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This question already has an answer here:

I've tried solving $\sqrt[3]{3} =a + b* \sqrt[3]{2}+c* \sqrt[3]{4}$, but there is no obvious contradiction, even when taking the norms/traces of both sides. I can't think of another approach. This is part of a proof, it seems obvious, but it's really bothersome I can't prove it.

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marked as duplicate by Watson, Paul Frost, Did, Botond, José Carlos Santos Nov 26 '18 at 12:51

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You write that there is no obvious contradiction even when taking traces/norms. That is false. There certainly is a contradiction that way. You simply have to take more than one trace! Instead of just taking the trace of $\sqrt[3]{3}$ down to ${\mathbf Q}$, look at the trace of $\sqrt[3]{3}$ and $\sqrt[3]{3}\sqrt[3]{2}$.

We will assume $\sqrt[3]{3}$ is in ${\mathbf Q}(\sqrt[3]{2})$ and get a contradiction. From that containment we get that $\sqrt[3]{6}$ is also in this field, and therefore ${\mathbf Q}(\sqrt[3]{2}) = {\mathbf Q}(\sqrt[3]{3}) = {\mathbf Q}(\sqrt[3]{6})$. Call this common field $K$. Let us assume $$ \sqrt[3]{3} = a + b\sqrt[3]{2} + c\sqrt[3]{4} $$ for some rational $a$, $b$, and $c$. Viewing $K$ as ${\mathbf Q}(\sqrt[3]{3})$, we have ${\rm Tr}_{K/{\mathbf Q}}(\sqrt[3]{3}) = 0$. Viewing $K$ as ${\mathbf Q}(\sqrt[3]{2})$, we have ${\rm Tr}_{K/{\mathbf Q}}(a + b\sqrt[3]{2} + c\sqrt[3]{4}) = 3a$. Thus $3a = 0$, so $a = 0$. Next, multiply the above equation on both sides by $\sqrt[3]{2}$, leading to $$ \sqrt[3]{6} = 2c + b\sqrt[3]{4}. $$ Viewing $K$ as ${\mathbf Q}(\sqrt[3]{6})$, we have ${\rm Tr}_{K/{\mathbf Q}}(\sqrt[3]{6}) = 0$. Viewing $K$ as ${\mathbf Q}(\sqrt[3]{2})$, we have ${\rm Tr}_{K/{\mathbf Q}}(2c + b\sqrt[3]{4}) = 3(2c)= 6c$. Therfore $6c = 0$, so $c = 0$. That makes the above equation $\sqrt[3]{6} = b\sqrt[3]{4}$, and multiplying both sides of this by $\sqrt[3]{2}$ tells us $\sqrt[3]{12} = b \in {\mathbf Q}$, which is false.

The book "Number Fields" by Marcus has an exercise using this idea to show $\sqrt{3} \not\in {\mathbf Q}(\sqrt[4]{2})$. It is exercise 16 of Chapter 2. You should try that exercise to check that you understand this method. (But once you learn more of algebraic number theory you should find that the ramification ideas mentioned by Pete are a more efficient way to see at a glance why $\sqrt[3]{3}$ can't be in ${\mathbf Q}(\sqrt[3]{2})$.)

Edit: Here is a proof using norms. If ${\mathbf Q}(\sqrt[3]{2}) = {\mathbf Q}(\sqrt[3]{3})$, then passing to Galois closures over ${\mathbf Q}$ implies ${\mathbf Q}(\sqrt[3]{2},\omega) = {\mathbf Q}(\sqrt[3]{3},\omega)$. Write this as $F(\sqrt[3]{2}) = F(\sqrt[3]{3})$, where $F = {\mathbf Q}(\omega)$. Then by Kummer theory $2 = 3\gamma^3$ or $2 = 3^2\gamma^3$ for some $\gamma \in F$. Taking norms of both equations from $F$ to $\mathbf Q$, $4 = 9c^3$ or $4 = 81c^3$ for some rational $c$. This is a contradiction.

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Another argument: The question is whether $\mathbb Q(2^{1/3})$ and $\mathbb Q(3^{1/3})$ are the same field. But $\mathbb Q(3^{1/3})$ embeds into $\mathbb Q_2$, while $\mathbb Q(2^{1/3})$ does not. Reason? The binomial series for $(1+2)^{1/3}$ is $2$-adically convergent; and $X^3-2$ is Eisenstein over $\mathbb Z_2$, so any root generates a cubic extension of $\mathbb Q_2$.

