# Neat method to show that $\mathbb{Q}(2^{1/3}) \ne \mathbb{Q}(3^{1/3})$? [duplicate]

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?

• 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 $\Bbb Q(\sqrt[3]{2})$ ($2=(2^{1/3})^3$) – user8268 Jun 7 '16 at 7:26
• 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 $\mathbb Q_2$ – user8268 Jun 7 '16 at 7:41
• Thank you. Both solutions are also good! – user29422 Jun 7 '16 at 13:45
• More general problem: math.stackexchange.com/questions/1657374 (basically the rational prime $2$ ramifies in the first field but not in the second one). – Watson Dec 17 '18 at 18:34

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.

• 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? – Jyrki Lahtonen Jun 7 '16 at 7:35
• 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. – Ege Erdil Jun 7 '16 at 7:37
• ... like the fact that conjugates of $3^{1/3}$ can only be of the form $\zeta_3^k3^{1/3}$. – Jyrki Lahtonen Jun 7 '16 at 7:37
• Yes, alternatively, use Vieta on $x^3 - 3$ :) – Ege Erdil Jun 7 '16 at 7:38
• Yup. I realized that possibility, too. +1 – Jyrki Lahtonen Jun 7 '16 at 7:38

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.