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We all know that $\sqrt[3]{3}\not\in \mathbb{Q}({\sqrt[3]{2}})$ intuitively. We even know that $\sqrt[3]{p}\not\in \mathbb{Q}(\sqrt[3]{q})$ for two distinct primes $p, q$. However, I don't know how to prove these things rigorously. In case of $\sqrt{p}\not\in \mathbb{Q}(\sqrt{q})$, we can just set assume $\sqrt{p}=a+b\sqrt{q}$ for some $a, b\in\mathbb{Q}$ and deduce a contradiction. However, in cubic case, expansion of $(a+b\sqrt[3]{2}+c\sqrt[3]{4})^{3}$ is very complicated and even I expand this I couldn't get any contradiction by this. To be specific, it gives us 3 diophantine equations

$$a^{3}+2b^{3}+4c^{3}+12abc=3$$ $$a^{2}b+2b^{2}c+2c^{2}a=0$$ $$ab^{2}+2bc^{2}+ca^{2}=0$$

and I don't know how to check solvability of this kind of diophantine equations (over $\mathbb{Q}$). Even if there exists a way to check solvability, I cannot convince that it can be used to prove $\sqrt[3]{p}\not\in \mathbb{Q}(\sqrt[3]{q})$.

So I tried to use Galois theory, but actually I don't know where to go. I assumed $\mathbb{Q}(\sqrt[3]{3})=\mathbb{Q}(\sqrt[3]{2})$ and take a Galois closure, i.e. $K=\mathbb{Q}(\sqrt[3]{3}, w)=\mathbb{Q}(\sqrt[3]{2}, w)$ where $w^{3}=1$ then tried to use field norm maps. We can check that $\beta=\sqrt[3]{3}\cdot\sigma(\sqrt[3]{3})\cdot\sigma^{2}(\sqrt[3]{3})\in \mathbb{Q}$ when $\sigma\in Gal(K/\mathbb{Q}), \sigma(\sqrt[3]{2})=\sqrt[3]{2}, \sigma(w)=w^{2}$. But I cannot get any information from this.

Edit : It would be great if there is a simple algorithm to check whether $\alpha\in K$ or not for a given algebraic number $\alpha$ and a number field $K$.

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    $\begingroup$ You could find the primitive element of $\mathbb{Q}(\sqrt[3]{3},\sqrt[3]{2})$, find its minimal polynomial and show that it is of degree larger than $3$. $\endgroup$ – Michael Burr Sep 16 '17 at 1:22
  • $\begingroup$ Algebraic number theory might be able to give a solution by considering discriminants of rings of integers - but I'm a bit rusty on that. $\endgroup$ – Daniel Schepler Sep 16 '17 at 1:27
  • $\begingroup$ @MichaelBurr Thanks! I think we can even solve the general problem with that idea and computers. However, I don't know how to do this by hands. It seems that $\sqrt[3]{2}+\sqrt[3]{3}$ is a degree 9 element, but I cannot convince it without any help of computer programs. $\endgroup$ – Seewoo Lee Sep 16 '17 at 1:34
  • $\begingroup$ $a$ cannot be divisible by $2$ (first equation). Therefore $b$ is divisible by $2$ (second equation). Therefore, $c$ is divisible by $2^2$ (third equation). Hence $b$ is divisible by $2^3$ (second equation). Keep repeating and you get that $b$ is divisible by $2^k$ for any $k$. For other pair of primes it is the same business. $\endgroup$ – Hellen Sep 16 '17 at 1:34
  • $\begingroup$ What is $\Bbb{Q}\left( \sqrt[3]{2} \right)$? $\endgroup$ – gen-z ready to perish Sep 16 '17 at 4:03
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Here’s another proof, making light use of $p$-adic theory:

First, note that Eisenstein tells us that both $\Bbb Q(\sqrt[3]3\,)$ and $\Bbb Q(\sqrt[3]2\,)$ are cubic extensions of the rationals, and thus either they are the same field or their intersection is $\Bbb Q$. Therefore, a question equivalent to the stated one is, “Why is $\sqrt[3]2\notin\Bbb Q(\sqrt[3]3\,)$?” I’ll answer the new question.

