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I am wondering how to show looking obvious $\mathbb{Q}(2^{\frac{1}{3}}) \ne \mathbb{Q}(3^{\frac{1}{3}}) $?

This question has appeared to compute the order of $\text{Gal}(\mathbb{Q}(2^{\frac{1}{3}},3^{\frac{1}{3}},\xi_3)/\mathbb{Q})$ where $\xi_3$ is a primitive root of unity.

I already know to show this by brutal force by assumimg $2^{\frac{1}{3}}=a+b\cdot2^{\frac{1}{3}}+c\cdot 2^{\frac{2}{3}}$ for some $a,b,c\in \mathbb{Q}$ and take 3rd power on both side and compare their coefficients. But its too boring.

Is there any neat method?

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    $\begingroup$ I guess it's not what you want, but $2$ is unramified in $\Bbb Q(\sqrt[3]{3})$ (the discriminant of $x^3-3$ is $-3^5$), and it's totally ramified in $\Bbb Q(\sqrt[3]{2})$ ($2=(2^{1/3})^3$) $\endgroup$ – user8268 Jun 7 '16 at 7:26
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    $\begingroup$ another solution: $x^3-3$ actually has a root in $\mathbb Q_2$, as $3^3=27=3+8\times 3$, but $x^3-2$ has certainly no root in $\mathbb Q_2$ $\endgroup$ – user8268 Jun 7 '16 at 7:41
  • $\begingroup$ Thank you. Both solutions are also good! $\endgroup$ – user29422 Jun 7 '16 at 13:45
  • $\begingroup$ More general problem: math.stackexchange.com/questions/1657374 (basically the rational prime $2$ ramifies in the first field but not in the second one). $\endgroup$ – Watson Dec 17 '18 at 18:34
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We can argue using the field trace. Consider $ K = \mathbb{Q}(2^{1/3}) $ and $ L = \mathbb{Q}(2^{1/3}, \zeta_3) $ where $ \zeta_3 $ is a primitive third root of unity. $ L $ is the normal closure of $ K $. Now, assume that we had $ 3^{1/3} \in K $, then we would have

$$ 3^{1/3} = c_0 + c_1 2^{1/3} + c_2 2^{2/3} $$

for some $ c_k \in \mathbb{Q} $. Let $ T = \textrm{Tr}_{L/\mathbb{Q}} $ denote the field trace, defined by

$$ T(x) = \sum_{\sigma \in \textrm{Gal}(L/\mathbb{Q})} \sigma(x) $$

The properties $ T(cx) = c T(x) $ for rational $ c $ and $ T(x) = 0 $ iff $ x = 0 $ for rational $ x $ are evident. Now, note that applying $ T $ to both sides of the relation yields $ c_0 = 0 $. Multiply both sides by $ 2^{1/3} $ to get

$$ 6^{1/3} = c_1 2^{2/3} + 2c_2 $$

and apply the field trace to both sides again to get $ c_2 = 0 $. These results imply that we must have $ (3/2)^{1/3} \in \mathbb{Q} $; impossible. Thus, $ 3^{1/3} \notin K $ after all.

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  • $\begingroup$ A good idea. But don't you need a few more properties of the trace to conclude that $T(3^{1/3})=0$? Your argument still works if you use transitivity of trace and go via $\Bbb{Q}(2^{1/3},3^{1/3},\zeta_3)$. Or may be I missed something? $\endgroup$ – Jyrki Lahtonen Jun 7 '16 at 7:35
  • $\begingroup$ We don't. If $ \mathbb{Q}(3^{1/3}) = \mathbb{Q}(2^{1/3}) $, then any $ \mathbb{Q} $-embedding of this field into its normal closure $ L $ will extend to a $ \mathbb{Q} $-automorphism of $ L $ by the isomorphism extension theorem; so $ 3^{1/3} $ has full orbit under the action of the Galois group. $\endgroup$ – Ege Erdil Jun 7 '16 at 7:37
  • $\begingroup$ ... like the fact that conjugates of $3^{1/3}$ can only be of the form $\zeta_3^k3^{1/3}$. $\endgroup$ – Jyrki Lahtonen Jun 7 '16 at 7:37
  • $\begingroup$ Yes, alternatively, use Vieta on $ x^3 - 3 $ :) $\endgroup$ – Ege Erdil Jun 7 '16 at 7:38
  • $\begingroup$ Yup. I realized that possibility, too. +1 $\endgroup$ – Jyrki Lahtonen Jun 7 '16 at 7:38
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Arithmetic tools (ramification, $p$-adics…) surely do the job, but a more « direct » Galois theoretic argument would be more in the spirit of the question. The introduction of $\zeta_3$ obviously hints at Kummer theory above $K = \mathbf Q (\zeta_3)$ . The hypothesis $\mathbf Q(\sqrt[3] {2}) = \mathbf Q (\sqrt[3] {3})$ would imply $K(\sqrt[3] {2}) = K(\sqrt[3] {3})$, hence, by Kummer theory, $2 = 3. x^3$, with $x \in K^*$. Norming down from $K$ to $\mathbf Q$ would give $4 = 9. y^3$, with $y \in \mathbf Q^*$, which obviously contradicts the factoriality of $\mathbf Z$ .

NB : The concluding step is equivalent to $({2/3})^{2/3}\notin \mathbf Q$ , which in turn, because 2 is invertible mod 3, is equivalent to the conclusion $({2/3})^{1/3}\notin \mathbf Q$ obtained by @Starfall. But using the norm instead of the trace to deal with the multiplicative structure was easier because more natural.

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