Yet another argument: look modulo the prime $31$, where we see that $2\equiv4^3$, in fact $x^3-2=(x-4)(x-7)(x-20)$ in $\mathbb F_{31}[x]$, so that $31$ splits completely in $\mathbb Q(2^{1/3})$. On the other hand, $3$ is a primitive root modulo $31$, and in fact $31$ is inert in $\mathbb Q(3^{1/3})$.

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I will make use of some basic algebraic number theory.

Let $K_2 = \mathbb{Q}(\sqrt[3]{2})$ and $K_3 = \mathbb{Q}(\sqrt[3]{3})$. As the polynomials $t^3-2$ and $t^3-3$ are both irreducible, both $K_2$ and $K_3$ are cubic number fields. Thus if $\sqrt[3]{3} \in K_2$ we'd have $K_3 = K_2$.

The discriminant of the polynomial $t^3-3$ is $-243 = -3^5$. The discriminant of the number field $K_3 = \mathbb{Q}(\sqrt[3]{3})$ divides this (in fact they are equal but in general this is a more delicate computation), so it is prime to $2$, and thus the prime $2$ does not ramify in $K_3$. On the other hand, the prime $2$ visibly ramifies in $K_2$, so $2$ does divide the discriminant of $K_2$. So $K_2$ and $K_3$ cannot be equal (or even isomorphic).

In some sense we got lucky to "see" that $2$ ramifies in $K_2$. The following gives a more general looking criterion along similar lines.

Variant: Let $K_1 = \mathbb{Q}[t]/(f_1)$ and $K_2 = \mathbb{Q}[t]/(f_2)$ be two number fields. Let $\tilde{d}_1$ be the discriminant of the polynomial $f_1$ and $\tilde{d}_2$ be the discriminant of the polynomial $f_2$. Let $d_1$ and $d_2$ be the discriminants of $K_1$ and $K_2$, and recall that $\tilde{d}_1 = d_1 \cdot a_1^2$ and $\tilde{d}_2 = d_2 \cdot a_2^2$ for some nonzero integers $a_1$ and $a_2$. Thus if there is a prime $p$ such that $\operatorname{ord}_p(d_1)$ is odd and $p \nmid \tilde{d}_2$, then $p \mid d_1$, $p \nmid d_2$ and thus $p$ ramifies in $K_1$ but not in $K_2$. It follows that $K_1$ cannot be embedded in $K_2$.

Unfortunately the variant does not quite apply here since the discriminant of $t^3-2$ is $-108 = -2^2 3^3$, so it was good that we got lucky.

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  • $\begingroup$ Probably in this case we could do something more elementary, but for the general question "Does this algebraic number lie in this algebraic number field?" the above gives a quick and easy sufficient criterion to be able to answer "no". $\endgroup$ – Pete L. Clark Apr 20 '13 at 3:38
  • $\begingroup$ Let me also say that several standard software packages have a built in command that will test whether one number field embeds in another. You can be sure that they do something more sophisticated than this! As usual for algorithmic algebraic number theory, Henri Cohen's GTM is the first place to look. $\endgroup$ – Pete L. Clark Apr 20 '13 at 3:41
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How 'bout this? Let $\alpha=\root3\of 3$ and $\beta=\root3\of 2$. If $\alpha\in\mathbb Q(\beta)$, then, as Pete says, $\mathbb Q(\alpha)=\mathbb Q(\beta)$. Then their Galois closures (i.e., the respective splitting fields) are equal. The Galois group would then be $S_3$, but would have far too many $\mathbb Q$-automorphisms in it—in particular, two subgroups of order $3$ coming from permuting the roots of each of the two irreducible cubics.

I bet there's a horrid, brute-force argument just cubing your original equation and using linear independence of $1$, $\beta$, and $\beta^2$.

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    $\begingroup$ If we assume Q(α)=Q(β), then we can express $\beta$ as a linear combination of the basis elements for $\alpha$. So I don't think we would have 2x as many automorphisms, the actions would already be defined by how they are related. Would your argument suggest $\sqrt[3]{9}$ is not in $\mathbb{Q}(\sqrt[3]{3})$? $\endgroup$ – Tahlor Apr 20 '13 at 4:27
  • $\begingroup$ OK, but then a more naive version of the trace argument will give a contradiction if we apply the automorphism $\sigma$ given by $\sigma(\beta)=\beta\omega$ to $\alpha$ and get either $\alpha\omega$ or its complex conjugate. $\endgroup$ – Ted Shifrin Apr 20 '13 at 4:51
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This method use Galois theory. The notation and theorem is follows from the book "Abstract Algebra: An Introduction" writed by Hungerford.

