field theory - $mathbf{Q}[sqrt 5+sqrt[3] 2]=mathbf{Q}[sqrt 5,sqrt[3] 2]$?

Is there a general or elegant way to approach this problem? One can show that $\sqrt 5+\sqrt[3] 2$ is a root of the hexic $x^6-6x^4-10x^3+12x^2-60x+17$, which should then be its minimal polynomial to get the result. I see no straightforward way to show this is irreducible, though. It has a root in $\mathbf{F}_4$, and to show it's irreducible over $\mathbf{F}_3$ already seems to require a lot of grunt work.

One thing I can do is rule out the possibility that $[\mathbf{Q}[\sqrt 5+\sqrt[3] 2]:\mathbf{Q}]=2$, because $[\mathbf{Q}(\sqrt 5,\sqrt[3] 2):\mathbf{Q}]=6$ and $\mathbf{Q}(\sqrt 5,\sqrt[3] 2)$ is the compositum of $\mathbf{Q}[\sqrt 5+\sqrt[3] 2]$ and $\mathbf{Q}[\sqrt 5]$, where the latter has degree two. But I don't seem to be able to play an analogous trick to rule out $3$, because I know of no general lower bound on the degree of a compositum.

Could there be some way to bring Galois theory into the picture? These extension aren't Galois, and so it's by no means clear to me how to reason from their normal closures, but perhaps that's the direction to go in. Suggestions are very much appreciated.

Answer

Here is an approach using a bit of Galois theory.

Let $K \subset L$ a Galois extension and subfields $K \subset K_1, K_2 \subset L$. We have $K_1 \supset K_2$ if and only if the following hold: whenever $g$ is in the Galois group of $L/K$ such that $g$ fixes any element of $K_1$, $g$ also fixes any element of $K_2$ ( therefore $K_1 \supset K_2$ if and only if $\text{Gal}(L/K_1) \subset \text{Gal}(L/K_2))$.

Consider the same extension $K\subset L$ and $\alpha_1$, $\alpha_2$ in $L$. From the above we conclude: $K(\alpha_1) \supset K(\alpha_2)$ if and only if: for every $g$ in $\text{Gal}(L/K)$ that fixes $\alpha_1$, $g$ also fixes $\alpha_2$.

Consider again $K\subset L$ and $u$, $v$, $w$ $\ldots$ elements in $L$. Let $R(u,v,w,\ldots)$ a rational expression in $u$, $v$, $w$ $\ldots$. Say we want to show that $K ( R(u,v,w,\ldots) )$ contains $u$. Let $u_1=u$, $u_2$ $\ldots$ in $L$ the conjugates of $u$,$\ \ $ $v_1= v$,$v_2$, $\dots$ the conjugates of $v$,$ \ \ $ $w_1 = w$, $w_2$ $\ldots$ ... of $w$. ( in fact it's enough to assume that $u_1=u$, $u_2$, $\ldots$ roots in $L$ of a polynomial with coefficients in $K$ and same for $(v_j)$, $(w_k)$ )

Assume that
$$R(u,v,w,\ldots) \ne R(u_i, v_j, w_k,\ldots )$$
if $i \ne 1$. Then
$$K ( R(u,v,w,\ldots) )\ni u$$

Indeed, let $g$ in $\text{Gal}(L/K)$. We have $g u = u_i$, $g v = v_j$, $g w = w_k$, $\ldots$ for some indexes $i$, $j$, $k$, $\ldots$ since $g u$ is a conjugate of $u$, $g v$ is a conjugate of $v$ and so on. Therefore we have
$$ g( R(u,v,w,\ldots) ) = R(u_i, v_j, w_k, \ldots)$$

Assume that $g( R(u,v,w,\ldots) )= R(u,v,w,\ldots) $. Therefore we have
$R(u_i, v_j, w_k, \ldots)= R(u,v,w,\ldots)$. From the condition above we conclude that $i=1$ and therefore $g u = u_1 = u$.

We have showed that $g( R(u,v,w,\ldots) )= R(u,v,w,\ldots) $ implies $g u = u$. Therefore $K ( R(u,v,w,\ldots) )\ni u$.

Application: Let $a_1$, $a_2$ $a_3$, $\ldots$ rational numbers $>0$,$\ $ $e_1$, $e_2$, $\ldots$ natural numbers $>0$. Then we have

\begin{eqnarray}
\mathbb{Q}( \sum \sqrt[e_s]{a_s}) = \mathbb{Q}( \sqrt[e_1]{a_1}, \sqrt[e_2]{a_2}, \ldots)

\end{eqnarray}

Indeed, $\sum \sqrt[e_s]{a_s}> \text{Real part}( \text{sum of any other set of conjugates} ) $ and so no other set of conjugates has the sum equal to $\sum \sqrt[e_s]{a_s}$ and so
\begin{eqnarray}
\mathbb{Q}( \sum \sqrt[e_s]{a_s}) \supset \mathbb{Q}( \sqrt[e_{s_0}]{a_{s_0}})
\end{eqnarray}
for all $s_0$.

Similarly we can also prove:
\begin{eqnarray}

\mathbb{Q}( \sqrt[3]{2} \cdot \sqrt{5}) = \mathbb{Q}( \sqrt[3]{2} , \sqrt{5})
\end{eqnarray}

Obs: the nonequality of certain expression involving algebraic numbers can be checked in $\mathbb{C}$ using "real methods" ( inequalities ).

Example: if $\sqrt[3]{2} \cdot \sqrt{5} = \zeta^{j-1} \sqrt[3]{2} \cdot (-1)^k \sqrt{5}$ then
$\zeta^{j-1}$ must be real and so $j=1$.