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I want to know the degree of the splitting field of $x^6-7$ over $\mathbb{Q}$, call him $\mathbb{E} = \mathbb{Q}(\sqrt[6]{7},\omega) = \mathbb{Q}(\sqrt[6]{7})(\omega)$. By the theorem "characterization of extensions", we have that $[\mathbb{E}:\mathbb{Q}] = \deg \text{Irr}_{\mathbb{Q}}^{\sqrt[6]{7} \omega}$ = 6 where $\omega$ is the 6th primitive root of unity, and to see irreducibility we can apply Eisenstein with $p=7$.

But when dealing with the subfields I have $[\mathbb{Q}(\sqrt[6]{7}):\mathbb{Q}]=6$ because its irreducible polynominomial is $x^6-7$, and also $[\mathbb{E}:\mathbb{Q}(\sqrt[6]{7})]=2$ beacause $x^2-x+1$ is the irreducible polynomial of $\omega$ in $\mathbb{Q}$ and also it is irreducible in $\mathbb{Q}(\sqrt[6]{7})$ as $\omega \in \mathbb{C} \setminus \mathbb{R}$.

Therefore by transitivity of degrees I get to a contradiction: $$ 6 = [\mathbb{E}:\mathbb{Q}] \neq [\mathbb{E}:\mathbb{Q}(\sqrt[6]{7})][\mathbb{Q}(\sqrt[6]{7}):\mathbb{Q}] = 2 \cdot 6 = 12 $$

I'll really appriciate any hint of where I'm mistaken. Thanks for your time, best regards.

Jyrki Lahtonen
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baristocrona
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  • The same problem as here. You have the special case of $x^n-a$ with $n=6$ and $a=7$. The degree is $\phi(n)n=2\cdot 6=12$. – Dietrich Burde Apr 21 '25 at 11:07
  • @DietrichBurde Yes, but in that problem they only apply degree transitivity. I want to know why the left hand side of my equality is wrong. – baristocrona Apr 21 '25 at 11:10
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    Because the left hand side is $\phi(n)n=2\cdot 6\neq 6$ for $n=6$. There is no contradiction. The degree of $\Bbb E$ over $\Bbb Q$ just is not $6$, see the duplicates. We have $\Bbb E=KL$, in my notation, and this doesn't have degree $6$. So your third line "By the theorem "characterization of extensions", we have that" is wrong. – Dietrich Burde Apr 21 '25 at 11:11
  • @DietrichBurde Why did you rollback my question? – baristocrona Apr 21 '25 at 11:55
  • Because you have changed it afterwards. If you are not satisfied, then ask what is not clear. In your original question, you cannot take as a primitive element $\alpha$ the product of $\omega$ with $\sqrt[6]{7}$. – Dietrich Burde Apr 21 '25 at 12:05
  • @DietrichBurde why can't I take the primitive element of $\omega$ with $\sqrt[6]{7}$ if they generate all roots of the polynomial and form a basis with its powers? I really don't see where your $\phi(n)$ arises. I mean when I put I add all the roots and only the roots to the extension then I certainly have the splitting field. – baristocrona Apr 21 '25 at 12:08
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    We have $[\Bbb Q(\zeta_n):\Bbb Q]=\phi(n)$. This is where $\phi(n)$ arises. In your case $n=6$. In the duplicate I explain why this cyclotomic extension cannot be ignored. – Dietrich Burde Apr 21 '25 at 12:37

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