Note that $\Bbb Q(\sqrt[3]3\,)$ may be embedded into $\Bbb Q_2$, the field of $2$-adic numbers, because $X^3-3\equiv(X+1)(X^2+X+1)\pmod2$, product of two relatively prime polynomials over $\Bbb F_2$, the residue field of $\Bbb Q_2$ (more properly the residue field of $\Bbb Z_2$). So Strong Hensel’s Lemma says that $X^3-3$ has a linear factor in $\Bbb Q_2[X]$, in other words, there’s a cube root of $3$ in $\Bbb Q_2$. But if there were a cube root of $2$ in $\Bbb Q(\sqrt[3]3\,)$, there’d be such a root in $\Bbb Q_2$, and there isn’t since it’d have to have (additive) valuation $v_2(\sqrt[3]2\,)=1/3$. No such thing, the value group of $\Bbb Q_2$ is $\Bbb Z$.

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I really like all of the 3 answers, and I think I found one more solution which uses Trace, not Norm. (I don't know why I didn't try to use trace map before.) So I'm going to write it down here.

Suppose $K=\mathbb{Q}(\sqrt[3]{2})=\mathbb{Q}(\sqrt[3]{3})$. Consider a trace map $$ \mathrm{Tr}_{K/\mathbb{Q}}:K\to \mathbb{Q}, \quad \alpha \mapsto \sum_{i=1}^{n}\sigma_{i}(\alpha) $$ where $\sigma_{1}(\alpha)=\alpha, \dots, \sigma_{n}(\alpha)$ is roots of minimal polynomial of $\alpha$. (We can consider trace map as a trace of a linear map $m_{\alpha}:K\to K, x\mapsto \alpha x$.) This is a $\mathbb{Q}$-linear map and we can check that $\mathrm{Tr}_{K/\mathbb{Q}}(\sqrt[3]{2})=\sqrt[3]{2}+\sqrt[3]{2}w+\sqrt[3]{2}w^{2}=\sqrt[3]{2}(1+w+w^{2})=0$ where $w=e^{2\pi i /3}$ and $\mathrm{Tr}_{K/\mathbb{Q}}(\sqrt[3]{3})=0$, too. If we assume $$ \sqrt[3]{3}=a+b\sqrt[3]{2}+c\sqrt[3]{4}\,\,\,\,(a, b, c\in \mathbb{Q}) $$ as above, by taking trace on both sides we get $0=3a$ and $a=0$. By multiplying $\sqrt[3]{2}$ and $\sqrt[3]{4}$ on both sides, we get $b=c=0$ and contradiction. I think this methods can be used to prove for higher power cases, such as $\sqrt[n]{3}\not\in \mathbb{Q}(\sqrt[n]{2})$ for $n\geq 2$.

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  • $\begingroup$ Great idea! I used it to prove that roots appear in root extensions only in "obvious" ways. $\endgroup$ – Orest Bucicovschi Sep 16 '17 at 6:53
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Let $\omega := e^{2 \pi i/3}$, and consider the splitting field $\mathbb{Q}(\sqrt[3]{2}, \omega)$ of $x^3 - 2$, and the unique automorphism $\sigma$ such that $\sigma(\sqrt[3]{2}) = \omega \sqrt[3]{2}$, $\sigma(\omega \sqrt[3]{2}) = \omega^2 \sqrt[3]{2}$, $\sigma(\omega^2 \sqrt[3]{2}) = \sqrt[3]{2}$. Suppose that $\sqrt[3]{3} \in \mathbb{Q}[\sqrt[3]{2}]$ is equal to $a + b \sqrt[3]{2} + c \sqrt[3]{4}$ for $a, b, c \in \mathbb{Q}$. Then $\sigma(\sqrt[3]{3})$ is also a root of $x^3 - 3 = 0$, which implies it's equal to either $\sqrt[3]{3}$, $\omega \sqrt[3]{3}$, or $\omega^2 \sqrt[3]{3}$.

