} \left( \alpha \right )$, so that $\mathbb{Q} \left( \sqrt{5} \right )$ is a subfield of $\mathbb{Q} \left( \alpha \right )$, $\mathbb{Q} \left( \alpha \right )=\mathbb{Q}\left( \sqrt{5}\right) \left( \alpha \right)$, and we have
\begin{equation}
\left[ \mathbb{Q} \left( \alpha \right ) : \mathbb{Q} \right] = \left[ \mathbb{Q} \left( \sqrt{5} \right)\left( \alpha \right ) : \mathbb{Q} \left (\sqrt{5} \right) \right] \left[ \mathbb{Q} \left( \sqrt{5} \right ) : \mathbb{Q} \right] .
\end{equation}
Now $\alpha$ is a root of the polynomial $g(x) \in \mathbb{Q} \left (\sqrt{5} \right) [ x ]$ given by $g(x) = x^3 – 4 – \sqrt{5}$. So to prove our thesis it is enough to prove that this polynomial is irreducible in $\mathbb{Q} \left (\sqrt{5} \right) [ x ]$. Being $g(x)$ of third degree, if it were not irreducible, its factorization would have at least one linear factor, so that $g(x)$ would have some root in $\mathbb{Q} \left (\sqrt{5} \right)$. Hence our problem boils down to show that there are no integers $m_0, m_1, n$, with $n \neq 0$, such that
\begin{equation}
\left( \frac{m_0}{n} + \frac{m_1}{n} \sqrt{5}\right)^3 = 4 + \sqrt{5},
\end{equation}
which gives
\begin{equation}
m_0^3 + 5 \sqrt{5} m_1^3 +3 \sqrt{5} m_0^2 m_1 + 15 m_0 m_1^2 = 4 n^3 + \sqrt{5} n^3,
\end{equation}
or
\begin{equation}
m_0^3 + 15 m_0 m_1^2 – 4 n^3 + \sqrt{5} \left( 5 m_1^3 +3 m_0^2 m_1 – n^3 \right)=0,
\end{equation}
which implies, being $\sqrt{5}$ irrational,
\begin{cases}
m_0^3 + 15 m_0 m_1^2 – 4 n^3 = 0, \\ 5 m_1^3 +3 m_0^2 m_1 – n^3 = 0.
\end{cases}
At this point I am stuck, because I do not know how to prove that this system admits the only integer solution $m_0 = m_1 = n = 0$.
Any help is welcome!
Best Answer
As requested by OP I am rewriting my comment as an answer. We will show that $[\mathbb{Q}(\sqrt[3]{4+\sqrt{5}}) : \mathbb{Q}(\sqrt{5})] = 3$ by showing that $f(x) = x^3 - (4+\sqrt{5})$ has no solution in $\mathbb{Q}(\sqrt{5})$.
Rather than the approach in the question we notice that $\operatorname{Nm}_{\mathbb{Q}(\sqrt{5})/\mathbb{Q}}(4 + \sqrt{5}) = 11$. In particular suppose that $\alpha$ is a root of $f(x)$ in $\mathbb{Q}(\sqrt{5})$, then \begin{align*} \operatorname{Nm}_{\mathbb{Q}(\sqrt{5})/\mathbb{Q}}(\alpha)^3 &= \operatorname{Nm}_{\mathbb{Q}(\sqrt{5})/\mathbb{Q}}(\alpha^3) \\ & =\operatorname{Nm}_{\mathbb{Q}(\sqrt{5})/\mathbb{Q}}(4 + \sqrt{5}) =11 \end{align*} a contradiction.
What's really going on under the hood here is that $f(x)$ is Eisenstein for the prime ideal $\mathfrak{p} = (4 + \sqrt{5})$.