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"Find the degree of $\mathbb{Q}(\sqrt[\leftroot{-2}\uproot{2}12]{2},\sqrt[\leftroot{-2}\uproot{2}8]{2})$ as an extension of $\mathbb{Q}$ and find basis for such an extension"

Let $L = \mathbb{Q}(\sqrt[\leftroot{-2}\uproot{2}12]{2},\sqrt[\leftroot{-2}\uproot{2}8]{2})$

The minimum polynomials of $K_1 =\mathbb{Q}(\sqrt[\leftroot{-2}\uproot{2}12]{2})$ and $K_2 = \mathbb{Q}(\sqrt[\leftroot{-2}\uproot{2}8]{2})$ to have degrees $12$ and $8$ respectively.

But as $gcd(8,12) = 4$ then $[L:\mathbb{Q}]=[L:K_1][K_1:\mathbb{Q}]<96$ (by the tower law)

That's as far as I've gotten. I tried to work out the basis (using a grid) and thus the degree and got 42, but that doesn't seem right at all.

My assumption is that I divide $96$ by $4$ giving a degree of $24$

What am I getting wrong?

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  • $\begingroup$ Your field must also contain $\dfrac{1}{2}(\sqrt[12]{2}\sqrt[8]{2})^5=\sqrt[24]{2}$. $\endgroup$ – David Peterson Oct 17 '18 at 23:47
  • $\begingroup$ All the fields in here have infinitely many elements so I replaced the finite-fields tag with a more appropriate one. $\endgroup$ – Jyrki Lahtonen Oct 18 '18 at 5:58
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call the field $F$. then:

$$ \mathbb{Q}(\sqrt[24]{2}) \subset F $$

because: $$ \sqrt[24]{2} = \frac{\sqrt[8]{2}}{\sqrt[12]{2}} $$

on the other hand: $$ F \subset \mathbb{Q}(\sqrt[24]{2}) $$

because $$ \sqrt[12]{2} = \bigg(\sqrt[24]{2} \bigg)^2 \\ $$ and $$ \sqrt[8]{2} = \bigg(\sqrt[24]{2} \bigg)^3 $$

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