However, in the first case, $$\sigma(\sqrt[3]{3}) = \sqrt[3]{3} \Rightarrow a + b \omega \sqrt[3]{2} + c \omega^2 \sqrt[3]{4} = a + b \sqrt[3]{2} + c \sqrt[3]{4}.$$ Rewriting $\omega^2 = - \omega - 1$ and using the linear independence of the basis $\{ 1, \omega, \sqrt[3]{2}, \omega \sqrt[3]{2}, \sqrt[3]{4}, \omega \sqrt[3]{4} \}$ of the splitting field, we get $b = c = 0$, which gives a contradiction since $\sqrt[3]{3}$ is irrational.

Similarly, in the second case, $\sigma(\sqrt[3]{3}) = \omega \sqrt[3]{3}$ would imply $\sqrt[3]{3}$ is a rational multiple of $\sqrt[3]{2}$, which is impossible since $\sqrt[3]{3/2}$ is irrational. And in the third case, $\sigma(\sqrt[3]{3}) = \omega^2 \sqrt[3]{3}$ would imply $\sqrt[3]{3}$ is a rational multiple of $\sqrt[3]{4}$, which is impossible since $\sqrt[3]{4/3}$ is irrational.

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HINT:

If $p$ a rational number, not a cube, and $a$, $b$, $c$ rational numbers so that

$$(a + b \sqrt[3]p + c \sqrt[3]{p}^2)^3 $$ is rational, then at most one of the numbers $a$, $b$, $c$ is nonzero.

To prove this, it's enough to show that if $(b,c) \ne (0,0)$ then $a=0$. Consider $\omega = -\frac{1}{2} + i \frac{\sqrt{3}}{2}$. We have $\omega^2 + \omega + 1=0$, and $\omega^3 = 1$.

From $$(a + b \sqrt[3]{p} + c \sqrt[3]{p}^2)^3 = q$$ we get $$(a + b \sqrt[3]{p}\omega + c\sqrt[3]{p}^2 \omega^2)^3=q$$ Since $(b,c)\ne (0,0)$ we conclude $$a + b \sqrt[3]{p}\omega + c\sqrt[3]{p}^2 \omega^2 = \sqrt[3]{q} \omega^{\pm 1}$$ and the conjugate equality $$a + b \sqrt[3]{p}\omega^2 + c \sqrt[3]{p}^2 \omega = \sqrt[3]{q} \omega^{\mp 1}$$

To these two equalities we add $$a + b \sqrt[3]p + c \sqrt[3]{p}^2= \sqrt[3]{q}$$ and we get $3 a = 0$, so $a = 0$.


$\bf{Added:}$

Details:

Assume $a + b \sqrt[3]{p}\omega + c \sqrt[3]{p}^2\omega^2 \ne a + b \sqrt[3]{p} + c \sqrt[3]{p}^3$. We get $$b\omega + c \sqrt[3]{p}\omega^2 = b + c \sqrt[3]{p}$$ and with $\omega^2 =-1-\omega$ we get $(b-c\sqrt[3]{p}) \omega = b + 2 c \sqrt[3]{p}$, and so $\omega = \frac{b + 2 c \sqrt[3]{p}}{b-c\sqrt[3]{p}}$, contradiction ( since, say, $\omega$ is not real). So $a + b \sqrt[3]{p}\omega + c \sqrt[3]{p}^2\omega^2$ must be one of the other two roots of the equation $x^3 = q$.

$\bf{Added 2:}$ We follow the beautiful idea of @See-Woo Lee: to use traces.

