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I'm studying (basic) Galois theory and I'm stuck in front of this exercise:

Find the degree of $\mathbb{Q}(\zeta_{169},\sqrt[169]{34})$ over $\mathbb{Q}$.

It seems a very standard exercise, anyway I cannot find a solution..maybe because I'm used to solve this kind of exercises directly by finding the minimal polynomial and now this is a little harder. Please, give me some hints or a solution.

Thank you in advantage.

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  • $\begingroup$ All I've managed to prove is that $169\le[\mathbb{Q}(\zeta_{169},\sqrt[169]{34}):\Bbb Q]\le 26364=156\cdot 169$ from the degree-$156$ and $169$ minimal polynomials for these elements. I too would like to see what approach one uses to attack this sort of question. $\endgroup$ – Mario Carneiro Feb 6 '15 at 19:03
  • $\begingroup$ This is the splitting field of the polynomial $x^{169}-34$ over $\mathbb{Q}$, in particular it is a Galois extension. Moreover that polynomial is irreducible by Eisenstein criterium. Since the degree of $[\mathbb{Q} (\zeta_{169}):\mathbb{Q}] = \phi (169) = 12 \cdot 13$, we have that the degree is at least $12 \cdot 13^2$, and divides $12 \cdot 13^3$. $\endgroup$ – Crostul Feb 6 '15 at 19:05
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I have some more thoughts:

If $\mathbb Q(\sqrt[169]{34},\zeta_{169})/\mathbb Q(\zeta_{169})$ would be of degree $13$ (we know $13$ or $13^2$, we want to rule out $13$ in the following), consider the action of $Gal(\mathbb Q(\sqrt[169]{34},\zeta_{169})/\mathbb Q(\zeta_{169})) \cong \mathbb Z/13\mathbb Z$ on the roots of $X^{169}-34$. The action is transitive on each irreducible factor. The lenghts of the orbits of each root is $1$ or $13$. But there is definitely no linear factor because $\sqrt[169]{34} \notin \mathbb Q(\zeta_{169})$, hence length $1$ can be ruled out. So $X^{169}-34$ splits into 13 factors of degree $13$ over $\mathbb Q(\zeta_{169})$.

Now let us multiply $13$ distinct roots of $X^{169}-34$. We get a number of the form $\zeta_{169}^m \sqrt[13]{34}$. On the other hand this is the constant term of an irreducible factor over $\mathbb Q(\zeta_{169})$. This shows $\sqrt[13]{34} \in \mathbb Q(\zeta_{169})$. But this is definitely a contradiction, since $\mathbb Q(\zeta_{169})/\mathbb Q$ is an abelian extension, hence admits only normal intermediate fields.

All in all we deduce that the degree of $\mathbb Q(\sqrt[169]{34},\zeta_{169})/\mathbb Q$ is $169 \cdot 12 \cdot 13$. And the galois group is the semidirect product of $\mathbb Z/169\mathbb Z$ and $(\mathbb Z/169\mathbb Z)^*$

Edit: Of course we can abbreviate the argument a little bit: If the degree of $\sqrt[169]{34}$ over $\mathbb Q(\zeta_{169})$ would be $13$, the minimal polynomial would be the product of $13$ linear factors of the polynomial $X^{169}-34$. The constant term (up to sign) has the form $\zeta_{169}^m \sqrt[13]{34}$, so we can deduce $\sqrt[13]{34} \in \mathbb Q(\zeta_{169})$, getting the contradiction above.

But i think the longer version somehow shows what galois theory is capable of, when everything comes together.

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This is a partial answer.

This is the splitting field of the polynomial $x^{169}-34$ over $\mathbb{Q}$, in particular it is a Galois extension. This means that the degree coincides with the order of the Galois group $G$.

The polynomial $x^{169}-34$ is irreducible by Eisenstein criterium, so $$[\mathbb{Q}(\sqrt[169]{34}) : \mathbb{Q} ] = 169 = 13^2$$

This means that $G$ has a subgroup of index $169$.

Moreover $$[\mathbb{Q} (\zeta_{169}):\mathbb{Q}] = \varphi (169) = 12 \cdot 13$$ so $G$ has a subgroup of index $12 \cdot 13$

Finally

$$[\mathbb{Q}(\sqrt[169]{34}, \zeta_{169}) : \mathbb{Q}] = [\mathbb{Q}(\sqrt[169]{34}, \zeta_{169}) : \mathbb{Q}(\zeta_{169})] \cdot [\mathbb{Q} (\zeta_{169}):\mathbb{Q}]$$ and clearly $[\mathbb{Q}(\zeta_{169})(\sqrt[169]{34}) : \mathbb{Q}(\zeta_{169})]$ divides $[\mathbb{Q}(\sqrt[169]{34}) : \mathbb{Q}] = 13^2$, so it is $13$ or $13^2$.

So there are two options: the answer is $12 \cdot 13^2$, or it is $12 \cdot 13^3$. But now I am not able to conclude.

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