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.

  • $\begingroup$ Check the polynomial. I got $-121 - 60 x + 75 x^2 - 4 x^3 - 15 x^4 + x^6$. $\endgroup$ – Jyrki Lahtonen Sep 20 '14 at 7:21
  • $\begingroup$ Yep, the orders of the radicals were switched at some point. $\endgroup$ – orangeskid Sep 20 '14 at 11:07
  • $\begingroup$ Thanks, @JyrkiLahtonen. I still can't show that's irreducible, so I'm glad to have a different approach. $\endgroup$ – Kevin Arlin Sep 20 '14 at 15:03
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    $\begingroup$ I entertained some hopes for using the fact that the constant term $-121=-11^2$. It is impossible for the two (putative) factors to have constant terms $\pm11$ for then the linear term of the product should also be divisible by eleven. Unfortunately the possibility that one factor has constant term $\pm121$ and the other $\mp1$ is not as tidy. $\endgroup$ – Jyrki Lahtonen Sep 20 '14 at 16:38

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 $j=1$.

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  • $\begingroup$ Ah, that's just excellent, thanks very much! $\endgroup$ – Kevin Arlin Sep 20 '14 at 15:00
  • $\begingroup$ No worries! I wonder how to get explicit expressions in an elegant way, for instance $\sqrt[3]{2}= P(\sqrt[3]{2}+ \sqrt{5})$ with $P$ coefficients in $\mathbb{Q}$ of $\deg \le 5$. Can do by brute force. $\endgroup$ – orangeskid Sep 20 '14 at 17:05

We don't need any sophisticated tools like Galois Theory here. It is possible to prove this using simple algebraic manipulation. It should be obvious that $\mathbb{Q}(\sqrt{5} + \sqrt[3]{2}) \subseteq \mathbb{Q}(\sqrt{5},\sqrt[3]{2})$ and proving that $\mathbb{Q}(\sqrt{5} + \sqrt[3]{2}) \supseteq \mathbb{Q}(\sqrt{5},\sqrt[3]{2})$ is not difficult. Let $c = \sqrt{5} + \sqrt[3]{2}$ and we need to establish that both $a = \sqrt{5}$ and $b = \sqrt[3]{2}$ are rational functions of $c$.

We have $$(c - a)^{3} = 2\Rightarrow c^{3} - 3ac^{2} + 3a^{2}c - a^{3} = 2$$ or $$c^{3} - 3ac^{2} + 15c - 5a = 2\Rightarrow a = \frac{c^{3} + 15c - 2}{3c^{2} + 5}\tag{1}$$ so that $a$ is a rational function of $c$. And further we have $a + b = c$ so that $b = c - a$ and therefore $b$ is also a rational function of $c$.

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It's not the most general, but I actually find the explicit approach to be relatively elegant for this problem:

Let $\alpha=\sqrt{5}+\sqrt[3]{2}$. Then $\alpha-\sqrt{5}=\sqrt[3]{2}$, so $(\alpha-\sqrt{5})^3\equiv \alpha^3-3\sqrt{5}\alpha^2+15\alpha-5\sqrt{5}=2$. In other words, $(3\alpha^2+5)\sqrt{5}=\alpha^3+15\alpha-2$ or $\sqrt{5}=\dfrac{\alpha^3+15\alpha-2}{3\alpha^2+5}$.

An analagous calculation should show that the result does generalize to the case $\mathbb{Q}[\sqrt{p}+\sqrt[n]{q}]$ for any relatively prime $p,q$ and any $n\geq 2$; the fact that any powers of the quadratic are either rational or a rational multiple of the quadratic itself make it straightforward to construct this sort of explicit expression.

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Apologies if this duplicates content in the nice answer of orangeskid. Here is another approach, also using Galois theory.

Let $K$ be the splitting field of the polynomial $(x^{2}-5)(x^{3}-2)$ over $\mathbb{Q}$, and let $\alpha = \sqrt[3]{2}+\sqrt{5}$. One can see that $K = \mathbb{Q}(\sqrt{5}, \sqrt[3]{2}, \xi)$, where $\xi$ is a primitive third root of unity, and hence is the normal closure of the subfield $L := \mathbb{Q}(\sqrt{5}, \sqrt[3]{2})$. Now, $[L:\mathbb{Q}] = 6$, since $\deg(\sqrt{5}) = 2$ and $\deg(\sqrt[3]{2}) = 3$ both divide $[L:\mathbb{Q}]$, and $\mathbb{Q}(\sqrt[3]{2}+\sqrt{5})$ is clearly a subfield of $L$. If we can show that $\deg(\alpha) = 6$, the conclusion follows.

Since $L \subset \mathbb{R}$ and $\xi \notin \mathbb{R}$, it follows that $[K:L] = 2$, whence $[K:\mathbb{Q}] = 12$. Then $G := \mathrm{Gal}(K/\mathbb{Q})$ has $12$ elements, and any element $\sigma \in G$ is determined by what it does to the generators $\sqrt{5}, \sqrt[3]{2}, \xi$ of $K$ over $\mathbb{Q}$. Any $\sigma \in G$ must satisfy $\sigma(\sqrt{5}) \in \{\pm \sqrt{5}\}, \sigma(\sqrt[3]{2}) \in \{\xi^{i}\sqrt[3]{2}\}_{i = 0}^{2}, \sigma(\xi) \in \{\xi, \xi^{2}\}$, so counting, there are at most 12 possible automorphisms, arising from these possible assignments on generators. Since we have already determined $|G| = 12$, every assignment on generators described above must give rise to an automorphism of $K$ over $\mathbb{Q}$. Hence, let $\sigma, \tau \in G$ be the automorphisms defined on generators by $$\sigma(\sqrt[3]{2}) = \xi\sqrt[3]{2}, \sigma(\sqrt{5}) = \sqrt{5}, \sigma(\xi) = \xi$$ $$\tau(\sqrt[3]{2}) = \sqrt[3]{2}, \tau(\sqrt{5}) = -\sqrt{5}, \tau(\xi) = \xi$$

Then the Galois orbit $G\alpha$ contains $\{\pm \sqrt{5} + \xi^{i}\sqrt[3]{2} \mid i = 0, 1, 2\}$, so $|G\alpha| \geqslant 6$. Since $|G\alpha| \leqslant 6$, we obtain the equality $|G\alpha| = 6$. Further, as the minimal polynomial of $\alpha$ over $\mathbb{Q}$ factors as $\prod_{s \in G\alpha} (X-s)$, it follows that $\deg(\alpha) = 6$, as desired.

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