The original question is to find the Galois extension of the polynomial $x^9-5$. There are multiple questions before the actual question. Assuming $F$ = $\mathbb{Q}(\zeta)$ where $\zeta$ is a primitive 9th root of unity and $L$ = $\mathbb{Q}(\gamma)$ where $\gamma$ is the real root of $x^9-5$. I have proved that the only non-trivial subextension of $L$ is $\mathbb{Q}(\gamma^3)$. I am unable to prove whether $\mathbb{Q}(\gamma^3)$ is equal to $\mathbb{Q}(cos(2\pi/9))$ (which is subextesion of $F$ ) or not. If yes then $F$ intersection $L$ is $\mathbb{Q}(\gamma^3)$ otherwise it is $\mathbb{Q}$. Any help would be appreciated
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Why do you need the intersection of these to find the splitting field? The splitting field is their composition, not their intersection. – Vercassivelaunos Aug 10 '20 at 07:52
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1For the Galois group of $x^n-a$ and intersection see this duplicate. So the Galois group is $\Bbb Z/9\rtimes (\Bbb Z/9)^{\times}$ of order $9\cdot \phi(9)=54$. – Dietrich Burde Aug 10 '20 at 08:04
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1This $\gamma$ is usually denoted by $\sqrt[9]{5}$. – tomasz Aug 10 '20 at 08:53
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@DietrichBurde Thanks for sharing the link and for the answer. – Bhargav Kale Aug 10 '20 at 09:14
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@DietrichBurde so the intersection will precisely be $\mathbb{Q}$ right? just clarifying – Bhargav Kale Aug 10 '20 at 09:23
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1Yes, indeed, it is just $\Bbb Q$. – Dietrich Burde Aug 10 '20 at 10:31
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$\Bbb{Q}(\gamma^3)$ is not a Galois extension of $\Bbb{Q}$ but $\Bbb{Q}(2\cos(2\pi/9)$ is. – Jyrki Lahtonen Aug 11 '20 at 18:52