# $[\mathbb{Q}(\sqrt [3] {2}+\sqrt {5}):\mathbb{Q}]$ [duplicate]

What is $[\mathbb{Q}(\sqrt [3] {2}+\sqrt {5}):\mathbb{Q}]$? A straight forward way will be to just set $x=\sqrt [3] {2}+\sqrt {5}$, take powers and reach at a polynomial in $x$ and show the polynomial is irreducible. But the method is very tedious and seems to be hopeless. Any idea?

Consider the tower of fields $\mathbb{Q} \subset \mathbb{Q}(\sqrt[3]{2}+\sqrt{5}) \subset \mathbb{Q}(\sqrt[3]{2},\sqrt{5})$. We know that $[\mathbb{Q}(\sqrt[3]{2},\sqrt{5}):\mathbb{Q}]=6$ so the only possibilities for $[\mathbb{Q}(\sqrt[3]{2}+\sqrt{5}):\mathbb{Q}]$ are 2, 3, and 6. It does not take long to verify that $\sqrt[3]{2}+\sqrt{5}$ is not a root of any degree 2 or 3 polynomial over $\mathbb{Q}$, so this implies that $[\mathbb{Q}(\sqrt[3]{2}+\sqrt{5}):\mathbb{Q}]=6$.

Note that $\mathbb{Q}[\sqrt[3]{2} + \sqrt{5}] \subset \mathbb{Q}[\sqrt[3]{2}, \sqrt{5}]$. Since $\mathbb{Q}[\sqrt[3]{2}, \sqrt{5}]$ is a degree $6$ extension over $\mathbb{Q}$, then it must be the case that $[\mathbb{Q}[\sqrt[3]{2} + \sqrt{5}]:\mathbb{Q}] = 2, 3, \text{or } 6$.

Could $\sqrt[3]{2} + \sqrt{5}$ be a root of a quadratic? Of a cubic?

If the degree were $6$, then it would be true that $\mathbb{Q}[\sqrt[3]{2} + \sqrt{5}] = \mathbb{Q}[\sqrt[3]{2}, \sqrt{5}]$.

We're already down to just three possibilities, so try some stuff out and see which ones you can further eliminate.

Denote $\alpha = \sqrt[3]{2}+ \sqrt{5}$.

Let $\zeta = - \frac{1}{2} + i \frac{\sqrt{3}}{2}$.$\$ The product \begin{eqnarray} P(X) =\prod_{0\le k \le 2, 0 \le l \le 1} ( X - (\sqrt[3]{2}\cdot \zeta ^k + \sqrt{5}\cdot(-1)^l)\ ) \end{eqnarray} is a polynomial with integer coefficients. We calculate: \begin{eqnarray} P(X) = X^6- 15 X^4 - 4 X^3 + 75 X^2- 60 X -121 \end{eqnarray} $P(X)$ is irreducible $\mod\!\! 7$ and therefore irreducible over $\mathbb{Q}$. It follows that $$[\mathbb{Q}(\alpha) \colon \mathbb{Q}] = 6$$

We can check by direct computation that \begin{eqnarray} \frac{-2275 + 7482 \alpha - 390 \alpha^2 - 1000 \alpha^3 + 9 \alpha^4 + 60 \alpha^5}{4054} = \sqrt{5} \end{eqnarray} and therefore \begin{eqnarray} \alpha - \frac{-2275 + 7482 \alpha - 390 \alpha^2 - 1000 \alpha^3 + 9 \alpha^4 + 60 \alpha^5}{4054} = \sqrt[3]{2} \end{eqnarray} We conclude that $$\mathbb{Q}(\alpha) = \mathbb{Q}( \sqrt[3]{2}, \sqrt{5})$$