$3^{1/7}\notin \Bbb{Q} (2^{1/7})$.

Question

I am trying to show that $3^{1/7}\notin \Bbb{Q} (2^{1/7})$ using Galois theory. I'm not sure whether or not I'm on the right track but I have considered the normal closure of $\Bbb{Q} (2^{1/7})$ by adjoining the 7th primitive root of $1$ . Then I considered the Galois group of $\Bbb{Q} (\zeta)/Q$ and trying to use the lift to the overall Galois group by mapping $\zeta \rightarrow \zeta^i$ for some i $\in (0,1,2, ..., 6)$ and $2^{1/7} \rightarrow 2^{1/7}\zeta^j$ , j$\in (0,1,2, ..., 6)$. Then I''m trying to get a contradiction by showing that we have, for each i and j, a fixed element by the map which is not actually in Q.

I'm not too sure if this works and would appreciate any help or easier ways of doing this Problem!

Edit: I think just considering $3^{1/7} \in Q(2^{1/7}, \zeta)$ and considering its Galois Group over $Q(\zeta)$, generated by $\tau$: $\zeta \rightarrow \zeta , 2^{1/7} \rightarrow 2^{1/7}\zeta$ , then, $$ \tau(3^{1/7})^{7} = 3 \implies \tau(3^{1/7}) = 3^{1/7}\zeta^i, i\in (0,1,2,...,6) $$ for each case we get a contradiction as say if we consider $\tau(3^{1/7})=3^{1/7}\zeta$, then $\tau((3/2)^{1/7}) = (3/2)^{1/7} \implies (3/2)^{1/7} \in Q(\zeta)$ $\implies Q((3/2)^{1/7})= Q(192^{1/7}) \subset Q(\zeta)$, contradiction as 7 does not divide 6 (Eisenstein). Similar for other cases.

Answer 1

Both extensions $\mathbf Q(\sqrt [17] 2)$ and $\mathbf Q(\sqrt [17] 3)$ have degree $17$ over $\mathbf Q$ by Eisenstein criterion. If one were contained in the other, they would coincide and this would carry over to the normal closures, i.e. $\mathbf Q(\zeta_{17},\sqrt [17] 2)=\mathbf Q(\zeta_{17},\sqrt [17] 3)$. But Kummer theory applies over $\mathbf Q(\zeta_{17})$, and the latter equality is equivalent to $2=3.x^{17}$ for a certain $x\in \mathbf Q(\zeta_{17})^*$. Since $\mathbf Q(\zeta_{17}/\mathbf Q)$ has degree $16$, norming down to $\mathbf Q$ would give $2^{16}=3^{16}.y^{17}$ : this contradicts the uniqueness of prime factorization in $\mathbf Z$. Note that this kind of argument generalizes to many similar problems.

Answer 2

Suppose by the absurd that $\sqrt[7]3\in\mathbb Q(\sqrt[7]2)$. Then $\sqrt[7]3+\sqrt[7]2\in\mathbb Q(\sqrt[7]2)$.

It follows that the degree of $\sqrt[7]3+\sqrt[7]2$ should be at most $7$ because the degree of the extension $\mathbb Q(\sqrt[7]2)$ is $7$ but it is not the case and the minimal polynomial of $\sqrt[7]3+\sqrt[7]2$ at first view has degree greater than $7$. It is actually equal to $49$, according to Wolfram this polynomial is $$x^{49} - 35 x^{42} - 71505 x^{35} - 5622715 x^{28} + 152278889 x^{21} - 966420105 x^{14} - 11026463 x^7 - 78125$$