$\bf{Fact:}$ Let $\alpha$ a real radical, that is $\alpha \in \mathbb{R}$, $\alpha^m\in \mathbb{Q}$ for some $m$. Assume moreover that $\alpha$ is irrational. Then $\operatorname{trace}^L_{\mathbb{Q}}\alpha= 0$ for any $L$ finite algebraic extension of $\mathbb{Q}$ containing $\alpha$.

Proof: We may assume $\alpha >0$. Let $m$ smallest so that $\alpha^m \in \mathbb{Q}$. Clearly $m>1$. Let's prove that the polynomial $X^m - \alpha^m$ is irreducible over $\mathbb{Q}$. It factors over $\mathbb{C}$ as $\prod_{i=0}^m(X- \alpha \omega^i)$. If several of these linear factors produced a polynomial with rational coefficients, we would have the free term $$\prod_{i \in I} (- \alpha \omega^i) \in \mathbb{Q}$$ Taking absolute values we would have $\alpha^k \in \mathbb{Q}$ for some $1\le k < m$, not possible.

Denote by $K = \mathbb{Q}(\alpha)$. Then we have $$\operatorname{trace}^K_{\mathbb{Q}}(\alpha) = \sum_{i=0}^{m-1} \alpha \omega^i = 0$$. Hence for every $L\supset K$ we get $$\operatorname{trace}^L_{\mathbb{Q}} \alpha = [L: K] \operatorname{trace}^K_{\mathbb{Q}}\alpha = 0$$

$\bf{Main\ result:}$ Let $\alpha_i$ real radicals. Assume moreover that the ratios $\frac{\alpha_i}{\alpha_j}$ for $i\ne j$ are irrational. The the $\alpha_i$'s are linearly independent over $\mathbb{Q}$.

Proof: Let $a_i\in \mathbb{Q}$ so that $$\sum a_i \alpha_i = 0$$ Fix $i$, $1\le i \le k$. We have $$a_i = -\sum_{k \ne i} a_k \frac{\alpha_k}{\alpha_i}$$

Let a finite extension $L$ of $\mathbb{Q}$ containing all the $\alpha_i$ Taking traces on both sides we get $$d \cdot a_i = \sum_{k \ne i} a_k \operatorname{trace}^L_{\mathbb{Q}} \left(\frac{\alpha_k}{\alpha_i}\right )=0$$, since $\frac{\alpha_k}{\alpha_i}$ is a real irrational radical. So all the $a_i$ are $0$.

In words: incommensurable real radicals are linearly independent over $\mathbb{Q}$.

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  • $\begingroup$ This looks great, but I don't get how to get from $(a + b \sqrt[3]{p} + c \sqrt[3]{p}^2)^3 = q$ to $(a + b \sqrt[3]{p}\omega + c\sqrt[3]{p}^2 \omega^2)^3=q$. Any tips? $\endgroup$ – Robert Lewis Sep 16 '17 at 3:32
  • $\begingroup$ @Robert Lewis: Thanks. Added some details. $\endgroup$ – Orest Bucicovschi Sep 16 '17 at 4:06
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Somehow, when manipulating pure radicals, I think that using identification or the trace map, i.e. essentially the additive structure, could lead to more complicated calculations than using the multiplicative structure and the norm map. It seems to be the case here. Introduce the quadratic field $k=\mathbf Q (\omega)$, where $\omega$ is a primitive cubic root of $1$. Since $\mathbf Q(\sqrt[3] 2)$ and $ \mathbf Q(\sqrt[3]3)$ have degree $3$ over $\mathbf Q$ by Eisenstein criterion, Kummer theory tells us that the extensions $k(\sqrt[3]2)/k$ and $k(\sqrt[3]3)/k$ are Galois cyclic of degree $3$, and moreover they coincide iff $2=3x^3$, with $x\in (k^{*})^{3}$. Norming down to $\mathbf Q$, we get the diophantine equation $4a^3=9b^3$ with coprime integers $a, b$ : impossible because neither $4$ nor $9$ are cubes. Note that the argument still works with other (coprime) parameters than $2, 3$.

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