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I have already proven that $[\mathbb{Q}(\sqrt{3}, \sqrt[3]{3}, \sqrt[5]{3}) \colon \mathbb{Q}] = 30$ and $\sqrt{3} \not\in \mathbb{Q}(\xi_5)$. Consequently, $[\mathbb{Q}(\xi_5,\sqrt{3}) \colon \mathbb{Q}(\xi_5) ] = 2$ and then $[\mathbb{Q}(\xi_5,\sqrt{3}) \colon \mathbb{Q}] =[\mathbb{Q}(\xi_5,\sqrt{3}) \colon \mathbb{Q}(\xi_5) ] [\mathbb{Q}(\xi_5) \colon \mathbb{Q}] = 8$, so I have been trying to show that $[\mathbb{Q}(\xi_3,\xi_5,\sqrt{3}) \colon \mathbb{Q}(\xi_5,\sqrt{3})] = 2$ or, equivalently, that $\xi_3 \not\in \mathbb{Q}(\xi_5,\sqrt{3})$. If that was true I can easily conclude that $[\mathbb{Q}(\sqrt{3}, \sqrt[3]{3}, \sqrt[5]{3}, \xi_3,\xi_5) \colon \mathbb{Q}] = 240$, but I am stuck in that part.

David
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    That's a good start. Can you use the fact that $\Bbb{Q}(\xi_5,\sqrt3)/\Bbb{Q}$ is a Galois extension with (an easy to figure out) abelian Galois group isomorphic to $C_4\times C_2$? That allows you to figure out how many quadratic intermediate field there are? – Jyrki Lahtonen Jun 03 '24 at 19:28
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    Or, a modification: Complex conjugation on $L=\Bbb{Q}(\sqrt3,\xi_5)$ is the square of another automorphism. Consequently it belongs to every subgroup of $Gal(L/\Bbb{Q})$ of order four. Therefore all the quadratic subfields of $L$ are real. A lot of details to be filled in :-) – Jyrki Lahtonen Jun 03 '24 at 19:36
  • Are $\xi_3$ and $\xi_5$ primitive roots of unity (often denoted $\zeta_3$ and $\zeta_5$)? – J. W. Tanner Jun 03 '24 at 20:11
  • Yes, they are primitive roots. I forgot to specify it. – David Jun 03 '24 at 20:12

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Consider the two subfields $K_1=\mathbb{Q}(\sqrt3, \sqrt[3]3, \sqrt[5]3)$ and $K_2=\mathbb{Q}(\zeta_3,\zeta_5)$ of $L=\mathbb{Q}(\sqrt3, \sqrt[3]3, \sqrt[5]3,\zeta_3,\zeta_5)$. It is easily seen that $K_1 \cap K_2 = \mathbb{Q}$. $K_2$ is a normal and separable extension of $\mathbb{Q}$, therefore $K_1, K_2$ are linearly disjoint (see here and here), and it follows that $|[L:\mathbb{Q}]|=|[K_1:\mathbb{Q}][K_2:\mathbb{Q}]|=30*8=240$.

nor
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