Consider the splitting field $K$ of $(x^3-2)(x^3-3)$ over $\Bbb{Q}$. That is, $K=\Bbb{Q}(\sqrt[3]{2},\sqrt[3]{3},\xi_3)$, where $\xi_3$ is a primitive root of unity. Then $K$ is a Galois extension of $\Bbb{Q}$.

We find all the automorphisms in $Gal_\Bbb{Q}K$. To determine an automorphism $f\in Gal_\Bbb{Q}K$, it is sufficient to determine $f(\sqrt[3]{2}), f(\sqrt[3]{3})$ and $f(\xi_3)$. If $f\in Gal_{\Bbb{Q}}K$, then $f(\sqrt[3]{2})$ must be a root of the minimal polynomial of $x^3-2$. That is, $f(\sqrt[3]{2})\in\{\sqrt[3]{2}, \sqrt[3]{2}\xi_3, \sqrt[3]{2}\xi_3^2\}$. We can find $f(\sqrt[3]{3})$ and $f(\xi_3)$ by a similar method.

Therefore, the Galois group $Gal_\Bbb{Q}K$ is generated by the following three automorphisms $\sigma, \tau$ and $\rho$. Furthermore, $Gal_\Bbb{Q}K$ is isomorphic to $S_3\times C_3$, where $S_3$ is the symmetric group of degree 3 and $C_3=\langle g\rangle$ is the cyclic group of order 3. $$S_3\times C_3\ni((123),1)\leftrightarrow\sigma\in Gal_\Bbb{Q}K, ~~\sigma:\left\{ \begin{array}{ll} \sqrt[3]{2}\to \sqrt[3]{2}\xi_3 \\ \sqrt[3]{3}\to \sqrt[3]{3} \\ \xi_3\to \xi_3 \end{array} \right.$$ $$S_3\times C_3\ni((12),1)\leftrightarrow\tau\in Gal_\Bbb{Q}K, ~~\tau:\left\{ \begin{array}{ll} \sqrt[3]{2}\to \sqrt[3]{2} \\ \sqrt[3]{3}\to \sqrt[3]{3} \\ \xi_3\to \xi_3^2 \end{array} \right.$$ $$S_3\times C_3\ni((1),g)\leftrightarrow\rho\in Gal_\Bbb{Q}K, ~~\rho:\left\{ \begin{array}{ll} \sqrt[3]{2}\to \sqrt[3]{2} \\ \sqrt[3]{3}\to \sqrt[3]{3}\xi_3 \\ \xi_3\to \xi_3 \end{array} \right. $$

Let $E=\Bbb{Q}(\sqrt[3]{2},\sqrt[3]{3})$. Then $Gal_E K=\langle ((12), 1)\rangle$. By the Fundamental Theorem of Galois Theory, $$[\Bbb{Q}(\sqrt[3]{2},\sqrt[3]{3}):\Bbb{Q}] =[E:\Bbb{Q}] =|Gal_{\Bbb{Q}} K:Gal_E K| =|(S_3\times C_3):\langle ((12),1)\rangle|=9. $$

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    $\begingroup$ Nice diagram and otherwise good stuff. But how did you conclude that all $3\cdot3$ combinations of $(\sigma(\root3\of2),\sigma(\root3\of3))$ occur? Assuming that those two images can be chosen independently from each other is already assuming that $[\Bbb{Q}(\root3\of2,\root3\of3):\Bbb{Q}]=9$. In other words, $\sigma(\root3\of2)$ determines $\sigma(\root3\of3)$ uniquely should it happen that $\root3\of3\in\Bbb{Q}(\root3\of2)$, and you should first prove that this is not the case. So I'm concerned that there is an element of circularity in your argument. $\endgroup$ – Jyrki Lahtonen Jul 16 '16 at 7:52
  • $\begingroup$ @JyrkiLahtonen Thanks for your comment. I will try to solve it. $\endgroup$ – bfhaha Jul 16 '16 at 8:06
  • $\begingroup$ I hope you can say something about it. The diagram of intermediate fields is nice to have! Even if you cannot add another way of looking at that problem, it is good to have a bit more complicated case described at this level of detail. $\endgroup$ – Jyrki Lahtonen Jul 16 '16 at 8:41

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